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# Chemical Structure CHEM 20A

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This 193 page Class Notes was uploaded by Michael Reilly on Friday September 4, 2015. The Class Notes belongs to CHEM 20A at University of California - Los Angeles taught by Staff in Fall. Since its upload, it has received 9 views. For similar materials see /class/177990/chem-20a-university-of-california-los-angeles in Chemistry at University of California - Los Angeles.

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Date Created: 09/04/15

Lecture Series 5 Chemistry 20A Professor Jim Heath These Lectures will cover Chapter 14 The major point of this material is to develop a simple description of the chemical bond Much of that description will be derived from our earlier lectures on the quantum mechanical nature of electrons and atoms First let s get a physical feeling for bonds by considering the lengths of some homonuclear diatomic molecules twoatom molecules made from atoms of the same element Recall that for the hydrogen atom we calculated a Bohr radius an of about 054 Angstroms or 054 A Thus we might guess that the molecule Hz would have a bond length that is about twice the Bohr radius or 2a0 11 A This would put us in the right ballpark 7 the actual answer is 075 A The fact that the bond distance is a little less than 2a0 is important and we will return to this But first let s consider a few of the other bonds we have already encountered Take the following molecules H3CCH3 CH length 11 A CC length 15 A HzCCH2 CH length 11 A CC length 13 A HCECH HC length 11 A CEC length 12 A We can state a few things already First bond lengths in these simple covalent molecules tend to hover around 1 7 15 A s Second by comparing the bond lengths of a triple double and single bond it is apparent that higher order bonds are characterized by shorter bond lengths If we had looked at a whole series of organic molecules we would have found that carboncarbon bonds only vary slightly from molecule to molecule but vary more strongly as bond order changes Furthermore if we had tabulated a series of MH bonds where M is just some element bound to hydrogen then we would have found that most MH bonds are about 1 A in length The Energy of a Chemical Bond The Bond Energy When two atoms come together to form a stable diatomic molecule a chemical bond is formed Why does the bond form The reasons are similar to what we used to describe the formation of a stable atom Recall that for the case of a hydrogen atom the electron and the proton were held together by a Coulombic potential When they were far apart the potential energy was 0 When A the came close together the potential energy was lt 0 The electron in a 13 orbital is in an orbital that is roughly the size of the Bohr radius a0 and it takes 136 eV to ionize that electron Thus we can represent the hydrogen atom with the graph in Figure 1 In a similar way there is a potential energy curve that describes how one 136 e QlQ2 lt 0 atom chemically binds to another atom It looks Attractive something like that shown in Figure 4 Note that in Figure 2 the XaXis is still length and the yaXis is still energy In fact the yaXis still represents potential energy Why Because when atoms are far apart from one another and those atoms can form a stable chemical bond then the potential energy is high 0 relative to when they are close together This is very similar to the electronproton argument However two atoms that are charge neutral for example two hydrogen atoms do 0 r a0 distance Energy Fig 1 The Coulombic potential curve that accounts for the electron associated with a proton the H atom not attract each other in the same way that two oppositely charged atoms or two oppositely charged particles do Why Because the Coulombic potential describes the interaction between two charged particles not neutral atoms Nevertheless the form of the attractive potential for two neutral atoms doesn t look so different Let s note a few things about Figure 2 First the potential energy curve is really a well not a curve When the atoms get close to each other small r then the well increases steeply to positive values of potential energy What this is saying is that two atoms are rather like a magnet and a piece of iron The magnet attracts the iron but the iron can t actually penetrate into the magnet It can only r bond length Bon approach the magnet Once the iron has Energy approached up to the surface of the magnet it will penetrate no further The cost in energy is Fig 2 The potential energy curve that too great describes a chemical bond We will come back to this picture several times However for now the point we want to make is that the bond energy is extracted from this picture in exactly the same way that the ionization energy is extracted from the electronproton picture of a hydrogen atom The bond length is extracted from the picture in the following way Note that a given negative potential energy value will be associated with two values of r since the curve is a well In Fig 2 a double arrow line is drawn across the well intersecting the well at two values of r This is a constant energy line The midpoint of this double arrow line is the equilibrium bond length This length is approximately equal to the value of r for which the potential energy well is a minimum The ionization energy of H is as stated above 136 eV In chapter 13 at the end of the chapter a discussion of periodic trends is presented and the ionization energies of the elements are graphed The ionization potential of He appears in Fig 1333 to be about twice that of hydrogen or about 25 eV while Li is about half that of hydrogen or 7 eV give or take an eV Thus the range of ionization energies in eVs is from about 5 to about 25 How about bond energies In your text on p 531 table l4l lists bond enthalpies The bond enthalpy is a more exact term but means that same thing as the term bond energy that we have been using Properly what the bond enthalpy tells us is how much energy do we get back when we form a chemical bond or how much energy does it take to break the chemical bond The values listed range from l50 ldmol to 1100 kJmol We can change this to eV s by dividing through by NA Avagadro s number and then converting kiloJoules to Joules and then to eV s It turns out that the net conversion factor is rather easy 7 you just divide through by 100 so that 150 kJmole equals about 15 eV s or 15 eV and 1100 kJmol equals about 11 eV s Thus the bond energies tend are of similar magnitude but about 50 smaller than ionization energies Thus even though the potential energy of interaction between two neutral atoms is not a Coulombic potential it is not too dissimilar The types of Bonds We have been discussing chemical bonds and rather loosely using terms such as single bonds double bonds covalent bonds and ionic bonds As we progress through this chapter we will try to develop a more indepth view of what these terms mean Let s de ne them a little here in a qualitative fashion and then move on to a more quantitative description afterwards Covalent Bond This type of bond may be a single double triple or even some fraction such as a 112 bond When we speak of the type of covalent bond we use the term bond order For a single bond the bond order is 1 For a triple bond the bond order is 3 And so on For all types of covalent bonds the electrons that are involved in forming the chemical bond are more or less shared equally by both atoms participating in the bond Ionic Bond This type of bond may be described in similar terms that are used for the Coulombic attraction between two oppositely charged particles In an ionic bond charge is separated meaning that one of the two atoms involved in the bond is positively charged and one of the atoms is negatively charged Thus the bond is really an attraction between two oppositely charged particles Inbetween covalent and ionic bond This type of bond is the most common All homonuclear diatomic molecules are characterized by purely covalent bonds However most other types of bonds are somewhere in between a covalent and an ionic bond How do we estimate whether the bond is ionic or covalent It turns out that the periodic table coupled with a concept known as electronegativity are the tools that we need Electronegativity The discussion in OXtoby is horrible Use these notes to understand this subject Electronegativity was de ned by Mulliken to be 12 Ionization Potential Electron Affinity For the case ofH atom the electron af nity is 75 kJmol while the IP is 1300 ldmol The average of these two numbers is about 700 kJmol Lets do a few more atoms Cesium is 12375 45 N 200 0 kJmol Fluorine is 12 1681 328 1000 kJmol Pauling devised a similar scale His electronegativity is more or less Mullikan s electronegativity divided by 250 so the the H atom has an electronegativity of about 25 the Cesium atom s value is 08 and the value for Fluorine is about 40 Figure 141 in your text gives these values In F1gure 3 The electronegat1v1ty trend in the periodic general one can get an overall table The electronegativity of the rare gases is usually feeling for electronegativity not considered so they are greyed out values by looking at the periodic table See Figure 3 So how does electronegativity help us when we try to rationalize bonding Well consider what it is Atoms that are difficult to ionize are usually easy to add electrons to and vice versa The periodic table can be utilized to rationalize such behavior 7 see the end of Lecture Series 4 Thus an atom such as Fluorine will readily accept an extra charge but will also be reluctant to give one up An atom such as Na however will readily give a charge up but is also not too enthusiastic about accepting a charge The Pauling electronegativity of sodium is about 1 and for Fluorine as mentioned above it is about 4 Consider the absolute value of the difference in electronegativity between the two atoms I ElecNeg F 7 Elec Neg Na I 3 This is a large difference and is a measurement of the charge separation that takes place in the chemical bond between a Na atom and a F atom Such a large difference indicates that the NaF bond is largely ionic and may be described as NaF39 One finds that this estimation of the charge separation gives widely varying numbers for chemical bonds It can range from 00 for homonuclear diatomics and other similar molecules to 30 for very ionic bonds How do we define an ionic bond If the absolute value of the difference in electronegativities between the two participating atoms is greater than 20 then the bond is very ionic If it is less than 05 then the bond is mostly covalent If it is between 20 and 05 then the bond has a mixed character 7 it is both ionic and covalent Estimating the Energy of an Ionic Bond In the discussion above we found that we could use electronegativities to determine whether a chemical bond formed between two elements was covalent ionic or mixed in character However we have said nothing about how strong is the bond that is formed For ionic bonds we can make a few simple approximations and come to an answer that is close to the true bond energy or bond enthalpy Consider the NaF molecule 7 which we have already shown to be held together by a mostly ionic bond If we describe NaF as NaF39 then we can begin describing the bond energy Let s separate the chemical reaction between a Na and a F atom into a few steps Na F 9 NaF Breaks down into Na 9 Na e39 Energy Ioniz Pot of Na 500 ldmol And F e39 9 F39 Energy Elec Aff ofF 328 kJmol Note that the equations above are charge balanced Thus it costs us 500 kJmol to prepare Na and we get back 328 kJmol when we make F Note that we write 7328 kJmol for the Fluorine electron affinity The book lists the value as 328 ldmol The reason behind this is that the book considers the reaction going in the direction of F39 9 F e rather than how we have stated it If we add these two values 500 7 328 then we find that it has still cost us 7170 ldmol to make this solid Thus if we only consider the above process we have spent l70 kJmol and got nothing back 7 the formation of the bond is only costing us energy What s going on here It turns out that we aren t quite done yet We still haven t formed a bond we have only formed a cation Na and an anion F39 We need an additional step 7 that is we need to bring the two ions together to actually form the bond Na Fquot a NaF This reaction looks suspiciously like two oppositely charged particles coming together attracted to one another by a Coulombic potential In fact that is exactly what it is It turns out that the energy we get back from forming the bond is related to the drop in potential energy when we bring the two charged particles together minus the energy cost associated with forming the two ions Thus we may estimate our bond energy as Qle 4118 or1 IPsodium 7 EAFluorine Bond Enthalpy Recall that for two oppositely charged particles the Coulombic potential yields a negative energy as the interparticle separation distance is reduced EXAMPLE Consider the ionically bound molecule NaF If this molecule has a bond length of 19 Angstroms then estimate the bond enthalpy First thing we need to do is to calculate the potential energy of interaction between two oppositely charged particles separated by 19 Angstroms Since both ions that form the molecule are characterized by single charges Z l e2411 0r391 l6el9 C2 43148854 el2 CZJ39lm391l9e10m391 l2e18 J Convert to k and then to kJmole by l2e18lld1000J6022e23 mol39l 730 kJmol So we gain 730 ldmol in potential energy when we bring these two charged atoms together Recall however that we spent 170 ldmol when we ionized the atoms Thus the total bond enthalpy is 730 kJmol 170 kJmol 560 ldmol 55 eV Why do we denote the bond enthalpy as a negative number Because when we make the bond this is the energy we get back How do we get it back Possibly as a photon or several photons or possibly as heat If we wanted to break the molecule apart we would have to supply 55 eV of energy per molecule Dipole Moments for Diatomic Molecules It turns out that just about every diatomic molecule except for homonuclear diatomics has a dipole moment Once again Oxtoby manages to present an opaque discussion of a relatively simple topic Pick any two atoms on the periodic table Let s call them atom A and atom B Now look at Figure 141 in your text p 534 Almost all combinations of atoms will have some slight difference in the Pauling electronegativity Exceptions include things such as Pd Ru Os At H and and Ir which all have an electronegativity value of about 220 or Hg and Po which have values of 200 Aside from these few exceptions all heteronuclear different atoms diatomic molecules will be characterized by atoms with differing values of electronegativity Look at the table to determine which atom has the greatest electronegativity That atom will contain a little more negative charge than the other atom when a bond is formed For example consider the following diatomics and the atomic electronegativities HF Electro Neg H 220 Electro Neg F 398 Al78 NaBr Electro Neg Na 093 Electro Neg Br 296 A 203 ClBr Electro Neg Cl 316 Electro Neg Br 296 A 020 CO Electro neg C 255 Electro neg O 344 A 089 CN Electro neg C 255 Electro neg N 304 A 049 Which atom has the highest value of electronegativity For HF it is F For NaBr it is Br For ClBr it is Cl For CO it is O For CN it is N We can represent the charge separation in these diatomics by placing a symbol over the molecule as we show in Figure 4 Such a separation of negative and positive charge in a I I molecule gives the molecule a dipole moment The magnitude othe dipole O Br moment depends upon both how ionic the bond is or how much charge is separated and how far apart the charge is held Note that the arrow points in the direction of the negative charge the most electronegative atom and the cross on the arrow looks more or less like a plus sign In the above table we denote the difference between the two electronegativities as A All you need to worry about is that the amount of ionic character in the bond is proportional to A For example NaBr about as ionic a molecule as there is and it has a large A of 203 ClBr however has a tiny A indicating it is almost entirely covalent It should be since chlorine and bromine are in the same column of the periodic table The dipole moment measured in units called Debyes after a guy named Peter Debye has units of charge X distance 1 Debye 33 X 103930 Coulomb meters Molecules have dipole moments also and we will try to return to them in a little bit For now however let s just use a few linear cyanoalkynes to make a simple point about dipole moments Consider the following series of molecules 2T T aBr 2T ET Fig 4 The dipole directions for a few diatomic molecules HCECCECCEN HCECCEN HCECCECCECCECCEN and HCEN In these molecules there is a little bit of negative charge piled up at the Nitrogen and a little bit of positive charge piled up at the hydrogen It is roughly the same amount of charge for all 3 molecules However the charge is separated by varying amounts The dipole moment in these molecules depends not only on the amount of charge that is separated but also on the distance over which the charge is separated and the dependence on charge separation is linear Let s assume that HCN has a dipole moment of 2 Debye HC3N is about 2 time longer than HCN so it dipole moment is 2 X 2 4 Debye HC5N is about 3 time longer and so it would have a dipole moment of about 6 Debye HC9N is about 5 times longer so it would have a dipole moment of 10 Debye The molecules are actually the most compleX molecules that have ever been observed in interstellar space and it is eXactly because they are characterized by large dipole moments that makes it possible to see them They are observed through a technique called radioastronomy which in a way images molecules with dipole moments If you watch that silly Jody Foster movie Contact that was out this past summer you will see a large array of satellite dishes That large array is a radio telescope The Shapes 0f Molecules VSEPR Theory We have pretty much covered the nature of an ionic bond Before we cover the nature of a covalent bond in detail let s return to our Lewis Dot Structures and try to elucidate molecular shapes from them Just as I stressed organic molecules when we were talking about Lewis Dot structures I will again stress them here However just as Lewis dot structures do apply to more than just organics so does the theory of this section It is called Valence Shell Electron Pair Repulsitm theory or VSEPR theory VSEPR theory is a very simple phenomenological model and that s all it is that is used to predict molecular structures It works surprisingly well but it also has severe limitations In order to utilize VSEPR theory first draw the Lewis dot structure of the molecule that you are interested in Let s do a few examples of simple organic molecules First I will give the structural formula and then the Lewis Dot structure If you are trying to learn Lewis dot structures of organic molecules which you should be then try to work out the Lewis dot structure yourself before you look the structure that I give The molecules we will work on here are CH3CH2CH3 CHCH CH3CHO and CH30H Their Lewis dot A B structures have to 39 39 be those shown in H H H Figure 5 Now that we have drawn H l IZ b3H H1 CZ39H these Lewis dot I I I structures we can H H H begin trying to back out structural information from C H them by using H VSEPR theory I H C 0 Notice a few things HCl C220 I about the structures I I H H First we have H H labelled all of the carbon atoms within Fig 5 Lewis Dot structures of the organic molecules mentioned in the a given structures text with a number This will allow us to refer to them later Second we have not included lone pairs Which lone pairs have we left out Now that we have the Lewis dot structures we can figure out the geometrical structures using VSEPR theory The first thing that we need to do is to count the number of things around a given atom in the molecule By things we mean the following Each bond whether it is single double or triple counts for 1 thing Each lone pair counts for 1 thing Thus in Structure A Carbon Atom 1 is bound to 3 hydrogens and 1 carbon and it has no lone pairs Therefore there are 4 things around carbon atom 1 Carbon atom 2 is bound to 2 hydrogens and 2 carbons and has no lone pairs It also has 4 things around it Carbon atom 3 is bound to 1 carbon atom and 3 hydrogens It also has 4 things around it Why did we group bonds and lone pairs together and for what reason It turns out that every bond and every lone pair represents a region of electron density For the bonds this corresponds to shared electron density between bonded atoms For lone pairs it is simply the electron density of a lled atomic orbital VSEPR theory states that these regions of high electron density will repel each other 7 simply because of Coulombic repulsion between the orbitals A slightly deeper interpretation comes from the Pauli Exclusion Principle PEP Recall that the PEP states that no two electrons can have the same set of 4 quantum numbers where those numbers are n l m l and ms Since two electrons in the same orbital can have different spin quantum numbers they can occupy the same space Imagine however they we hypothesized the following bonding environment shown in Figure 6 about carbon atom 1 from Figure 5 A Now here we have all four things C1 occupying a very small fraction of the volume about carbon 1 Since the electrons involved in these bonds are occupying nearly the same physical space they are very nearly all occupying the same orbitals and the implication is that the Fig 6 structure is more or less a violation of the PEP So in order to be consistent with the PEP and to minimize Coulom repulsions between the various regions f high L Fig 6 One wrong possibility for the bonding environment surrounding carbon 1 from Figure 5A electron density we need a different chemical environment around that central carbon atom than what is shown in Figure 6 In fact we have already seen this chemical environment in our previous Lecture Series 4 notes It is shown in Figure 7 It turns out that if you want to arrange 4 objects around a central point and keep those 4 Figure 7 The correct molecular objects as far apart from each other as possible structure surrounding carbon 1 in then this is the arrangement that you would use Fig 5 A Carbon 1 is at the center of This arrangement is called a tetrahedral the box structure and we say that carbon 1 is tetrahedrally bound to 3 hydrogens and one other carbon Note that we could have arranged the three hydrogens and carbon 2 around carbon 1 in any order As long as the bonding things Structure One Lone Pair TWO LOW P air S around A 1800 B A B 2 linear trigonal planar 10957 B tetrahedral Figure 8 Expanded structure table to account for how lone pairs affect the structure and the naming of the structures environment around carbon 1 is tetrahedral all arrangements are equivalent Now we can make some simple rules for local molecular structure These rules are based on the following Tally up the number of things surrounding the atom you are interested in Arrange those things so that they are as far apart from each other as possible This is the repulsion part of VSEPR The structures in Figure 8 satisfy this criteria It turns out that these structures are all you will ever need to know to be able to assign geometries to organic molecules Let s return to Figure 5 A All three carbons are bonded to 4 atoms and there are no lone pairs Thus the environment around each of the carbons is that of AB4 in Fig 8 or tetrahedral Figure 5B is the structure of HCCH In this molecule each carbon is bonded to one other carbon and one hydrogen There are no lone pairs Thus each carbon atom represents a BAB structure in Figure 8 and so the molecule is linear Figure 5C is the structure of CH3CHO Carbon 1 is bound to 3 hydrogens 1 other carbon and there are no lone pairs Thus its chemical environment is AB4 or tetrahedral Carbon 2 is bound to 1 carbon 1 hydrogen and an oxygen and there are no lone pairs Thus its chemical environment is AB3 and its structure is trigonal planar Finally Figure 5D is the structure of CH30H methanol The carbon is bound to 3 hydrogens and one oxygen so it is in an AB4 or a tetrahedral environment How about the oxygen It is bound to a carbon and to a hydrogen However it also has two lone pairs Be sure you can show this to yourself Thus it is in an AB4 where two B s are lone pairs or tetrahedral chemical environment but the geometrical structure is bent Note that there are two types of bent molecules AB3 where one B is a lone pair and AB4 where 2 B s are lone pairs Note also that I have denoted the bond angles for the AB3 bent structure as ltl200 and for the AB4 bent structure as lt 10950 Why is this It turns out that the lone pairs tend to occupy a little more space around the central atom than does a chemical bond What this means is the repulsion between a bonding pair of electrons and a lone pair is a little greater than between two bonding pairs The net result is that the chemical bonds are squeezed together a little more than they would be if there were no lone pairs Thus the AB4 structure that has one lone pair trigonal pyramidal would be expected to have a bond angle of about 108 while the AB4 structure with 2 lone pairs bent would have a bond angle of about 106 The AB3 structure with one lone pair bent would probably have a bond angle of about 1180 or so Here are two sample problems for you to work out yourselves These problems involve Lewis dot structures and VSEPR theory 1 For the molecule H3CCH2CHCHCOCHN2 draw the Lewis dot structure label all formal charges that you nd and for each nonhydrogen atom use VSEPR and Figure 8 to assign a local bonding geometry Estimate the bond angles 7 watch out for lone pairs 2 Repeat the various parts of problem 1 but for FObenzeneCHzCOzH Note that the benzene structure will be C5H4 instead of C5H5 with two any two it doesn t matter of the hydrogens replaced one by the FO group and the other by the CHZCOZH group 0th er types of VSEPRdetermined structures nonorganics For any ABz AB3 or AB4 molecule regardless of whether it is organic or inorganic the above structures in Fig 8 apply when trying to determine local bonding geometries For example BF3 is trigonal planar NH3 is trigonal pyramidal and water is bent However there are a few other structures that are very important for other types of inorganic compounds For 00 1 Zonepaz39r 2 Zonepaz39rs Figure 9 AB5 Structures predicted by VSEPR theory Notice that for the case of2 lone pairs the lone pairs are as far apart as possible 7 the repulsion between the lone pairs is strong The structure given by Oxtoby has both lone pairs at the equator Such a structure seems wrong to me but I have seen it before I am actually not sure which one mine or Oxtoby s is correct and so I will never ask you about molecules that are AB3 2 lone pairs AB5 systems the various structures as predicted by VSEPR theory are given in Fig 9 In the AB 5 structure with no lone pairs three B s are arranged about the equator of the molecule with trigonal symmetry Can you determine the bond angles in this molecule In the AB4 one lone pair molecule the axial B s and the equatorial B s are distorted 7 they are pushed back by the strong repulsion with the equatorial lone pair In the AB32 lone pair structure the B s are again in a trigonal alrangement about the equator and the lone pairs are arranged axially 7 as far apart from each other as possible Lecture Series 3 Chem 20A James Heath Recall that we concluded our discussion of the discoveries by Planck Einstein and Bohr by stating that the single theme that ran through those discoveries was that at very small length scales various physical phenomena were no longer continuous but rather discrete or quantized Light as well as matter also comes in discrete units 7 the photon Electrons which are discrete particles of matter orbit a nucleus in discrete orbits that are characterized by discrete angular momenta Now there is one more piece of information that we need before we can proceed to a more applicable description of quantum mechanics When we described electromagnetic radiation light we described it as both waves and as discrete packets of energy photons This waveparticle duality is not unique to light In fact it is very general Some time after the three paradoxes were confronted by Planck Bohr and Einstein other scientists began doing experiments on electrons and finding some very puzzling results It was well known from Newton s time that if light was sent through a very small slit it would diffract as shown in Fig 1 This is how diffraction gratings work and also why a rainbow is a rainbow Some scientists began doing this same experiment with electrons They would make a beam of electrons send it Fig 1 When a single color light beam passes through a Slit and 100k at a Phosphoroscem through a slit it creates a diffraction pattern a screen placed behind the Slit What they series 0f bright and dark Spots 0139 rings on the found was rather surprising They observed White Screen behind the Slit diffraction patterns just as if they had done the experiment with a beam of light This implied that electrons 7 which had previously been considered to be particles 7 also had wavelike characteristics Based on these results and the other work that was being done by Planck Louis de Broglie wrote a very brief doctoral thesis in which he postulated that all matter was characterized by a wavelength and that wavelength could be calculated from the following equation it hmv Planck s constant divided by the momentum of the particle This equation was in agreement with the observed diffraction measurements Example Say that we had a beam of electrons travelling with a kinetic energy of 10000 eV s 1 eV 16x103919J What is the wavelength associated with those electrons By the way eV s are a very convenient unit to use here If we accelerate an electron through an electric field of x volts the that electron has a kinetic energy of x eV So if kinetic energy 1x 104 eV 12 mvz then the momentum mv We know that m me 91 x 103931 kg so we can solve for v and hence momentum 104 eV 16 x 103919JeV 7 1291 X103931kgv2 v 7 5 x107 ms A 7 663x103934 J s 1x104 eV16x103919 JeV5 x107 ms391 12 x 103911 m 7 01 Angstroms Ifwe had used protons instead of electrons 7 would 3 X 103913 m 7 a quantity that is essentially not measureable As objects get more and more massive their momentum increases and the wavelengths associated with them get smaller and smaller 7 quickly becoming too small to measure We know that mass and light each have wavelike and particlelike characteristics and that electrons in atoms are quantized 7 they have discrete energies or equivalently they have discrete values of the angular momenta We have so far presented only the facts Now we try to ask why Our first question Why does an electron become quantized Quantum mechanics is not a particularly intuitive subject and so analogies to real world phenomena are a little dangerous Nevertheless here is one that works a little bit Imagine that we have a guitar string tied down at either end as if it were on a guitar If we pluck the string then we hear a note and this note is the resonance frequency of the string This is of course the basis for all stringed instruments One selects an appropriate string puts an appropriate tension on that string and stretches it across some length These things we do to the string will give it a desired resonant F39 2 A 39t t 39 d 39t lgure gm H S nng an 1 S frequency 7 say an A note Now 1t turns out that when resonant frequencies Notice that all frequencies are integral you pluck the string you not only hear the A note but if multiples of the fundamental you listen very closely you will hear an A note at the frequency Shown at the bottom next octave up and even the 2d octave up as well and so 0f the drawmg39 on and so forth These are harmonics and their frequencies are integral multiples of the first primary note or the fundamental frequency This is shown graphically in Fig 2 In the language of physics let s restate the various phenomena that control the resonance frequency of the string 0 First we have the mass of the string 7 larger mass translates to lower frequencies Second we have the tension on the string When we stretch a string we are actually using it as a spring 7 ie if we let loose of both ends it will retract back to its initial length When we pull or compress a string or a spring this is the same as increasing its potential energy When we compress or expand a spring we store energy into the spring in the form of potential energy Thus the amount of potential energy stored in the string is important More potential energy tighter string corresponds to higher p I 0 Third we have the length of the string that is resonating Shorter lengths lead to higher frequency resonances This is why frets are on a guitar Note that the Vibrating string is a one dimensional system 7 ie you can represent the string itself as an xaxis So if we are going to make an analogy between a Vibrating string and a quantum mechanical entity we need to choose a ldimensional quantum mechanical system There is a theoretical model for such a system known as the particleina box and it represents one of the most famous problems in quantum mechanics Although this model teaches us many important things about what to expect from electrons in atoms it is just a model 7 it does not represent any real experimental system We note that this system actually more resembles a particle on a string rather than a particle in a box We are not going to solve the particle in a box problem exactly in these notes although it is possible to do so yonr book does it seep 499500 In our attempts to make an analogy to a string system we are going to neglect some things and include others that we shonldn t Nevertheless onr incorrect approach to this problem actually comes up with the correct solution The particleinabox is represented by the drawing in Fig 3 The particle in our case is an electron We keep it in the box by putting in nitely high energy barriers at These circles represent energy barriers that keep the e39 con ned to the region of the line between the circles w electron exists in this region Fig 3 Particle in a ldimensional box The potential energy has a value of 0 between the green dots either end of the box If we think of the electron as a particle then we imagine a particle scooting back and forth between the two barriers However if we think of the electron as a wave then we are back to Fig 2 7 our guitar string In fact electron waves are going to appear more or less just like the resonant frequencies of the guitar string However there are differences Recall that 7 hmV the de Broglie relationship Let s look again at our guitar string resonance and its overtones redrawn in Fig 4 Notice that the overtones are characterized by shorter wavelengths than is the primary resonance x 203 ig 4 Same as Fig 2 but we have noted the relationship between the various harmonics and the fundamental frequency frequency the fundamental frequency We didn t think too much about it when we were discussing guitar strings but here it becomes important According to the deBroglie equation if the wavelength is made shorter then the momentum of the particle and hence its kinetic energy must increase Thus we have an electron trapped in a 1 dimensional box and it has a resonance frequency that is determined by the size of the box However the various overtones of the resonance frequency are at successively higher electron kinetic energies Furthermore this energy separation is necessarily discrete not continuous 7 simply because the overtones are integral fractions of the fundamental wavelength Let s work this out a little bit If we have a box of length L then according to Figs 2 and 4 the fundamental frequency has a wavelength of 2L Notice how only 12 of the fundamental wavelength ts into the box 2L hmv so mv h2L and so V h2Lm hth Energy kinetic energy potential energy However if we keep our particle within the boundaries of the box then the potential energy 0 so the total Energy the kinetic energy 12 mvz So if E 12 mv2 the energy of the fundamental is given by E 12 m h24L2m2 12 hz4L2m h28mL2 Let s go to the 1st overtone 7t L hmv so v hmL E 1 W 12 mvz 12 m hzmsz hZszz 4h28mL2 And for the 2d overtone 7t 23 L hmv so v 32 hmL and E2d W 12 mvz 12 m 9h24m2L2 9h28mL2 So we have a series of states with En n2h28mLz where the fundamental frequency has n l the 1st overtone has n 2 the 2d overtone has n 3 etc Here n is the quantum number that denotes a quantum state of the system En is the energy of the n3911 state There is a function that describes the 3 waves shown in Fig 4 and all the other ones that aren t shown and that function is called the wave function of the system Each of the separate waves is a quantum state of the system and each state is characterized by a quantum number The wave function is often denoted by the symbol 1 or psi Notice that the quantum number determines the shape of the wave function If we had really set up a differential equation and solved this problem as your book does we would have also found that En h2n28mL2 It turns out that the reason our approach works is because this is an unrealistic but very simple system the potential energy contribution to the total energy of the system was negligible Thus other than using the potential energy to de ne how big our box was we neglected it when we calculated the energy of the system Usually we can t do this E3 So let s go back to the factors that affected the resonance frequency of the guitar string mass of string potential energy length of string and ask how those factors affect the E E2 properties of our quantum mechanical waves a First however we need to adopt a slightly different representation for our waves Since we E1 know that shorter wavelengths correspond to higher kinetic energies we can represent the fundamental frequency and it harmonics as length shown in Fig 5 Fig 5 Now we redraw our previous figure that described the resonances of a String to represent the resonances of a o F1rst we had the mass of the string 7 larger confined electron The dots on the mass translated to lower resonance length3X15 31396th thS 0f Fig 32N0te frequencies Here we have 7 fixed by L the the enemv Snacmgs Increase as n 39 length of the box to which the electron is con ned However we also still have the relationship 7 hmv Thus if 7 is xed and we increase mass then the velocity of the particle decreases Recall once again our favorite equation for energy E 12 mvz If the product mv stays the same since the wavelength doesn t change yet the mass is increased then velocity must proportionately decrease The energy only varies linearly with mass but it varies as the square of the velocity Thus a small increase in mass plus a small change decrease in velocity translates into decrease in kinetic energy Increasing mass reduces the enerng of the wave Second we had the tension on the string More potential energy tighter string cm A J to higher f 39 Our situation is similar In the particleina box the potential energy is what keeps the particle restricted to a small part of the axis If we remove the potential energy barriers remove the dots in Fig 3 then the electron is no longer con ned to a particular part of the axis and it can have all ranges of wavelengths associated with it including very long ones 0 Third we had the length of the string that was resonating Shorter lengths led to higher frequency resonances In fact if we reduce the width of the boxI we shorten the wavelength 7 just like in the case of a guitar But we also increase the energy of the electron If 7 is reduced then momentum mv is increased and hence kinetic energy of the electron is increased One Last Background Point The Heisenberg Uncertainty Principle There is one more thing we need to cover before we get to a real quantum mechanical system the atom and it is called the Heisenberg Uncertainty Principle This principle is fundamental to quantum mechanics 7 it is not strictly provable but like the rest of quantum mechanics it has withstood the test of time admirably In any measurement that we can imagine doing we will always have a little bit of uncertainty in our answer 7 that uncertainty is governed by our instrumentation or by the nature of the thing we are trying to measure Let s say that we are trying to measure the position of an electron The only way we can do this is to scatter something off of the electron 7 like a photon When we do this we may get a very accurate measurement of where the electron was at the time of the scattering event but we have also changed the system Imagine for example that the electron is travelling along with some velocity V in some direction giving it a momentum p mass x velocity If we scatter a photon off of the electron in order to measure its position then we change its momentum We don t really know how we change it but we change it So now if we try to measure its momentum which we can do it is not the same momentum that it had when we measured its position 7 confusing huh What Heisenberg showed is that if we try to measure both the position and the momentum of an electron then the precision of our position measurement times the precision of our momentum measurement must be greater than h4TE We can write this as Apr Z h4TE where the A is read as the uncertainty in Let s do an example Let s say that we have done a measurement that shows that an electron is located right within the nucleus of an atom Assume the size ofa nucleus is given by dn103913m then how well do we know the kinetic energy of the electron Answer Since we know that the electron is within the nucleus then we know its position to a Ax of 103913 m Apr Z h4 r 663x103934 J s 47 so Ap Z 52x103922 kg m s39l So if Amv Z 52x103922 and we know me9lx103931 kg we can calculate the uncertainty in the velocity to be AV 2 58x108 ms So at best we can only pin down the velocity to somewhere between 0 and 58x10 ms Obviously if our velocity is 0 then the kinetic energy of the electron is 0 However if the velocity is 58x108 ms then our kinetic energy is 15x103913J For 1 mole of electrons this corresponds to an uncertainty of 9x1010 Jmol This is a huge uncertainty in the energy If instead we had stated that we knew the location of the electron to within the dimensions twice the Bohr radius lO39lOm we would have an uncertainty in the momentum and hence the kinetic energy that is much more reasonable Show this to yourself This is why the electron doesn t collapse into the positively charged nucleus as would be predicted by the Coulombic interaction potential energy curve If it collapsed into such a small volume its kinetic energy would possibly be tremendous T has the size of an atom is dictated more or less by the Heisenberg uncertainty principle Real Systems The Hydrogen Atom When we solved the problem of the particle con ned to a 1 dimensional box we were able to generate a single quantum number n for that particle If we had solved for a particle in a 3dimensional box we would have found that the solution to the problem involved 3 different quantum numbers 7 one describing how the particle behaved in the xdirection one for Z the ydirection and one for the z direction This has to do with the degrees of freedom of the particle If the particle is confined to a 1 dimensional box it has 1 degree of freedom and l quantum number In a 3dimensional box it has 3 degrees of freedom and 3 quantum numbers With this in mind what do we expect from an atom In an atom the electron exists within a particular volume surrounding the Figure 6 The three degrees of freedom that correspond to Dueleus Because the electron has an electron moving about a nucleus Here we have shown wave1ike properties we describe an xyz coordinate system for reference However we can the Space occupied by the electron also descrlbe the pos1tlon of the electron by us1ng the as the electronic wave function distance r and the angles theta e and phi 4 to describe Where the wow points Because th1s space 1s 3d1mens1onal it is a volume surrounding the positive nucleus we know that there will be 3 quantum numbers describing the electron 7 one quantum number corresponding to each degree of freedom Thus we already see one of the problems with Bohr s atom 7 it only has one quantum number So if we need one quantum number per degree of freedom then what are those degrees of freedom If we had a particle in a 3dimensional box we mightjust call them X y and z and get quantum numbers nx ny and n2 However that is not what we have Instead we have a particle the electron held within some region of space by a centrosymmetric Coulombic potential the positive nucleus One obvious degree of freedom that we have is a measurement of how far out from the nucleus is the electron 7 ie what is its radial distance The quantum number corresponding to this is the radial quantum number If we decide on this as one of our quantum numbers then what are the other two Look at Figure 6 Here we show that we can describe the position of the electron by r its distance from the origin and by the angles theta 9 and phi 1 These are related to the xyz coordinate system because 9 is the angle of r within the xy plane and I is the angle that the vector r makes with the with respect to the zaXis Let s not worry about this too much other than that we need one degree of freedom to describe the distance of the electron from the origin and two degrees of freedom that describe angles Thus we have one radial quantum number and two angular quantum numbers The radial quantum number will dictate how the wavefunction spreads out from the nucleus and the two angular quantum numbers will determine the shape of the wavefunction Quantum numbers of electronic orbitals The wave function of an electron is usually denoted by 1 For the case of the electron orbiting around the nucleus as is shown in Fig 6 1 is a function of r 9 and I The energy levels of the electron are described by the three quantum numbers n the principal or radial quantum number I or the angular momentum quantum number and m which is called the magnetic quantum number We ll return to these quantum numbers in a minute We write 1 as 1 n1mr9 and we find that we can separate 1 into a radial component Rn1r and x1m9 so that we can write LIJnlmraeal Rnlr We also find that there are certain relationships between the quantum numbers and we ll describe those in a minute I just list this information so that you see that there are three quantum numbers and three degrees of freedom What do those quantum numbers mean 7 We don t know yet If we had set up and solved an equation that described Fig 6 which is a hydrogenlike atom we would have been able to derive all of the stuff that is being presented here This is a very difficult thing to do and is usually not attempted until the last year of an undergraduate chemistry program or the first year of a graduate chemistry program Nevertheless what we would have found would have been just a more complicated version of the quantum guitar string problem we solved earlier Just like the guitar string the energy levels would have been described by an energy E where n is the principal quantum number En Z2e4me8802n2h2391 Let s de ne the constants in this energy expression Z is the nuclear charge For H atom it is simply I for Liz it is 3 and for He it is 2 Why is this H has one proton so one positive charge Li2 has 3 protons so a positive nuclear charge of 3 and He has 2 protons We don t worry about neutrons since they are charge neutral Also all of these atoms are hydrogenlike 7 meaning that there is just one electron orbiting about a positive charge Even though the magnitude of the positive charge differs the physical solutions work out to be the same they are just sealed for the different charges You should understand this concept Going back to the equation for EH 7 we have e4 which is the fundamental unit of charge to the 43911 power we have me which is the mass of the electron In the divisor we have 802 the constant that corresponds to the permittivity of a vacuum squared We have hz which is Planck s constant squared and we have n which is the primary or radial quantum number squared Thus just like our quantum guitar string the energy of a given quantum state is related to only one quantum number n However unlike the solution to the quantum guitar string the energy depends inversely on n scaling as n39z This means that if we plot the energy levels of this system as a function of n they will look something like what is shown in Fig 7 The energy levels that describe our hydrogenlike system do not depend upon the other two quantum numbers I and m However a given En describes the energy of one or more quantum states of the system Those quantum states are more or less electronic orbitals and they have a shape It is the shape of those wavefunctions which is related to the orbital angular momentum that is determined by the quantum numbers I and m We could have solved for the shapes and angular momentum of these various states and we would have found that n l and m have certain relationships with one another 1 0 12 11 m 1 41 42 0 1 2 1 1 So for example if n 3 what are the possible combinations of other quantum numbers that we could have What are the allowed quantum numbers if we have n l n 4 You should be able to figure this out yourself When we want to describe a particular orbital there are some nomenclature rules First we consider the principal quantum number n Then we consider the angular momentum quantum number I Ifl 0 it is an sorbital Ifl 1 it is a porbital and ifl 2 it is a dorbital Other less commonly encountered orbitals include forbitals 13 and g orbitals 14 Returning to the case of n 3 we have 10lor 2 These combinations yield 3s 3p and 3d orbitals So how about m It turns out the number of values for m describe the number of 3s 3p or 3d orbitals For example for a 3s orbital m can only be 0 Thus there is only one combination n3 10 and m0 For n3 and 11 m can take on the values of 71 0 or 1 Thus we have 3 3p orbitals each with a different m quantum number For n3 and 12 m has the values of72 l 0 l 2 ithus there are 5 3d orbitals In Fig 7 we plot the various orbitals 114 113 112 Is 111 as a function of energy Recall from the energy expression for hydrogen like atoms that the energy depends only on the principal quantum number n and that energy scales as n We have hinted that the angular quantum numbers describe the shape of an orbital and that the principal or radial quantum number describes the radial extension of the orbital Let s look a little more at what we mean by this First we consider the s orbitals For all sorbitals both I and m are 0 What this means is that the I I I 1 2 contains all of the m values for that orbital 0 AngllarMnI 39lnann lnlnNanerd Fig 7 The various orbitals are plotted as a function of energy Note that since E goes as n2 the levels get increasingly compressed in energy as n increases At the right of each orbital notation is a shaded box that 393 orbital does not really have much angular shape 7 it is simply a sphere radiating outward from the positive nuclear core But how about n This quantum number affects the radial part of the wavefunction the 39d Ener length Figure 8 The quantum guitar string but only the first two energy states are drawn The point where the wave function passes through the dashed lines is called a node function Rn1r at the top of this section As we move from n0 to nl to n2 etc the principle quantum number is changing so the radial part of the wavefunction must change In fact this change is very similar to what we observed for our quantum guitar string Recall that as we increased n we moved from the fundamental frequency to higher harmonics Let s look at that graph one more time and discuss terms In Fig 8 we have redrawn the lowest two states of the quantum guitar string with a couple of additions We have drawn dashed lines through the strings indicating the nodes in the function When a wave function passes through the dashed line it changes sign We call the point at which it crosses a node Consider a sin function sin0 0 sinlo gt 0 and sin 1 lt 0 Thus when sinangle passes through 0 it passes through a node Now return back to one of the earlier figures of these waves 7 one that shows the first three states Notice how the first state has 0 nodes the second state has 1 node and the 3rd state has 2 T15 o gt through 0 YZS node 0 r FSS nodes r Fig 9 The radial wavefunctions of ls 2s and 3s orbitals Note how the number of nodes increases from 1 to 2 to 3 As the wavefunction passes through a node its sign changes just like a sin function passing nodes It turns out that the n3911 state would have n 1 nodes The radial wavefunction that describes the electronic orbits is very similar For n1 it has 0 nodes For n2 it has 1 node For n 3 it has 2 nodes and so on and so forth However unlike the quantum guitar string the radial function it not symmetrically bound by in nite potential barriers so it looks different The n1 2 3 radial functions for l 0 are shown in Fig 9 If one is given a radial wavefunction then just count the number 0fn0des add 1 and you have the principle quantum number Something that is implied by this statement is that a wavefunction with more nodes is a higher energy wavefunction Since these wavefunctions are spherically symmetric the situation for l and m both 0 they look like concentric rings of opposite sign radiating out from the positive nuclear core In your text you will nd graphs of the 1s 2s and 3s orbitals shown to highlight these nodes So how about if I 1 or something other than 0 This implies that we have shape to our wavefunction 7 ie its structure is a little more interesting than just a spherical orbital Notice that the radial part of the l K Fig 10 2p orbitals wavefunction the equation given early in this section is given as Rn1r so it depends on both n and 1 For n 2 we know that Rn1r will have 1 node If I 1 then it turns out that the node is at the nucleus not some distance out from the nucleus such as is shown for the 2s orbital in Figure 9 For n2 11 we have porbitals and we have the possible values for m of 71 0 or 1 It turns out that the fact that l 1 means that the porbitals are going to have the shape of two stacked eggs aligned along some axis These orbitals are shown in Figure 10 The three values of m de ne which axis the porbital is aligned along and we will arbitrarily say that for m 1 the porbital is aligned along 2 If m 0 it is aligned along x and ifm 1 it is aligned along y Since the wavefunction of a 2p orbital changes sign as it passes through the nucleus one ofthe two eggs will have one sign ie is positive while the other egg has another sign ie is negative For example in the 2pZ orbital at the top of Fig 10 ifthe top egg is positively signed then the lower egg is negative Just like the radial functions drawn in Fig 9 if the wavefunction passes through a node then it always changes sign For n3 and 12 we again have dorbitals 7 the 3d orbitals We aren t going to worry about these too much in this class Nevertheless there are 5 3dorbitals and they each have two nodes Pictures of them are in your text Filling the Orbitals Let s list the various orbitals for n l 2 and 3 in a slightly different way but once again with energy as the yaxis Fig 11 This is similar to Fig 7 except that n we have replaced the individual labels of 35 3p 3d the m quantum number with just a series of empty spaces underline marks Each empty space stands for an unoccupied n ZS 2p orbital and so Fig 11 represents a system a with no electrons So how do we fill these s orbitals up with electrons There are L several rules Let s state two now I fill lowest energy orbitals first 2 each orbital can hold two electrons So let s start filling For the first two 39 39 39 electrons its easy We fill in the following way and then We call these two con gurations ls1 and lsz respect vely The superscript at the right of these notations stands for the number of electrons in the orbital Note that when we fill these orbitals we are denoting the individual electrons with a little arrow either pointing up or down We ll get back to that in a minute So far we have built the H atom and the He atom Now we move down to the next atom which is Li We are faced with a choice Apparently all of the 2s and 2p orbitals are degenerate meaning that they are at the same energy level Thus where do we put the electrons It turns out that all of the orbitals of same principle quantum number are degenerate only for hydrogen and hydrogenlike atoms Once we begin putting electrons into the orbitals we find that the degeneracy is lifted and that the orbitals are no longer isoenergetic What this means is that if we put an electron into the 2s orbital then its energy is lowered relative to the 2p orbital In other words if we put an electron into the 2s orbital we gain a little bit of energy especially when compared to putting that same electron into a 2p orbital Thus the 2s orbital gets filled first and then the 2p orbital So now we have the 3rd rule 3 In general when we fill orbitals that are characterized by the same principle quantum number n we fill the lowest l value orbital first all the way up to the highest 1 ivalue For example we first fill the 3s orbital then the 3p orbital then the 3d orbital It is not obvious why this should be so iespecially from what you have learned so far Yet it is so and because of this we can explain many of the periodic properties of the elements So let s continue First we can redraw our energy diagram to re ect our new ordering Let s only do it for n l and 2 levels 7 see Fig 12 With our new ordering we can not only make Li but we can make Be and B atoms as well We show 21 the Boron atom electronic structure in Fig 12 It is labeled as ls2 2s2 2p1 Now note something about this arrangement We stated that this is the lowest 2 energy way to arrange the electrons We call the H n lowest energy con guration the ground electronic state However what if we had put one electron into a 2s orbital and two into the 2p orbitals the ls I 2s1 2p2 con guration Or what if we moved an 1 electron from the ls orbital and put it into a 2p orbital a ls1 2s2 2p2 con guration Certainly this isn t against the law 7 it just costs more energy Such con gurations that are higher energy con gurations than the ground state are called excited electronic states Note that there are many excited electronic states Is there more than one ground state For some atoms no For example for the Be atom ls2 2s2 or 1 less electron that the B atom shown in Fig 12 there is only 1 way to con gure the ground electronic state However for the B atom we could have put the 2p electron in any one of 3 orbitals Thus the ground state of B is 3fold degenerate Recall the de nition of degenerate that was given above Now we come to yet another dilemma if we want to add one more electron to make a carbon atom Where do we put the electron In an unoccupied porbital or into the same porbital that already has 1 electron in it To answer this we need to de ne the little arrows that we have used to denote electrons What do those symbols mean It turns out that we have actually one more quantum number that we have neglected 7 the spin quantum number Fortunately it is a simple one 7 it can have values 12 or 7 12 We denote the 12 value as spin up up arrow and the 7 12 value as spin down Wolfgang Pauli stated the Pauli Exclusion Principle which is sort of like rule 2 listed above It states No electron in an atom will have the same quantum numbers as any other electron Why is this like rule 2 Well as a consequence of the Exclusion Principle each orbital can only hold two electrons 7 one with spin up and one with spin down If it held any more then we would have a violation of the Pauli exclusion principle It turns out that spin is a real physical quantity but all we need to know is that two electrons can exist in the same orbital only if they have opposite spin So how does the spin quantum number guide us when we put together a carbon atom This gives us rule 4 2 n1 lz Fig 12 The filled orbitals of the ground state of a B atom 4 When lling orbitals that have the same n and l quantum numbers but different m rst put one electron into each orbital keeping all electron spins the same Once all the orbitals are lled with one electron each with the same spin then nish lling the orbitals by pairing those electrons CHEM 20A winter 2001 LECTURE NOTES eighth set HYDROGEN LIKE ATOMS We turn next to the study of the hydrogen atom the simplest of atoms because it contains only a single electron Actually we shall consider hydrogenlike atoms by which we mean atoms and ions that contain only a single electron Thus He Li2 and Be3 in addition to H are hydrogenlike Explain this In many ways we can envisage the hydrogenlike atoms as examples of a particle electron in a box but a rather more complex box than the one discussed in the last section Potential energy First we examine the potential energy 1 Ze2 Ur F807 1 where Ze is the charge on the nucleus r is the distance between the nucleus and the electron and so is a constant already introduced Problem Derive Eq 1 Plot Ur versus r for Z l and Z 2 Note that Ze2 411780 is a constant the electron position is not xed so that r is a variable Total energy We now present a partial derivation of the expression for the energy ie for the energy levels of the hydrogenlike atom This is a derivation built upon the fact that the energy E is conserved i e a constant and upon the principles of dimensional analysis Dimensional analysis is based upon the concept that both sides of an equation must have the same dimensions The energy is the sum of the kinetic and potential energy The kinetic energy depends upon the mass and the velocity the mass is a constant and the velocity a variable The potential energy depends upon the constant Ze241t80 and the variable r Since the expression for E is quantum mechanical it must contain the constant h Thus the constant energy E must depend upon the constants Ze24nso h and the mass of the electron me Let us write the most general expression we can for E in terms of these constants C Z6241tso hlimeY 2 where at 5 y are dimensionless unknown exponents The factors in brackets all have dimensions whereas C is just a dimensionless number The reason that we use a minus sign is because we are interested in the energy for a bound particle which in this case must be negative we discuss this below With M being mass L being length and t being time the energy E has dimensions E gtML2t392 Mass me has dimensions me gt M and h has dimensions h gtML2t391 Prove this To determine the dimensions of Ze24nso look back at Eq 1 note that the dimensions of r are L and that the dimensions of U are MLZt39Z Thus the dimensions of Ze24nso are Ze24n80 gt ML3t392 Prove this We can then reformulate Eq 2 as MLZt39Z CML3t3920 ML2t1l3MY The power of each dimension must be the same on both sides of this equation so M1 MocBv 1 0c y L2 L3002l3 2 30c2 t 2 t JoeB 2 2x We have three equations so we can solve for the three unknowns 0c 2 l 2 y 1 Thus Eq 2 becomes E C Ze24nso2h392me1 3a which can be rewritten as Z2e4me E 2 3b 161T280h2 This is a rather complicated expression which we have derived without even knowing the Schroedinger equation in our derivation we have merely used dimensional analysis the fact that h always enters into quantum mechanical expressions and the fact that the energy is a constant But dimensional analysis cannot be used to determine dimensionless quantities such as the constant C in Eq 3b for that we need a more detailed solution of the Schroedinger equation Here we merely state the result C 2121 1392 where the quantum numbern l234 Thus Z2e4me L n 7 883112 n2 n 1234 4 Often one writes this in the simpler form En ZZRHLZ n 1234 5 n where the constant RH is known as aRydberg and RH Mme883 h2 6 Energy levels Next we examine Eq 4 We note that as with the particle in a box the energies are quantized As with the particle in a box the quantum numbers n are positive integers but whereas the energies for the particle in a box vary as n2 for the hydrogen atom they vary as n392 This indicates that for the particle in a box the energy levels get further apart the higher the energies whereas for the hydrogenlike atoms they get closer together One turns again to graphs for greater insight Plot the potential energy Ur in Eq 1 versus the distance r between the electron and the nucleus Note that this potential energy is always negative but of greater importance is the fact that the associated forces are negative 139e attractive binding the electron to the nuclear atom Be sure you understand this The potential energy looks much like a potential energy box or more descriptively a potential well On the same graph indicate the total energies En as given by Eq 4 these are horizontal lines on the negative energy axis which intersect the potential energy curve which is a hyperbola Indicate the n values for each of the energy levels These horizontal energy lines constitute an energy level diagram much like those encountered for the particle in a box but with different trends in the energy gaps as indicated above The total energy En is a constant i e it has the same value regardless of the position r of the electron relative to the nucleus at the same time En is the sum of the kinetic and potential energies Classically the kinetic energy cannot be negative because it equals mu22 which means that En cannot be less than Ur i e that the particle must remain bound within the potential well as long as En is negative If however the energy E is positive the particle can roam as far from the nucleus as it wants because E is always greater than the potential energy and the particle will always have kinetic energy Summarizing some of the results of the last paragraph we see that for the bound states of the hydrogenlike atom the energy En is negative quantized and given by Eq 5 For the unbound states ie those for the ionized electron the energy E is positive and unquantized continuous These are quite general features as already discussed in this course For the hydrogenlike atoms but not for all systems there are an in nite number of quantized states the gaps between the levels getting so close for large n that even the very excited bound states form a virtual continuum Check this out More about energy levels The n 1 level corresponds to the ground state energy When the particle is in the potential well this energy is greater than the potential energy and consequently the electron has kinetic energy even in the ground state Thus the electron always has kinetic energy ie it is always in motion even in the ground state As indicated in the last set of notes this concept of zero point energy is strictly quantum mechanical The energy needed to ionize an electron that starts in the ground state is known as the ionization energy Illustrate it on the graph constructed above Thus the ionization energy of a hydrogenlike atom is I z2RH 7 m Obtain this expression for the ionization energy Compare the ionization energies of H He Li2 and Be3 Can you give a qualitative explanation of these trends Problem If a photon of frequency v is absorbed by a hydrogenlike atom and if the frequency is more than enough to ionize the atom show that the kinetic energy of the ionized electron is given by KE hv 1 Compare this with the photoelectric effect Problem Depict the absorption spectrum from the ground state from the first excited state from the second excited state How will the spectrum of H differ from that of He Degeneracies amp spin The energy levels for the hydrogenlike atoms are highly degenerate For a particle in a hyperbolic well such as that depicted above for hydrogen like atoms the n 1 ground level is nondegenerate as expected but the n 2 first excited level is 4fold degenerate the n 3 second excited level is 9fold degenerate and the n 4 third excited level is 16fold degenerate A 4fold degeneracy means that there are four distinct wave functions which yield the same energy These results are obtained by solving the Schroedinger equation something we shall not do here Problem Indicate the degeneracies on the energylevel diagram Electrons are special so special that the degeneracies indicated above are not correct and should in fact be doubled Electrons have a very special property called spin Whereas the wave functions we have discussed depend upon the position of the electrons they have not had any dependence upon spin However they should depend upon spin The electron spin is itself quantized and can take on two values corresponding to two distinct states These values are usually depicted by a spin quantum number ms which can take on the values 12 or l2 One often talks ofthe n1S 12 state as the quotspin upquot and the ms l2 state as the quotspin downquot state The spin of the electron does not affect the energy Since each of the quantum states described above can have quotspin upquot or quotspin downquot the degeneracies listed above should be doubled Problem Explain this and explain why the actual degeneracies for the hydrogen like atoms are 2 for the n 1 level 8 for the n 2 level 18 for the n 3 level What is this spin Why is it important How can one decide whether it is important Why does one care about degeneracies whether with or without spin These are all questions we shall address Magnetism and Spin Electric elds forces can act on electrical charges either pulling them or pushing them in one direction or another Electrical forces have no effect upon spins On the other hand magnetic fields forces can have a direct effect upon spins If a hydrogen atom in its ground state n l is subjected to a magnetic field its energy level splits into a doublet ie two separated energy levels the doublet separation on the energy level diagram is proportional to the strength of the magnetic eld the stronger the magnetic field the greater the energy splitting This indicates that the ground state energy level of the hydrogen atom in the absence of the magnetic eld is at least 2fold degenerate and that the magnetic eld removed the degeneracy Problem Show this on an energy level diagram The lower one of the ground state doublets is the n1S 12 state and the upper one the n1S l2 state One way to detect the doublet splitting of the ground state in the presence of a magnetic eld is to study the absorption spectrum an absorption line should be observed at very low frequency corresponding to the transition from the n1S 12 to the mS l2 state Furthermore the absorption frequency should increase with increasing magnetic eld Problem Why is the frequency low Why is the frequency proportional to the strength of the magnetic eld Draw a graph of energy versus strength of magnetic eld This kind of spectroscopy is known as magnetic resonance spectroscopy in particular electronspin magnetic resonance called esr Problem Look back at the particle in a ldimensional box Suppose the particle is an electron with spin What are the degeneracies of the levels in the absence of magnetic elds Draw the energy level diagrams for the electron in a box subjected to a weak magnetic eld The rst six or so levels will do Now do the same for an electron in a 2dimensional square box If you can39t do this problem go back and review the material in both this and the last set of notes The Hatom is a 3dimensional problem so it can be considered as an electron in a 3dimensional potential well The potential energy in Eq 1 is a function only of the distance r between the nucleus and the electron and does not depend upon direction i e upon angles that speci direction The properties of a sphere are the same in all directions and depend only upon the distance from the center ie upon the radius r Thus any property that depends only on the radial distance r is said to have spherical symmetry or to be spherically symmetric Quite generally atoms are spherical whereas molecules are not Problem Explain this Thus although when plotting the potential Ur versus r one might envisage a ldimensional potential well it is actually 3dimensional For the particle in a box we found that in 2dimensions we could nd degeneracies ie energy levels corresponding to more than one state Even more degeneracies are possible in 3dimensions Thus we are not surprised to nd high degeneracies for the electron in hydrogenlike atoms as indicated above Problem Look back at how we handled the problem of the particle in a 2 dimensional square box and obtain an expression for the energy levels of a particle in a 3dimensional cubic box It has 3 quantum numbers Draw an energylevel diagram for the rst six or so levels and indicate the degeneracies of each of these levels The quotmotionquot of the electron about the nucleus can be envisaged as radial motion in and out along r and angular motion with changing directional angles The kinetic energy can be separated into radial kinetic energy and angular kinetic energy the total energy involves both of these It is thus possible for an electron in the atom to have a given energy En speci ed by the total quantum number n but to have different angular and radial energies This is how degeneracies build up i e there can be two states with the same total energy but with different angular and radial components Remember that for the particle in a 2dimensional square box the degeneracies occurred because a particle could have a given energy but could have most of it either in the x or in the y direction To specify the distinctiveness of degenerate states we need more than one quantum number For the hydrogenlike atom one has the total quantum number n which as we have seen speci es the total energy We have already mentioned and discussed the spin quantum number ms There are also two additional quantum numbers land m1 that are used to specify the angular motion we discuss these below Note that n does not speci merely the radial energy but the total energy The wave functions for the hydrogenlike atoms are therefore denoted as Tn 1mth although the energy in the absence of a magnetic field is dependent only upon the quantum number n as depicted in Eqs 4 and 5 e wave functions I mlmlms for the single electron in a hydrogenlike atom is known as a quotspin orbitalquot The wave function 13me for which the spin has been neglected is known as an quotorbitalquot The quantum numbers for the orbitals and spin orbitals take on the following values The total quantum number n as we indicated above takes on the values 8a The quottotal angular momentum quantum numberquot 1 takes on the values 10123n1 8b Angular momentum is a vector orbital motion can take place about any axis and the motion about some arbitrary zaxis is indicated by the quantum number m1 where m1 0ili2i3 lil 8c And as indicated above the spin quantum number mS has the values mS i 12 8d In this course we merely state that these are the allowed values for the quantum numbers but it can be shown by solving the Schroedimger equation for the hydrogen atom that these results are indeed correct Problem Above we indicated that the n 1 level was 2fold degenerate the n 2 level was 8 fold degenerate the n 3 level was 18 fold degenerate Prove that this follows from Eqs 4 and 5 and from the allowed values of the quantum numbers listed above The wave functions or orbitals are often labeled somewhat differently than indicated above So orbitals with l 0 are called s functions the value of n precedes the s and the functions are denoted nsfunctions i e ls 2s 3s etc The wave functions with l l are called p functions more specifically np functions ie 2p 3p 4p etc The wave functions with l 2 are called d functions more specifically 3d 4d etc And finally one has f functions with l 3 ie 4f 5f etc Problem Why are there no 1p 2d 3f functions When the quantum number 1 0 there is no angular energy i e all the energy is radial in and out Thus in the ground state with n l and l 0 all the energy is radial The fact that there is a ground state is interesting classically there is no ground state and the electron collapses into the nucleus thereby destroying atomic structure and nearly everything else in the universe So the existence of the ground state is an important feature of quantum mechanics Note that even in the ground state the system has kinetic energy ie it has zero point energy much like that of the particle in a box This means that even in the ground state the electron moves in and out The fact that 1 can assume the value zero indicates that the angular kinetic energy can be zero the fact that the largest value that 1 can assume is nl indicates that the radial kinetic energy can never be zero ls functions The ground state functions are ls functions n l l 0 m1 0 In this discussion we neglect the dependence on spin leaving that for later In solving the Schroedinger equation it is found that these functions are spherically symmetric ie LP100r or I lsr As indicated above if it depends only upon r it is spherically symmetric It then follows that the probability density l I 100rl2 is spherically symmetric ie that there is equal probability of finding the electron in all directions However there is not equal probability of finding the electron at all distances from the nucleus In fact the wave function has the form Nexpra0 where quotexpquot indicates an exponential and where N and a0 are constants This function is a maximum at the nucleus r 0 and drops off as the distance r from the nucleus increases The same is then true of the probability density l I l srl 2 N2exp 2rZa0 9 The constant N is the normalization constant which is adjusted so that the total probability of nding the electron somewhere in space is one Go back to our earlier discussions if you no longer remember this The constant aO must have the dimensions of a length why it is called the Bohr radius and is about 0053 nm aoZ is a measure ofthe size of the atom Problem Plot l I 15rl 2 versus r Why is aoZ a measure of the atomic size The atomic size aJZ is clearly smaller for HeJr than for H Explain physically why this is so and why it is so based upon Eq 9 Problem It has been shown that aO 80h211mee2 10 Show by dimensional analysis that any characteristic length for the hydrogenlike atom must be proportional to SOhZmeZeZ Think about the concept of a quotcharacteristic lengthquot The probability density for the hydrogenlike atom is the same as the electron distribution ie the distribution of the electron over space about the nucleus This concept together with Eq 9 illustrate some important points the electron is not a point but an quotelectron cloudquot distributed in space A characteristic feature of all sfunctions i e for all values of n is that s functions have nite probability of being at the nucleus in contrast as we shall see all other orbitals pdf have nodes at the nucleus i e zero probability density at the nucleus The probability density given in Eq 9 is the probability density per unit volume Sometimes we are interested in a somewhat different quantity the probability that the electron will be found in a shell between r and r5r where Sr is a small increment in the radius It can readily be shown that this probability is 411er Plsrl Sr 11 Problem Plot l I 15rl2 and 411r2l 1 1srl2 versus r and interpret these quantities in terms of probabilities Note that l P15rl 2 has a maximum at r 0 while 411r2l 1 1srl2 vanishes at r 0 and has a maximum at r aOZ Make sense of all this Look at graphs in Munowitz ns function By solving the Schroedinger equation for hydrogenlike atoms one can obtain the functional forms of all the ns functions Although we shall not solve the Schroedinger equation we shall examine the resulting wave functions All the ns functions re spherically symmetric i e they depend only upon the radial distance r All the ns functions have finite probability density at the nucleus The ns functions have nl nodes See the diagrams in Munowitz It then follows that all the probability densities have nl nodes these nodes are spherical surfaces ie nodal surfaces Problem Be sure you understand these nodes These spherical nodal surfaces indicate that the electron can be on either side but not on the nodal surfaces Recall that the nodes are the consequence of destructive interference of waves We have already encountered such nodal effects for the particle in a box up nd and nf functions These functions too can be obtained by solving the Schroedinger equation although we shall not do so here These wave functions are not spherical but rather directional ie they favor certain directions in space From this one can anticipate molecular structure with specific bond angles and molecular shapes Problem Study these functions in Munowitz All these functions have nodes at the position of the nucleus as well as increasing numbers of nodal planes with increasing n and increasing 1 Tunneling Plot Ur versus r and draw the horizontal line representing the E1 energy On a separate graph plot the l P15rl2 of Eq 9 versus r making use of the same scale for r on both graphs We have already seen that classically the electron could not go further from the nucleus than the ra where Ura E1 because that would mean that the kinetic energy had turned negative But quantum mechanically there is no such restriction it is clear that l P15rl2 diminishes but does not vanish for r gt ra This phenomenon the possibility of having Ur gt E is called tunneling it is a structly quantum mechanical phenomenon attributed to the wave nature of quantum mechanics Despite the fact that the electron can go beyond the point ra where Ur gt E as long as E lt 0 the electron is bound and it is only if E gt 0 that the electron is free At very large distances Ur gt0 and so unless E gt 0 the particle will ultimately return closer to the nucleus Hybrid functions A very important property of degenerate states is that although they are distinct solutions of the Schroedinger equation so too are all linear combinations of degenerate solutions Thus T25 L112p and T35 1mm are also good solutions of the Schroedinger with the same associated energies E2 and E3 respectively These linear combinations of degenerate state functions are called hybrid functions or hybrid orbitals We shall make much use of them later Although any linear combination of the orbitals is as good as any other the number that are independent i e that contain independent information is fixed as indicated above For deeper understanding consider these problem Lecture Series 9 Covered for final In this chapter we will learn a little about the science that underlies many of the analytical techniques that chemists use to identify molecules In addition these techniques are the major tools that chemists use to elucidate the structure that characterizes both the bonding and the electronic con guration of molecules This chapter also seems a little haphazard and so I will try to cover it in a way that is consistent with what you have already learned Spectroscopy Recall that we discussed that light comes in packets of energy or quanta called photons and that the energy of a single photon is related to its wavelength by the following relation ship worked out by Planck E hck the wavelength of the photon where h is Planck s constant c is the speed oflight and 7 is Now it turns out that molecules and solids interact with most forms of radiation Long wavelength radiation probes very low energy processes while short wavelength radiation will probe highenergy processes Why does this happen Imagine that we have the 5 839 839 8 L1 F FL1 gtlt r H gtK 49m lt Ult t1me gt Fig l The rotation ofa polar molecule LiF and the AC electric electric eld that is generated by that rotation 8 eld molecule LiF moving around freely in space We show this in Figure 1 Now it takes a certain amount of energy to rotate this molecule and as you might suspect the molecule can only be rotated at certain frequencies ie the rate at which it spins is quantized So let s assume that the molecule is rotating from LiF to F Li and back again at some frequency LiF is as we know a largely ionic molecule with a large dipole moment The dipole moment represents the direction of an electric eld within the molecule So now if we rotate this molecule we are rotating an electric eld Since the molecule will rotate only at some particular frequency because of the quantization the electric eld is some AC alternating current eld with a characteristic frequency that is the rotational frequency of the molecule Consider the voltage coming from a wall socket The electricity that is supplied to the wall socket is supplied at 60 Hz meaning that 60 complete cycles of the electric eld are completed every second In other words the AC voltage coming out of a wall socket has a period not unlike the AC voltage generated by this rotating molecule Now recall that light is electromagnetic radiation The electric part of the electromagnetic radiation is an AC eld with a period de ned by the wavelength of the light and the velocity of light according to Planck s relationship If the period of the molecular rotation and the period of an incident beam of light are the same then the molecule is in resonance with the light beam and it can absorb photons Now let s consider a different type of motion with the same molecule LiF Recall that the dipole moment charge X distance and that the magnitude of the dipole moment is related to the strength of the electric eld that is generated by the molecule If we stretch the molecule a little bit then the dipole moment increases since the distance between the separated charges increases If we vibrate the molecule so that it is stretching relaXing stretching etc then we get another type of alternating electric eld 7 one that alternates between a large eld molecule stretched to a small eld molecule compressed Once again the molecule can only vibrate at certain frequencies 7 ie its vibrations are also quantized So if light is resonant with those frequencies then the molecule can absorb photons The alternating eld that is produced by a vibrating 5 5 5 539 LiF molecule is shown in LiF Figure 2 I Now there are lots of other things that a molecule can do Vlbl atlng LIF that will generate an alternating eld 7 some of them are easy to show A pictorially and some of them are more dif cult For example if we take the MO tlme gt diagram of LiF and move one of the electrons from an occupied orbital to an unoccupied orbital we are changing the distribution of electric charges in the molecule and so we are once again changing its dipole moment We have already shown that the energy levels of LiF are quantized and so we eXpect that moving electrons from one state to the next will only occur at certain energies Once again we generate a varying electric eld within a molecule that has some characteristic Figure 2 The AC eld generated by a vibrating polar molecule frequency If a light beam incident upon the molecule has that same characteristic frequency then it can be absorbed Anything that a molecule can do that can potentially generate an AC eld can in principle be probed by light When we probe such molecular phenomena with light we say that we are doing spectroscopy where spectroscopy is de ned as the interaction of light with matter Wavelength of Light Molecular Interaction Spectroscopy Name Millimeter wavelengths Figure l Radiowaves Nuclear spins ip from up Nuclear magnetic 01 meter and longer to down or from down to resonance wavelengths up NMR Information about local chemical environments as well as global structure Microwaves Molecules rotate Molecular rotation spectroscopy Information about dipole moments and bond lengths 6 l m1crometer 10 meters and angles Infrared Molecules Vibrate Molecular vibrational l 7 40 micrometers Figure 2 spectroscopy Information about the functional groups present on a molecule the symmetry of the molecule the structure of the molecule Visibleultraviolet 1000 to 100 nanometers Electrons move from one electronic configuration to another Electronic Spectroscopy Information about the molecular orbitals It turns out that all of these techniques listed in the table are very important for chemistry and within the chemistry department at UCLA there are a dozen or so instruments for each type of spectroscopy Although NMR spectroscopy is perhaps the single most important tool that chemists use we won t discuss it any more here because it involves nuclear spin 7 something that we haven t covered in this course Instead we will spend a little bit of time discussing microwave and infrared and Visibleultraviolet spectroscopy since they deal with molecular phenomena that should be a little familiar Return for a moment to Figure 1 In this Figure we show the LiF molecule rotating and we draw out the AC field that accompanies this rotation 7 a field that originates because the molecule has a permanent dipole moment According to the table above we can use microwave spectroscopy or molecular rotational spectroscopy to look for characteristic absorption by this molecule as it rotates What if the molecule were F2 or Liz Would we get this AC field when the molecule rotates No There is no dipole moment so there is no AC field generated by rotation Thus if we are going to do molecular rotational spectroscopy we can only do it on a molecule that has a permanent dipole moment Therefore other larger molecules will behave similarly Molecules like SF6 or CCl4 both of which have no permanent dipole moment will not generate an AC eld when they rotate and so their rotational spectra cannot be interrogated using microwave absorption spectroscopy Molecules like CH3Cl H20 or NH3 each of which have a permanent dipole moment h ow does it point will eXhibit a characteristic microwave absorption spectra When we say that the molecule eXhibits a characteristic rotational spectra what do we mean It turns out that the microwave absorption spectra of a rotating polar molecule will be a ngerprint of that molecule and it will also contain a fair amount of information about molecule structure We will only discuss the very simplest type of rotational spectroscopy here 7 the rotational spectra of a heteronuclear why not homonaclear diatomic molecule Rotational Energy Levels of a Diatomic Molecule When we think about a rotating mass one way to characterize it is by its moment of inertia Consider a child playing with a ball that is attached to a string figure 3 If the child spins the mass child 2 Inc ball around by holding on to one end of the string and swinging the ball then the ball has a moment of length of string r inertia which is mass of the mass of the b 11 2 ball times the l l lcl l lb a mb length2 of the M 39 string Actuall mcmb the moment of y Figure 3 A child spinning a ballonastring inertia is really just a little bit more complicated but we shall show that it may be estimated by the above description For the child spinning the ball in Figure 3 we can calculate out u the reduced mass which is defined in the Figure If the child weighs much more than the ball then the reduced mass is essentially that of the ball sh ow that this is so However to be exact we say that the moment of inertia for this childball system I urz where u is the reduced mass For a diatomic molecule such as LiF the situation is similar and we get a moment of inertia that is 1LRez where u is the reduced mass of the molecule Example Calculate the moment of inertia for LiF given the following values mass Li 7 gmol mass ofF l9 gmol Bond length 22 A pl 719719 51 gmol 51 x 10393 kgmol602 x 1023 mol39l 853 x 103927 kg 1853e27 kg22e10 m2 41 x 103946 kg m2 As we indicated above the way that a molecule can rotate is dictated by quantum mechanics and so it can only rotate a discrete quantized frequencies Those frequencies are given by Eh28721 JJ1 J012 J is the rotational quantum number In microwave spectroscopy one can observe these various rotational states by monitoring the wavelength dependence of microwave absorption by the molecule of interest Example For LiF what are the first 3 rotational energy levels Eh2872 41x 103946 kg m2l J J1 forJ 12 and3 For a 1 E 27 x 103923 J 2 E 80 x 103923 J 3 E 16 x 103922 J WWW What wavelengths do these energy levels represent We can use Planck s relationship albeit a little inverted to calculate this Using 7 hcE Fl 7 74 millimeters F2 7 25 millimeters F3 7 12 millimeters Recall from the previous table that microwave radiation was characterized by millimeter wavelengths In fact that is what we see here so this answer makes sense Note that if we had measured these three wavelengths and we knew that we were measuring the microwave spectrum of LiF then we could have calculated the bond length of LiF by reversing the order of these calculations and solving for the bond length r In fact this is exactly how bond lengths are measured for diatomic molecules It turns out that it is possible to measure bond lengths to extreme accuracy the millionth of an Angstrom However such measurements represent the average bond length to a high accuracy not the bond leng t Why As an aside a microwave oven works by exciting the rotational states of water Because microwaves are millimeter wavelength however the water mu st be contained in something relatively large like a glass If you live in a notsoclean house or apartment you might notice that ants can walk around inside a microwave oven while it is on The ants are too small to interfere with the microwaves and so no radiation is absorbed Infrared or Molecular Vibrational Spectroscopy Perhaps the simplest way to detect if there is a functional group on an organic molecule ie an alcohol or a ketone or an aldehyde or an ene etc is to measure the infrared absorption spectra of the molecule As was shown in Figure 2 if the molecule generates an alternating eld when it vibrates then it will interact with electromagnetic radiation and we can observe photon absorption in the infrared Chemical bonds may be modeled as two atoms held together by a spring It turns out that such a system will have a resonance frequency and that resonance frequency is determined by the stiffness of the spring ie the stiffness of the chemical bond and the weights of the two atoms that are bonded together We can quantify the spring stiffness by using a spring constant k If the spring is stiff k is large If the spring is oppy k is small We can once again model the mass of the two atoms using the reduced mass u If we do this then we will get resonance frequency in s39l of v l2nldu12 Problem What are the units of k Equating both sides of the resonance equation we find that sl ldkg12 So s392 ldkg and kg Squot2 k Newton m391 Because the spring constant has units of Forceunit length we often call it a force constant Problem Based on the above equation and the fact that a CH has a characteristic resonance wavelength of vibration 3 micrometers estimate the resonance frequencies of C0 OH N H and CC single bonds I urge you to do this problem I will provide the answers in my office hours and perhaps in a later Lecture Series I won t hold you responsible for knowing much beyond this for rotational and vibrational spectroscopy Example 151 in the book is a fine problem to try to understand however Electronic Spectroscopy Let s review what we have learned and apply it to a new type of spectroscopy electronic spectroscopy Rotational spectroscopy requires millimeterwave radiation and vibrational spectroscopy requires infrared or micrometerwave radiation The implication is that it takes a lot less energy to rotate a molecule than it does to excite molecular vibrations How do these energies relate to the potential energy wells that we have used many times in this class to describe a molecular bond In Figure 4 we show rotational states Energy c ground state potential well of molecule Figure 4 The relationship that exists between rotational state energies vibrational energies labelled vl v2 etc for the various quanta of vibrational excitation and the ground state potential energy well of a diatomic molecule Note that as we excite higher vibrational quanta the average bond length the midpoint of the line that is drawn across the potential well is increasing This means that as we excite vibrations we are beginning to pull the molecule apart It typically takes the excitation of many many vibrations to do this as many as 40 or so and so this picture here actually is not in correct proportion to the true energy level separations The vibrational spacings are a little smaller and the rotational level spacings are much smaller However except for the scale this picture is correct p6 55 Fig 5 The MO diagram for the gound state of C2 these various energies in relationship to each other The ground state potential well refers to the ground electronic state of the molecule 7 ie the lowest energy state If you were to put together the MO diagram for a chemically bonded diatomic molecule the lowest energy con guration would have a potential well that would look something like what is shown in Figure 4 Within the potential well there exists a manifold of vibrational states each corresponding to some number of vibrational quanta We label that number v Within each vibrational state there is a manifold of rotational states Each rotational state also corresponds to various quanta of rotational energy J similar to the vibrational manifold Note that if we could measure all of the vibrational frequencies and if we could determine the average bond length for each of those frequencies by doing rotational spectroscopy we might be able to exactly determine the shape of the potential well In fact this is actually how such wells are experimentally determined Why should we do such an experiment It turns out that if we can determine the shape of the ground state potential energy well then we can determine values like the bond dissociation energy It turns out that each electronic state is going to be accompanied by its own potential energy well Consider for example the MO diagram in Figure 5 for the diatomic molecule C2 This electronic con guration is a triplet state How can we tell this Count up the number of unpaired and aligned spins In this molecule there are 2 Give each unpaired aligned spin a value of 12 and add all the spins together to give s Then calculate the number S 2s 1 In this case S l The first excited electronic state of this molecule will be at nearly the same energy as this triplet state except that all the spins P39G pc5 1345 pTE 2f 2f p 1375 C p r C P395 P395 Figure 6 Two lowenergy singlet states of C2 will be paired and it will look something like that shown in Figure 6 Note that in this case we have both of the highest energy electrons are antialigned with each other They are paired in the same orbital for the case to the right For these two MO diagrams there are not unpaired aligned spins and so s0 giving S 1 Thus these are singlet states These states are slightly higher in energy than the ground triplet state Probably the state shown in Fig 6 left is the first excited state ie the state that is next highest in energy above the ground state and the state in Fig 6 right is the second excited state Both of these states as well as the ground state are characterized by a bond order 2 and they are all probably very close in energy to one another Thus both the ground electronic state and the first two excited singlet states are very likely to have potential energy wells that look like one another How about if we excite one of the p r bonding electrons to the first unoccupied M0 the p r MO We would call this a 757239quot transition To cause such a Figure 7 The first excited triplet state of C2 transition we wou1d need to shine visible or ultraviolet light on this molecule which would be a way of doing electronic spectroscopy The excited state MO diagram is shown in Fig 7 where we show the first excited triplet state of C2 This state involves reducing the bond order in the molecule to 1 since the electron in the p r MO is cancelled by the electron in the p r MO Thus we expect that the potential energy well for this excited state is going to look very different from the ground electronic potential energy well For one thing it will not be nearly as deep since we CHEM 20A1 winter 2001 LECTURE NOTES fourth set CLASSICAL MECHANICS Mechanical variables Consider the behavior of a particle of mass m moving in ldimension At a given time t let the position of the particle be given by the variable x which is a function of t ie xt The velocity u of the particle is the rate of change of position Ax over a very short change of time At Thus Ax dx u og 3 Of course the change of position Ax must be very small if measured over a very short time interval At but the quantity AxAt need not be either very large if At were very small and Ax not so small or very small if Ax were very small and At not so small Problem Explain this last sentence Why in de ning the velocity do we take At to be very small Note that the velocity might change as the particle moves from one position to another ie the velocity can be a function of position ux Furthermore since the position is a function of time xt the velocity can be expressed as a function of time ut The acceleration a of the particle is the change of velocity Au over a very short change of time At Thus Au du A l A 3397 AtlgoAt T dt Note that the acceleration might change as the particle moves from one position to another ie the acceleration can be a function of position ax Furthermore since the position is a function of time xt the acceleration can be expressed as a function of time at Force amp energy Well over three hundred years ago Isaac Newton formulated the laws of motion which are referred to as Newton s laws or the classical laws of motion In particular Newton described the effect of a force F on a quotbodyquot of mass m This effect can be summarized by the following equivalent equations du F ma ma It follows that the force is a function of position ie Fx this means that the force may be different at different positions in space Problem Why is this so The motion of a moving object can be described in terms of its kinetic energy KE which is defined as KE 12mu2 Large fastmoving objects have great kinetic energy while small slowmoving objects have little kinetic energy The potential energy UX is de ned by the relation dngl dx We say that the force is the negative gradient of the potential energy We shall see momentarily why the potential is a useful quantity But note rst of all that it is a bit strange to de ne a quantity through its derivative Note secondly that UX has the same dimensions and can therefore have the same units as the kinetic energy KE We shall discuss dimensions and units soon Note that U can be a function of X Fx Ifwe combine the de nition of UX with Newton39s law we obtain dU du 39dx mdt We can multiply the right side by dXdX l dU dudx dudx du dKE 39dx mdt dx mdx dt mdx u dx This can be rewritten as dgKEU1 dX This last equation is equivalent to stating that KEU is a constant we express this as KE U E where E is a constant known as the total mechanical energy Problem Be sure you understand how to obtain this result but even if you have trouble doing so the important point is to remember and understand the result that the sum of the kinetic and potential energies ie the total energy is a constant This means that the total energy is conserved that it can be neither created nor destroyed This is another veg important conservation law Keep in mind however that the energy of a system is conserved if the system is left to its own devices but that one can always subject it to additional forces thereby changing its energy So for example an object of mass m held at rest at a height h above the oor is known to have a potential energy of mgh where g is the acceleration of gravity and its kinetic energy is also zero if it is released and falls its velocity and kinetic energy increase as its potential energy decreases but at all times its total energy remains a constant E mgh Therefore when it hits the oor the potential energy is zero and KE mgh However if the particle is not simply released but pushed downward with an initial velocity uo then its total energy is E mghl2mu02 Problem Work all of this out and calculate the velocity with which the object strikes the oor if its initial velocity is uO 10 cm S391 and the height at which it is initially held is h 2 m Potential energy Recall that UX is a function of X and consequently since E is a constant KEX must also be a function of X Changes in KEX must be compensated for by changes in UX Because of the intimate connection between UX and FX each of these quantities contains very much the same information Although the force FX is most directly related to the equation of motion we focus attention almost exclusively on the potential energy UX Therefore we spend a few moments discussing the relationship between these quantities Iftwo particles exert forces on each other we say they are interacting If they pull together we say the force is negative because the distance between them decreases and attractive Ifthey push apart we say the force is positive and repulsive The potential energies that describe these are likewise denoted attractive and repulsive interaction potential energies But it is the slope of UX at a point X and not its sign that determines whether it is attractive or repulsive To show the connections between UX and FX consider a plot of UX versus X and consider only positive values of X If the UX increases as the curve goes from left to right the slope is positive ie dUdX is positive and consequently the force is negative this means that the object is pushed to smaller values of X and if the potential energy arises from interactions among particles the particles are pushed closer together This means the interactions are attractive Similarly if the UX decreases as the curve goes from left to right the slope is negative the forces positive and repulsive Note that whether the potential energy UX is positive or negative is irrelevant what counts is its slope Ifthe UX versus X curve goes through a minimum or a maXimum at these points the slope of the curve is zero ie dUdX 0 Consequently the force is zero If the force is zero and the particle is at rest 139e the velocity u 0 then the particle will remain at rest So these eXtremum points minima and maXima are very special and are known as equilibrium points They are however very different the minimum is a point of stable equilibrium and the maXimum a point of unstable equilibrium If the particle is displaced a bit X is changed a bit from the minimum the forces will then always act to restore it back to the minimum ie to the point of stable equilibrium regardless of whether the displacement is to larger or smaller values of X However slight displacements from the maXimum result in forces that drive the particle away from the maXimum and inevitably towards a minimum Problem Prove that these statements are true Ifthe potential energy has a minimum at X X0 and the UX consequently swings upward for X both greater and lower than X0 one talks about a potential energy quotwellquot If UX describes the potential energy of a particle one can think of the particle as trapped or bound within the well as it moves up either wall it is forced back to the bottom This is very much like the behavior of a particle in a physical well in which the solid walls cannot be breached and the gravitational forces are at work The analogy is useful and we shall make use of it but remember that we will not be discussing gravitational forces but electrostatic forces between particles Potential amp total energy In describing the particle in relation to its energy we note that the curve we have described is the potential energy curve The total energy E of the particle is a constant and so if we think of the graph as energy versus position we can draw both the potential energy curve and the total energy curve the latter is just a horizontal line which cannot lie below the potential energy curve My Thus if the slope of the Ux curve is everywhere positive the interaction attractive the total energy curve must be restricted to regions of x where E 2 Ux the particle is therefore said to be bound The higher the energy E given to the particle the more delocalized it becomes ie the greater the region of space over which it can exist Problem Show that if the slope of the Ux curve is everywhere negative the particle is unbound or free at all energies E If the potential energy curve has regions of positive and negative slope and both a minimum and a maximum then the particle is bound in the region about the minimum provided the total energy E is less than Ux at the maximum if the energy E is greater than Ux at the maximum then the particle is unbound or free Thus at low energies particles can be trapped or bound in a potential energy well but if given enough kinetic energy so that E is sufficiently large the particle can escape ie become unbound or free Quite generally one can state that the lower the energy of the particle the more tightly bound it will be and consequently the more stable the system This description fits an electron that is bound to an atom but that can be ionized if given enough energy the ionization energy or ionization potential or work function Similarly the energy lost by an incoming electron that is attached to an atom trapped in its well is the electron af nity Problem Prove the above Make use of diagrams graphs These are very important concepts that we will use over and over again both in studying atoms and molecules Each time we run into these concepts we will cover them at a deeper level and at each encounter your understanding of them should increase Of course we are actually interested in 3dimensional problems but the generalizations from 1 to 3 dimensions is very straightforward Classical versus quantum mechanics Here we stress that the discussion above is restricted to classical mechanics the mechanics that describe the phenomena that we observe in daily life About one century ago it became obvious to scientists that this description did not apply in the atomicmolecular world Classical mechanics break down on such small length scales and quantum mechanics then become the correct way to describe such phenomena Consequently we will next study quantum mechanics In classical mechanics the energy is continuous ie it can take on any sequential value whereas in quantum mechanics it can take on only certain discrete or quanti ed values much like the mass In classical mechanics the potential energy cannot be less than the total energy but in quantum mechanics this is not quite so as a consequence a trapped or bound particle is not truly trapped but can tunnel to freedom However even in quantum mechanics the mechanical energy is conserved and this fact remains fundamental to all that we discuss in chemistry APPLICATION 0f CLASSICAL CONCEPTS When discussing high energy phenomena high compared to what is encountered in chemistry relativistic effects must be incorporated into the quantum mechanics and modifications must be made to the conservation laws even the energy conservation law But this need not concern us here We take the conservation of energy as absolute Example of Forces Isaac Newton determined that the attractive gravitational force between two spherical objects one of mass m1 and the other of mass m2 is given by the relation where r12 is the distance between the centers of the objects and G is a universal measured constant known as the gravitational constant You can look up the value of G but this constant and in fact gravitational phenomena need not concern us here The potential energy corresponding to this force is U Gmlmz r12 Problem Prove that this U and F correspond to the same interaction The electrostatic analog to the gravitational forcelaw is Coulomb39s law 1112 r122 F l4739E80 where q1 and q2 are two point charges or charges distributed spherically and r12 is the distance between the points or the centers of the spherical distributions 4739E80 is a universal measured constant which we shall examine later The potential energy corresponding to this force is U 147ceo The Coulombic interactions differ from the gravitational ones in that they can be either attractive or repulsive depending upon the sign of q1q2 attractive if q1q2 lt 0 and repulsive if q1q2 gt 0 Problem Plot Ur12 for both attractive and repulsive forces APPLICATION t0 ATOMS Electrons in an atom are bound to the nucleus by attractive Coulombic interactions On the other hand the electrons within the atom repel each other via repulsive Coulombic interactions The fact that the atom actually holds together with these competing effects is by no means obvious and is worthy of study Most of this course is devoted to the study of electrons in atoms and molecules and most of the focus is on the energies involved Exact potential energy for atoms The potential energy of an atom containing N electrons can be expressed as 2 2 UNzi 12NJe J rJ Jrll where Z N rj is the distance from the nucleus to the jth electron and rij is the distance between the ith and the jth electron Problem Derive this expression Shielding If one sits on a given electron and studies all the attractive forces originating from the nucleus and all the repulsive forces originating from the other electrons the net effect is very complex However because the nuclear charge is Ze and the total charge of all the other electrons Zle one might expect the net force on the given electron to be attractive as indeed it is But this net attractive force will certainly be less than what would be expected if the given electron were the sole electron in the atom and was not repelled by all its neighbors This reduction of the nuclear attractive force by the presence of the neighboring electrons is called quotshieldingquot the nuclear attractive force is shielded or diminished by the repulsive forces exercised by the other electrons One can envisage this shielding as diminishing the effective nuclear charge ie diminishing the Coulombic attraction in much the same way that a reduction in the nuclear charge Ze would reduce the interaction How much diminishing of the effective nuclear charge the actual charge always remains Ze occurs is a complicated issue that we discuss later but the general concept of shielding should be simple to understand The electrons in an atom are distributed quotsphericallyquot about the nucleus some of them are in close to the nucleus and some are farther out the former the quotinner electronsquot form the quotcorequot of negative charge about the nucleus while the outer ones are known as valence electrons The inner electrons are very effective in shielding the outer ones from the nuclear attraction whereas the outer ones provide relatively little shielding for the inner ones Consequently the inner electrons have lower energy and are bound more tightly to the nucleus than are the outer ones Problem Explain why this is so Write down the potential energy of a Li atom It is the outer or valence electrons which are held less tightly that can be most easily shifted from atom to atom during a chemical reaction the core electrons remain relatively unchanged Thus chemistry is largely the study of the behavior of the valence electrons UNITS amp DIMENSIONS The term quotdimensionsquot can be used to refer to the dimensionality of space but here we use it to describe a basic physical property The four dimensional properties to which we refer are mass M length L time T and charge Q Some physical quantities have complex dimensions for example velocity has the dimensions LT acceleration has the dimensions LTZ force has the dimensions MLTZ and energy has the dimensions MLZTZ Problem Show that this is so It is essential that the dimensions on both sides of an equation be the same or else one cannot have equality Furthermore if separate terms are added or subtracted they each have the same dimensions So for example in the equation E KEU all three terms EKEU must have the same dimensions Problem Prove that they do Can both sides of an equation be dimensionless Problem Determine the dimensions of the universal constants G and 80 introduced above One check on the correctness of a formula is to make sure that at the very least the dimensions are the same term by term and on both sides of the equation Furthermore one can make sure that the dimensions are those of the quantity being sought so for example if you are calculating a force it should have the dimensions MLT39Z Quantities with dimensions must be measured relative to a set of units We make use of the socalled SI units which are based upon the metric system Length is measured in meters m time in second s mass in kilograms kg and charge in Coulombs C Thus velocity is measured in units of m S39l acceleration in units of m 8392 force in units of Newtons N where l N 1 kg m 8392 and energy in Joules J where l J 1 kg m2 8392 Other units can be used but one should be careful ultimately to make sure that each term contains only a single choice and that each term contains the same choices Of particular interest still within the metric system is the dyne as the unit of force and the erg as the units of energy 1dyne 1g cm S392 10395N l erg l g cm2 S392 10397erg Problem Prove this CHEM 20A1 winter 2001 LECTURE NOTES eleventh set THE CHEMICAL BOND Atoms bind together to form molecules The bond between atoms within a molecules is known as the chemical bond or an intramolecular bond or an interatomic bond Why do chemical bonds form That is the first question we address Ionic and Covalent Bonds Atoms are electrically neutral The positive nucleus lies at the center of spherically distributed negative electron An observer far from the atom senses no electrical charge because the negative charge of the spherically distributed electrons acts as though it were concentrated right at the nucleus thereby ensuring that all electrical forces are canceled And indeed atoms far separated from each other do not interact However when they get close to each other the situation is altered As their outer shells begin to overlap the shielding of an atomic nucleus by its own electrons is not perfect and consequently the outer electrons on the second nucleus can be affected by the incompletely shielded positive charge Two extreme we denote them quotlimitingquot situations can occur the outer electrons on the two atoms are equally affected in which case a covalent bond is formed the outer electrons on one atoms are preferentially pulled towards the other atom in which case an ionic bond is formed An example of a covalent bond is that in H2 139e the HH bond There is no reason why the electron on one H should be preferentially pulled towards the other An example of an ionic bond is that in CsF 139e the CsF bond It turns out that the F atom pulls so hard on the Cs electrons that it nearly pulls one electron off the Cs atom thus very nearly becomes a CsJr ion and the F atom very nearly becomes an F39 ion In this case one can understand the ionic bond as due to the electrostatic attraction between the CsJr and F39 ions Understanding the covalent bond is a more subtle challenge Sometimes to indicate an ionic bond we place a positive sign on the atom that loses some of its electron cloud and a negative sign on the one that gains electron density e g CslF39 however this does not necessarily mean that a full electron has transferred from the vicinity of the Cs atom to that of the F atom only that there has been substantial transfer In summary an ionic bond is formed when substantial uneven electron transfer from one atom to the other takes place covalent bonds are bonds that form without such substantial transfer Most bonds can be envisaged as mixtures of the two quotlimiting casesquot The CsF39 ionic bond forms because the ionization energy ofF is much higher than that of Cs thus the electron readily moves away from Cs but not from F Furthermore the chemical affinity of F is high whereas that of Cs is low so that the F readily accepts an extra electron whereas Cs does not The basic idea is that the single outer electron on Cs is so shielded that the effective nuclear charge holding it is only slightly greater than e whereas even an added electron to the outer 2p electrons of F is held by a positive nuclear charge of approximately 15e Look back at the discussion of the periodic table in the last set of notes The HH and FF bonds are covalent as indicated above because there is no preference for electrons to reside on one atom or the other On the other hand the rather low ionization energy of H suggests that it is not too hard to distort the electron distribution about each of the atoms and indeed this is what happens with the consequent formation of a covalent bond For FF the situation is different the high electron affinity suggests that it is not too hard to distort the electron distribution about each of the atoms and indeed this is what happens with the consequent formation of a covalent bond On the other hand the noble gases do not readily form bonds because they have both a very high ionization energy and a very low chemical affmity Stability and Bonding When why and how do chemical bonds form Systems are more stable when they have lower energies When two atoms are brought near each other near enough that there is some overlap of their electron clouds they interact with each other the attraction can be attractive in which case they will bond or it can be repulsive in which case they repel each other If they attract each other and form a bond the energy of the electrons electronic energy decreases below that of the sum of the electronic energies of the separated atoms If they repel each other and do not bond the electronic energy increases over that of the separated atoms Since energy is conserved there must be compensation for the decrease or increase of the electronic energy upon atomicatomic interaction electronic energy must be either discarded or retained and redistributed we shall not delve into this matter but there are other relevant sources and sinks of energy such as those associated with other atoms and bonds in molecules translational motion of the atoms and molecules rotational and vibrational motions of the molecule and absorption or emission of photons But the bottom line is that bonds are formed if the formation of a bond lowers the electronic energy Furthermore electronic systems always adjust themselves so as to lower their energy the lower the energy the more stable the system and systems adjust themselves so as to maximize stability One can therefore think of the formation of chemical bonds as the selfstructuring 0f the system so as to maximize stability A simple way by which to conceive of a chemical bond is in terms of an effective potential energy Ur written as a function of the distance r between the two nuclei If the atoms do indeed form a stable bond they must do so at an interatomic separation r0 at which the forces pushing them together or apart are zero Why Thus the interatomic potential energy must have a minimum at r r0 Why When the atoms are very far apart they interact very feebly which means that the interatomic force is nearly zero which in turn means that at large r the interatomic potential energy versus r curve is a nearly horizontal line On the other hand it is difficult to push the atoms together much closer than the quotequilibrium distancequot r0 at small r ie for r lt re the interatomic force is repulsive in fact very repulsive This means that as r decreases below re the potential energy increases rapidly Thus the interatomic potential energy versus r curve has a distinctive shape nearly horizontal at very large r curving downward slowly as r decreases curving downward more rapidly as r approaches r0 reaching a minimum at r r0 increasing as r decreases below re and then increasing extremely rapidly as r decreases further Thus the Ur versus r plot has the appearance of a potential energy well with a minimum at r r0 with a steep wall at smaller r and with a more moderate wall at larger r a wall that ultimately attens into a plateau Problem Plot this Ur versus r Earlier we wrote down the exact potential energy for molecules For a diatomic molecule it is the sum of the attractive interactions of all the electrons with both nuclei of the repulsive interactions of all electrons from both atoms with each other and the repulsive interaction of the two nuclei with each other Problem Write down the exact potential energy of the diatomic molecules This potential energy is quite complicated and does not look at all like the simple potential energy well described above The exact potential energy depends upon all the coordinates of all the electrons and nuclei whereas the effective potential energy discussed depends only upon the internuclear separation The effective potential energy represents an approximation to a partial solution of the true physical problem such effective potential energies constitute models that provide effective means for understanding the actual physical problem Let us focus on the approximate effective intermolecular potential energy for the two atom system This two atom system also has kinetic energy and a total constant energy En the total energy is quantized and the allowed energy levels are indicated by the quantum number n Note that each pair of atoms has its own set of En s much as a particle in a box and as a hydrogenlike atom or other atomic species have their own distinctive En39s On an energy versus r plot we can place both the potential energy Ur and the allowed energy levels En the latter being a series of horizontal lines The ground state will be indicated by n 0 so that the ground state energy is E0 We could start numbering the states at n l or n l or wherever we wish E0 must be higher than the minimum of the potential energy at r r0 Problem Draw this double graph of both EO and Ur versus r Problem Explain why the zeropoint energy equals EOUr0 As you will recall the atoms are bound together in this ground state because if they move too far from r r0 whether to larger or smaller r the potential energy would exceed the total energy and the kinetic energy would become negative Although quantum mechanically such negative kinetic energy is possible ie by means of tunneling the atoms nevertheless remain bonaleal However if the twoatom system absorbs sufficient energy perhaps through irradiation to boost its energy above Uoo then the molecular bond is broken and the two atoms have enough kinetic energy to y apart as free atoms Problem Be sure you understand this We can summarize the picture outlined above of bonding in a diatomic molecule The molecule in its ground state is stable with an interatomic separation of about r r0 the atoms are bound by a chemical bond If it absorbs energy in excess of UooE0 the molecule dissociates and the atoms are free The dissociation energy is defined as dissoc energy Uoo E0 it is the energy required to break the molecule apart We also can define this quantity as the bond energy bond energy Uoo E0 A strong bond is one with a large bond energy i e it is one that cannot be broken without the input of a large amount of energy A weak bond is one with a small bond energy i e it is easily broken with the input of very little energy Not all pairs of atoms form bonds in fact some may repel each other In this case the effective interatomic potential energy is not a quotwellquot but a barrier Ur is large at very small r and decreases as r increases rapidly at first and ultimately attening out into a horizontal line Problem Plot this repulsive potential energy Discuss the behavior of two atoms interacting with this potential energy and headed on a collision path Energy Minimization An insightful way to envisage the formation of a chemical bond is to think of it as the tendency of a system of atoms to stabilize itself by lowering its potential energy If the formation of bonds reduces the potential energy then bonds form if bond formation does not reduce the potential energy ie if it leaves it unchanged or increases it then bonds do not form Of course this discussion need not be restricted to a system of two atoms such as H F HF but may involve several atoms H O O H20 atoms and molecules HCl F HF C1 or several molecules H2 F2 2HF The tendency to reduce the potential energy not only helps understand whether bonds form or not but it helps understand what bonds form and consequently what structure the resulting molecule will assume So for example carbon dioxide ali means two and oxide indicates an oxygen compound though written as C02 has a linear structure OCO rather than the linear structure OOC or a bent structure on the other hand water written as H20 has a bent OHO structure as does sulfur dioxide 02 The reason that the atoms combine to form a molecule with a particular structure is that in so doing they minimize the potential energy Therefore to determine the strength of bonds bond energies or dissociation energies and to determine the molecular structure we must determine how best to minimize the potential energy of the molecule as a whole We have been a bit cavalier with the concepts above It is the total electronic energy rather than the potential energy that we must minimize We must therefore find the minimum ground state electronic energy for the system If we look back at the diagrams above with both allowed energy levels and potential energy on the same plot we see that the minimum ground state electronic energy anal the minimum potential energy differ only by the zero point energy which is o en not great Problem Untangle all these concepts Valence Electrons When atoms bond to form molecules it is largely the outer or valence electrons that participate in the bond formation The inner or core electrons are held very tightly anal very closely to the nucleus and so they are not much a ecteal by the approach ofother atoms Thus a simplifying model of chemical bonding is one in which the core electrons are unaltered as the bond is formed we therefore need only focus on the outer or valence electrons although the core electrons still play a very big role in shielding the outer electrons This model does not yield perfect results ie results that agree exactly with experiment but it is a pretty good model good enough for all but the most precise studies And it is clear that this model greatly simplifies the work of studying molecular structure For H the single 1s electron is an outer electron For He the two 1s electrons form a complete shell and although they must by de nition be outer electrons completed shells do not participate readily in reactions We discussed this above and concluded that He was very unreactive Electronegativity amp Dipole Moments For ionic bonds and even for covalent bonds the outer electrons move to some degree from one atom towards the other for perfectly covalent bonds the motion is identical for the electrons on each of the atoms Ifone atom borrows electron density from the other it is said to be more electronegative than the other the greater the electronegativity the more it tends to attract electrons The greater the electron affinity and the smaller the ionization energy the greater the electronegativity Explain In fact one could define electronegativity as the sum of ionization energy plus af nity It therefore follows that the electrons are distributed more densely about the more electronegative atom and that the bond forms a dipole moment A dipole moment is a neutral object in which the equal positive q and negative q charges are separated by a distance d the dipole moment u is given by Hqd For a bond dipole moment the charge q may be far less than a full electron charge lei it may correspond to a very slight shift in the electron clouds about the nuclei The distance d is often taken as the distance between the nuclei In any case whether or not one actually knows the magnitude of q and d one can measure the dipole moment u when u is large one expects a largely ionic bond whereas when u is small the bond is likely to be mostly covalent Ionic bonds are likely to form when the electronegativities are very diiTerent for the combining atoms Problem Which of these would you expect to have no dipole moment small dipole moments large dipole moments H2 HF F2 N0 N2 02 FCl C02 H20 802 BF3 03 which forms atriangle with two equal sides Some Chemistry For Li the 2s electron is an outer electron shielded by the ls2 core and most of the chemical properties of Li can be described in terms of the wellshielded 2s electron If a Li atom is brought in contact with a F atom with its high electronegativity the Li atom readily loses its valence electron to the F thereby becoming a LiJr ion with a ls2 core and no outer electrons The F atom becomes a F39 ion with a ls22s22p6 core and no outer electrons The LiJr and F39 ions are similar in structure to He and Ne respectively and therefore are not readily altered Why Thus Li tends to lose one electron and F to gain one electron none of the other electrons contribute much to the chemical reaction Why For Be both 2s electrons are outer electrons that participate in chemical reactions If a Be atom is brought in contact with an F atom the Be loses an electron and the F gains one and one might expect a BeF39 bond to form Why The BeJr ion still has an outer 2s electron which can be ionized without too much difficulty Why The F39 ion cannot add another electron the exclusion principle would require that it be placed in a 3s orbital which has high energy and is almost completely shielded by the ls22s22p6 core However if a second F atom is brought in contact with the BeF39 it can pull the second outer electron from Be forming F39Be2F39 The Be2 and the two Fquots are very stable and they neither wish to add or lose electrons we conclude that F can readily form one bond but Be can readily form two bonds ie the valency of Be is two that of F one This model is somewhat oversimpli ed the electrons do not fully leave the Be and do not fully reside on the F we must keep this in mind when writing formulas such as F39Be2F39 We come back to this later in discussing oxidation states and formal charges Suppose a Li atom is brought up to the much more electronegative oxygen atom with its 2s22p4 outer electrons The O atom steals the electron from the Li forming Li O39 The LiJr cannot do more but the O39 with its 2s22p5 outer electrons can still attract another electron So if LiO39 is brought up to another Li atom it can withdraw the latter s outer electron and LiJr0239Li4r can be formed The two LiJr ions and the 0239 are very stable and they neither wish to add or lose electrons we conclude that Li can readily form one bond but 0 can readily form two bonds ie the valency of O is two that of Li one Once again this model is somewhat oversimpli ed the electrons do not fully leave the Li39s and do not fully reside on the 0 Important problem Discuss the chemistry of Na Mg Cl and S And also of H Think about the wonders of the periodic table What happens when two Li atoms are brought together The outer 2s electrons on the two Li atoms overlap each other and are altered by this overlap Attracted by the other Li atom they may move more into the region between the atoms or repelled by the other electrons they may be pushed out of the region between the atoms If they are pushed out of the region between the atoms the two nuclei are poorly shielded from each other and will not form a bond This is called antibonding If they are pulled into the region between the atoms then they serve the double function of shielding the two nuclei from each other and because each of the participating electrons are attracted simultaneously to both nuclei of linking the nuclei Thus the LiLi molecule is formed This is the origin of the covalent bond whatever shi s occur in the electron distribution must be the same for both Li atoms Why Problem Plot the interaction potential energy of LiLi as lnction of the internuclear separation r for both bonding and antibonding What if two F atoms are brought together Of course since they have the same electronegativity the charge is symmetrically distributed and there is no dipole moment Because the ionization energy is quite high there is no tendency for the electron to wander off too far on the other hand the electron affinity is high so that each F atom would gladly accept another electron So there is some delocalization of electron density between the two F atoms ie the electrons are less restricted to a single atom and tend to spread out somewhat over the two reacting atoms This leads to lowering of the energy and formation of a bond If this delocalization leads to a build up of electron density between the nuclei it helps shield the nuclei from each other thereby decreasing the repulsive forces between the nuclei this helps form a bond At the same time the electrons between the nuclei can be pulled towards both nuclei simultaneously and this leads to bond formation And so the stable F2 molecule can be formed The FF bond is quite strong ie F2 has a large dissociation energy Quantum Mechanics and the Chemical Bond The fact is however that if the atoms were classical objects brought into close proximity they would never form substantial chemical bonds Classically the electrons would spiral into the nuclei and the neutral collapsed atoms would not bind to each other So ultimately we must turn to quantum mechanics to understand chemical bonding We have of course already implicitly implied the use of quantum mechanics in all that we have said about bonding because the very structure of the atom depends as we have shown upon quantum mechanics But we move on from there We take the properties of atoms as discussed previously as given properties and we ask why starting with these atoms should molecules be formed We have already given some answers to this by considering the ionization energies affinities and electronegativities With these in mind we can explain why and how the electron densities shift between atoms and how all this contributes to the lowering of energy and consequent stabilizing of the molecule To obtain more profound answers within the context of quantum mechanics we focus on two concepts the exclusion principle and delocalization Delocaliz ation By quotdelocalizationquot we mean that the electron cloud of a valence electron on an atom spreads out onto its neighbor or neighbors the wave functions and probability densities extend over a greater volume than do those of the isolated atoms We have already indicated how this can promote bonding because the electron now has a better chance of being attracted to both nuclei thereby linking these nuclei But this link is not great and the consequent reduction in energy is not great However there is another reason that delocalization reduces the energy and stabilizes the bond This can be understood by noting that the more delocalized the electron the lower the energy and therefore the more stable the bond This is best seen for the particle in a box the ground state energy decreases as L39Z where L is the size of the box For a ls electron in a hydrogenlike atom the ground state energy varies as Z392 and the size of the atom varies as Z39l Problem Check these out Why do these results indicate that delocalization lowers energy We therefore conclude that electron delocalization decreases energy and increases stability In a large molecule with more than two atoms it might be possible to delocalize the electron over several atoms this might result in considerable lowering of the energy and signi cant increase of molecular stability This sometimes occurs So for example the benzene molecule C6H6 is a planar equalsided hexagon formed by the six C atoms and one H atom also in the plane is attached to each C atom It turns out that some of the outer electrons on each of the C atoms are delocalized over the entire ring and move quite freely as though they were in a hexagonal wire Note the similarities with the particle in a 1 dimensional box This extreme delocalization is a very important aspect of the bonding in many molecules we shall study this in some detail later In conclusion we nd that all bonding electrons 139 e outer electrons participating in a bond are delocalized in the sense that they are spread over at least two atoms In some molecules the delocalization is much greater stretching over many molecules Since delocalization reduces energy there is always a propensity for electrons in molecules to delocalize as much as possible But there are other considerations that keep in from doing so and in each molecule a balance is reached Usually the term delocalized is restricted to electrons spread over more than two atoms A local bond which refers to most bonds in chemistry is one in which the bonding electrons are constrained to remain in the vicinity of two neighboring atoms the binding electrons in a delocalized bond are spread over more than the two neighboring electrons bove we examined Li as an atom as a covalently bonded Li2 molecule as ionically bonded with the much more electronegative F to form LiF and as LiZO with very electronegative 0 But one can also form solid Li in which many Li atoms are bonded together The reason a Li atom bonds to other Li atoms in this way is that the 2s electrons are all highly delocalized so delocalized that they spread over the entire solid macroscopic sample and by so doing lower the energy and bind the metal together The ls2 atoms in the core of each Li are tightly bound and do notjoin in the bonding process A solid with such highly delocalized electrons is called a metal The fact that the electrons can move so easily over the entire system is why metals have very high electrical conductivity 139e why electrical current the ow of charge or electrons is unimpeded A metal is the ultimate example of electron delocalization Valence We saw that Li tends to form bonds by giving up an electron Li readily loses a single electron a 2s electron and it is this 2s electron that participates in bond formation Li does not readily lose a second electron The reason that Li loses the 2s electron readily is that the 2s electron is forced by the exclusion principle to lie outside the ls2 core where it is heavily shielded and so weakly held Thus one electron from each Li participates in bond formation Be also tends to form bonds by giving up electrons but it can readily lose two outer electrons and so as we saw two electrons from each Be atom can participate in bond formation Said differently Be can form two bonds The reason that Be gives up two electrons readil is that the two outer 2s electrons are forced by the exclusion principle to lie outside the ls core where they are heavily shielded and so weakly held On the other hand we saw that F tends to form bonds by attracting a single electron it cannot readily attract a second electron because the second one would because of the exclusion principle be forced into a 3s orbital which would require a great deal of energy and leave it heavily shielded and hence detached from the nucleus This would destabilize the system Freshman Chemistry 20A Winter Quarter Lecture Week of Jan 12 Chemistry and the Periodic Table The point of this course is to learn how to think about chemical reactions the periodic table and the relevance of these things to the world around us Consider the periodic table The elements at the far right are essentially inert The elements at the far left are amongst the most reactive of all elements Yet the difference between the two elements is that the ones on the far right are characterized by 1 less electron and 1 less proton than the corresponding element 1 row down on the far left If you don t know what an electron or a proton is just wait a bit We ll get to that Let s take one of the simplest reactions possible 2Na Br 9 2NaBr There are several things to note about this reaction First Na is a covalent solid and it is a metal 7 which means that it is very shiny like a silvered mirror In fact in terms of its electrical conductivity it is one of the best metals there is Recall also that by nature of its position at the far left of the periodic table it is a very reactive element Br is a molecule and it is a gas and it is red in color It turns out that Br is only slightly reactive NaBr is an ionic solid 7 which implies that it is not a metal but rather an insulator In fact it is one of the best insulators that there is In addition it is nearly completely inert or nonreactive and it is clear and colorless So we combine a highly reactive covalent metallic solid with a quasireactive red gas and we get an inert product that is an ionic insulating clear colorless solid Let s look at another reaction Na HO 9 NaOH 12 H In this reaction we are taking a reactive metallic shiny covalent solid and reacting it with a clear colorless molecular liquid how is the liquid held together and producing a colorless white reactive ionic solid and a slightly reactive colorless molecular gas We have only written down two apparently simple and common chemical reactions and we have already introduced a tremendous amount of complexity Other questions that we haven t brought up yet include How fast are these reactions Do they proceed to completion And others With this complexity already introduced how can we every hope to understand truly complicated chemical systems such as living organisms or the earth s atmosphere or the chemical processes that are involved in fabricating electronics or plastics etc In this class we will only begin to shed light on these issues and we will do it slowly one step at a time We will first introduce the language of chemistry and we will follow this with a highly simpli ed description of what makes up an atom This description will necessarily involve some quantum mechanics Chapter 13 We will discuss the types of experiments that have led to various descriptions of chemistry including spectroscopy which is the interaction of light with matter After we discuss the atom we will move onto the structure of simple molecules and how simple models may be used with caution to predict such structures Finally we will increase the complexity of the systems we are trying to describe and include organic molecules polymers and transition metal complexes For now let s back up a bit and just try to understand the language of the reactions we have already mentioned and discuss why they were written they way they were First let s return to the reaction 2Na Br 9 2NaBr What we are saying when we write this reaction down is that 2 units of sodium atoms react with 1 unit of bromine gas to produce two units of sodiumbromide The numbers in front of the chemical species are intended to balance the chemical reaction We could have just as easily written the reaction as Na 12 Br 9NaBr Notice that this works out just as well if we are trying to conserve mass In fact that is just what we are trying to do when we balance a chemical equation In all chemical reactions mass is conserved This is a principle that was first stated by Lavoisier early in the last century Notice that we did something similar when we wrote down our second reaction of the day Na H20 9 NaOH 12 Hz However this reaction is a little more complicated 7 2 reactants to produce 2 products Balancing equations never involve more than algebra but nevertheless the algebra can get a little complicated Molecular Unitsa Molesa and the Periodic Table When we say that 1 unit of sodium plus 1 unit of water reacts what do we mean by units Chemists use all types of units to describe quantities of chemicals Below we have listed a few Mass Volume Grams g liters l 02 gallons roughly milligrams mg 0001 g milliliter ml 0001 I kilograms kg 1000 g 22 pounds microliter ul 10396 1 However when we are dealing with reactions such as those described above we need to work in units that are related to the number of molecules or atoms that are reacting or are produced We could certainly just say that 1 atom of sodium reacts with 12 molecule bromine to produce 1 molecule of sodiumbromide However this is not very much material For example 1 atom of sodium only weighs 4 x 103923 grams The most sensitive mass scales measure only a few nanograms 10399 grams or so and so even those scales are 14 orders of magnitude off 103923 grams is so very small that it is incomprehensible to most people Thus we would like to work with amounts that are comprehensible The mole is such an amount 1 mole ofthings is 6022x1023 ofthose things so 1 mole of sodium atoms weights about 23 grams This is much better 6022x1023 is called Avogadro 3 number and is represented in shorthand by NA How did someone ever come up with the de nition of a mole as 6022x1023 It turns out that 1 mole of carbon atoms weighs 12 grams exactly It is good to remember that 1 mole of things is a tremendously large number of things 1 mole of seconds is longer than the age of the universe However 1 mole of grams is approximately the mass of water on the earth We will nd that many things in chemistry are characterized by masses charges velocities etc that are of a scale that makes them very dif cult to comprehend Things like N A are used to make such physical quantities a little more manageable Scienti c notation Notice that above we have written that 1 ul 10396 Z We could have written 1 ul 0000001 1 however such notation becomes rather painful and unmanageable Imagine writing out Avagadro s number This type of preferred notation is called scienti c notation and if you have any problems with it you should review the supplement to the course text and the appendices in the text Now that we have a number the mole with which to deal with chemical quantities how do we relate that to our metric system units the gram and the liter In the front inside cover to your textbook you can nd a periodic table Let s consider one of the elements listed in that table Its entry looks something like There are three parts to this box At the top is the number 11 1 1 which is often shorthanded by Z and means the atomic number of the element In the middle is the abbreviation for the element N This particular box is for sodium Why is the symbol Na a Well sodium was one of the rst elements discovered and it was initially given the Latin name Natrium hence Na There are 22 0 other elements like this For example Tungsten is shorthanded W standing for Wolfram However most of the abbreviations of the elements make a little more sense It is an excellent idea to memorize the symbols for the first 3 rows of the periodic table I won t ever ask you to regurgitate this information but it will make your life easier Finally at the bottom of the box is listed the number 22990 This is the atomic weight of sodium and means that 1 mole of sodium atoms weighs 22990 grams or that 1 atom of sodium weighs 22990 atomic mass units or amu s Note that it is always an atomic weight For example in the above equations we referred to Br and H which don t exist as atoms but rather as diatomic molecules Their weight as given in the periodic table is for the atoms not the molecules So what are the molecular weights of Br or H They are simply 2 times the atomic weights ofthose respective elements For H this is 2 x 1 g 2 g and for Br it is 2 x 799 g 1598 g Molecular weights of more complicated molecules are calculated in the same way Let s use atomic weights and molecular weights to gure out how much material we would have to measure out to do our favorite reaction Na 12 Br 9NaBr Assume that we want to end up with 10 gram of product The molecular weight ofNaBr is 229 799 1028 gramsmole Therefore 1 gram of NaBr 1g1028g mole39l 97x10393 moles That means that we will need 97 mmoles 1 mmmole 103 moles ofNa metal and 972mmoles of Br gas So we need 97x10393 moles x 229 gmole weight Na 0222 g Na And 1297 x 10393 moles Br1598 gmol 0775 g Br Let s check our calculations We add 022 g Na to 077 g Br to get 0997 g NaBr I A Si i cant Figures 0 39 s and Wrong Answers It turns out that this number 099 g is the same as 10 g to the accuracy that we specified at the start of the problem As the matter of fact it should actually be reported as 10 g since we really don t know this number any more accurately than that One thing that calculators will give you is lots and lots of numbers to the right of the decimel place However we must constantly ask ourselves how well do we know the numbers that we are calculating This is not the same thing as how many figures that the calculator gives you At the start of this problem I specified that I wanted to have 10 grams of product I specified this number to the nearest tenth of a gram This means that any number between 095 and 104 grams would have been just fine 7 they are all the same to me since 095 rounds up to 10 and 104 rounds down to 10 Therefore even though a calculator will turn the 077 g Br listed above into 077503 Don t be fooled and lose points Your answer should be quoted only to as many significant figures as you know it Certain numbers that are used in algebraic equations like this are going to be absolutely known or known to a very high accuracy In this case we of course could look up the atomic weights of Na and Br to a tremendous accuracy and the 12 is known exactly as is the stoichiometry of the reactants and the products However since we asked for 10 gram of product we only need an accuracy of a tenth of a gram and we should only quote our answer to a tenth of a gram This is a way that many students lose credit on homework and test problems Signi cant gures Signi cant gures are so important that Ithink that we should do another example Assume that we have a compound of chemical formula C8H1102 and we want to burn it which means react it with oxygen However we only have 25 grams of oxygen How much of the compound can we burn The equation for this process is C8H1202 02 9 C02 H20 First let s balance this equation A good way to approach this problem would be to rst balance the carbons and the hydrogens on the left and the right sides of the equation After doing that then you will only need to adjust the relative amount of 02 to make the reaction stoichiometrically balanced Since we have 8 CS on the left and 12 H s on the left then we can do the following C8H12O2 XO2 9 8CO2 6 H2O Note that now the carbons and the hydrogens are balanced on both sides The amount of oxygen is not correctly balanced yet At left we have 2 2X 0 s and on right we have 16 6 24 O s Therefore X 11 and so the balanced equation is C8H12O2 1002 9 8 CO2 6 H2O Now recall that we only had 25 grams of 02 to burn 25 g 32 gmol 078 moles of 02 That means we can only burn 1100078 moles ongH1202 or we can burn 0078 moles Now the molecular weight ofthis compound is 81201gl2l01g 21600g 1402 gmole So l4ll gmol x 0078 moles 10936 g compound 11 grams ofcompound The underlined number is the one that should be quoted as an answer Make certain that you understand why Unitsa Units Systems and Algebra Notice in the above calculation that the units work out correctly This always has to be the case If the units don t work out then you most certainly have the wrong answer You can always use the units as a check to see if you have set up the algebraic equation correctly Although just having the units work out doesn t mean that the answer is correct it is certainly a necessary condition for getting the correct answer Lets work out an example with just units Recall from high school physics that kinetic energy E 12 massvelocity2 l2mv2 an equation that is hopefully familiar to you According to our units above we can use kilograms for mass Velocity is in units of distance per unit time If we use kg s for mass then we should use meters for distance and seconds for time This is the MKS units system also called SI units MKS stands for Meters Kilograms Seconds Another unit system is the cgs centimeters grams seconds system and energy in that system is in units of ergs So E 12 kgms2 Do these units describe an energy In the MKS units system the unit for energy is the Joule The de nition of the Joule is in fact kg m2 s39z Note that if we had used cms for velocity or some other measurement of distance over time we would not have ended up with a Watt If we had used cms for velocity then we probably should have calculated out the energy in the cgs centimeters grams seconds unit system Lets convert l J to an energy in the cgs system lJ 1 kg m2 s392 now multiply by the conversion factor for meters to cm s lJ 1 kg m2 s392100cmm2 104 kg cm2 s392 Now convert kg to g 104 kg cm2 s3921000 gkg 107 g cm2 s392 107 ergs There is a name for this unit and it is ergs 7 ie the erg is the energy unit in the cgs system Therefore 1 erg 10397 J There are other units for energy including the electron volt eV the kiloWatt hour kW h the british thermal unit BTU etc The units that you use depend upon the problem that you are solving Densities Now let s return to our favorite chemical reaction Na 12 Br 9NaBr Recall that in this reaction we are reacting a solid metallic element with a gas We have already discovered how to think about this reaction in terms of moles and how to convert from moles to grams and thus calculate the masses of equivalent amounts of reactant and product Often times chemists will not just consider the masses of the chemicals but their volumes also To do this we need to understand the concept of density Density is a massperunit volume and is usually described in the units of gcm3 Note that cm3 is a volume A cm is about as long as a trimmed fingernail so a cm corresponds to the volume of a box with each side about the size of a ngernail Let s list the densities for a few common substances Here we are not going to be exact but just provide rough estimates Br gas 006 gcm3 Water HO l gcm3 Organic materials 07 7 09 gcm3 Iron Fe 8 gcm3 Copper Cu 9 gcm3 Silver Ag 11 gcm3 Gold Au 19 gcm3 Let s say we had a set of 4 immiscible they don t mix but rather separate materials that we put into a closed container These materials were Br benzene liquid formula C5H5 water and Iron After we had prepared this rather bizarre mix we can use the densities to predict how it would appear on our benchtop The iron has the highest density so it sinks to the bottom Although the densities of benzene and water are very close to one another the benzene has a slightly lower density so it oats on top of the water For another example of an organicaqueous mix consider cooking oil on water or an oil slick Finally the density of the Br gas is much lower than any other component and so it fills the volume above the liquid surface Note that in the table listed above the densities of Cu Ag and Au are given and that their density increases in the direction of Au Now note their positions on the periodic table We will return to this later when we come to atomic structure So now let s convert densities of some substance into moles of that same substance Let s assume that benzene s density is in the middle of the range listed 08 gcm3 How many moles in 1 cm3 1 mole of benzene weighs 6 C atoms times 12 gmol 6 H atoms times 1 gmol 78 gmol 1 cm 08 g benzene so 08 g cm3 78 gmol 10 mmol cm3 or 001 moles How about for Br 1 cm3 006g Br so 0006 g cm393 160 gmol 38 umol cm393 4 x 10395 moles Note that equivolumes of a liquid contain approximately 103 times more molecules or atoms that for a corresponding gas Typically the densities of a liquid are only slightly more 20 or so than for the corresponding solid of the same material water is a very notable exception However when a substance is converted from a liquid to a gas its volume increases 103 Avogadro s Hypothesis 0n gases This brings us to a fundamental point concerning gases It turns out that equivolumes of two gases contain just about the same number of moles of those gases This equality was first noted by Avogadro Avogadro s hypothesis is Equal volumes of alz39jfkrent gases at the same temperature anal pressure contain equal numbers of particles Let s de ne this a little further by the following example Assume that you were going to burn hydrogen with oxygen to make water and you were doing this reaction at a temperature and pressure so that the reactants and products were gases The reaction would be Hz12029H20 Assume that this reaction goes all the way to products It does This reaction by the way is a very famous one 7 it was the reaction responsible for ending the age of the Zeppelin s ie this reaction was responsible for the Hindenburg disaster Anyway we have 15 units particles on the left and 1 unit of particles on the right According to Avogadro s hypothesis the volume of the product will then be 23 the volume of the reactants It turns out that this is indeed the case So we find that Avogadro s definition of a particle is either an atom or a molecule Does Avogadro s hypothesis hold for liquids and solids With the information thus far provided you should be able to prove that it does not However similar to gases volume is not conserved when liquids or solids react The structure of the Atom Recall that we defined 1 mole of C atoms as weighing 12 grams Let s check this definition by looking at the periodic table under C Wait We see that the mass of Hydrogen is listed as 12011 gmol not 120000 gmol Did we screw up Well let s consider the various components of an atom of carbon or an atom of any element for that matter It turns out that atoms are made up of protons positively charged particles neutrons no charge but roughly the same mass as protons and electrons negatively charged particles and much lighter than protons or neutrons Let s tabulate the masses and charges on these particles Particle mass charge Electron n1e 911 X 103931 kg 549X10394u 16 X 103919 Coulombs Proton mp 167 x 103927 kg 10087 u 16 x 1039 Coulombs Neutron mn 168 X 103927 kg 10078 u no charge How do we know these quantities Let s imagine setting up an eXperiment to measure these separate components of an atom Recall that Kinetic Energy 12 mv2 Now imagine that we could somehow smash apart some atoms and then we could give each fragment an identical amount of kinetic energy then we could determine the masses of those fragments by measuring their velocities Consider the eXperiment described in the picture below In this eXperiment an atom A is disassembled by hitting it with an imaginary hammer Remember we don t really have to do this eXperiment we can just think about it This is one of the luXuries of theory Now if we smack atom A while atom A is in the presence of an electric field then any charged components of the atom will be accelerated by the electric field In fact the electric field imparts the same amount of kinetic energy to each charged component that is produced This is just what we need So all we have to do is put detectors some distance away hammer smacks apart from where we smashed the atom atom A into Protons and we can measure how long it 7 Ctrons and Clemons took for the fragments to get to 39 the detector This is therefore a measurement of the fragment s A velocity and thus its mass The first thing that we would notice is that some of the particles are accelerated in one direction by the electric eld and some fragments Voltage are accelerated in the other applied here direction This tells us that we have two different charges present neutrons electrm travel 7 a negative and a positive charge promns remain faSt to ght We would also find that travel n e the positively charged particles SIOWIy p n gt 8 travel much more slowly than the to left P n e negatively charged particles and therefore they must be much Fi ure 2 Ex eriment to determine the various com onents g p p heav1er Finally we would add of an atom up our collected masses ofthe charged particles and we would compare the sum of those masses to the weight of our original atom We would find that we are missing some weight This means that some fragments were produced that were not affected by the electric field and thus those fragments are neutral By doing a series of careful experiments on a series of atoms we would eventually obtain numbers for the charge and mass that are associated with the electron the proton and the neutron Ithink that one of the most amazing things to consider is that it has only been within the past century that the electron was discovered by J J Thomson So now let s return back to the question of why Carbon apparently weighs 12000 gramsmol from the definition of Avogadro s number while it weighs 12011 grams according to most periodic tables Let s say that we could do the experiment discussed above on a whole bunch of carbon atoms looking at one atom a time What we would find is that the vast majority of carbon atoms would contain 6 electrons 6 neutrons and 6 protons However occasionally about 1 out of every 100 measurements we would measure a carbon atom with 6 electrons 7 neutrons and 6 protons This is called an isotope of carbon Recall that we discussed what the various numbers and symbols mean for a given entry in the periodic tableand that the number at the top of each entry was Z or the atomic number The atomic number refers to the number of protons or electrons in a charge neutral atom of the element described by the entry However many actually most elements have a number of different isotopes Hydrogen for example only has 1 proton and 1 electron If you consider the atomic weight of hydrogen you should be able to figure this out However there are two other isotopes of hydrogen 7 deuterium with on neutron and tritium with two neutrons Deuterium weighs 2 amu hence the deu in the name while tritium weights 3 amu and so the tri in the name Most isotopes found in nature are stable although some are not and will decay by the emission of neutron by fissioning in to new lighter elements or by other processes associated with radioactive decay Now let s return to carbon It has has three isotopes 12C 13C and 14C When one is writing out elemental symbols and specifying particular isotopes then at the top left of the element symbol one writes the total number of neutrons protons contained by that element The molecular weight quoted in most periodic tables is the isotopically averaged molecular weight For carbon for example this means that it is going to be something like 99 X 120000 1 X 13000 001 or something X 14000 which turns out to be 12011 amu The isotope 14C is of particular interest because it is taken up by living organisms but it is not taken up by dead organisms It is also a radioactive isotope with a halflife of I m guessing 6000 years This means than in 6000 years half ofthe element will have decayed into something else 7 usually something that is not radioactive Therefore the amount of 14C that one measures in a fossilized lifeform yields information about when that lifeform was alive The process of dating fossils by measuring the ratio 14C 12C is called radiocarbon dating and was invented by Bill Libby who was on the faculty in this department and who won the Nobel Prize in Chemistry for discovering this technique in 1961 Atomic Structure So now that we know the ingredients of an atom what is the structure The structure of an atom turns out to be a wonderful and surprising thing Knock on a piece of wood or your notebook or your desk Feels pretty solid huh That is eXactly how most scientist thought of the structure of atoms up to the turn of this century Atoms were solid billiard balls packed together in a tightly configured lattice Fig 3 closestpacked billiard balls Consider the Figure 3 Notice that there is not much dead space between the billiard balls and if one stacked the billiard balls to be many layers thick then there would be no way to shoot a bullet through the lattice and not hit a ball even if the bullet were infinitely small However Rutherford had a good idea He decided to take a very thin metal foil and send very small bullets in the form of electrons through the foil In this way he could confirm the atomic billiard ball model The electrons should be re ected back or should bounce off at random angles rather than passing through the lattice However what he found is that by far the vast o o majority of electrons passed through the metal film as if nothing was there The very few that didn t pass through the metal film were sharply de ected It was as if the electrons either hit nothing and this was most of the electrons or they hit a very solid billiard ball This indicated that matter was not made up of continuous stuff but was very unevenly Fig 4 7 the nuclear model ofthe 2mm distributed Most of the mass was concentrated into very tiny regions much smaller than individual atoms Thus the model that was presented for the atom was the nuclear model given in Figure 4 In this model the neutrons and protons are all concentrated in the middle while the electrons ll the space between the nuclei This drawing greatly over exaggerates the size of the nucleus relative to the atomic radius The nucleus is really really small How small7 we don t really care It is just small Remember that Why is this important for chemistry What this means is that when atoms encounter one another it is really clouds of electrons encountering one another In fact protons and neutrons are relatively unimportant in chemistry and we will really not mention them very much the rest of this course All of chemistry is really determined by how diffuse clouds of electrons encounter one another For this reason we will spend quite a bit of time discussing just what is the nature of those diffuse couds of electrons Are they random or is there some order to those couds etc You will note that the book launches right into chemical structure and bonding and periodic trends relevant to the periodic table We aren t going to do that Ionic and Covalent Bonding Mostly ionic and part of Chapter 13 Before we move on to the details relating the electronic structure of atoms let s finish up our chemistry review by discussing the various types of chemical bonds Once again consider the reaction Na 12 Br 9NaBr Na metal is a covalent solid Brz gas is a covalently bound diatomic molecule and NaBr is an ionic solid In a covalent bond electrons are shared between the atoms and each atom has the same number of electrons associated with it that it does as a neutral isolated atom In an ionic solid or molecule however one or more charges are donated by an atom or a part of the molecule to another atom or another part of the molecule The molecule remains charge neutral overall but charge within the molecule is separated For example NaBr might also be written NaBr39 Note that the number of positive charges in this molecule is equal to the number of negative charges 7 the molecule is charge balanced We are a little bit away from being able to rationalize this behavior However we can use the principle of charge balancing within a molecule to predict how various ions will combine with one another Take the following examples of ions At right is shown how the negative and positive ions will combine with each other to form a charge neutral ionic molecule Cation Anion Charge Balanced Molecule Nat 0239 NaZO NaCl NazCO339 NaBr Kt Cl39 K20 KC1K2CO3 KBr Ca2 Cng39 CaO CaClz CaC03 CaBrz Fe3 Br39 FezO3 FeCl3 FezCO33 FeBr3 In an ionic molecule the two counterions are held together by Coulombic forces 7 which means forces that are related to interacting charges If we consider bringing two oppositely charged ions together then one can imagine that they will attract each other opposites attract and if we bring likecharged ions together they should repel one another It turns out that this energy of attraction or repulsion scales as 1r where r is the distance that separates the ions A more accurate equation looks like E Q1Q241T801 where Q1 and Q2 are the charges on ions 1 and 2 and so is a fundamental physical constant called the permittivity of the vacuum It is equal to 8854 X 103912 CZJ39lm39l The charge on an ion is Ze where Z is the number of charges it has a sign and e is the fundamental unit of charge 7the charge on a single electron e 160 X 1039 9 C We use meters for the separation between the ions r So now just plugging in units we have E CoulombsCoulombsCoulombszl39lm391m Joules an energy Good Notice that if Q1 and Q2 are of opposite sign as for an anion and a cation then E is negative This means that the energy of the system is reduced when the ions are brought together However if Q1 and Q2 are both of the same sign two cations or two anions then E increases as the ions are brought near each other Let s plot the energy for these two cases de ning E0 as the case for when the ions are very far apart from each other Thus when oppositely charged ions are brought close to R3 one another the energy of the system is decreased However if those ions are of like charge then the energy system is increased This energy is called the potential energy Imagine the following analogy Let s take a spring As we compress the spring we put energy into it and that energy is stored 7 ready to cause the spring to stretch back to its initial configuration The case is similar here If we compress likecharged ions together it is similar to compressing the spring As soon as we remove our tweezer or whatever it is we are using to press the ions together then the ions will quickly move away from one another 7 turning that potential energy into kinetic energy Similarly if we have opposite ions that are far apart then they have a large amount of potential energy This is like an expanded spring As soon as we release the ions they will quickly move toward one another and the potential energy of the system is lowered Thus the two curves drawn here are called potential energy curves Now there is something interesting that we can do with this graph Let s say that we put a cation and an anion at a distance corresponding to the dot at r ri on the attractive potential energy curve We know that if we release them then they will begin moving toward each other and so as we hold them we need to exert a force to keep them apart How much force is necessary It turns out that if we move the ions just a little bit ie we change ri by a small amount Ar then the potential energy will change by a small amount AE Now if we calculate AEAr we are measuring the slope of the potential energy curve in the vicinity of the dot r ri Some of you may recognize this as evaluating dEdr at ri or taking the derivative of the potential at ri This slope AEAr X QlerE Energy a Qle lt 0 Attractive Fig 5 7 The Coulombic Potential Energy for ionic interactions Lecture Series 4 Chemistry 20A Professor Jim Heath The discussions in Oxtoby Chapter 13 137 relating to periodic trends are pretty good Although I cover them in class I won t add anything on them in our notes You can skip the little bit on Interstellar Space p 521 Chapter 2 Sections 22 and 23 Take a look the column labelled IV of the periodic table on p 52 of your book We have the elements C Si Ge Sn and Pb meaning carbon silicon germanium tin and lead Now forget about lead Pb We are going to point out some relationships between these elements and lead turns out to be an exception to the rule In many ways the chemistry of the elements is rationalized in a way that is similar to grammatical rules for using English The most important rules of chemistry a lumped into the periodic table However like grammar there are always exceptions to the rule In the following discussion of the structure of the Group IV elements Pb doesn t fit Before we get into the structure however let s recall what we know about these elements Based on the Aufbau principle and our atomic wavefunctions we can pretty much guess what the electronic configuration of these elements is going to be Carbon is lsz 2s2 2p2 Si is lsz 2s2 2p6 3s2 3p2 Ge is 1s2 2s2 2p6 3s2 3p6 4s2 3d 4p And Sn is 1s2 2s2 2p6 3s23p64s23d10 4p6 5s2 4d 4p2 2 All four of these elements form stable structures in which 1 atom is bonded to 4 others as is shown in Fig 1 Why should their bonding patterns be so similar The lattice shown is called the diamond lattice by the way It turns out that even through the electronic structure of these atoms differs substantially 7 Sn contains 50 electrons while C only contains 6 7 it is the outer or valence electrons that really matter when it comes to bonding We can recategorize the electronic structure of the atoms as C He2s2 2p2 3 Si Ne3s23pz quot Ge Ar4sz3d104p2 Sn Kr5sz4d105p2 4 Fig 1 7 the structure of the diamond lattice The central atom is bound to 4 others arranged as shown at the corners of a square box All atoms are equivalent in this structure Bonding How to predict The best clue to trying to predict bonding is to consider a rare gas atom 7 take Argon for example Argon has a completely closed electronic shell It doesn t react with anything It is a happy atom So you might ask how does an inert atom tell us anything about chemical reactivity and bonding It turns out that Ar and the other rare gases 7 also called inert gases are the ideal supreme atom beings of the elemental world They have reached electronic nirvana What does it mean to reach electrical nirvana For the element He it means to have 2 electrons or a completely lled ls shell For Ne it means to have a completely lled 2s and 2p shell including of course a lled ls shell Neon has 8 valence electrons 7 2 selectrons and 6 pelectrons Ar also has 8 valence electrons 7 a completely lled 3s and 3p shell The next one up Kr has 18 valence electrons 2 4s electrons 6 4p electrons and 10 3d electrons Thus we may de ne nirvana for an atom to be a shell of 8 or 18 valence electrons depending on where it is on the periodic table Going back to our diamond lattice system 7 each of the atoms involved had 4 valence electrons or 14 if we count count the delectrons for Ge and Sn To reach elemental nirvana these atoms would like an additional 4 electrons or they would like to get rid of the four electrons that they have It turns out that in the diamond lattice solids they sort of do both things Each atom shares its four electrons with 4 other atoms and it borrows l electron from each of those 4 atoms at the same time This is kind of confusing but it is major issue behind what are called Lewis Dot Structures Consider the carbon atom and its four electrons One way to draw this concept of an atom is as In this gure we denote each of the valence electrons as a single dot uniformly surrounding the atom Now if we were to surround this carbon atom with four other likedotted carbon atoms we would have the electronicequivalent of graphite although you may note that there is really no true structural information in these graphs yet We ll get to that In the case of a carbon atom surrounded by four others we would draw a new structure that would look like In this 5 carbon atom representation we have tried to point out the central carbon atom and its C new shell of both the dots and electrons by changing the color of q I C the other carbon 0 the elemental symbol from that of 7 atoms Now notice that the central C surrounded by 8 electrons 7 it is in However does it now have a carbon atom is elemental nirvana charge of74 No Each of 8 electrons is a shared electron meaning that it is part of the central carbon O atoms shell and it is also part of one of the neighboring atom s shells Note however that all of the electrons surrounding the central atom are colored the same In fact there is no way to distinguish these electrons from each other Once we form a solid the electrons are truly shared We no longer can identify which ones were initially associated with the neighboring carbon atoms and which ones were initially associated with the central carbon atom Now how about the neighboring carbon atoms They appear to have 5 electrons 7 needing 3 more to complete their respective trips to nirvana In fact each carbon atom in the diamond lattice is surrounded by 4 other atoms Thus as we alluded to before we have shown an incomplete structure All carbons have a full octet of electrons The point here is that carbon has 4 electrons If it can somehow nd and share 4 more then it will have an octet Where can it get those extra four Well we already see that it can get them from 4 other carbon atoms It can also get them from four atoms from the following series H F Cl and Br Notice that all of these atoms lack just 1 electron to have a complete shell H is lsl while F Cl and Br are nsznp5 valence atoms If H gets its extra electron it is isoelectronic with He If F Cl and Br get their extra electrons they become isoelectronic with Ne Ar and Kr respectively Since each of these atoms can get that extra electron by forming a bond with a carbon atom each of those atoms will Fig 3 Two representations of CH methane The Lewis enter elemental nirvana When the dot structure at left shows that both carbon and hydrogen bond is formed For example in have fully satisfied valences The structure at the right is Figure 3 we represent both the grawn t show the similarity between methane and Structural picture and the Lewis 1amon dot drawing of methane The structural representation that we have used the Lewis dot structure also tells us a little about what we can expect in terms of chemical structure We have already alluded to the fact that for the case of the molecules and solids we have already discussed a tetrahedrally coordinated 4bonded carbon atom sits at the center of a box as shown in Fig 3 For the system shown in Fig 3 we say that the coordination number of the carbon atom is 4 The coordination number of each hydrogen atom is 1 since they each form 1 bond Now each bond represents a region of heightened electron density 7 ie there is increased electron density along the line connecting the carbon atoms to one of the hydrogen atoms At the four comers of the box that don t have a hydrogen atom the electron density is lower We know from our previous work that electrons or any particles of like charge repel each other through a Coulombic repulsion Thus the four bonds want to be as far separated from one another as possible The optimal solution is then to arrange the 4 hydrogen atoms at the corners of the box as shown The angle that describes any one ofthe HCH bond angles is 10950 and such an angle is indicated in Fig 3 We will come back to these geometries later but let s do more Lewis dot structures Consider the molecule NH3 which is ammonia Can we make a Lewis dot structure of this molecule that satisfies all of the valences Yes and this is it r L Note that the number of electrons adds up here to be that of 3 H iquot N H hydrogens 1 each to give 3 total and one nitrogen 5 to give 8 electrons for the molecule How about the extra two electrons on the H nitrogen atom that aren t involved with any hydrogens We call those electrons lone pairs In terms of the octet rule each lone air counts for two electrons Let s take another familiar molecule H0 or water 39 Note that water has two lone pairs and it still has an octet of electrons Let s move one further down the periodic table and make the molecule HF or hydro uoric acid It looks like HF has 3 lone pairs and only 1 hydrogen If we tally up the electrons on water and HF we nd that they like ammonia are equal to the sum of electrons from oxygen or uorine plus 2 or 1 electrons from hydrogen atoms Hydrogen by the way doesn t need a full octet but it has a satis ed valence anyway just like the N O and F atoms in these structures With these three structures we can discuss the issue of formal charge Formal charge is the amount of charge on an atom and may be tallied up in the following way For each shared pair of electrons each atom involved in sharing that pair gets 1 of those two electrons For each lone pair the atom that has that lone pair gets 2 electrons For example we tally up the electrons on the C N O and F atoms in the methane ammonia water and HF structures presented here TO get the formal charge we molecule shared lone total formal subtract the total number of electrons e s pairs e S charge from the atomic number Z Recall CH 4 4 0 4 0 that the atomic number tells us how NH3 3 1 5 0 many electrons are associated With a H O 2 2 6 0 given atom In the case of this series 2 of molecules each atom has a formal HF 1 3 7 0 charge of 0 and each atom has a satisfied valence How about the molecule NH4 If we make a Lewis dot structure this PI is what we get since it shares Note that this nitrogen now has a formal charge of l 4 electrons has no lone pairs and Z 5 The hydrogen atoms each only have 1 electron and so their formal charge is 0 H Thus the sum of the formal charges is 1 and NH4 is a cation or a positively charged ion and we call it the ammonium ion Note that even though the formal charges do not balance to 0 this molecule is characterized by atoms with completed octets or valences and thus we expect that it will be a relatively stable cation Let s move away from these relatively simple molecules and look at the Lewis dot structures of a few more molecules Consider the molecule 80 Can we draw a Lewis dot structure for this guy Yes but it is substantially more com licated than the structures we have been drawing up to this point We show it at right Note three things here 1 We have a new type of bond 7 one that involves 4 electrons not just two 2 We have drawn two exactly equivalent structures 3 All atoms have completed octets Let s talk about point 1 first In order to generate satisfied valences on all of the atoms we introduce a new type of bond 7 the double band This bond is similar to the previous bonds in which just a single pair of electrons were shared and so we now call that previous bond a single band In terms of terms of the octet rule we treat the double bond similarly to the single bond all electrons involved in the double bond add to the octet around the involved atoms In terms of tallying up formal charges the rules are again similar The electrons involved in the double bonds are split evenly between the two atoms For the case of a single bond that means each atom gets 1 electron and for the case of a double bond that means that each atom gets 2 electrons counted towards its formal charge There are also such things as triple bonds consisting of 3pairs of shared electrons The molecule HCzH has one of these Work out the Lewis dot structure to convince yourself that this is so Bonds up to hextuple bonds 6 pairs of shared electrons have actually been observed but you will really only have to deal with single double and triple bonds at most Now we get to point two There are two ways to draw the SO structure 7 in the top structure we put a formal charge of 7l on the left oxygen atom and on the bottom structure that charge is on the right oxygen atom Be sure that you can generate the formal charges that are indicated on the figure Both structures are exactly equivalent 7 we couldn t do an experiment that would distinguish between them In reality the structure of the molecule will be continuously moving between these two structures and we say that the two structures are in resonance with each other and so they are resonance structures It turns out that the molecule will generally get a little bit of stabilization from having resonance structures Finally we get to point 3 7 all atoms have completed octets You might be wondering when you would predict that molecules would have resonance structures double or triple bonds etc It seems like a lot to do to memorize all of the molecular structures In fact you shouldn t have to memorize anything Just work out the Lewis dot structures keeping in mind that you are trying to satisfy the octet rule for every atom involved As long as you allow for single double and triple bonds you will naturally generate the correct structure with a few exceptions discussed below Once you have the structure tally up the formal charges on the atoms Check for the possibility of resonance structures Returning to the molecule 80 there is another possibility for drawing it Since all the atoms involved have the electronic configuration of nsznp4 they all have 6 valence electrons An alternative pair of structures would be those shown on the right These molecules are structural isomers of the molecules shown above In addition these two structures do not represent resonance structures Note that in the OSO structures if we take one of the resonance structures and rotate it by 180 then we generate the other resonance structure That is not the case here If we rotate the SOO structure by 180 then we are still left with the same structure it is just rotated ie the double bond is still between the oxygens not the sulfur and the central oxygen Thus both of the 00 structures are structural isomers of each other When we move the double bond from the 00 and make it an 80 double bond we are changing the structure of the molecule Let s look at a few more Lewisdot structures Take for example N03 Is there anyway to predict which anion negative ion will be most stable ie what is the value of n Yes What we need to do is to work out Lewis dot structure possibilities for this ion The ion that has the least amount of net charge and has a completed set of octets is going to be the winner From now on we will denote a double bond with an symbol we will denote a single bond or a lone pair with an symbol and we will still denote an unpaired electron with a dot In Fig 4 A we shown a Lewis dot structure of N03 that appears to be perfectly satis ed Each atom has an octet of electrons Each oxygen atom is equivalent to every other one 7 characterized by a single bond to the N atom and by a 7l redrawn with bars charge The ion is a triam39tm B called so because it has a net charge 1 1 of3 In Fig 4 B we present a 0 1 391 1 0 second possibility In this case there is also a satis ed valence of 8 I I electrons associated with each 1 1 atom However there is also one F394V39 trturfNO 39Tht A double bond Since that double 1g anouss uc es or janlon 6 Op structure represents a trianion 3 charge while the bond can be drawn as any one Of bottom structure B is a monoanion 1 charge the three NO bonds there are three Two resonance structures are shown for B a third resonance Structures only two resonance structure is not shown such structures are known So which anion is the most stable 7 N03339 or N03 The answer is N03 for a couple of reasons Both anions have satis ed octets but the amount of charge on the N03 anion is less Lower charge is better In addition the N03 anion has 3 resonance structures 7 resonance structures generally denote some sort of enhanced stabilization Breakdown of the octet rule p 6263 of the text is good for this Lewis Dot structures of Organic Compounds We have already discussed one of the simplest organic compounds methane Organic compounds are really where Lewis dot structures work best The following bonding principles should be kept in mind 1 All atoms always have satis ed valences octets for everything except H 2 Carbon 7 carbon bonds can be CC CC or CEC Carbon 7 hydrogen oxygenhydrogen and nitrogenhydrogen bonds are always CH OH or N H Carbonoxygen bonds are either CO or CO and carbonnitrogen bonds are either CN CN or CEN Carbonhalogen a halogen is either F Cl Br I are always CX where X denotes the halogen You should be able to apply these simple rules to assign the Lewis Dot structure of just about any organic molecule that you might see The chemical formulae of organic molecules can be quite long and complex but the structures turn out to be much more predictable With these simple rules in mind let s assign the Lewis dot structure to the following molecule shown in Figure 5 Try drawing this structure in another way and you will nd that you can t and still obey the above rules These types of structures are Chemical Formula CH3COCHCHOCHC1CHNH Lewis Dot Structure no lone pairs drawn Figure 5 The chemical formula and the Lewis dot structures of an organic molecule very important for organic chemistry and for biochemistry as well as for materials science Much of the structure is implied by the way I have written out the chemical formula If I hadn t and instead had written C5H8NOZ you would have been able to produce many possibilities For example consider the molecule that has the chemical formula C5H5 Various possibilities are worked out in Figure 6 Notice that in all of these cases Figure 5 and 6 that the formal charge on the carbon atoms is always 0 regardless of how it is bonded to other atoms Also notice in Fig 6 that the top I 1 HCCECCECcH H H Fig 6 Various structural isomers that are satisfied Lewis dot structure versions of the molecule C6H6 Note that the top isomer benzene has a second resonance structure that isn t shown structural isomer benzene has an identical resonance structure in which the double bonds are rotated by 60 You shouldn t have to worry about naming these structures and the detailed nature of the chemical bonds involved here will be discussed in another section However if I give you a formula for an organic molecule you should be able to generate a satisfied Lewis dot structure If I give you a formula such as CxHyOZ meaning that the molecule contains X C atoms y H atoms and z 0 atoms you will probably be able to generate a few of Lewis dot structures depending on the molecule If I give you a more structural formula such as that given in Figure 5 then you will probably only be able to generate a single satisfied Lewis dot structure Supplement to Lecture Series 4 Periodic Trends We breezed through periodic trends and so now we will backtrack and try to cover them in a little more detail The idea with any periodic trend is to take some atomic property and plot its value vs atomic number Z and then to try to rationalize the plot you have made with the periodic table It turns out that this type of plot coupled with the knowledge that we already have on the Aufbau principle atomic orbitals etc can give tremendous insight into how chemistry works Let s actually put a plot together but do it backwards In other words we will use the periodic table and our knowledge of what atomic orbitals are filled for various atoms to generate the plot We will plot the ionization potential vs Z and we will start armed with knowledge from Chapter 13 concerning atomic orbitals and we will also know our rst point on the graph The IP of Hydrogen Zl is 136 eV The reasons that the IP is such a good property to correlate with the periodic table is because it is a measurement of the stability of an atom A very stable atom such as a rare gas has a very high Ionization Potential 7 if an electron is removed then it loses the stable closed shell of electrons that it had as a neutral species Let s start lling our chart with IP s The rst element H has an IP of 136 eV Hydrogen is one electron shy of a closed ls shell and so it is not a particularly stable atom The next atom is Helium which is of course a rare gas atom It is very stable and it should take a lot of energy to remove an electron Thus we predict that its ionization potential will be much higher than that of hydrogen Let s guess that it is just about but not quite 2 times higher Why do we guess this Well the Coulomb potential that describes the electron s interaction with the positive nucleus now has to account for a nuclear charge of 2 Thus if we assume that the Coulomb potential is the only thing that we care about this implies that the electron will be bound twice as tightly as it is for hydrogen However the other electron the one we aren t going to remove is shielding this 2 charge a little bit so that the electron we are going to remove probably sees an effective nuclear charge that is a little less than 2 Since twice the IP of H is 27 eV let s estimate the IP of He 25 eV Now we come to the atom Li Li has one electron more than the closest inert gas shell He so it should be easy to ionize Is it as easy to ionize as H It is probably easier The electron we are going to remove is in the 2s shell which extends farther out from the nucleus than does the ls shell Does this electron see a nuclear charge of 3 If so then it should be very dif cult to ionize In fact it only sees a nuclear charge of 1 Two of the 3 positive nuclear charges are completely cancelled out by the lled ls shell Thus we have an electron orbiting a single positive charge at a distance that is greater than the case for hydrogen The electron should be easier to remove Let s say that the ionization potential is 8 eV Now we come to Be Be is not a closed shell in the same way that a rare gas atom is but it does have a lled 2s shell something we will refer to as a minor shell closing Thus it will be a little harder to ionize than Li Let s say that the ionization potential is 10 eV Now we come to B If Boron loses l electron it has a lled 2s orbital 7 a minor shell closing It should actually be a little easier to ionize than Be Let s call the ionization potential 9 eV Now we come to the series C N O and F As we add more and more electrons we work our way closer and closer to a closed shell In addition each ofthese atoms sees a slightly increasing nuclear charge For example carbon is 2s2 2p2 Each 2s electron completely shields 1 unit of nuclear charge Since Z6 there are 6 protons in the nucleus 2 of the are shielded by the ls electrons and 2 more are shielded by the 2s electrons so we are left with 2 remaining Now there are 2 2p electrons and they do shield each other from the nuclear charge but not completely Thus we expect a relatively high IP for carbon For oxygen we have a net 4 nuclear charge after we account for all of the ls and 2s electrons This 4 charge is attracting 4 2p electrons Lecture Series 8 Covered for Final Descriptive Chemistry of Organic Molecules Macromolecules and Polymers Remember how I have stressed time and time again that I didn t want you to have to memorize anything Unfortunately you will probably have to memorize some stuff here I am not really certain how to enhance what is given to you in Chapter 23 other than to perhaps highlight the stuff that I feel is the most important H H H I I I npropane H C C C H H H H npentane H H H H H H C C C C C H H H H H H Let s rst try to make a little sense of some of the stuff that Oxtoby gives you For example in the rst few pages especially Figure 235 and the related discussion it is stated that organic molecules that are larger are characterized by higher boiling points than smaller ones provided that the types of organic molecules being compared are very similar For example take the following two molecules npropane and npentane It turns out that npropane is a gas propane stoves for example while npentane is a liquid that boils A van der Waals 0 interaction distance I is 35 Lu Bond Energy chemical bond distance Figure 2 A van der Waals potential compared to a normal bonding potential Note that vdW minimum is at a larger internuclear separation than for the chemical bond The depth of the vdW curve ie the strength of the interaction is exaggerated here around 65 0C If we made noctadecane which isjust like the two molecules at left but contains 18 carbon atoms all arranged in a row we would have a waxy solid that boils around 200 0C So what we want to explain is why does higher molecular weight ie larger molecules translate into higher boiling points It turns out that molecules attract each other via weak interactions A bonding interaction or a Coulombic interaction is a strong interaction However things that don t bond to one another still attract each other through what are called dispersion forces or van der Waals interactions How strong are these interactions It depends on the system of course but they are typically 100 times weaker than a chemical bond These are the forces that hold a solution of pentane molecules together or a solid of octadecane molecules together After all the solution doesn t y apart spontaneously We compare a van der Waals potential energy curve of attraction with that of a chemical bond in Figure 1 Note that both curves have a repulsive part that is about equally steep This makes sense Consider a liquid of molecules held together by van der Waals attractions and a solid of atoms held together by ionic or covalent chemical bonds Neither material is particularly compressible That is why the repulsive part of the potential appears the same for both systems Recall that we discussed that we could eject an electron from an atom ionization of the atom by supplying energy 7 possibly the form of a photon We could also supply energy in the form of heat but we would have to get an atom pretty hot before it lost electrons It turns out that because van der Waals interactions are relatively weak we only have to heat up a solution of molecules a relatively small amount to break it apart 7 or in other words to get it to boil It turns out that larger molecules are characterized by stronger van der Waals attractions than small but similarly structured molecules Why Think of the following eXperiment Take a small 1 cm2 square of glass and put it on top of a second square with a drop of water sandwiched in between Now pull don t slide the two pieces apart off They will stick a little bit but you should be able to separate them without too much trouble Now take a large 1 m2 plate of glass and put it on another plate of glass once again sandwiching a small amount of water in between the two glass panes Try to pull the two plates apart now You probably won t be able to do it You may even break the glass before you get two plates separated Obviously the interactions between the various pieces of glass are the same there are just more of them when the glass pieces are larger This is the same situation for molecules The van der Waals interactions per unit lengl h of the molecule are nearly identical for two interacting propanes or two interacting pentanes or two interacting octadecanes However the interactions add up for the longer molecules Thus a solution of octadecane is held together more strongly than is a solution of pentane than is a solution of propane The net result is larger molecules boil and melt at higher temperatures The molecules we discussed above are nalkanes The n pre x indicates that they are linear and the ane suf x indicates that all carbon atoms are sp3 hybridized which means that there are no double bonds Obviously there are many different types of organic molecules of which alkanes are the simplest Even if we just stay with the two elements C and H we can generate several different types of organic molecules We have already discussed linear alkanes There are also branched alkanes Once again all carbon atoms are sp3 hybridized but they are not bonded in a linear arrangement A really painful thing that we are going to have to do now is to develop a nomenclature so that when we refer to an organic molecule the name that we call it tells us what its structure is Consider the molecule shown in gure 3 which is named 36 dimethyl octane To name this molecule we count the longest continuous carbon network that we Figure3 3ethyl 36dimethyloctane H H C H E I HCHH C H n HIJEDH H H c 32 H H H H H H H H IC H H C H IH Figure4 can regardless of how the molecule is drawn When we do this we nd that we come up with an uninterrupted chain of 8 carbon atoms Since there are no double bonds and the only atoms present are carbon and hydrogen it is an alkane The 8 carbon atom chain makes it an octane If we number the carbon atoms as we count them then we can locate the positions along the carbon backbone where the extra hydrocarbon branches originate We nd that a methyl group 1 carbon atom 3 hydrogens originates at carbon 3 and a second one originates at carbon 6 Since there are two methyl groups and they are at positions 3 and 6 we call it 36dimethyloctane Let s do one more 7 the molecule shown in gure 4 Here we once again have a branched octane molecule We name this molecule in the following way we number the carbon atoms starting at the terminal carbon that is closest to the rst branch In this case it doesn t matter since carbon 3 from both directions is the rst to have a branch Notice that the numbering is now reversed from Figure 3 although either direction would have been ne Now we refer rst to the most compleX branch which is the ethyl group at the 3position Then we refer to the 2 methyl groups at positions 3 and 6 and then we call it an octane Thus the final name is 3ethyl 36 dimethyloctane Isn t this exciting Hold on to your seats It gets even better Rather than go through a tedious list of nomenclature rules I will simply de ne some of the terms that you need for the problems that you are assigned If molecule that contains hydrogen and carbon has at least 1 double bond and no triple bonds it is an alkene If such a molecule has at least one triple bond it is an alkyne Alkenes and alkynes don t occur naturally in great abundance they they are very useful for fuels and as starting materials for making other organic molecules Alkanes are pretty darn inert However it is possible to react alkanes at very higher temperatures and break them down This process is called cracking Cracking When a long linear or branched alkane is broken into smaller components the process is called cracking For example it is often desirable to have relatively short branched alkanes and alkenes for use as fuels Such cracking is done by heating the alkane to very high temperatures several hundred degrees Celsius in the presence of some sort of catalyst When an alkane is cracked the result is a shorter alkane plus a shorter alkene For example if we crack hexane into methane and a 5 carboncontaining molecule we get the following reaction H3CCH2CH2CH2CH2CH3 9 CH4 CH CHzCHzCHzCH3 Note that the number of carbons and hydrogens is not changed by the cracking process Now try to draw the Lewis dot structure for the molecule at the far right It turns out that you can t do it You will have to rearrange the hydrogens a little bit and even then you will have to include a double bond The product molecule looks like CH2CHCH2CH2CH3 So if you crack an alkane branched or straight you will get an alkane and an alkene Cyclic Alkanes and alkenes It turns out that there is a third type of structure for a molecule that contains only hydrogen and carbon and that is a cyclic structure It is actually possible to make cyclic structures from hydrocarbons that contain 3 or more carbon atoms However for n 3 4 or 5 such structures are pretty strained Because of the strain cyclopropane C3H6 is very unstable cyclobutane C4H8 is unstable cyclopentane C5H10 is sort of unstable and cyclohexane C6H12 is stable Why are they strained Well recall that for an sp3 hybridized carbon each atom that is bonded to the carbon is separated from every other atom by a bond angle of 10950 All of the cyclic molecules mentioned above are sp3 hybridized carbons However cyclopropane has to have bond angles of 120 cyclobutane has to have bond angles near 900 it can distort from a square geometry Cyclopentane is a better 7 it s bond angles aren t too different from 10950 but it is still a little strained Only cyclohexane can satisfy the requirement of having 10950 bond angles These cyclic alkanes and alkenes also have a nomenclature associated with them We will just do one here problem 6a to give you a feeling for the subject In this molecule we number the carbons around the ring in a clockwise direction starting with the first carbon that is involved in the double bond Thus there are two methyls at the 2 and 3 positions respectively and l i double bond starting at the lposition 2 3 1 CyCIObutene This gives the molecule its formal name CH3 CH3 Table 233 in your text lists some of the most important functional groups in organic chemistry including alkyl and aryl halides alcohols phenols ethers aldehydes ketones carboxylic acids esters amines and amides You will undoubtedly become more and more familiar with these functional groups as the course progresses One of the things that should stand out here is the versatility of the carbon atom in terms of its chemistry No other element will form such a broad range of compounds with such a large number of other atoms No other element even comes close This is largely due to the fact that carbon readily forms double and triple bonds 7 something that no other element does with any regularity Figure 5 The cyclobutene structure referred to in the text Note that we have begun to utilize a shorthand notation for referring to this molecule All of the bonds sticking out from the cyclobutene are bonds to Hatoms except where the methyl groups are shown All vertices ie the 4 comers of the tetrahedron are carbon atoms This type of shorthand notation is very common in organic chemistry and we will use it more and more throughout this course A Brief Survey of Some Organic Chemical Reactions For many of the chemical reactions that are important in organic chemistry a catalyst plays an important role We already referred to the fact that a catalyst was present in the cracking of alkanes In that case the catalyst is a solid piece of alumina or aluminum oxide A1203 You may have encountered this material before in the form of sapphire 7 a clear colorless gem What does a catalyst do Let s make an analogy with something that we covered in our last material 7 the formation of an ionic bond Recall that if we wanted to form a molecule AB that was characterized by an ionic bond so that it was AB we first had to do the following two reactions A9 A e cost the ionization potential of A B e 9 B we get back the electron affinity of B Preparing the ions always cost a certain amount of energy We got this energy back when we formed the ionic bond as the potential energy gain from the Coulombic interaction of the two ions was more than sufficient to overcome the energy we had spent preparing the ions Thus we could represent that reaction in the way shown in Figure 6 In this diagram we show the sequence of steps as proceeding from right to left At far right we have our atomic precursors As we A B proceed to the left we generate the IPEA lt cation and the anion from A and B and the energy of the system is A B raised since the IPEA is a 0 positive number At the far left 3 we have brought the two atoms 0 COUIOHIbiC together and formed an ionic bond g Interaction There is an energy gain and so the L1 energy energy of the system drops to that of the bound molecule We show AB the various energies of the system Bond by making the yaXis an energy energy scale Now let s consider something that looks like a very Figure 6 Energy diagram for forming an ionic bond different process and that is the reaction of H20 with CH2CH2 to produce H20 gt H OH G CHZOHCH3 This reaction is called the hydrolysis of an alkene and is discussed in I I your text on p 813 Ifwe H C C gt C 39 C took this reaction in reverse I I so that we started with H ethanol and ended up with I O ethyene and water we would I I call it the dehydration of C C oOH gt C C ethanol Now consider one I I I I possibility for this hydrolysis Figure 7 the stepwise hydrolysis of ethene Ef gterE ctsigzv ig 62656 1 y favorable which means that H20 CH2CH2 is less stable than CH3CHZOH However to get the reaction to proceed at least according to the reaction mechanism listed in Fig 7 we need to break apart the water molecule first 7 a step that costs a lot of energy This is similar to the formation of the ionic bond that we went through In order to form the ionic bond we first need to ionize the participating atoms creating a cation and an anion Thus similar to Fig 6 we can draw an energy pathway that describes the mechanism of Fig 7 We present the energy pathway in Figure 8 In the pathway we describe just two reaction steps First we need to break apart the water molecule Second we add the water molecule across the CC double bond to produce ethanol Note that we have shown the product ethanol molecule as being more stable than the reactants ethene and water This indicates that the reaction is energetically favorable Note however that the intermediate step the breaking of the water bond means that the reaction pathway has to A H OHgtCC Energy of Intermediates Activation Energy A H O C 20 Energy of 1 V 2 C Reactants H A LL I I tEnergy gain romreaction C C v Energy of 1 Products Figure 8 The energy diagram of the reaction of water ethene to form ethanol proceed through a energy barrier or an energy of activation before the reactants can form the product If this energy barrier were not there then this reaction would occur spontaneously However the presence of the energy barrier means that the reaction will probably happen very slowly if at all The role of a catalyst is to lower the energy barrier of a chemical reaction For example in the hydrolysis of ethene the catalyst will do the job of breaking apart the water molecule Although even the catatalytically assisted reaction may have some energy of activation that energy of activation is likely to be much less There is one more characteristic of the catalyst and that is that it participates in the reaction but emerges at the end unchanged from its initial form In other words consider the role of the catalyst here Assume it is a piece of NiO nickel oxide NiO grabs onto a water molecule breaking it apart into H and OH An ethene molecule removes the H and the OH from the chunk of NiO thereby forming ethanol In the end the NiO looks exactly like it did at the start of the whole process We don t even need it when we write a balanced chemical reaction But it certainly helped speed things along Let s write the reaction out and see this HZO NiObulk 9 H on NiO OH on NiO CH2CH2 H on NiO 9 CH3CH2 radical CH3CH2 radical OH on NiO 9 CH3CHZOH NiObulk Now add all of these steps together canceling out components that are listed on both left hand sides and right hand sides of the equation and you will get H20 CH2CH2 e CH3CH20H This hydrolysis reaction when spelled out in detail hints at some of the complexity involved in chemistry Although the nal balanced chemical reaction looks very simple the process by which it occurs can be quite complex Polymerization synthetic Polymers The vast majority of chemical systems that we have thus far encountered are small molecules like the water ethylene and ethanol molecules in the above hydrolysis reaction However there is another very important class of organic molecules called polymers Polymers are generally classi ed as either naturally occurring biological polymers or synthetic polymers We will talk about synthetic polymers rst Examples include te on polyurethane polystyrene styrofoam nylon rayon and others Other there are examples of inorganic polymers they are rare and we won t discuss them Many polymerization reactions occur through mechanisms that involve free radicals Free radicals are molecules that are characterized by having one unpaired electron For example when we broke water apart into H OH in the above hydrolysis reaction we OH is a radical Since OH has 7 valence electrons 6 from oxygen 1 from hydrogen it simply can t have all of its electrons paired It has to be a radical It turns out that as a class radicals are just about the most reactive chemical species that exist Furthermore imagine what happens when a single radical reacts with a stable molecule Radical odd number of electron molecule even number of electrons 9 new species odd number of electrons Since an odd number plus an even number produces an odd number it is not surprising that radical reactions can quickly run out of control 7 ie reacting a radical with a molecule will often produce another radical Obviously we can terminate one of these reactions by reacting two radicals with themselves 7 producing a product that has an even number of electrons Thus if we are doing chemistry in which radicals are the major reacting species then we can classify our reaction mechanism into the following types of steps Create a radical Initiation step radical molecule 9 larger radical propagation step radical radical 9 molecule termination step Note that we can have just one initiation step followed by many propagation steps and nally end it all with a single termination step In many cases the propagation steps result in the growth of a very large molecule Let s say for example that we have a molecule AA and we can use ultraviolet light to split this molecule into two radicals 2A0 We can imagine the following sequence of events 1 AA photon 9 2A 2 A AA 9 AAA 3 AAA AA 9 AAAAA 4 Step 3 repeats many times and nally 5 AnA A 9 An2 Termination Step Initiation Step propagation Step propagation stepspolymer growth Note Any step that produces a radical is an initiation step Any step that has a radical reacting with something to form another radical is a pr0p0gati0n step and any step that gets rid of radicals is a termination step Let s look at a real example of polymer growth from monomeric units We are going to grow a polymer made from the monomer shown at right In this drawing we just show the skeletal framework of the monomer In this case we have a functionalized benzene ring In situations like this the benzene ring is often referred to as a phenyl group It of course has 5 hydrogens and one alkyl group coming off the 6 vertices We can initiate this polymerization process by taking a peroxide ROOR and splitting it in half with ultraviolet light photons that are slightly higher in energy than visible light That will give us two radicals ROOR hv 9 2 R0 The reaction of this radical with vinyl benzene is almost identical to the hydrolysis reaction discussed above In this case the radical adds across the alkene double bond creating a new radical and so this is a propagation step We show this new radical at right Now this guy will proceed to react with more monomer and we will quickly react with more monomer and we will eventually get a monomer 7 styrene or C6H5CHCH2 R O M Product of R00 monomer RO Fig 9 A short chain polymer 7 actually a piece of polystyrene polymer that looks something like what we show in Figure 9 This is the polymer polystyrene Spend a little time trying to gure out how the monomeric units added up to make this guy As you can see one end of the polymer contains the RO group that initiated the polymerization process and the other end of the polymer is still ready to grown by adding more monomer This polymer will eventually quit growing by nding another RO radical or by nding another piece of growing polystyrene When that happens we have a chain termination reaction and polymer growth stops Note that if we wanted to make this polymer really long we would put just a very small amount of free radical initiator the peroxide molecule into the pot with the polymer monomer It just gets things started Too much of it and we will have too many radicals around that will quench the growth processes Molecular weights of polymers often reach 106 amu or more For the polystyrene shown here each monomer weighs 100 amu its molecular weight is 100 gmol and so a polymer with a molecular weight of 1 million amu would have 104 monomer units Biopolymers Biopolymers are naturally occurring polymers and include such things as DNA RNA polypeptides etc One term that will come up several times with biopolymers is the fact that they often contain chiral units A chiral molecule is rather different from any type of structure that we have thus far discussed Consider your left hand and your right hand Hold your left hand up to a mirror and in the re ection you will see a right hand In other words your two hands are mirror images of each other but they are not superimposable No matter how hard you try don t hurt yourself you can not rotate or twist your left hand to make it a right hand Your hands are chiral Micky Mouse s hands which look a bit like large white doughy ngers are not chiral because they don t obviously have a front and a back Because of that Mickey Mouse s right hand could be rotated so that it was a left hand Think about this question If your hands consisted of a single finger in the middle and two thumbs symmetrically arranged on either side would they be chiral What types of molecules are chiral Consider the two Figure 10 Enantiomers molecules shown below in Figure 10 Are these two molecules the same It turns out that they aren t Note that they are mirror images of each other Now try and take one of the two structures and superimpose it on the other structure You will find that you can t do it These two molecules are chiral isomers or enantiomers of each other More appropriately the central atom is a chiral center How can you tell if an atom is a chiral center First is it sp3 hybridized If not it has no chance of being chiral Second are there 4 different groups including lone pairs that are bonded to the central atom If there are 4 CH2 different groups not 4 different CH3 CH3 C1 I atoms then the central atom 1s a F ch1ral center Let s do a couple H examples shown in Fig 11 Each molecule has 1 chiral center On 4 the cyclobutene molecule it is the carbon3 and on the molecule at H3C the right it is the carbon that is bound to the chlorine atom Make sure you understand why I will give you chirality questions like Fig l 1 Some chiral centers atom 3 on left atom with Cl bound to it on rioh this on the nal One of the unique things about biopolymers is that they are usually not made up of repeating units but rather they are made up of strings of peptides or amino acides etc The length of biopolymers some can be more than a millimeter long coupled with the way fact that they can be constructed from a dozen or so building blocks arranged in arbitrary order means that biological polymers are remarkably varied I won t go into them too much more than this Cistrans isomers There is one last thing that this chapter covers that I want to include here and that is cis and transisomers So far we know about resonance isomers structural isomers and stereoisomers enantiomers or chiral pairs Cis and transisomers are actually structural isomers but very subtle C1 H ones Enantiomers are CC 1200 also structural isomers and they are also very subtle In Figure 12 we show tran s dichloroethene Note that in the trans version the Cl atoms are opposite each other across the molecule In the cis version the molecule looks like Cl 01 CC 5 120 H Now why are these two structures different It appears that one could just rotate the CC bond and generate one from the other In fact that is the catch You can t do it If all we had was the obonds Fig 12 middle we could rotate the bond But we also Fig 12 Transdichloroethene The molecule at top the G bonded frameworkmiddle and the nframework bottom have a rtbond and if we rotate the CC we break the rtbond Note also that the trans version doesn t have a molecular dipole moment While the cisversion does Make sure that you understand why this is so These two structures are actually a little different energyWise I believe that the transversion is a little more stable but that should not be obvious to you End lecture series 8 CHEM 20A1 winter 2001 LECTURE NOTES tenth set SHIELDING In the last set of notes we studied systems consisting of many noninteracting electrons Here we recognize that the electrons in an atom are compressed into tight quarters and that they therefore not only interact with each other but interact very strongly Why do they interact strongly when crowded together Although the effect of the electrostatic interactions among electrons is great the effect of the exclusion principle on the electronic structure of atoms and molecules is even greater this is why the periodic features outlined in the last section persist even when the electronelectron repulsions are considered Stability and Shielding We have already discussed to some extent the effect of the electronelectron repulsions The lower the energy the more stable the system and since the electron electron repulsions increase the energy of the atom they make the atom less stable than would be the case for fictitious noninteracting electrons Why Secondly because any given electron is subject simultaneously to the attractive force of the nucleus and the repulsive forces due to all the other electrons the net attractive force holding the electron within the atom is less than that due to the nucleus alone This reduction in the net attractive force that is attributed to the nucleus is called quotshieldingquot the other electrons can be envisaged as partially shielding any given electron from the full attractive force of the nucleus Think this one over We focus on shielding There is atheorem in the study of electrostatics which states that if charges are distributed spherically and observed from a large distance they act as though they were all situated at the center of the sphere Thus in an atom the electrons are distributed spherically about the nucleus and they therefore act as though they were all concentrated at the nucleus Since the nucleus has a plus charge equal to the negative charge due to all the electrons the atoms as a whole acts as though equal positive and negative charge were concentrated at the nucleus Consequently the atom is neutral The electrons have completely shielded the positive charge on the nucleus and no electrical force is exerted at the distant observation point However note that the analysis above holds only if observed from a large distance if observed from inside the atom only those charges that lie inside the observation point can exert an electrical force This means that outer or valence electrons are well shielded from the nucleus by the core electrons but that the core electrons are not well shielded from the nucleus by the outer electrons This is the concept of quotshieldingquot In the simplest of pictures each core electron is totally effective in shielding outer electrons thus each electron in an inner shell has the effect of reducing the attractive force on an electron in an outer shell by reducing the effective nuclear charge from Ze to Zle At this level of approximation shielding of inner electrons by outer electrons is neglected as is shielding by electrons in the same shell The following examples are worth considering Problem Neglecting shielding we previously estimated the ionization energy of H Li and Na to be RH 9RH4 and lZlRH 9 respectively With the rough shielding model given above one might expect the ionization energies to be RH RH4 and RH9 respectively Why Problem Neglecting shielding we estimated the ionization energy of F and C1 to be 81RH4 and 2899 respectively With the rough model of shielding given above one might expect the ionization energies to be 49RH4 and 49RH9 respectively Why But this model of shielding is too extreme The electrons are distributed as described by the probability densities given by the square of the orbitals Outer electrons penetrate the core to some extent and the core electrons leak out beyond the outer electrons The major effect of this is that an electron in a given shell is somewhat effective but not completely effective in shielding the other electrons in that shell This has the effect of reducing most of the ionization energies estimated in the two problems above Why m Compare the ionization energy of O with Z 8 and con guration ls22s22p4 with that of F with Z 9 and configuration ls22s22p5 The ls electrons do quite an efficient shielding job thereby reducing the effective charges to just slightly over 6 and 7 for O and F respectively Why One of the outer n 2 electrons the one to be ionized is then partially shielded by the other ve n 2 electrons in O and by the six in F If one were to estimate the partial shielding as 07 per electron in the same shell then the effective charge on the electron to be ionized would be about 25e for O and about 28e for F It follows that the ionization energy for O and F would be approximately 252RH 4 and 282RH4 respectively Be sure you understand this NOTE This model is slightly different than that proposed in class in that here the average shielding of one electron by another in the same orbital energy level is taken as about 07e whereas in lecture it was taken as about 05e This model is closer to the quottruthquot but is still quite rough We shall not try to do better here There are however a few other aspects of shielding that we investigate Energy level diagram for atomic orbitals For hydrogenlike atoms the 2s and 2p orbitals are degenerate Because of shielding ie because of electronelectron repulsions this is not true for other atoms In fact the orbital energies for the ns state lie slightly lower than those for the np states This can be understood by examining the shapes of the wave functions and hence of the probability densities Specifically the ns functions have finite probability densities at the nucleus whereas the np functions have nodes at the nucleus This implies that ns electrons tend to get quotinsidequot the np electrons consequently the ns electrons shield the np electrons more than the np electrons shield the ns electrons Since shielding cuts down the attractive forces of the nucleus it raises the potential energy This explains why the us levels lie below the up levels Problem It does explain it doesn t it Similar reasoning leads to the conclusion that nd levels lie above np levels and nf levels lie above nd levels With the information above one can draw an energy level diagram for atomic orbitals At the bottom one finds the ls level which can accommodate two electrons Then with avery large gap comes the 2s level which can accommodate two electrons The gap is large for two reasons If there were no shielding the gap would be a large ZZRH34 furthermore whereas there is little shielding ofthe ls electrons the ls electrons provide considerable shielding for the 2s electrons thereby increasing the energy of these Next after a small gap comes the 2p level which can accommodate six electrons The gap is small because its existence is entirely due to the fact that the 2p electrons are more shielded by the 2s than are the 2s by the 2p electrons There is then quite a large gap before one finds the 3s level which can accommodate two electrons The gap is quite big because ifthere were no shielding it would be a rather large ZZRH536 and shielding increases it further Explain ext after a small gap comes the 3p level which can accommodate two electrons Explain At this point things get complicated One might expect the 3d levels to come next but it is the 4s that come next After that it is indeed the 3d levels that are found but curiously as electrons are placed into the 3d levels the 3d levels slip back below the 4s level consequently the 4s electrons remain the outer electrons The 3d level can accommodate ten electrons Next come the 4p electrons and then the 5s electrons We shall not look any higher But we might ask why there is growing confusion after the 3p levels The basic problem is that the shell separation ie the difference in energies for different n values is so large between n l and n 2 and even between n 2 and n 3 that the effect of shielding can almost be ignored But the difference is small between the n 3 and n 4 levels and even smaller between the n 4 and n 5 levels and this means that the shielding effects which become more complex with the growing number of electrons start to dominate This gives rise to complexity Summary of Atomic Structure We now have a reasonably valid reasonably simple model with which to describe and understand the properties of atoms To understand the discussion of the very important chemistry that follows be sure you understand the material in the last section thoroughly The atomic model can be summarized as follows The atoms consists ofa positively charged nucleus Ze which has a mass very close to AMp where Mp is the mass of a proton or a neutron Z is known as the atomic number and A as the mass number Chemical properties are determined primarily by the electrostatic interactions between the positively charged nucleus and the surrounding negatively charged electrons Thus the chemical properties are determined largely by the value of Z and indeed each element and its properties are specified by the value of its Z The individual electrons in an atom can be envisaged as existing in independent quantum states described by oneelectron wave functions known as orbitals 1m1r these states are quantized and have quantized allowed orbital energies snl These orbital energy levels are subject to a high degree of degeneracy associated with the high spherical symmetry of the atom The collection of electrons within an atom can be described by a product of the orbitals and the energy of the atom can be described as a sum of the orbital energies The orbitals are quotoccupiedquot by the electrons subject to the exclusion principle which states that at most two electrons can occupy a given orbital To understand the exclusion principle one must incorporate the property of spin in which case one can describe a spinorbital which describes the state of an electron including its spin and only one electron can occupy any given spin orbital In describing the energy of the atom by means of the quotoccupancyquot of orbitals one must take account of the degeneracy of the orbital energy levels The electronelectron repulsive interactions increase the energy and reduce the stability of the atom This effect can be represented as shielding where each electron is envisaged as partially shielded from the attractive force of the nucleus by the repulsive forces of all the other electrons This shielding effect can be described in terms of reduction in the effective charge Ze of the nucleus Shielding removes the degeneracy between the ns np nd and nf levels so that ans lt snp lt and lt Snf Because of the nature of the degeneracies the atoms develop a shell structure with inner core and outer valence electrons The core electrons are less shielded are held very tightly by and very closely to the nucleus and they are not altered much in chemical reactions The outer electrons are more shielded held less tightly and less closely to the nucleus and it is these that participate in chemical reactions Because of the shell structure of atoms similar chemical properties are expected and found associated with electronic structures outside each shell core This gives rise to the periodicity of properties summarized by the periodic table The Periodic Table Tacking account of the exclusion principle and the concept of shielding one can get a good understanding of the observed periodicity in the properties of the atoms as well as an understanding of their chemical reactive properties I shall not detail these here but will discuss them in lecture Your text also gives information on these properties In particular we can examine ground state electronic structure con gurations ionization energies electron affmities atomic sizes and reactivity The alkali metals Li Na K Rb Cs Fr all lie in the same column of the periodic table H also lies in this column The ground state con gurations of all these elements are similar in that they have a single outer electron only a single ns electron outside a fully occupied core H is slightly different in that it has no core The outer ns electron is thus heavily shielded and the effective nuclear charge holding it is not much greater than le ie similar to that for H Since the orbital energy for this outer ns electron should vary as n39Z as one moves down the column one might expect the ionization energy to decrease but because the shielding is not perfect one might expect the effective nuclear charge holding the electron within the atom to increase slightly and counteract the decrease in ionization energy As a consequence the ionization energies for the alkali metals should be rather low and decrease but not much as one goes down the column And this means that these elements and H as well can readily lose an electron and form positive ions or cations HLi Na K Rb Cs Fr In chemical reactions it is usually the ease with which these elements give up an electron that controls their roles The halogens F Cl Br I At all lie in the same column of the periodic table The ground state configurations of these elements are similar in that they consist of five np electrons outside a core and outside a two member ns2 subshell Even the ns2 subshell does a pretty good shielding job so the inner electrons reduce the effective attractive charge of the nucleus on the np electrons to about 5e Why Additionally the np electrons shield each other to some extent let us say about 70 each This means that any single np electron feels an effective nuclear charge of about 22e Why This is much greater than the effective nuclear charge felt by the lone ns outer electron in the alkali metals furthermore the orbital binding energy depends upon the square of the effective nuclear charge Consequently we expect the outer electron on a halogen and it can be any one of the five np electrons to be much more tightly bound than on its alkali metal brethren Thus the halogens have very high ionization energies and do not loose electrons very easily On the other hand they can gain an electron quite easily If an extra electron is added it interacts repulsively with the other electrons thereby increasing the energy and destabilizing the atom However since the shielding is not perfect the added electron still feels an attraction to the nucleus To understand this note that there are now six np electrons and that any one of these is shielded by the core electrons and by the other five np electrons the effective nuclear charge acting on any one of the six np electrons is then approximately 15e Why This is less than the effective nuclear charge acting on an outer electron in the atom but it is still sufficient to give rise to a considerable attractive force Thus the extra np electron is readily accommodated This means that the halogens readily form negative ions or anions F39 Cl39 Br39 I39 At39 In chemical reactions it is usually the ability of these elements to attract an additional electron that determines their reactivity and chemistry The energy of attraction of an extra electron is known as the electron af nity The noble gases He Ne Ar Kr Xe Rn are all placed in the same column They are all very inert chemically This means that they do not lose electrons easily have high ionization energies they do not have electron affinities they do not add electrons easily and it is difficult to change the state of their electrons no low energy excitations They all have complete np subshells 139e they have six outer np electrons except for He which has a complete n l shell If an extra electron is added without violating the exclusion principle it must be placed outside the completed shells and so the attractive force due to the nucleus is almost completely shielded out Consequently there is nothing holding the extra electron to the atom 139e there is no significant electron affinity and it is difficult to form negative ions At the same time they have large ionization energies and it is not easy to remove an electron and form positive ions To understand this recognize that although the inner electrons shield the nuclear charge almost completely the outer np electrons only shield each other partially so that one might expect an effective charge of about 25e on any given outer np electron Why This is a very large effective charge and so the electrons are tightly bound and it takes much energy to ionize them The reasoning for why He has a high ionization differs slightly but is quite analogous Explain It is now le to you to explain the following Problem Why on the whole do you expect ionization energies to increase as one moves from left to right on the periodic table Problem Why on the whole do you expect electron affinities to increase as one moves from left to right on the periodic table Problem Why do you expect elements on the left to form positive ions and those on the right to form negative ions Problem Why are second ionization energies higher than first ionization energies Problem Why might you expect the alkali earth elements Be Mg Ca Sr Ba Ra to form doubly positive ions Bez Mg2 Ca2 Sr2 Ba2 Ra2 whereas the alkali metals form only singly charged positive ions Problem Why might you expect 0 S to add two electrons without too much difficulty whereas the halogens can only add one Problem Discuss the relative sizes of the atoms as one moves across a row ofthe periodic table and as one moves down a column For an atom to be reactive 139e for it to participate in a chemical reaction it must be susceptible to change If it readily loses an electron and becomes positive 139e if it has a low ionization energy it can react readil If it readily adds an electron and becomes negative 139e if it has a high electron affmity it can react readily If its ground state con guration can readily be changed by excitation 139e by addition of very little energy then it also has a good chance of being reactive Such excitation changes are associated with the least energy required to move an outer electron into a higher orbital Problem Why would you expect the excitation energies in the noble gases to be high Problem Why would you expect the chemistry of all the elements in a given column to be quite similar Why would you expect them to be not quite identical The elements on the left of the periodic table tend to be metallic Metals have high electrical conductivity 139e they conduct electricity well This means that electrons can flow easily throughout the metal For the elements on the left of the periodic table the electrons are not very tightly held and so they can ow rather easily Problem Think back about the photoelectric effect Would you expect the maximum wavelength of radiation that is needed to drive electrons off a piece of K to be greater than needed to drive them off a piece of 1 Addition of Spins Each electron has a spin that can have one of two values ms 12 and l 2 For two electrons the spin states can be m51m52 where msl and msz indicate the spin states of electron 1 and 2 respectively The spins add so that the net spin for the two electrons is given by MS mslm52 Thus MS l for the l2l2 state MS l for the l2l2 state and MS 0 for both the l2l2 and l2l2 states One reason that this is interesting is that if MS 0 there is no net spin and the electrons are not affected by a magnetic field while if Ms 0 there is a net spin and the electrons are affected by a magnetic eld as desc1ibed above The electrons in all fully occupied orbitals do not contribute to the spin if the orbital is fully occupied it means that it contains two electrons one with spin up and one with spin down Furthermore it is possible for atoms with even numbers of electrons to have a net spin of MS 0 because the electrons could fill orbitals in pairs It is impossible for an atoms with an odd number of molecules to have MS 0 Why But it is possible for atoms with even numbers of electrons to have Ms 0 Why It follows from the above that the spins of electrons in atoms with MS 0 are not affected by the application of magnetic fields whereas those in atoms with MS 0 are affected by magnetic fields The latter are called paramagnetic substances If the atoms or molecules have an odd number of electrons they must be paramagnetic such substances are called free radicals Free radicals play a major role in chemical reaction We shall come back to this subject later Lecture Series 3 Chem 20A James Heath Recall that we concluded our discussion of the discoveries by Planck Einstein and Bohr by stating that the single theme that ran through those discoveries was that at very small length scales various physical phenomena were no longer continuous but rather discrete or quantized Light as well as matter also comes in discrete units 7 the photon Electrons which are discrete particles of matter will orbit a nucleus in with atomic orbitals that are characterized by discrete angular momenta Now there is one more piece of information that we need before we can proceed to a more applicable description of quantum mechanics When we described electromagnetic radiation light we described it as both waves and as discrete packets of energy photons This waveparticle duality is not unique to light In fact it is very general Some time after the three paradoxes were confronted by Plancls Bohr and Einstein other scientists began doing experiments on electrons and finding some very puzzling results It was well known from Newton s time that if light was sent through a very small slit 1 it would diffract as shown in Fig 1 This is how diffraction gratings work and is related to why a rainbow is a rainbow Some scientists began doing this same experiment with electrons They would make a beam of electrons send it through a slit and look at a phosphoroscent screen placed behind the slit What they found was rather surprising They Fig 1 When a single color light beam passes Observed diffraaion Patterns juSt as ifthey had through a slit it creates a diffraction pattern a done the experiment With a beam of light This series 0f bright and dark Spots 0139 rings on the implied that electrons 7 which had previously been White Screen behind the Slit considered to be particles 7 also had wavelike characteristics Based on these results and the other work that was being done by Planck Louis de Broglie wrote a very brief doctoral thesis that was almost turned down in which he postulated that all matter was characterized by a 39 39 and that 39 439 could be 39 39 J from the following equation 7M hmv Planck s constant divided by the momentum of the particle This equation was in agreement with the observed diffraction measurements Example Say that we accelerated a beam of electrons through a 10keV eld so that they were now travelling with a kinetic energy of 10000 eV s 1 eV 16x103919J What is the wavelength associated with those electrons Note that eV s are a very convenient unit to use here Ifwe accelerate an electron through an electric eld of x volts then that electron has a kinetic energy of x eV So if kinetic energy 1x 104 eV 12 mvz then the momentum mv We know that m n1e 91 x 103931 kg so we can solve for v and hence momentum 104 eV 16 x103919JeV 1291 x103931kgv2 v 7 5 x107 ms 1 7 663x103934 J s 1x104 eVl6x103919 JeV5 x107 ms391 7 12 x103911m 7 01 Angstroms Ifwe had used protons instead of electrons 7 would 3 x 103913 m 7 a quantity that is essentially not measureable As objects get more and more massive their momentum increases and the wavelengths associated with them get smaller and smaller 7 quickly becoming too small to measure What if we con ned a neutron We know that mass and light each have wavelike and particlelike characteristics and that electrons in atoms are quantized 7 they have discrete energies or equivalently they have discrete values of the angular momenta We have so far presented only the facts Now we try to ask why Our rst question Why does an electron become quantized Quantum mechanics is not a particularly intuitive subject and so analogies to real world phenomena are a little dangerous Nevertheless here is one that works a little bit Imagine that we have a guitar string tied down at either end as if it were on a guitar If we pluck the string then we hear a note and this note is the resonance frequency of the string This is of course the basis for all stringed instruments One selects an appropriate string stretches it across some length and then puts an appropriate tension on that string These things we do to the string will tune it to a desired resonant frequency 7 say an A note Now it turns out that when you pluck the string you not only hear the A note but if you listen very closely you F39 2 A 39t t 39 d 39t lgure gm H S nng an 1 S w111 hear an A note at the next octave up and even the 2d octave resonant frequencies Notice that all frequencies are integral up as well and so on and so forth These are harmonics and their multiples of the fundamental frequencies are integral multiples of the rst primary note or the frequency Shown at the bottom fundamental frequency This is shown graphically in Fig 2 of the drawmg39 In the language of physics let s restate the various phenomena that control the resonance frequency of the string First we have the mass of the string 7 larger mass translates to lower 7 ie thick strings give lower notes Second we have the tension on the string When we stretch a string we are actually using it as a spring 7 ie if we let loose of both ends it will retract back to its initial length When we pull or compress a string or a spring this is the same as increasing its potential energy When we compress or expand a spring we store energy into the spring in the form of potential energy Thus the amount of potential energy stored in the string is important More potential energy tighter string corresponds to higher f 39 0 Third we have the length of the string that is quot Shorter lengths lead to higher frequency resonances This is why frets are on a guitar Note that the vibrating string is a one dimensional system 7 ie you can represent the string itself as an xaxis So if we are going to make an analogy between a vibrating string and a quantum mechanical entity we need to choose a ldimensional quantum mechanical system There is a theoretical model for such a system known as the particleina box and it represents one of the most famous problems in quantum mechanics Although this model teaches us many important things about what to expect from electrons in atoms it is just a model 7 it does not represent any real experimental system Note that there is actually nothing boxlike about this model 7 the particleinabox actually more resembles a particle on an xaxis rather than a particle in a box We are not going to solve the particle in a box problem exactly in these notes alth ongh it is possible to do so yonr book does it seep 54 7550 In our attempts to make an analogy to a string system we are going to neglect some things and include others that we sh onldn t Nevertheless onr incorrect approach to this problem actually comes up with the correct solution and givse as some physical insight These circles represent energy baniers The Pa ieleinabOX iS represented by the that keep the e con ned to the region drawing in Fig 3 The particle in our case is an of the line between the circles electron We keep it in the box by putting infinitely high energy barriers at either end of the box Ifwe think of the electron as a particle then we imagine a particle scooting back and forth between the two barriers However if we think of the electron as a wave then we are back to Fig 2 7 our guitar string In Vquot fact electron waves are going to appear more or less electron exists in this region just like the resonant frequencies of the guitar string Fig 3 Particle in a 1dimensiona1box The However there are differences Recall that 7 hmv potential energy has a value of 0 between the de Broglie relationship Let s look again at our the green dOtS guitar string resonance and its overtones redrawn in Fig 4 Notice that the overtones are characterized by 7b bro3 shorter wavelengths than the primary resonance frequency the fundamental frequency We didn t think too much about it when we were discussing guitar strings but here it becomes important According to the deBroglie equation if the wavelength is shorter then the momentum of the particle and hence its kinetic energy must increase Thus we have an electron trapped in a ldimensional box and it has a resonance frequency that is determined by the size of the box However the various overtones of the resonance frequency are at successively higher electron kinetic energies Furthermore this energy separation is necessarily discrete not ig 4 Same as Fig 2 but we continuous 7 simply because the overtones are integral fractions of have noted the relationShiP the fundamental wavelength between the varlous harmomcs and the fundamental frequency Let s work this out If we have a box of length L then according to Figs 2 and 4 the fundamental frequency has a wavelength of 2L Notice how only 12 of the fundamental wavelength ts into the box 2L hmv so mv h2L and so V h2Lm hkm Energy kinetic energy potential energy However if we keep our particle within the boundaries of the box then the potential energy 0 so the total Energy the kinetic energy 12 mv So if E 12 mv2 the energy of the fundamental is given by Ef 12 m h24L2m2 12 hz4L2m h28mL2 Let s go to the 1st overtone 7 L hmv so v hmL E 1 W 12 mvz 12 m hzmsz hZszz 4h28mL2 And for the 2d overtone 7 23 L hmv so v 32 hmL and Em 12 mvz 12 m 9h24m2L2 9h28mL2 So we have a series of states with En n2h28mL2 where the fundamental frequency has n l the ISI overtone has n 2 the 2d overtone has n 3 etc Here n is the quantum number that denotes a quantum state of the system En is the energy of the n3911 state There is a function that describes the 3 waves shown in Fig 4 and all the other ones that aren t shown and that function is called the wave function of the system Each of the separate waves is a quantum state of the system and each state is characterized by a quantum number The wave function is often denoted by the symbol l called psi and pronounced si Notice that the quantum number determines the shape of the wave function If we had really set up a differential equation and solved this problem as your book does we would have also found that EI1 h2n28mL2 It turns out that the reason our approach works is because this is an unrealistic and simple system the potential energy contribution to the total energy of the system was negligible as long as the particle stayed between the barriers Thus other than using the potential energy to de ne how big our box was we neglected it when we calculated the energy of the system Usually we can t do this Also notice that we force fed quantum mechanics into this solution by way of introducing the de Broglie wavelength of the particle If we had actually set up and solved the differential equation we would also have had to force feed quantum mechanics into the problem in order to get the correct solution E3 So let s go back to the factors that affected the resonance frequency of the guitar string mass of string potential energy length of string and ask how those factors affect the properties of our 3 E E2 quantum mechanical waves First however we need to adopt a slightly different representation for our waves Since we know that shorter wavelengths El correspond to higher kinetic energies we can represent the fundamental frequency and its length harmonics as shown in Fig 5 Fi 5 Now we redraw our previous figure that described the resonances of a string to represent the resonances Ofa con ned 0 First we had the mass of the string 7 larger electron The dots on the lengthaxjs are the mass translated to lower resonance frequencies dots of Fig 3 Note the energy spacings increase as n2 Here we have 7 xed by L the length of the box to which the electron is con ned However we also still have the relationship 7 hmv Thus if 7 is xed and we increase mass then the velocity of the particle decreases Recall once again our favorite equation for energy E 12 mvz Ifthe product mv stays the same since the wavelength doesn t change yet the mass is increased then velocity must proportionately decrease The energy only varies linearly with mass but it varies as the square of the velocity Thus a small increase in mass plus a small decrease in velocity translates into decrease in kinetic energy Increasing mass reduces the energy of the wave Second we had the tension on the string More potential energy tighter string J J to higher 3 Our situation is similar In the particleina box the potential energy is what keeps the particle restricted to a small part of the axis Ifwe remove the potential energy barriers remove the dots in Fig 3 then the electron is no longer con ned to a particular part of the axis and it can have all ranges of wavelengths associated with it including very long ones LUI Third we had the length of the string that was quot Shorter lengths led to higher freguency resonances In fact if we reduce the width of the boxz we shorten the wavelength 7 just like in the case of a guitar But we also increase the energv of the electron If 7 is reduced then momentum mv is increased and hence kinetic energy of the electron is increased A couple of more things about the wavefnnctions The sign and the probability Sign of Pl l 2 l 3 Probability of I l I 2 I 3 There are a few E3 E3 more things that we V can extract from the plots of the wavefunctions We 5 now label our g E 3 wavefunctions as a 2 E2 l 1 l z etc The l notes that it is a wavefunction and E1 E the integer subscript l a 39 indicates which 1 th energy level we are eng length talking about Figure 6 This is essentially Figure 5 replotted with the addition of Figure 7 Here we have taken the In Flgures 6 notations that indicate sign of the Sqlla e 0f the three Wavgfmlcnqns39 and 7 we have wave mc on Whenever a wave This is the probability distribution replotted the function passes below the blue line When the wave mcnglls touCh the Wavefunaions Fig we say it is negative valued Else it blue lmes the Pmbablllty 9f ndmg 6 indicates that the is Posi ve valuedr an electron at those pomts 1s 0 Sign of the p u can actually be either negative or positive and the higher the energy level the more times that the wavefunction changes sign Recall that a wavefunction is just described by sin andor cosine functions and those can adopt negative values so this should not be too surprising Nevertheless this is a very important concept for thinking about bonding and we will return to it several times In Figure 7 we have plotted l I nlz for n 123 This is called the probability amplitude of the wavefunction At the points that the wavefunction changes sign touches the blue line the probability of nding our particle there is 0 This is called a node We see that higher energy wavefnnctions have more nodes 7 a point we will come back to again because it will be a very important issue to consider when we begin to think about chemical bonding One more point on the probability amplitude l I nlz If an electron is in one of the states described by a wavefunction I n then if we look all throughout the entire space occupied by that wavefunction we should be able to nd the electron This sounds like an obvious statement and it is However you can also make this statement mathematically in perhaps a less obvious way If the chance of nding the electron in the state I n is unity then ol I H 2 dx 1 Here the unusual appearing integral sign indicates that we are summing up the probability over the entire volume of the wavefunction and that the summed probability l One Last Background Point The Heisenberg Uncertainty Principle There is one more thing we need to cover before we get to a real quantum mechanical system the atom and it is called the Heisenberg Uncertainty Principle This principle is fundamental to quantum mechanics 7 it is not strictly provable but like the rest of quantum mechanics it has withstood the test of time admirably In any measurement that we can imagine doing we will always have a little bit of uncertainty in our answer 7 that uncertainty is governed by our instrumentation or by the nature of the thing we are trying to measure Let s say that we are trying to measure the position of an electron The only way we can do this is to scatter something off of the electron 7 like a photon When we do this we may get a very accurate measurement of where the electron was at the time of the scattering event but we have also changed the system Imagine for example that the electron is travelling along with some velocity v in some direction giving it a momentum p massvelocity If we scatter a photon off of the electron in order to measure its position then we change the electron s momentum We don t really know how we change it but we change it So now if we try to measure the momentum of the electron which we can certainly do it is not the same momentum that it had when we measured its position 7 confusing huh What Heisenberg showed is that if we try to measure both the position and the momentum of an electron then the precision of our position measurement times the precision of our momentum measurement must be greater than h4TE We can write this as ApAX 2 h4TE where the A is read as the uncertainty in Let s do an example Let s say that we have done a measurement that shows that an electron is located right within the nucleus of an atom Assume the size of a nucleus is given by dn103913m then how well do we know the kinetic energy of the electron Answer Since we know that the electron is within the nucleus then we know its position to a AX of 103913 m ApAX 2 h4TE 663XlO3934 J s 47 so Ap 2 52103922 kg m s39l So ifAmv 2 52X103922 and we know me9 leO3931 kg we can calculate the uncertainty in the velocity to be Av 2 58XlO8 ms So at best we can only pin down the velocity to somewhere between 0 and 58x108ms or six hundred million meters per second Obviously if our velocity is 0 then the kinetic energy of the electron is 0 However if the velocity is 58x108 ms then our kinetic energy is 15x103913J For 1 mole of electrons this corresponds to an energy uncertainty of 9x1010 Jmol This is a huge uncertainty in the energy If instead we had stated that we knew the location of the electron to within the dimensions of twice the Bohr radius 103910m we would have an uncertainty in the momentum and hence the kinetic energy that is much more reasonable Show this to yourself This is why the electron doesn t collapse into the positively charged nucleus as would be predicted by the Coulombic interaction potential energy curve If it collapsed into such a small volume its kinetic energy would possibly be tremendous Since total energy PE KE then the uncertainty in Potential energy is also tremendous Thus we would have no way to draw our electron on our Coulombic potential graph This sounds like a bizarre bit of circular reasoning but it is worth thinking about for awhile Thus the size of an atom is dictated more or less by the Heisenberg uncertainty principle Real Systems The Hydrogen Atom When we solved the problem of the particle con ned to a 1 dimensional box we were able to generate a single quantum number n for that particle If we had solved for a particle in a 3dimensional box we would have found that the solution to the problem involved 3 different quantum numbers 7 one describing how the particle behaved in the xdirection one for the y direction and one for the zdirection This Z has to do with the degrees of freedom of the particle If the particle is confined to a 1 dimensional box it has 1 degree of freedom and hencel quantum number In a 3dimensional box it has 3 degrees of freedom and hence 3 quantum numbers With this in mind what do we expect from an atom In an atom the electron exists within a particular volume surrounding the nucleus Because the electron has wavelike properties we describe the space occupied by the electron as the electronic wave function Because Figure 6 The three degrees of freedom that correspond to this space is 3dimensional it is a volume an electron moving about a nucleus Here we have shown surrounding the positive nucleus we know an xyz coordinate system for reference However we can that there will be 3 quantum numbers also descrrbe the pos1tron of the electron by us1ng the I d b the electron 7 one uamum d1stance r and the angles theta 9 and phi 1 to descrrbe escrl mg q Where the wow points number corresponding to each degree of freedom Thus we already see one of the problems with Bohr s atom 7 it only has one quantum number So if we need one quantum number per degree of freedom then what are those degrees of freedom Ifwe had a particle in a 3dimensional box we might just call them x y and z and get quantum numbers nx ny and nz However that is not what we have Instead we have a particle the electron held within some region of space by a centrosymmetric Coulombic potential the positive nucleus One obvious degree of freedom that we have is a measurement of the distance separating the electron and the nucleus 7 ie what is the radial distance For this degree of freedom we have the radial quantum number If we decide on this as one of our quantum numbers then what are the other two Look at Figure 6 Here we show that we can describe the position of the electron by r its distance from the origin and by the angles theta 9 and phi 1 These are related to the xyz coordinate system because 9 is the angle of r within the xy plane and I is the angle that the vector r makes with the with respect to the zaXis Let s not worry about this too much other than that we need one degree of freedom to describe the distance of the electron from the origin and two degrees of freedom that describe angles Thus we have one radial quantum number and two angular quantum numbers The radial quantum number will dictate how the wavefunction spreads out from the nucleus and the two angular quantum numbers will determine the shape of the wavefunction Quantum numbers of electronic orbitals The wave function of an electron is usually denoted by T For the case of the electron orbiting around the nucleus as is shown in Fig 6 l is a function of r 9 and The energy levels of the electron are described by the three quantum numbers n the principal or radial quantum number I or the angular momentum quantum number and m which is called the magnetic quantum number We ll return to these quantum numbers in a minute We write l as l n1mr9 and we find that we can separate l into a radial component Rn1r which describes how far out from the nucleus the wavefunction extends and X1m9 which describes the shape of the wavefunction Thus we can write LIlnlmraeal Rnlr We also find that there are certain relationships between the quantum numbers and we ll describe those in a minute I just list this information so that you see that there are three quantum numbers and three degrees of freedom What do those quantum numbers mean 7 We don t know yet If we had set up and solved an equation that described Fig 6 which is a hydrogen like atom we would have been able to derive all of the stuff that is being presented here This is a very difficult thing to do and is usually not attempted until the last year of an undergraduate chemistry program or the first year of a graduate chemistry program Nevertheless what we would have found would have been just a much more complicated version of the quantum guitar string problem we solved earlier Just like the guitar string the energy levels would have been described by an energy E where n is the principal quantum number EH 7 Z2e4me8802n2h2391 Let s define the constants in this energy expression Z is the nuclear charge For the H atom it is simply I for Li it is 3 and for He it is 2 Why is this H has one proton so one positive charge Li2 has 3 protons so a positive nuclear charge of 3 and He has 2 protons We don t worry about neutrons since they are charge neutral Also all of these atoms are hydrogenlike 7 meaning that there is just one electron orbiting about a positively charged nucleus Even though the magnitude of the positive charge differs the physical solutions work out to be the same they are just scaled for the different charges You should understand this concept Think about how the Coulombic potential graph would change for these three hydrogenlike systems Going back to the equation for EH 7 we have e4 which is the fundamental unit of charge to the 43911 power we have me which is the mass of the electron In the divisor we have 802 the fundamental constant that corresponds to the permittivity of a vacuum squared We have hz which is Planck s constant squared and we have n2 which is the primary or radial quantum number squared Thus just like our quantum guitar string the energy of a given quantum state is related to only one quantum number n However unlike the solution to the quantum guitar string the energy depends inversely on n scaling as n39 This means that if we plot the energy levels of this system as a function of n they will look something like what is shown in Fig 9 The energy levels that describe our hydrogenlike system do not depend upon the other two quantum numbers I and m However a given En describes the energy of one or more quantum states of the system Those quantum states are more or less electronic orbitals and they have a shape It is the shape of those wavefunctions which is related to the orbital angular momentum and that is determined by the quantum numbers I and m We could have solved for the shapes and angular momentum of these various states and we would have found that n l and m have certain relationships with one another J combinations of other quantum numbers that we could have 5 a I 4 IE 4 396 m M Q 3 0 2 nl 2 g m l l1 l2 0 2 l 1l 1 1 1 n 4s 4p 04d 04fo So for example 1fn 3 what are the poss1ble 1 2 1 1 1 2 0 3 1 n312m 21012 Q n3llml0l E n73170m7 Hn2 25E What are 1 the allowed quantum E2 numbers n1 15 IE v node if we have I I I I g n 1 n 0 1 2 3 m 4 You Angular Momentum Quantum Number I Should be Fig 9 The various orbitals are plotted as a E able to illincition1 of energy Not that since Edgoes as n39z 1 t e eve s et increasm com resse 1n ener as 0 quot 2131th n increasesg At the riglftbf each orbital notatigz is W H a shaded box that contains all of the m values for yourse 39 that orbital Flgure 10 The quantum guitar When we String but only the rst tWO want to describe a particular orbital there are some energy States are dfaWII The 130111t nomenclature rules First we consider the principal Where the wave fun ftion Passes quantum number n Then we consider the angular through the daShed llnes ls called a momentum quantum number I If I 0 it is an sorbital If 16 z 1 it is a porbital and ifl 2 it is a dorbital Other less commonly encountered orbitals include forbitals 13 and gorbitals 14 Returning to the case ofn 3 we have 10 lor 2 These combinations yield 3s 3p and 3d orbitals So how about m It turns out the number of values for m describe the number of 3s 3p or 3d orbitals For example for a 3s orbital m can only be 0 Thus there is only one combination n3 10 and m0 For n3 and 11 m can take on the values of7l 0 or 1 Thus we have 3 3p orbitals each with a different m quantum number For n3 and 12 m has the values of 72 l 0 l 2 7 thus there are 5 3d orbitals In Fig 9 we plot the various orbitals as a function of energy Recall from the energy expression for hydrogenlike atoms that the energy depends only on the principal quantum number n and that energy scales as n39z We have hinted that the angular quantum numbers describe the shape of an orbital and that the principal or radial quantum number describes the radial extension of the orbital Let s look a little more at what we mean by this First we consider the sorbitals For all s orbitals both I and m are 0 What this means is that T the orbital does not really have much angular shape 7 it is simply a sphere radiating outward from the positive nuclear core But how about n This quantum number affects the radial part of the wavefunction the function Rn1r at the top of this 0 gt section As we move from n0 to nl to n2 etc the principle quantum number is changing so the radial part of the wavefunction must change In fact this change is very similar to what we observed for our quantum guitar string Recall that as we increased n we moved from the fundamental frequency to higher harmonics Let s look at that 0 graph one more time and discuss terms In Fig 10 r we have redrawn the lowest two states of the 1135 quantum guitar string with a couple of additions We node have drawn dashed lines Z through the strings T 2 pl indicating the nodes in the function When a wave gtY 0 function passes through the g r dashed line it changes sign I Fig 11 The radial wavefunctions of ls We call the point at which it nodes 2s and 3s orbitals Note how the crosses a node Consider a 2 py number of nodes increases from 1 to 2 Sin function 5mmquot 0 to 3 As the wavefunction passes Sinao gt 0 and Sin1 lt 0 through a node its Sign changesust Thus when Sinangle passes like a sin function passing through 0 through 0 it passes through a node Now return back to one of the earlier figures of these waves 7 2 pX one that shows the first three states Notice how the first state has 0 I nodes the second state has 1 node and the 3rd state has 2 nodes It turns out that the n3911 state would have nl nodes The radial wavefunction that describes the electronic orbits is very similar For nl it has 0 I nodes For n2 it has 1 node For n 3 it has 2 nodes and so on and Fig 12 2F orbitals so forth However unlike the quantum guitar string the radial function it not symmetrically bound by in nite potential barriers so it looks different The nl 2 3 radial functions for l 0 are shown in Fig 9 If one is given a radial wavefunction then just count the number of nodes add 1 and you have the principle quantum number Something that is implied by this statement is that a wavefunction with more nodes is a higher energy wavefunction Since these wavefunctions are spherically symmetric the situation for l and m both 0 they look like concentric shells of opposite sign radiating out from the positive nuclear core In your text you will find graphs of the ls 2s and 3s orbitals shown to highlight these nodes So how about if I l or something other than 0 This implies that we have shape to our wavefunction 7 ie its structure is a little more interesting than just a spherical orbital Notice that the radial part of the wavefunction the equation given early in this section is given as Rn1r so it depends on both n and 1 For n 2 we know that Rn1r will have 1 node Ifl 1 then it turns out that the node is at the nucleus not some distance out from the nucleus such as is shown for the 2s orbital in Figure 9 For n2 11 we have porbitals and we have the possible values for m of 71 0 or 1 It turns out that the fact that l 1 means that the porbitals are going to have the shape of two stacked eggs aligned along some axis These orbitals are shown in Figure 12 The three values of m define which axis the porbital is aligned along and we will arbitrarily say that for m l the porbital is aligned along 2 If m 0 it is aligned along x and if m 1 it is aligned along y Since the wavefunction of a 2p orbital changes sign as it passes through the nucleus one of the two eggs will have one sign ie is positive while the other egg has another sign ie is negative For example in the 2pZ orbital at the top of Fig 12 if the top egg is positively signed then the lower egg is negative Just like the radial functions drawn in Fig 11 if the wavefunction passes through a node then it always changes sign For n3 and 12 we again have dorbitals 7 the 3d orbitals We aren t going to worry about these too much in this class Nevertheless there are 5 3dorbitals and they each have two nodes Pictures of them are in your text Filling the Orbitals Let s list the various orbitals for n l 2 n3 E 33 l and 3 in a slightly different way but once again with energy as the yaxis Fig 13 This is similar to Fig 9 except that we have replaced the individual labels of the m quantum number with just a series of empty spaces underline marks Each empty space stands for an unoccupied orbital and so Fig 13 represents a system with no electrons So how do we fill these orbitals up with electrons There are several rules Let s state two 1 Is now I I I I ll lowest energy orbitals rst 0 1 2 Figure 13 2 25 2p Energy 5 2 each orbital can hold two electrons So let s start lling For the rst two electrons its easy We ll in the following way and then I I 1 s 1 s C We call these two cor 39 1s1 and 1s2 respectively The superscript at the right ofthese notations stands for the number of electrons in the orbital Note that when we ll these orbitals we are denoting the individual electrons with a little arrow either pointing up or down We ll get back to that in a minute So far we have built the H atom and the He atom Now we move down to the next atom which is Li We are faced with a choice Apparently all of the 2s and 2p orbitals are degenerate meaning that they are at the same energy level Thus where do we put the electrons It turns out that all of the orbitals of same principle quantum number are degenerate only for hydrogen and hydrogenlike atoms Once we begin putting electrons into the orbitals we nd that the degeneracy is lifted and that the orbitals are no longer isoenergetic What this means is that if we put an electron into the 2s orbital then its energy is lowered relative to the 2p orbital In other words if we put an electron into the 2s orbital we gain a little bit of energy especially when compared to putting that same electron into a 2p orbital Thus the 2s orbital gets lled rst and then the 2p orbital So now we have the 3rd rule 3 In general when we fill orbitals that are characterized by the same principle quantum number n we fill the lowest l value orbital rst all the way up to the highest livalue For example we rst ll the 3s orbital then the 3p orbital then the 3d orbital It is not obvious why this should be so iespecially from what you have learned so far Yet it is so and because of this we can explain many of the periodic properties of the elements So let s continue First we can redraw our energy diagram to re ect our new ordering Let s only do it for n 1 and 2 levels 7 see Fig 14 With our new ordering we can not only make Li but we can make Be E L and B atoms as well We show the Boron atom electronic 2s 2p n Z structure in Fig 12 It is labeled as Is2 2s2 2p1 Now note R something about this arrangement We stated that this is h the lowest energy way to arrange the electrons We call the 0 lowest energy con guration the ground electronic state l However what if we had put one electron into a 2s orbital and two into the 2p orbitals the 1s2 2s1 2p2 con guration 1 1s Or what if we moved an electron from the 1s orbital and I I put it into a orbital a 1s1 2s2 2p2 con guration 0 1 Certainly this 1sn t against the law and we won t get arrested for doing this 7 it just costs more energy Such con gurations that are higher energy con gurations than the ground state are called excited electronic states Note that there are many excited electronic states Is there more than one ground state For some atoms no For example for the Be atom 1s2 2s2 or 1 less electron that the B atom shown in Fig 12 there is only 1 way to con gure the ground electronic state However for the B atom we Fig 12 The lled orbitals of the ground state of a B atom could have put the 2p electron in any one of 3 orbitals Thus the ground state of B is 3fold degenerate Recall the de nition of degenerate that was given above Now we come to yet another dilemma if we want to add one more electron to make a carbon atom Where do we put the electron In an unoccupied porbital or into the same p orbital that already has 1 electron in it To answer this we need to de ne the little arrows that we have used to denote electrons What do those symbols mean It turns out that we have actually one more quantum number that we have neglected 7 the spin quantum number Fortunately it is a simple one 7 it can have values 12 or 7 12 We denote the 12 value as spin up up arrow and the 7 12 value as spin down Wolfgang Pauli stated the Pauli Exclusion Principle which is sort of like rule 2 listed above It states No electron in an atom will have the same quantum numbers as any other electron Why is this like rule 2 Well as a consequence of the Exclusion Principle each orbital can only hold two electrons 7 one with spin up and one with spin down If it held any more then we would have a violation of the Pauli exclusion principle It turns out that spin is a real physical quantity but all we need to know is that two electrons can exist in the same orbital only if they have opposite spin So how does the spin quantum number guide us when we put together a carbon atom This gives us rule 4 4 When lling orbitals that have the same n andl quantum numbers but different m rst put one electron into each orbital keeping all electron spins the same Once all the orbitals are lled with one electron each with the same spin then nish lling the orbitals i i i i i 7 by pairing those electrons n4 In Fig 15 we show how this is done for IS much of the 2p series Assume in this n3 sip gure that the 35 atongs already have u 1L 1L Ne a ls 2s con guration that i f i is not shown We 11 2 f p are just worrying T O gt S about the 2p p on con guration 2 In Fig 14 N g we show for the n 1 2 and 3 and some of 4 how the T 1 15 orbitals line up in 2 B I I I energy You 0 1 2 should be able to F39gbliilFllllflg the por 1 s us1ngru e Figure 14 The energy ordering of the atomic orbitals ufe 1118 dlal39grartn t d 4 for the first 4 rows of the periodic table Note that the p us 6 ru es S a e 3d orbitals are at lower energy than the 4p orbitals here to predwt the ground State eleCtromc con guration of pretty much any atom in the rst 4 rows of the periodic table Lecture Set 2 The Paradoxes of Classical Mechanics and the Origins of Quantum Theory If we are to ever understanding chemical bonding then it is important that we first understand the structure of the atom Unlike most of the princ39lples involving conservation of mass volumes of gases etc the structure of the atom is a truly 201 century discovery As the matter of fact about 100 years ago many scientists felt that they pretty much understood the physical universe 7 there were only one or two nagging problems and those problems would soon work themselves out 7 no doubt about it The solutions to those nagging problems turned out to completely revolutionize both science and society Let s go over these problems as they existed 100 years ago and try to understand how they were solved In the latter half of the 19111 century James Clarke Maxwell uni ed the study of electricity with the study of magnetism Although Oersted had shown in 1820 that electricity and magnetism were related it was Maxwell who really uni ed the two phenomena What Maxwell showed is that an oscillating electric magnetic eld must produce a complimentary and orthogonal magnetic electric eld The term orthogonal refers to a magnetic eld that is at right angles to an electric eld For example a zaxis is orthogonal to both the x and yaxis To understand his model let s rst describe what a wave is This Figure is a periodic anction If we were watching some phenomena such as waves at an ocean beach or ifwe were listening to some particular musical note then we might receive a signal such as this one The xaxis here is time and the yaxis represents the strength of the signal There are three properties of this wave that we need to worry about First is the amplitude 7 or how strong is the Amplitude or strength of signal signal Second is the frequency of the wave in s39l denoted by the symbol v 7 ie how many crests pass by a xed point in a given period of time Finally we have the wavelength of the wave which we denote by the symbol 7 7 is de ned as the velocity of the wave in ms for example divided by the frequency of the wave and so its units are those of length For example assume that we have an acoustic wave travelling through some material that is characterized by a speed of sound of l x 103 ms Let s lrther assume for the above graph the entire xaxis spans 01 second We have about 85 complete waves that pass us in 01 second Now we can calculate frequency and wavelength v 85 waves01 second 85 s39l or 85 Hz 7t 1 X 103 ms85 s39l 115 m 7 a pretty long wavelength Let s try to develop a feeling for waves Waves at the beach Audible Sound 7 20 m v 01 s39l velocity 2 ms a in air 16 mto 016 m v 7 20 s391 to 20000 s39l velocity 331 ms in air at 25 C 1500 ms in H20 at 25 C Light or electromagnetic radiation 7 103916 to 108 m spanning yrays to radio waves v 1024 s391 to 100 s391 8 100 1 by the way velocity 3 X 10 ms in vacuum Velocity in air is very similar Visible Light 70750nmto 400nm1nm10399m Note that electromagnetic radiation spans wavelengths covering 24 orders of magnitude This is a lot We can break this down into yrays Gamma Rays 7 103916 to 103911 m Xrays 7t 103911 to 10398 m UV ultraviolet 7t 10398 to 4 X 10398 m Vis visible a 7 4 x 10398 to 75 x108 m IR infrared a 7 75 x 10398 to 10393 m Microwave 7 10393 to 100 m Longer wavelengths are FM the AM radio waves and the Long radio waves which you may have heard before referenced as kband or Xband or some such type of band A major diiTerence between all these waves is that acoustic waves ocean waves and similar phenomena propagate through some definite medium but electromagnetic EM waves can propagate through vacuum This was a rather bizarre concept around the turn of the century 7 many people searched for the ether which was the supposed omnipresent medium that supported electromagnetic wave propagation The are a couple of other diiTerences between EM and other waves as well Above we stated Magniftic eld Figure 2 A current loop generating a magnetic field The arrows indicate the direction of current flow and the direction of the generated field will learn about this in physics that J C Maxwell had found the connection between electricity and magnetism At left is shown this connection 7 a loop of current generates a magnetic eld at right angles orthogonal to the current ow The direction of the magnetic eld is given by a cross product which means that the magnetic eld direction is that shown in the gure Conversely if we just had a wire loop and we didn t hook it up so that there was no supplied current but rather we put the loop into a magnetic eld such as that shown in Fig 2 the eld would eXert an electromotive force voltage more or less on the wire and thus generate a current One eld does not eXist without the other You We have referred to the electromagnetic properties of radiation Now we make the connection between these terms and Maxwell s discoveries Imagine an alternating electric eld ie a sinusoidal eld passing through vacuum The existence of this alternating eld means that orthogonal to the eld there will be a magnetic eld alternating at the same frequency We could have made this statement the other way around This is shown in Figure 3 Note that the magnetic eld one of the two colors and the electric eld the other color are at right angles to one another just as in Figure 2 This is one example of what we mean by orthogonal So now we have a picture for light 7 it is an electromagnetic wave travelling at a velocity given by c where c represents the speed of light 30 x 108 ms and it is characterized by a wavelength or a frequency that can vary over many many orders of magnitude Paradox 1 Black Bod Radiation and Planck This picture of light was understood at the turn of the century However there were glitches Let s assume that we can describe a wave by the following equation Wave A Sinv7lt Now look at the expression Wit V is a constant the frequency of the wave and t is a variable Plot this lnction 7 you will get a wave Why Now the energy carried by this wave of electromagnetic radiation is proportional to A 7 or the square of the amplitude of the wave We haven t said much about the amplitude of the wave yet However more intense light equals a higher amplitude wave Fig 3 An electromagnetic wave Note that the blue and red NSWV f we apply hear to some waves are travelling at right angles to one another One mammal 1t W111 emlt energy m the form represents an electric eld and the other represents a magnetic Ofradiation Think Of SOldierS crawling field around at night with infrared night vision goggles on Those soldiers are looking at this radiation that comes off the warmest structures in their vision path In any case since we can get the energy of this wave of light from the above wave equation there should be a very straightforward way to calculate the energy given off by some substance at some temperature T It turns out that this energy was calculated to be proportional to TNl Let s not worry too much about just how this calculation was done It is going to turn out to be an incorrect result so we don t really care too much The major point is that when the light coming off a heated object was characterized by measuring the intensity of that light as a lnction of wavelength very dilTerent behavior than Tk4 was observed This was the glitch that bothered people 100 years ago Heated objects are o en called black body radiation emitters Let s go back to some of the implications that the above wave equation carries Wave A Sinv1lt One implication is that the energy of the wave is tunable continuously from 0 to in nity All we have to do is to turn the knob that controls A To see how the prefactor A controls the energy of a wave consider ocean waves at Hawaii as compared to Lake Tahoe The frequency velocity and hence wavelength of those waves may be very similar However the waves in Hawaii are simply bigger Each wave packs a more powerful punch 7 it has more energy An equation describing Lake Tahoe waves would have a relatively small A compared to a similar equation describing waves in Hawaii The energy carried by the waves would be proportional to 2 Planck in trying to understand the glitch in the theory of light made a tremendous leap He postulated that the energy of the wave was not ini nitely tunable Rather it was only tunable in discrete amounts For an electromagnetic wave of a given frequency he postulated that the minimum energy that such a wave could carry was equal to hv This implies that there is a minimum unit of light called a photon One photon equals one quanta of light In MKS or SI units h 662608 x 103934 J s This is the famous waveparticle duality that is so prevalent in quantum mechanics So now we come to a rather confusing point If something is heated and it gives off energy in the form of light then which energy do we care about We now have two energies of relevance here The first is related to our original wave equation and is the prefactor A A more intense wave means more energy in the wave The second energy is related to our description of a photon and is E hv The answer is that we care about both I won t hold you responsible for knowing the following equations but I will give them to you anyway Planck showed that using his assumption and what is known as a Boltzmann distribution he could reproduce the intensity and wavelength distribution of light coming from a thermal source His equation was introduced in October of 1900 and was Radiance A clX4expczLTl391 c1 2nc2h and c2 hck where c is the velocity of light and k is Boltzmann s constant You will encounter Boltzmann s constant throughout chemistry and physics k is the atomic equivalent of R in the famous ideal gas law PV nRT Divide R by Avogadro s number and you have k Let s consider radiance and Planck s equation a little bit and make sure that the units work out The units of Radiance are defined to be Watts m392 J s391 m392 kg m2 s392 s391 m392 kg s393 E6m Intensity Units of R0 in Planck s equation mzs39zJ s m39 Recall that a J kg m2 s392 4 E6 111 Wavelength Ole 6 m So R m2 s392 kg m2 s392 s m394 3 Fig 4 A plot ofemitted radiation from a heated mass This is the curve kg S that Planck successfully explained in 1900 A plot ofthe radiation coming from a heated source is shown in Figure 4 One useful thing about Boltzmann s constant is that it converts Temperature into an Energy by the equation E kT This is a very important relationship Thus if temperature can be considered to be an energy then it is not surprise that a black body held at a xed temperature will emit photons with a characteristic energy distribution that is related to the temperature ofthe box The value ofk is kl38066x103923 J K391 We can make some connections here between energy Temperature and photon wavelength The temperature of this radiation is 3000 Kelvin We can convert this into an energy by using the equation EkT and show that 3000 K is equal to 4 l4x103920 J Now what is the energy of a photon with a wavelength of 18x10396 m We can get this by E hv Converting wavelength into frequency we get a photon of energy E 6266x103934 J s3x108msl8x10396m 104E19 J We note that the energy that corresponds to 3000K and the energy that corresponds to a 18x10396 m photon are not the same although they are reasonably close These energies are not the same because the relationship between the energy curve and the wavelength is not a straight line relationship or a linear relationship in other words However they are only about a factor of 25 di erent If we were to heat up the mass to temperatures above 3000K the curve would shi to shorter wavelengths higher energy photons and if we were to cool it down then the curve would shi to longer wavelengths lower energy photons So what Planck did is to state that light came in discrete packets and that the energy of such a discrete packet was proportional to its frequency with a proportionality constant equal to h now known as Planck s constant One question that bothered me for a long time had to do with the equation that Planck used to describe black body radiation How did he come up with this It turns out that your text has a really wonderful discussion of this on pages 527529 and you should read it It is a wonderful example of scienti c reasoning Paradox 2 Einstein and the Photoelectric Effect The was another glitch in the classical model of the universe and it was the following Shine light on a metal surface As we discussed above the more intense the light you shine onto the surface the more energy you put into the metal If you put energy into the metal then you can presumably activate some chemical or physical process 7 and one such process is that you can ionize the metal 7 ie you supply enought energy so that electrons are ejected from the surface of the metal Let s assume that the energy required to eject l electron from the metal surface is 35 eV and we ll label this the Ionization Potential or the IP For bulk materials the IP is also called the work function The eV is an energy unit we haven t encountered yet We use it here because we want a manageable unit of energy For example if we listed this energy in Joules it would be 56x103919J Joules are useful when we discuss quantities such as moles For a single molecule or single electron energies the eV is much better lJ 62x1018eV We will use eV s a lot So now if we shine 10 Watts of infrared light on the sample in one second we are putting 10 or 62x1019eV s of energy into the sample However no electrons come of What s going on here Recall that we have two energies that we are concerned with here 7 the intensity of the light beam measured in Watts or J s39l and the energy of an individual photon hv We can increase the energy of a photon by increasing the frequency decreasing the wavelength of the light source What we nd is that when hv gt 35 eV we suddenly get a shower of electrons from the surface of the metal As we increase the photon energy beyond 35 eV then the electrons coming from the surface become more and more energetic and it turns out that the energy carried by the photon was hv 7 IP 12 mev2 the kinetic energy of the ejected electron This is what Einstein explained He borrowed Planck s ideas for the energy of a photon and stipulated that if hv was above the ionization threshold for the metal then electrons would come off now matter how intense the light beam was More intense light beams produce more electrons but even a single photon can produce an electron For hvIP lt 0 no electrons are emitted from the metal regardless of how intense the light source is Einstein won the Nobel prize in physics for explaining the photoelectric effect Let s do an example here Say that we have a H atom and we want to ionize it Its IP is 136 eV What is the longest wavelength photon that will emit an electron If we have a photon with k50 nm then what is the velocity of the emitted electron 136 eV x 16x10 19 JeV 7 hv 7 6626x10 34 J s 3x108 nusxm39l 7 91x10398 n1I or 91 nm light is necessary So if we have 50 nm light what is the velocity of the emitted electron 12 mev2 Kinetic Energy of electron hv k50 nm 7 hv k9l nm 7 6626x1039343xlOSmsl50x10399l91x10399l8x103918 J So 12 mv2 7 059 1 1x103931kgV2 7 18x10 18 J and v 7 198x106 ms Paradox 3 The structure of the Atom Recall from our previous lectures that we discussed that the structure of the atom appeared from Rutherford s experiments to consist of a very dense and tiny nucleus surrounded by a dilute and relatively large cloud of electrons Recall also that we discussed Coulombic potentials 7 ie the potential energy curves that describe the interactions between charges There was a nagging question here that we didn t answer If like charges repel each other as we know that they do and opposite charges attract then 0 Why are all the positive charges segregated from all the negative charges and 0 Why doesn t the atom collapse ie why don t electrons get sucked into the core of an atom Good thing that this doesn t happen by the way It turns out that the actual answer to this dilemma is related to what is known as the Heisenberg Uncertainty Principle 7 one of the most nonintuitive concepts in all of chemistry and physics However there are other more intuitive models that steered people toward the right pathways and the adoption of these models was aided not only by Rutherford s experiments but also by a set ofrecently collected emission spectra of various atoms To understand what an emission spectrum is just look at a neon sign As electric current is run through the neon gas the neon gas will convert that electric current into light 7 uorescent lamps work in the same way People had collected emission spectra from a whole number of atoms and ions including H He and My an isoelectronic set of atoms and ions Isoelectronic means that they each have the same number of electrons 7 exactly 1 each in this case What they had found was very surprising If emission intensity were plotted vs wavelength they would see something like what is shown in Fig 5 Intensity of Emitted Radiation Wavelength Figure 5 What the emission spectra of some arbitrary atom might look like So there were two clues to atomic structure The electrons existed as dif lse clouds surrounding tiny positively charged neuclei and the emission spectrum of the atoms consisted of discrete lines Niels Bohr rationalized these observations in the following way the electrons and the nuclei had a relationship with each other that was similar to that of the sun and the planets Notice that the gravitational attraction between the sun and the earth is tremendous However the earth doesn t collapse into the sun but rather it orbits around it 7 a planet orbiting the sun is sort of like a child spinning an airplane around on a string The child is pulling on the string but the airplane doesn t turn inward and collide with the child It has its on angular velocity that keeps it spinning around the child in balance with the child s pull In a similar way Niels Bohr imagined the electrons rotating around the nuclei The discrete lines in the emission spectrum resulted when an electron in one orbit switched to a di erent orbit 7 one that required a less energetic electron The change in energy between the two orbits was then given off as a photon This model turns out to be quite wrong Nevertheless it was the rst time that discrete orbits of the electrons were proposed a correct notion and it was the rst model that could correctly account for the electronnuclei model of the atom Force Slope of Energy vs r curve at r ri Energy O Fig 6 The Coulombic potential Angular momentum l l rmVsinG rmV for 9900 CHEM 20A1 winter 2001 LECTURE NOTES fth set WAVES We next turn to a discussion of waves and oscillators There are two reasons for this we are interested in electromagnetic radiation visible light being one form and we are interested in quantum mechanics Electromagnetic radiation can most effectively be described in terms of waves and quantum mechanics is a wavebased theory so much so that it is often called quotwave mechanicsquot Sinusoidal waves Consider the function Vx Vx Asinkx l where x is the position variable and A and k are constants A is the quotamplitudequot the maximum value that the function Vx can attain k is known alternatively as the propagation 0r wave vector The argument of a sine is dimensionless so k must be a reciprocal length presumably the units of k should be the reciprocal of those used for x the standard units for k are cm39l The function Vx is known as a wave function There are many wave functions and this one is known as a plane wave or sinusoidal wave This sinusoidal wave has a number of interesting properties in particular it oscillates periodically with amplitude A ie it increases to A decreases to A increases again to A etc The periodic nature is evident by the fact that the distance between peaks maximum values of Vx is always Ax 27Ek This constant repetition distance is denoted the wavelength and is represented by the Greek letter quotlambdaquot 7 7M ZTEk 2 One could then rewrite the sinusoidal function as Vx Asin27cx7t 3 It makes no difference whether one focuses on k or on X in some applications one refers to k in others to l The wavelength M has the dimensions of a length Problem Make sure you understand everything that has been stated above It is all very important Problem Show that the function Vx Acoskx 4 has similar properties to that in 1 Problem Show that the function W Asinkx 1 5 where the phase angle Q is a constant has similar properties to that in l and 4 Show that the function 5 becomes that in 1 when Q 0 and becomes that in 4 when Q 75 2 Important Problem Plot Vx versus x for l 4 and 5 with Q TE4 Indicate the meaning of A and M on these plots The points where the curves cross zero are known as nodes Another property of the sinusoidal wave is that it is in nite in extent 139e it extends from xQoo to x oo Wave of this kind can be used to describe all types of wavephenomena such as light waves and sound waves Such waves carry energy and the energy is proportional to a quantity known as the intensity and the intensity I is proportional to lAlZ Ioc AP 6 Thus the amplitude is a measure of the associated energy but it is lAlZ not simplyl that determines the energy and intensity Problem What is meant by quotproportional toquot Note that the symbol quotccquot means quotproportional toquot Problem Although we shall not say more in this course about the exact de nition of quotintensityquot and why it is proportional to lAlZ you may wish to pursue this question further It is not a very complicated or dif cult question but we do not have time to cover it in lecture Come and speak to the instructor about it if you wish Interference amp superposition We next consider the problem of the addition of waves So for example we can specify a new wave function l x which is the sum of two simple component sinusoidal waves l x A1sink1xQ2 A2sink2xQ2 6 where A1 A2 Q1 Q2 are constants Let s however take a somewhat simpler case l x Asinkx AsinkxQ 7 If Q 0 the two component waves are identical and one can write l x 2Asinkx 8a What is of interest here is that the amplitude has doubled but that the intensity quadrupled IxMAR m If however Q TE then l x 0 9a and 1 cc 0 9b Finally if Q TE2 then l x Asinkxcoskx 212AsinkxTE4 10 and Ice 2W2 11 Eq 10 can be obtained by means of simple trigonometric identities either trust this result or make use of high school trigonometry to prove it Problem All this can be understood most readily by means of graphs For each of the three cases plot the two component waves and the composite wave formed by the sum of the two Then explain the results above graphically When I 0 we say that the two component waves are in phase in this case they interfere constructively which is why the amplitude doubles and the intensity quadruples When i T we say that the two components are out of phase in this case they interfere destructively which is why the amplitude cancels out and the intensity vanishes When I TE2 we say that the component waves are 900 out of phase in which case the waves interfere somewhat constructively and the intensity merely doubles Problem Go back to the graphs in the last problem to explain the discussion in the last paragraph This property of adding wave functions giving rise to interference is known as the principle of superposition If wave functions with different wave vectors or wavelengths are superposed as suggested in Eq 6 then the resulting wave function l X need no longer be sinusoidal Problem Prove this graphically by adding two sinsoidal functions with different k s or X39s A reverse generalization of this result is that a large class of mathematically wellbehaved functions l X can be expressed as a sum not limited to two terms of component sinusoidal wave functions This is known as Fourier decomposition of a function Temporal waves In the above sections we considered wave functions that were functions of position X ie functions that show periodic behavior in space But one can also envisage periodic behavior in time i e oscillatory motion In particular Vt Asin03t 12 where A the amplitude and D the angular frequency are constants D is the Greek letter quotomegaquot The angular frequency has dimensions of reciprocal time why and the units used are often S391 sometimes called radians per second One often introduce the frequency V defined by the relation v N27 13 where V is the Greek letter quotnuquot Then one can also write Vt Asin27wt 14 It makes no difference whether one uses 0 or V but one must be clear which one is being used because they differ by 27 unfortunately sometimes 0 is carelessly called frequency in which case there is confusion about whether 0 or V is being used The only concern in all this is to keep track of the 27E s The units ofV are often quotcycles per secondquot or quotcpsquot this unit has also been denoted a quotHertzquot or quotHzquot Occasionally one introduces the period Tp defined as the reciprocal of the frequency Tp IN 15 Then Vt Asin27rtTp 16 Problem Compare Eqs 3 and 16 and the roles of and Tp Traveling waves Waves can propagate i e they can move Suppose we attach a very long cord at one end and oscillate the other end with frequency V If a short time later we photograph a section of the wave we see what appears much like a sinusoidal wave with peaks nodes and low points The distance between the peaks is a measure of the wavelength X If we photograph it a very short time after this we see a similar sinusoidal wave but with the peaks advanced somewhat This then is a quottraveling wavequot The speed v with which the peak moves forward is the velocity or speed of the wave It can readily be shown that v V This is a remarkable relationship that provides an interrelation between the three quantities v V 11 so that only two of them can be independent In many physical situations the speed of the wave is quite constant and known so that Eq 17 is a relationship between V and 7M Note that if X is increased then V is decreased and if X is decreased V is increased Problem Show that v km 18 Problem Eq 17 does not appear by magic it can readily be derived from the simple properties of sinusoidal traveling waves One can show that for a traveling wave VXt AsinkX03t l9 and that this leads to Eq 17 We will not pursue this problem here but if it interests you come by to discuss it Interfering traveling waves Suppose two traveling waves with the same frequency and same phase travel different distances and end up at the same spot what can one say about the interference at that spot The relevant superposition of the two waves is then l t Asinkr1ot Asinkr203t 20a where r1 and r2 are the distances traveled to the final point Straight forward trigonometric manipulations lead to the expression l t 2Acoskr2r1 sinkr1r2Dt 20b The derivation of this expression although indeed straightforward and well within what should be your knowledge of trigonometry need not concern us here What does concern us is the fact that the amplitude is 2Acoskr2r1 so that the intensity is given by Ice 4A2cos22nr2r17V 21a where we have converted from k to wavelength by means of Eq 2 If one sits at a given final spot for some time the wave function l t given in 20b oscillates and averages out to zero Why is this On the other hand the square l t2 averaged over time does not vanish Problem Why is this Estimate the value of l t2 graphically An alternative way of expressing 21a is then I cc timeaveraged l I tl2 2 lb What one sees from the above is that if the difference in path length for the waves is an integral number of wave lengths i e if r2r1 n7 22a where n is an integer then we observe fully constructive interference and I cc 4A2 22b while if the difference in path length is r2r1 2nlk2 23a then I 0 23b Problem Show that this is so Explain it graphically What happens for other values of r2r1 CLASSICAL THEORY 0f LIGHT Wave theory of light The classical theory of light has its origins in the 17th century work of the Dutch scientist Huygens The essence of the theory is that light can be represented as traveling waves In the 19th century it was established that light traveling in vacuum has a constant speed which is often represented by the symbol c The speed of light has been measured quite well and its approximate value is c 3x108 m S39l 24 This is a very great speed in fact according to the wellaccepted theory of relativity the speed of light is not only a constant in vacuo but it is the greatest speed at which matter or energy can be transported It then follows that Eq 17 takes the form 7W c 25 where c is a universal constant This is a very important relation Electromagnetic spectrum Light of different color is specified by its wavelength X or equally well by its frequency V We have already emphasized that increasing 7M corresponds to decreasing V The wave lengths M of visible light light that we can see extends from about 750 nm down to 450 nm the long wavelength 750 nm light being Fall the shorter wavelength 450 nm light being m The reason the different wavelengths frequencies appear to have different colors is that our eye distinguishes the different frequencies and passes this information to the brain which processes the information as different colors When light of all frequencies or wavelengths over the visible range are superimposed one sees white light Light consists of traveling waves in which the oscillating quantities are electrical anal magnetic fields quantities closely related to electric and magnetic forces Consequently visible light is described as electromagnetic radiation Electromagnetic radiation exists with a very large range of wavelengths frequencies a much wider range than observed by the eye Thus visible light covers only a small fraction of the total range known as the spectrum of electromagnetic radiation Above the long wavelength red end of the visible spectrum above 750 nm one has the near infrared IR radiation followed by far infrared radiation followed by microwave radar radiation and finally the very long km radiofreguency rf radiation Below the short wavelength blue end of the visible spectrum below 450 nm one has the ultraviolet UV radiation followed by the vacuum ultraviolet and then very short wave length m radiation at 7 S 10399nm Keep in mind that the speed c of the radiation is the same for all wavelengths so that if one knows 7M one can calculate V by means of Eq 25 and that short 7M means high V as in xrays The eye can detect electromagnetic radiation only in the small visible range but detectors exist for detecting electromagnetic radiation throughout a very wide spectral range Diffraction of light Since electromagnetic radiation can be described as waves one should be able to detect interference effects And indeed one can This is best seen in the famous Y oung39s two slit experiment performed around 1805 Light from a quotpoint sourcequot travels towards a screen in which two narrow slits equally distance from the point source have been cut a distance d apart The screen blocks all the light except that traveling through the slits and the light passing through the slits travels to a second screen a large distance R away ie R gtgt d What one sees on the second screen are light and dark lines ie lines where high and low intensity respectively of light is incident This is a classic interference pattern known as a diffraction pattern One can describe this phenomenon in terms of two separate sinusoidal waves emanating from each of the two slits the distance traveled to the second screen by the light from each of the two slits is different except along one particular line on the screen At points on the second screen where the difference in distance traveled from the two slits is r2r1 n7 where n is an integer one encounters constructive interference and a bright line wherever the difference in distance traveled from the two slits is r2r1 2nlk2 one encounters destructive interference and a dark spot or line Problem Explain this Remember our discussion of interference above The bright and dark lines are associated with large and small values respectively of the intensity ie with maximum and minimum values respectively of the timeaveraged WC Other waves The theory of sound sound waves is analogous to that of electromagnetic waves except that the speed of sound is much slower than that of light and the speed of sound varies considerable from substance to substance Actually when passing through matter as contrasted with vacuum the speed of light also varies diminishes but not very much By dropping a stone on a quiet surface one creates surface waves and these too can be described as indicated above QUANTUM THEORY 0f LIGHT Photons In 1903 Albert Einstein suggested that in certain experiments it was preferable to think of light as composed of particles rather than continuous waves The particles were denoted photons The photons are massless but they travel with the speed of light c though massless they have energy 8 and momentum Einstein postulated that the energy 8 of the photon is proportional to the frequency V of the classical electromagnetic wave a hv 26 where h is a constant denoted the Planck constant and subsequently measured The approximate value of h was determined experimentally to be h 662 x 1034 J s 27 Problem Check that the units given for h are acceptable 139e that the dimensions are correct It turns out that the presence of h is a sure sign that we are in the quantum world all 39 quot 39 39 relevant to the quantum world contain h Note that one can rewrite Eq 26 as 8 hck 28 What is interesting in the comparison of 26 and 28 is that the energy carried by a photon is proportional to V but inversely proportional to 11 Since in chemistry we are almost always primarily interested in energy this suggests that we focus more on frequency than on wavelength Photoelectric effect Why did Einstein make the assumption about photons He was trying to interpret the photoelectric effect that had recently been studied by Hertz It was known that when a beam of electromagnetic radiation struck the surface of certain metals and metal oxides electrons were driven off Einstein reasoned that the radiation had to transfer energy to the metal in order to drive off the electrons But it turned out that if the frequency of the radiation was too low electrons were not driven off even if the incident radiation beam carried a great deal of energy 139e even if it was very intense He reasoned that if the radiation consisted of photons and that the expulsion of each electron was due to the transfer of energy to that electron from a single photon then either the photon had enough energy to drive off an electron or it didn t If a single photon cannot knock off a single electron it makes no difference how many photons strike the metal no electrons will be emitted Einstein then determined that if the photons had sufficient energy to drive off electrons the kinetic energy meuZZ of each of the electrons driven off would be given by meu22 hv lt1 29 where 1 represents the minimum energy that a photon must transfer to an electron in order to drive it off Problem Be sure you understand why this is obvious So if the frequency of the incident beam is too low i e if V lt IDh then no electrons will be driven off if V gt IDh then the kinetic energy of the emitted photoelectrons will increase as V is increased as the photon energy increases This is exactly what is observed Since then similar effects have been observed in a wide range of phenomena in which electromagnetic radiation interacts with matter ie in spectroscopy The energy I is called the work function and is different for different materials Problem Think through all this Problem Glance back at our discussion of ionization and draw parallels between the photoelectric effect and ionization Intensity amp number of photons There is still another aspect of the photoelectric problem Suppose V gt IDh so that the radiation does indeed drive off electrons how many electrons are driven off Since each photon drives off one electron the number of electrons driven off per unit time must equal or at least be proportional to the number of photons per unit time that strike the metal surface We know from classical wave theory that the energy in a beam is dependent upon the amplitude squared and the energy in the beam must be equal to the number of photons in the beam times the energy hV of a photon Why Thus the intensity ie the timeaveraged l t2 of a beam with a given frequency V must give a measure of the number of photons in the beam Problem Explain the changes in the observed photoelectric effect for a given substance as the frequency and the intensity of incident radiation beam is changed What happens as one changes the substance being irradiated PROBABILITY DENSITY Let us return to the twoslit experiment The timeaveraged value of l l tl2 is large wherever bright lines are found and small wherever dark lines are found The bright lines are where many photons fall and the dark lines are where no or very few photons fall This experiment can be carried out so that only one photon at a time comes out and strikes the second screen We can ask where the photon strikes We do not know in advance but we can record where it strikes After hundreds of photons have been emitted one at a time we find the diffraction pattern on the second screen with exactly the same interference pattern as observed when many photons are emitted at once From this we conclude that each photon interferes with itself Although we cannot predict exactly where any given photon will strike the second screen we can say something about the probability of it striking certain places on the screen We know where the dark and bright spots arise and so we know that there is a higher probability of a photon striking where there are bright spots than where there are dark spots But that is all we can say So we conclude that we can make only probabilistic statements about where any given photon will strike This probabilistic interpretation is the consequence of melding a particle photon and a wave interference picture This probabilistic picture not only arises in the theory of electromagnetic radiation but also as we shall see in the quantum theory of matter More speci cally we know that a bright or dark spot is determined by the time averaged l wl tl2 at that spot and that this quantity determines the number of photons striking the spot Therefore the timeaveraged l l tl2 must determine the probability of a given photon striking that spot The major conclusion here is then that the timeaveraged l l tl2 is related to a probability In fact the probability that a photon will strike within a small area written as 5A around a given point is proportional to the timeaveraged l l tl25A Problem Why should the probability be proportional to SA Special discussion Ifthe photons are emitted one by one which slit do they pass through We don t know If we try to nd out by having it set off an alarm each time it passes through we will indeed know which slit it is passing through But much to our amazement there will then no longer be a diffraction pattern The conclusion is that if we want a diffraction pattern we cannot know which slit the photon passes through So in some sense it passes through both slits if we determine through which slit the photon has passed this knowledge or the way in which the knowledge is obtained destroys the diffraction pattern So the diffraction pattern is evidence of uncertainty uncertainty as to which slit the photon has passed through This is not a consequence of a clumsy experiment but is an inherent property of waves and their probabilistic interpretation This phenomenon is a manifestation of the well known Hisenberg uncertainty principle Above we concluded that the probability around a given spot on the screen was proportional to the timeaveraged l l tl25A Let the proportionality constant be given by NQ so that the timeaveraged Nzl l tl25A is the probability not merely proportional to the probability of a photon striking within a given small surface area 5A If we sum this probability over all area elements 5A in a plane then this sum would equal the probability of a photon striking someplace on the plane Since the photon must strike the plane this summed probability must be 1 Why We can represent this by Ztimeaveraged Nzl l tl25A l 30 This is known as the normalization condition and the constant N which is adjusted so that Eq 30 holds is known as the normalization constant One often introduces a quotnormalized wave functionquot I which we define by I N l 31 Eq 30 then becomes Ztimeaveraged lCIgttl25A l 32 The reason the wave functions are normalized is precisely so that the timeaveraged lltDtl25A is a probability and not merely proportional to a probability Lecture Series 3 Chem 20A James Heath Recall that we concluded our discussion of the discoveries by Planck Einstein and Bohr by stating that the single theme that ran through those discoveries was that at very small length scales various physical phenomena were no longer continuous but rather discrete or quantized Light as well as matter also comes in discrete units 7 the photon Electrons which are discrete particles of matter will orbit a nucleus in with atomic orbitals that are characterized by discrete angular momenta Now there is one more piece of information that we need before we can proceed to a more applicable description of quantum mechanics When we described electromagnetic radiation light we described it as both waves and as discrete packets of energy photons This waveparticle duality is not unique to light In fact it is very general Some time after the three A were f J by Planck Bohr and Einstein other scientists began doing experiments on electrons and nding some very puzzling results It was well known from Newton s time that if light was sent through a very small slit it would diffract as shown in Fig 1 This is how diffraction gratings work and is related to why a rainbow is a rainbow Some scientists began doing this same experiment with electrons They would make a beam of electrons send it through a slit and look at a phosphoroscent screen placed behind the slit What they found was rather surprising They Fig 1 When a single color light beam passes Observed diffraaion Patterns lUSt as if they had through a slit it creates a diffraction pattern a done the experiment With a beam of light This series 0f bright and dark Spots 0139 rings on the implied that electrons 7 which had previously been White Screen behind the Slit considered to be particles 7 also had wavelike characteristics Based on these results and the other work that was being done by Planck Louis de Broglie wrote a very brief doctoral thesis that was almost turned down in which he postulated that all matter was characterized by a 39 39 and that 39 39 could be 39 39 J from the following equation 7M hmv Planck s constant divided by the momentum of the particle This equation was in agreement with the observed diffraction measurements Example Say that we accelerated a beam of electrons through a 10keV eld so that they were now traveling with a kinetic energy of 10000 eV s 1 eV 16X103919J What is the wavelength associated with those electrons Note that eV s are a very convenient unit to use here If we accelerate an electron through an electric eld of x volts then that electron has a kinetic energy of x eV So if kinetic energy 1X 104 eV 12 mvz then the momentum mv We know that m me 91 X 103931 kg so we can solve for v and hence momentum 104 eV 16 x103919JeV 1291 X103931kgv2 v 7 5 x107 ms 1 7 663x103934 J s9lX103931kg5 x107 ms391 7 14 X103911m 7 01 Angstroms If we had used protons instead of electrons 7 would have been signi cantly smaller Calculate how much smaller Remember you can not just replug in the mass of the proton but you have to recalculate the velocity of the proton as well As objects get more and more massive their momentum increases and the wavelengths associated with them get smaller and smaller 7 quickly becoming too small to measure What if we found a way to accelerate a neutron We know that mass and light each have wavelike and particlelike characteristics and that electrons in atoms are quantized 7 they have discrete energies or equivalently they have discrete values of the angular momenta We have so far presented only the facts Now we try to ask why Our rst question Why does an electron become quantized Quantum mechanics is not a particularly intuitive subject and so analogies to real world phenomena are a little dangerous Nevertheless here is one that works a little bit Imagine that we have a guitar string tied down at either end as if it were on a guitar If we pluck the string then we hear a note and this note is the resonance frequency of the string This is of course the basis for all stringed instruments One selects an appropriate string stretches it across some length and then puts an appropriate tension on that string These things we do to the string will tune it to a desired resonant frequency 7 say an A note Now it turns out that when you pluck the string you not only hear the A note but if you listen very closely you F39 2 A 39t t 39 d 39t lgure gm H S rmg an 1 S w111 hear an A note at the next octave up and even the 2d octave resonant frequencies Not1ce that all frequencies are integral up as well and so on and so forth These are ha1monics and their multiples of the fundamental frequencies are integral multiples of the rst primary note or the frequency Shown at the bottom fundamental frequency This is shown graphically in Fig 2 of the drawing In the language of physics let s restate the various phenomena that control the resonance frequency of the string 0 First we have the mass of the string 7 larger mass translates to lower 7 ie thick strings give lower notes 0 Second we have the tension on the string When we stretch a string we are actually using it as a spring 7 ie if we let loose of both ends it will retract back to its initial length When we pull or compress a string or a spring this is the same as increasing its potential energy When we compress or expand a spring we store energy into the spring in the form of potential energy Thus the amount of potential energy stored in the string is important More potential energv tighter string corresponds to higher 11 0 Third we have the length of the string that is resonating Shorter lengths lead to higher frequency resonances This is why frets are on a guitar Note that the vibrating string is a one dimensional system 7 ie you can represent the string itself as an xaxis So if we are going to make an analogy between a vibrating string and a quantum mechanical entity we need to choose a ldimensional quantum mechanical system There is a theoretical model for such a system known as the particleina box and it represents one of the most famous problems in quantum mechanics Although this model teaches us many important things about what to expect from electrons in atoms it is just a model 7 it does not represent any real experimental system Note that there is actually nothing boxlike about this model 7 the particleinabox actually more resembles a particle on an xaxis rather than a particle in a box We are not going to solve the particle in a box problem exactly in these notes although it is possible to do so yonr book does it seep 54 7550 In our attempts to make an analogy to a string system we are going to neglect some things and include others that we shonldn t Nevertheless onr incorrect approach to this problem actually comes up with the correct solution and givse as some physical insight These circles represent energy barriers that keep the e39 con ned to the region of the line between the circles w electron exists in this region Fig 3 Particle in a ldimensional box The potential energy has a value of 0 between the green dots 7tA3 jig 4 Same as Fig 2 but we have noted the relationship between the various harmonics and the fundamental frequency The particleinabox is represented by the drawing in Fig 3 The particle in our case is an electron We keep it in the box by putting infinitely high energy barriers at either end of the box If we think of the electron as a particle then we imagine a particle scooting back and forth between the two barriers However if we think of the electron as a wave then we are back to Fig 2 7 our guitar string In fact electron waves are going to appear more or less just like the resonant frequencies of the guitar string However there are differences Recall that 7 hmv the de Broglie relationship Let s look again at our guitar string resonance and its overtones redrawn in Fig 4 Notice that the overtones are characterized by 0 shorter wavelengths than the primary resonance frequency the fundamental frequency We didn t think too much about it when we were discussing guitar strings but here it becomes important According to the deBroglie equation if the wavelength is shorter then the momentum of the particle and hence its kinetic energy must increase Thus we have an electron trapped in a ldimensional box and it has a resonance frequency that is determined by the size of the box However the various overtones of the resonance frequency are at successively higher electron kinetic energies Furthermore this energy separation is necessarily discrete not 1 7 simply because the overtones are integral fractions of the f 39 wavelen th Let s work this out If we have a box of length L then according to Figs 2 and 4 the fundamental frequency has a wavelength of 2L Notice how only 12 of the fundamental wavelength fits into the box 2L hmv so mv h2L and so v h2Lm hkm Energy kinetic energy potential energy However if we keep our particle within the boundaries of the box then the potential energy 0 so the total Energy the kinetic energy 12 mv So if E 12 mv2 the energy of the fundamental is given by E 12 m h24L2m2 12 hz4L2m h28mL2 Let s go to the 1st overtone 7 L hmv so v hmL E 1 W 12 mvz 12 m hzmsz hzszz 4h28mL2 And for the 2d overtone 7 23 L hmv so v 32 hmL and Ezd W 12 mvz 12 m 9h24m2L2 9h28mL2 So we have a series of states with En n2h28mL2 where the fundamental frequency has n l the ISI overtone has n 2 the 2 overtone has n 3 etc Here n is the quantum number that denotes a quantum state of the system En is the energy of the n3911 state There is a function that describes the 3 waves shown in Fig 4 and all the other ones that aren t shown and that function is called the wave function of the system Each of the separate waves is a quantum state of the system and each state is characterized by a quantum number The wave function is often denoted by the symbol l called psi and pronounced si Notice that the quantum number determines the shape of the wave function If we had really set up a differential equation and solved this problem as your book does we would have also found that EI1 h2n28mL2 It turns out that the reason our approach works is because this is an unrealistic and simple system the potential energy contribution to the total energy of the system was negligible as long as the particle stayed between the barriers Thus other than using the potential energy to define how big our box was we neglected it when we calculated the energy of the system Usually we can t do this Also notice that we force fed quantum mechanics into this solution by way of introducing the de Broglie wavelength of the particle If we had actually set up and solved the differential equation we would also have had to force feed quantum mechanics into the problem in order to get the correct solution E So let s go back to the factors that affected the resonance frequency of the guitar string mass of string potential energy length of string and ask g how those factors affect the properties of our E E2 quantum mechanical waves First however we need to adopt a slightly different representation for our waves Since we know that shorter wavelengths correspond to higher kinetic energies we can represent the fundamental frequency and its length harmonics as shown in Fig 5 Fig 5 Now we redraw our previous figure that described the resonances of a string to represent the resonances of a confined electron The dots on the lengthaxis are the dots of Fig 3 Note the energy spacings increase as n First we had the mass of the string 7 larger mass translated to lower resonance frequencies Here we have 7 xed by L the length of the box to which the electron is con ned However we also still have the relationship 7 hmv Thus if 7 is xed and we increase mass then the velocity of the particle decreases Recall once again our favorite equation for energy E 12 mvz If the product mv stays the same since the wavelength doesn t change yet the mass is increased then velocity must proportionately decrease The energy only varies linearly with mass but it varies as the square of the velocity Thus a small increase in mass plus a small decrease in velocity translates into decrease in kinetic energy Increasing mass reduces the enerng of the wave corresponded to higher Second we had the tension on the string More potential energy tighter string Our situation is similar In the particleina box the potential energy is what keeps the particle restricted to a small pa1t of the axis If we remove the potential energy barriers remove the dots in Fig 3 then the electron is no longer con ned to a particular pa1t of the axis and it can have all ranges of wavelengths associated with it including very long ones Third we had the length of the string that was resonating Shorter lengths led to higher frequency resonances In fact if we reduce the width of the boxI we shorten the wavelength 7 just like in the case of a guitar But we also increase the energv of the electron If 7 is reduced then momentum mv is increased and hence kinetic energy of the electron is increased A couple of more things about the wavefunctions The sign and the probability Sign of I v I ZH S E3 V 51 3 E2 a E1 o 1 length Figure 6 This is essentially Figure 5 replotted with the addition of notations that indicate sign of the wavefunction Whenever a wave function passes below the blue line we say it is negative valued Else it is positive valued Probability of I l I z I 3 E3 Ener m N E1 length Figure 7 Here we have taken the square of the three wavefunctions This is the probability distribution When the wavefunctions touch the blue lines the probability of nding an electron at those points is 0 There are a few more things that we can extract from the plots of the wavefunctions We now label our wavefunctions as I l I z etc The I notes that it is a wavefunction and the integer subscript indicates which energy level we are talking about In Figures 6 and 7 we have replotted the wavefunctions Fig 6 indicates that the sign of the iuu can actually be either negative or positive and the higher the energy level the more times that the wavefunction changes sign Recall that a wavefunction is just described by sin and or cosine functions and those can adopt negative values so this should not be too surprising Nevertheless this is a very important concept for thinking about bonding and we will return to it several times In Figure 7 we have plotted l I nlz for n 123 This is called the probability amplitude of the wavefunction At the points that the wavefunction changes sign touches the blue line the probability of nding our particle there is 0 This is called a node We see that higher energy wavefnnctions have more nodes 7 a point we will come back to again because it will be a very important issue to consider when we begin to think about chemical bonding One more point on the probability amplitude l I nlz If an electron is in one of the states described by a wavefunction I n then if we look all throughout the entire space occupied by that wavefunction we should be able to nd the electron This sounds like an obvious statement and it is However you can also make this statement mathematically in perhaps a less obvious way If the chance of finding the electron in the state I n is unity then ll l n 2 Here the unusual appearing integral sign indicates that we are summing up the probability over the entire volume of the wavefunction and that the summed probability l dxl One Last Background Point The Heisenberg Uncertainty Principle There is one more thing we need to cover before we get to a real quantum mechanical system the atom and it is called the Heisenberg Uncertainty Principle This principle is fundamental to quantum mechanics 7 it is not strictly provable but like the rest of quantum mechanics it has withstood the test of time admirably In any measurement that we can imagine doing we will always have a little bit of uncertainty in our answer 7 that uncertainty is governed by our instrumentation or by the nature of the thing we are trying to measure Let s say that we are trying to measure the position of an electron The only way we can do this is to scatter something off of the electron 7 like a photon When we do this we may get a very accurate measurement of where the electron was at the time of the scattering event but we have also changed the system Imagine for example that the electron is travelling along with some velocity v in some direction giving it a momentum p massvelocity If we scatter a photon off of the electron in order to measure its position then we change the electron s momentum We don t really know how we change it but we change it So now if we try to measure the momentum of the electron which we can certainly do it is not the same momentum that it had when we measured its position 7 confusing huh What Heisenberg showed is that if we try to measure both the position and the momentum of an electron then the precision of our position measurement times the precision of our momentum measurement must be greater than h4TE We can write this as ApAX 2 h4TE where the A is read as the uncertainty in Let s do an example Let s say that we have done a measurement that shows that an electron is located right within the nucleus of an atom Assume the size of a nucleus is given by dn103913m then how well do we know the kinetic energy of the electron Answer Since we know that the electron is within the nucleus then we know its position to a AX of 103913 m ApAX 2 h4TE 663XlO3934 J s 47E so Ap 2 52XlO3922 kg m s39l So ifAmv 2 52X103922 and we know me9 leO3931 kg we can calculate the uncertainty in the velocity to be Av 2 58x108 ms So at best we can only pin down the velocity to somewhere between 0 and 58x108ms or six hundred million meters per second Obviously if our velocity is 0 then the kinetic energy of the electron is 0 However if the velocity is 58x108 ms then our kinetic energy is 15x1039 J For 1 mole of electrons this corresponds to an energy uncertainty of 9x1010 Jmol This is a huge uncertainty in the energy If instead we had stated that we knew the location of the electron to within the dimensions of twice the Bohr radius lO39lOm we would have an uncertainty in the momentum and hence the kinetic energy that is much more reasonable Show this to yourself This is why the electron doesn t collapse into the positively charged nucleus as would be predicted by the Coulombic interaction potential energy curve If it collapsed into such a small volume its kinetic energy would possibly be tremendous Since total energy PE KE then the uncertainty in Potential energy is also tremendous Thus we would have no way to draw our electron on our Coulombic potential graph This sounds like a bizarre bit of circular reasoning but it is worth thinking about for awhile Thus the size of an atom is dictated more or less by the Heisenberg uncertainty principle Real Systems The Hydrogen Atom When we solved the problem of the particle con ned to a 1 dimensional box we were able to generate a single quantum number n for that particle If we had solved for a particle in a 3dimensional box we would have found that the solution to the problem involved 3 different quantum numbers 7 one describing how the particle behaved in the xdirection one for the y direction and one for the zdirection This Z has to do with the degrees of freedom of the particle If the particle is confined to a 1 dimensional box it has 1 degree of freedom and hencel quantum number In a 3dimensional box it has 3 degrees of freedom and hence 3 quantum numbers With this in mind what do we expect from an atom In an atom the electron exists within a particular volume surrounding the nucleus Because the electron has wavelike properties we describe the space occupied by the electron as the electronic wave function Because Figure 6 The three degrees of freedom that correspond to this space is 3dimensional it is a volume an electron moving about a nucleus Here we have shown surrounding the positive nucleus we know an xyz coordinate system for reference However we can that there will be 3 quantum numbers also descrlbe the pos1tlon of the electron by us1ng the I describing the electron 7 one quantum dlstance r and the angles theta 9 and ph1 1 to descrlbe number corresponding to each degree of where the vector r points freedom Thus we already see one of the problems with Bohr s atom 7 it only has one quantum number So if we need one quantum number per degree of freedom then what are those degrees of freedom If we had a particle in a 3dimensional box we might just call them x y and z and get quantum numbers nx ny and nz However that is not what we have Instead we have a particle the electron held within some region of space by a centrosymmetric Coulombic potential the positive nucleus One obvious degree of freedom that we have is a measurement of the distance separating the electron and the nucleus 7 ie what is the radial distance For this degree of freedom we have the radial quantum number If we decide on this as one of our quantum numbers then what are the other two Look at Figure 6 Here we show that we can describe the position of the electron by r its distance from the origin and by the angles theta 9 and phi 1 These are related to the xyz coordinate system because 9 is the angle of r within the xy plane and I is the angle that the vector r makes with the with respect to the zaXis Let s not worry about this too much other than that we need one degree of freedom to describe the distance of the electron from the origin and two degrees of freedom that describe angles Thus we have one radial quantum number and two angular quantum numbers The radial quantum number will dictate how the wavefunction spreads out from the nucleus and the two angular quantum numbers will determine the shape of the wavefunction Quantum numbers of electronic orbitals The wave function of an electron is usually denoted by T For the case of the electron orbiting around the nucleus as is shown in Fig 6 l is a function of r 9 and I The energy levels of the electron are described by the three quantum numbers n the principal or radial quantum number I or the angular momentum quantum number and m which is called the magnetic quantum number We ll return to these quantum numbers in a minute We write l as l n1mr9 and we find that we can separate l into a radial component Rn1r which describes how far out from the nucleus the wavefunction extends and X1m9 which describes the shape of the wavefunction Thus we can write LIlnlmraeal Rnlr We also find that there are certain relationships between the quantum numbers and we ll describe those in a minute I just list this information so that you see that there are three quantum numbers and three degrees of freedom What do those quantum numbers mean 7 We don t know yet If we had set up and solved an equation that described Fig 6 which is a hydrogen like atom we would have been able to derive all of the stuff that is being presented here This is a very difficult thing to do and is usually not attempted until the last year of an undergraduate chemistry program or the first year of a graduate chemistry program Nevertheless what we would have found would have been just a much more complicated version of the quantum guitar string problem we solved earlier Just like the guitar string the energy levels would have been described by an energy E where n is the principal quantum number En Z2e4me8802n2h2391 Let s define the constants in this energy expression Z is the nuclear charge For the H atom it is simply I for Liz it is 3 and for He it is 2 Why is this H has one proton so one positive charge Li2 has 3 protons so a positive nuclear charge of 3 and He has 2 protons We don t worry about neutrons since they are charge neutral Also all of these atoms are hydrogenlike 7 meaning that there is just one electron orbiting about a positively charged nucleus Even though the magnitude of the positive charge differs the physical solutions work out to be the same they are just scaled for the different charges You should understand this concept Think about how the Coulombic potential graph would ch ange for these three h ydrogenlike systems Going back to the equation for EH 7 we have e4 which is the fundamental unit of charge to the 43911 power we have me which is the mass of the electron In the divisor we have 802 the fundamental constant that corresponds to the permittivity of a vacuum squared We have hz which is Planck s constant squared and we have n2 which is the primary or radial quantum number squared Thus just like our quantum guitar string the energy of a given quantum state is related to only one quantum number n However unlike the solution to the quantum guitar string the energy depends inversely on n scaling as n39 This means that if we plot the energy levels of this system as a function of n they will look something like what is shown in Fig 9 The energy levels that describe our hydrogenlike system do not depend upon the other two quantum numbers I and m However a given En describes the energy of one or more quantum states of the system Those quantum states are more or less electronic orbitals and they have a shape It is the shape of those wavefunctions which is related to the orbital angular momentum and that is determined by the quantum numbers I and m We could have solved for the shapes and angular momentum of these various states and we would have found that n l and m have certain relationships with one another nl234 3 10l2n 2 2 m l ll l2 0 2 l ll 1 1 1 4 45E 4p 4d In 0 So for example 1fn 3 what are the poss1ble 1 2 1 1 combinations of other quantum numbers that 1 2 we could have n3 35E 3p 3d 0 3 n312m 2 1012 Q 1 1 n3llml0l g n73170m7 Hn2 25E What are 1 the allowed quantum E2 numbers n1 15 IE wode if We haVe I I I I g n n 0 1 2 3 m 4 You Angular Momentum Quantum Number I Should be Fig 9 The various orbitals are plotted as a E able to function of energy Note that since E goes as n39z 1 the levels et increasin 1 com ressed in ener as 0 quot 2131th n increasesg At the 1ingth each orbital notatiogyn is length i If a shaded box that contains all of the m values for yourse 39 that orbital Figure 10 The quantum guitar When we string but only the rst tWO want to describe a particular orbital there are some energy States are drawn The POint nomenclature rules First we consider the principal Where the wave fun Ktion Passes quantum number n Then we consider the angular through the daShed hues ls called a momentum quantum number I If I 0 it is an sorbital If 16 1 1 it is a porbital and ifl 2 it is a dorbital Other less commonly encountered orbitals include forbitals 13 and gorbitals 14 Returning to the case ofn 3 we have 10 lor 2 These combinations yield 3s 3p and 3d orbitals So how about m It turns out the number of values for m describe the number of 3s 3p or 3d orbitals For example for a 3s orbital m can only be 0 Thus there is only one combination n3 10 and m0 For n3 and 11 m can take on the values of 7l 0 or 1 Thus we have 3 3p orbitals each with a different m quantum number For n3 and 12 m has the values of 72 l 0 l 2 7 thus there are 5 3d orbitals In Fig 9 we plot the various orbitals as a function of energy Recall from the energy expression for hydrogenlike atoms that the energy depends only on the principal quantum number n and that energy scales as nquot We have hinted that the angular quantum numbers describe the shape of an orbital and that the principal or radial quantum number describes the radial extension of the orbital Let s look a little more at what we mean by this First we consider the sorbitals For all s orbitals both I and m are 0 What this means is that pls the orbital does not really have much angular shape 7 it is simply a sphere radiating outward from the positive nuclear core But how about n This quantum number affects the radial part of the wavefunction the function Rn1r at the top of this 0 gt section As we move from n0 to nl to n2 etc r the principle quantum number is changing so the l radial part of the wavefunction must change In fact 25 this change is very similar to what we observed for our quantum guitar string Recall that as we increased n we moved from the fundamental frequency to higher harmonics Let s look at that 0 graph one more time and discuss terms In Fig 10 r we have redrawn the lowest two states of the y35 quantum guitar string with a couple of additions We have drawn dashed lines Z through the strings T 2 pl indicating the nodes in the function When a wave 7 9 0 function passes through the X r dashed line it changes sign I Fig 11 The radial wavefunctions of 1s We call the point at which it node nodes 2s and 3s orbitals Note how the crosses a node Consider a 2 py number of nodes increases from 1 to 2 Sin function 5mmquot 0 to 3 As the wavefunction passes Sinai gt 0 and Sin10 lt 0 through a node its sign changes just Thus when Sinangle passes like a sin function passing through 0 through 0 it passes through a node Now return back to one of the earlier figures of these waves 7 2 pX one that shows the first three states Notice how the first state has 0 1 nodes the second state has 1 node and the 3rd state has 2 nodes It turns out that the n3911 state would have nl nodes The radial wavefunction that describes the electronic orbits is very similar For nl it has 0 Fig 12 2p orbitals nodes For n2 it has 1 node For n 3 it has 2 nodes and so on and so forth However unlike the quantum guitar string the radial function it not symmetrically bound by in nite potential barriers so it looks different The nl 2 3 radial functions for l 0 are shown in Fig If one is given a radial wavefunction then just count the number of nodes add 1 and you have the principle quantum number Something that is implied by this statement is that a wavefunction with more nodes is a higher energy wavefunction Since these wavefunctions are spherically symmetric the situation for l and m both 0 they look like concentric shells of opposite sign radiating out from the positive nuclear core In your text you will find graphs of the ls 2s and 3s orbitals shown to highlight these nodes So how about if I l or something other than 0 This implies that we have shape to our wavefunction 7 ie its structure is a little more interesting than just a spherical orbital Notice that the radial part of the wavefunction the equation given early in this section is given as Rn1r so it depends on both n and 1 For n 2 we know that Rn1r will have 1 node If I 1 then it turns out that the node is at the nucleus not some distance out from the nucleus such as is shown for the 2s orbital in Figure 9 For n2 11 we have porbitals and we have the possible values for m of 71 0 or 1 It turns out that the fact that l 1 means that the porbitals are going to have the shape of two stacked eggs aligned along some axis These orbitals are shown in Figure 12 The three values of m define which axis the porbital is aligned along and we will arbitrarily say that for m l the porbital is aligned along 2 If m 0 it is aligned along x and if m 1 it is aligned along y Since the wavefunction of a 2p orbital changes sign as it passes through the nucleus one of the two eggs will have one sign ie is positive while the other egg has another sign ie is negative For example in the 2pZ orbital at the top of Fig 12 ifthe top egg is positively signed then the lower egg is negative Just like the radial functions drawn in Fig 11 if the wavefunction passes through a node then it always changes sign For n3 and 12 we again have dorbitals 7 the 3d orbitals We aren t going to worry about these too much in this class Nevertheless there are 5 3dorbitals and they each have two nodes Pictures of them are in your text Filling the Orbitals Let s list the various orbitals for n l 2 n E 33 l and 3 in a slightly different way but once again with energy as the yaxis Fig 13 This is similar to Fig 9 except that we have replaced the n ZS 23 individual labels of the m quantum number with just a series of empty spaces underline marks Each empty space stands for an unoccupied orbital and so Fig 13 represents a system with no electrons So how do we fill these orbitals up with electrons There are several rules Let s state two n1 Is now 0 1 i I ll lowest energy orbitals rst Figure 13 2 each orbital can hold two electrons So let s start lling For the rst two electrons its easy We ll in the following way and then I I l s l s We call these two con gurations 1s1 and lsz respectively The superscript at the right of these notations stands for the number of electrons in the orbital Note that when we ll these orbitals we are denoting the individual electrons with a little arrow either pointing up or down We ll get back to that in a minute So far we have built the H atom and the He atom Now we move down to the next atom which is Li We are faced with a choice Apparently all of the 2s and 2p orbitals are degenerate meaning that they are at the same energy level Thus where do we put the electrons It turns out that all of the orbitals of same principle quantum number are degenerate only for hydrogen and hydrogenlike atoms Once we begin putting electrons into the orbitals we nd that the degeneracy is lifted and that the orbitals are no longer isoenergetic What this means is that if we put an electron into the 2s orbital then its energy is lowered relative to the 2p orbital In other words if we put an electron into the 2s orbital we gain a little bit of energy especially when compared to putting that same electron into a 2p orbital Thus the 2s orbital gets lled rst and then the 2p orbital So now we have the 3rd rule 3 In general when we fill orbitals that are characterized by the same principle quantum number n we fill the lowest l value orbital rst all the way up to the highest l evalue For example we rst ll the 3s orbital then the 3p orbital then the 3d orbital It is not obvious why this should be so iespecially from what you have learned so far Yet it is so and because of this we can explain many of the periodic properties of the elements So let s continue First we can redraw our energy diagram to re ect our new ordering Let s only do it for n l and 2 levels 7 see Fig 14 With our new ordering we can not only make Li but we can make Be n2 E L and B atoms as well We show the Bozronzatorln electronic 21 structure 1n F1g 12 It 1s labeled as Is 2s 2p Now note ZS something about this arrangement We stated that this is h the lowest energy way to arrange the electrons We call the d lowest energy con guration the ground electronic state l However what if we had put one electron into a 2s orbital and two into the 2p orbitals the ls2 2s1 2p2 con guration 1 1s Or what if we moved an electron from the 1s orbital and I I put it into a 2p orbital a Is1 2s2 2p2 con guration 0 1 Certainly this isn t against the law and we won t get arrested for doing this 7 it just costs more energy Such Flg 12 The lled orbltals of the con gurations that are higher energy con gurations than ground State Ofa B atom the ground state are called excited electronic states Note that there are many excited electronic states Is there more than one ground state For some atoms no For example for the Be atom ls2 2s2 or 1 less electron that the B atom shown in Fig 12 there is only 1 way to con gure the ground electronic state However for the B atom we could have put the 2p electron in any one of 3 orbitals Thus the ground state of B is 3fold degenerate Recall the de nition of degenerate that was given above Now we come to yet another dilemma if we want to add one more electron to make a carbon atom Where do we put the electron In an unoccupied porbital or into the same p orbital that already has 1 electron in it To answer this we need to de ne the little arrows that we have used to denote electrons What do those symbols mean It turns out that we have actually one more quantum number that we have neglected 7 the spin quantum number Fortunately it is a simple one 7 it can have values 12 or 7 12 We denote the 12 value as spin up up arrow and the 7 12 value as spin down Wolfgang Pauli stated the Pauli Exclusion Principle which is sort of like rule 2 listed above It states No electron in an atom will have the same quantum numbers as any other electron Why is this like rule 2 Well as a consequence of the Exclusion Principle each orbital can only hold two electrons 7 one with spin up and one with spin down If it held any more then we would have a violation of the Pauli exclusion principle It turns out that spin is a real physical quantity but all we need to know is that two electrons can eXist in the same orbital only if they have opposite spin So how does the spin quantum number guide us when we put together a carbon atom This gives us rule 4 4 When lling orbitals that have the same n andl quantum numbers but different m rst put one electron into each orbital keeping all electron spins the same Once all the orbitals are lled with one electron each with F4 W7 quotTaquot 45 iii n3 p 35 7271077 2 n 25 gt 61 25 s m 1 Is I I I 0 1 2 Figure 14 The energy ordering of the atomic orbitals for the rst 4 rows of the periodic table Note that the 3d orbitals are at lower energy than the 4p orbitals the rst 4 rows of the periodic table the same spin then nish lling the orbitals by pairing those electrons In Fig 15 we show how this is done for much of the 2p series Assume in this gure that the atoms already have a ls2 2s con guration that is not shown We are just worrying about the 2p con guration In Fig 14 we show for the n 1 2 and 3 and some of 4 how the orbitals line up in energy You should be able to use this diagram plus the rules stated Ne P 1421 0 LLL N 2p L2 B porbitals using rule 4 here to predict the ground state electronic con guration of pretty much any atom in Lecture Series 10 Transition Metal Chemistry Chapter 18 This is the last material that will be covered on the nal Chapter 18 covers a huge amount of material in a very few number of pages Not only is there a lot of material in this chapter the nature of the material is nearly completely new 7 at least on the surface There is simply no way that we can cover much of this chapter 7 nor would we want to It is simply too much and I only have a small amount of time in which to lecture on this stuff Nevertheless there are a few important topics that we will cover This chapter concerns itself with the elements that exhibit the chemistry of the dorbitals These are called the transition metals Look for a moment at the fourth row of the periodic table which contains the elements shown in gure 1 3D 4P Figure l The 43911 row of the periodic table The transition metals are the elements that range from Sc Scandium to Zn Zinc VaScTiVCrMnFeCoNiCuZnGaGeAsSeBrKr 4S These elements exhibit a range of oxidation states that are unparalleled in the S and P elements We brie y touched on oxidation states in previous sections but here is a formal definition The oxidation state of an atom is simply the charge on the atom For example Copper can exist as Cu39 Cuo Cu and Cu For these atoms and ions oxidation states are 71 0 1 and 2 respectively Often times students think that oxidation states must have something to do with the element oxygen This is not true Similarly an oxidation process also doesn t necessarily involve oxygen If an element is oxidized then its charge is increased in the positive direction Examples of oxidation processes include Cu39 9 Cu e Cu0 9 Cu e Mn3 9 Mn7 4e A reduction process is the opposite of an oxidation process If an element is reduced then its charge is increased in the negative direction Any of the above examples of oxidation processes would be reduction processes if they were written in reverse such as for the rst equation listed Cu e 9 Cu39 This is related but is not quite the same as ionizing an atom such as we discussed back in Chapter 15 with the photoelectric effect In an oxidationreduction collectively called redox process if an atom or a molecule is oxidized then something else is reduced so that charge is balanced in the end Likewise if something is reduced then something else is oxidized Let s consider for a moment the electronic structure of Cu and try to understand these various oxidation states Ifwe used the Aufbau principle Hund s rule Pauli principle etc we would say that the electronic structure of Cu0 is 482 3D9 Now if we add one more electron we can get a closed shell or 482 3D10 This would be Cu39 and we would expect it to be fairly stable it is after all a closed shell anion Now imagine if we removed one electron from Cuo We would then have 482 3D8 This arrangement of electrons doesn t look especially remarkable 7 and so one might not expect it to be very stable However try moving both of the 4S electrons into the D shell and you get 480 3D10 This is a closed shell cation and the electronic con guration now does look rather special In fact Cu is 480 3D10 and 1 is a common oxidation state for Cu Now let s consider Cu which would have an electronic con guration of 482 3D7 or of 4S0 3D9 depending on how we think about the 2 oxidation state Note that here I have been rather casual about moving the 4S electrons back and forth between the 3D orbitals In fact this is a ne thing to do and it is very common for transition metals to grab the S electrons for the dorbitals when some special stability can be gained However for this Cu species it is rather dif cult to understand why either of these two electronic con gurations would be particularly stable In fact Cu does exist but it is not as common as the other oxidation states From our simple arguments we would predict that the three most common oxidation states of Cu would be Cu39 Cu and Cu Cu would less common This is what is observed The preceding discussion of the oxidation states of copper highlights both the beauty and frustration of inorganic chemistry It would be great if we could just take the electronic con gurations of the transition metal atoms in their various oxidation states and predict which ones will to be observed which ones will be the most stable etc The fact is that we can t do this However we can use the electronic con gurations as a nge to which oxidation states will be most likely However lots of possibilities may be observed 7 possibilities that we don t think of when we just look for magic electronic con gurations What do we mean by magic electronic con gurations Here are a few with accompanying explanations The ones highlighted in red are especially stable 4S0 3D0 rare gas con guration ex Sc Ti Mn 432 3D0 closed Sshell ex TiII van MnV 4s0 3D5 12 lled Dshell ex Fe Mn 4S1 3D5 6 unpaired aligned electrons example Cro V39 Fe 482 3D5 closed Sshell aligned and unpaired electrons in D shell 2 3 ex Co N1 4s0 3D10 closed DShell Cuo Ago Au 4S2 3D10 closed Sshell closed DShell ex Hg Cdo Zno Note that we have in a few cases referred to the oxidation states of the various elements using Roman numerals such as MnV This is equivalent to Mn and is actually a more common way of representing the 5 oxidation state of Manganese We will use Roman numerals throughout this discussion to represent positive oxidation states of transition metals Note also that in some instances the system that represents the magic con guration is a neutral atom such as Cr0 for 48 3D In fact in those cases the bare metal is very stable Recall that Cr provides a shiny metallic plating on many automotive parts or that mercury Hg0 682 5D10 is a stable metal that is used in thermometers and elsewhere Note also that there are many examples of magic con gurations that are not found You can gure this out by comparing the above list with Figure 184 p 663 of the text For 481 3D5 one might expect Mn In fact Mn1 is not listed on 184 and so it is probably not very stable You ask Why not I am not really sure but I believe that it is because the MnII state is so stable that any MnI that you might form is immediately oxidized to MnII Transition metal compounds Recall from our VSEPR theory that we had a number of shapes such as octahedral and square planar that simply didn t shown up in our description of organic chemistry Those shapes show up quite commonly in transition metal chemistry In fact the most common transition metal compounds are characterized by the stoichiometry of ML5 octahedral or ML4 square planar Here M refers to the transition metal atom and L refers to ligand A ligand is something that forms a bond with a transition metal atom Some of the most common ligands are the halide anions Cl39 Br39 F39 the oxide anion Oz39 hydroxyl OH39 cyano CN39 thiocynato SCN39 and isothiocyanato NCS39 Charge neutral ligands include the ammine group NH3 the aqua group OHZ and the carbonyl group CO Here we name these according to how they are named when they are used in transition metal chemistry For example NH3 is normally called ammonia although it is an ammine ligand OH is water but it is called an aqua ligand If you can remember these various neutral and negatively charged ligands then you can usually determine the oxidation state of the transition metal that is bonded to them I won t make you remember their names Let s do several examples naming the molecules and identifying the oxidation states and the electronic con gurations of the transition metals in them Consider the compound NiCO4 This compound is a clear colorless gas that is highly toxic The CO ligands are charge neutral the compound is tetrahedral and Ni is in the Ni0 oxidation state NiCO4 is tetrahedral because it has a steric number of 4 Nio is essentially a closed shell so it gets 8 electrons from the 4 lone pairs on the CO ligands giving it a steric number of 4 This compound would be called tetracarbonylnickel The tetra implies the number 4 referring to 4 carbonyl groups on the nickel atom Note that Ni is SOD PtClzNH3z is called dichlorodiamminoplatinum is square planar and Pt exists in the PtII oxidation state Note that PtII is SIDS It is square planar because it has a steric number of 6 Of its 6 valence electrons 2 are used up in the PtCl bonds and 4 are used up in two lone pairs at the top and bottom of the octahedral bonding geometry Note that none of the PtII atom s electrons are used up in the amminobonds Why Consider the NH3 molecule It has a lone pair and it is that lone pair that is used up in the Pt NH3 interaction Since there are already two electrons in the lone pair the bond doesn t need any more electrons from the PtII metal atom Fe203 is called ironHIoxide Let s not worry about the structure FeIII is CrO4239 has a CrVI atom and is called the chromate ion Note that CrVI is SODO Cr207239 also has a CrVI atom and is called the dichromate ion There is no reason that you should be able to name these since the names don t really make a lot of sense The one thing that is important in these names is that they end in ate This means that the whole molecular complex is negatively charged Let s look at another negatively charged complex CoNH3zCN439 is called diamminotetracyanatocoboltateIII Do you understand why the Co is CoIII Consider the fact that the entire complex has a net charge of 71 and then consider the various groups that are bound to it NH3 is charge neutral but CN39 is 71 charged CoIII is SIDS and this structure is octahedral Again we end the name of the compound with ate to denote the fact that it is negatively charged If your purchased this compound or if you prepared it in a form that allowed you to put it into a bottle it would generally come as a chargeneutral compound such as KCoNH3zCN4 The name for this compound would be Potassium diamminotetracyanatocoboltateIII Note that this is written as two words There are a lot of other names for J on p 668 and in the surrounding pages Although it is important for you to d understand these names it is much more 3 xy important for you to be able to 391 the 391 39 state of the metal atom in those 1 the geometric structure of those 1 and the electronic configuration of the metal atom Are all these metal atoms characterized bv a magic electronic configuration Electronic structure of transition metal complexes In order to understand the electronic structure of transition metal complexes it is important to understand just what the d orbitals look like We will consider the d Figure 2 These three d0rbitals 100k orbitals as just two types of orbitals One type very Similar to each other They each is shown in gure 2 and those orbitals are contain 4 lobes here Ihave colored the Squareshaped With same39Signed 10bes negative lobes with in one way and the oriented diagonally to each other Note that positive signed lobes in another way each of these orbitals lies in the plane that is and they each lie in one ofthe planes speci ed by the name of the orbital The de ned by two out ofthe three axis orbitals also lie in between the cartesian Note that the orbitals bisect rather than coordinate aXiS 11m along the aXiS lie along the actual axis The second class of orbitals is indicated in Figure 3 Note that for these two atomic orbitals the actual lobes of the orbitals lie along the cartesian coordinate axis Zd Z2 Figure 3 the other 2 dorbitals Note that these are aligned along cartesian axis Now based on our previous experience with making chemical bonds our rst inclination would be to make a variety of sigma and pi bonds between these dorbitals and the orbitals of the ligand molecules We would then get a correlation diagram that contained 039 5 TE TE and nonbonding orbitals If we wanted to explain transition metal bonding in terms of molecular orbital theory MO theory then that is exactly what we would do In fact such an approach would eventually work and we would be able to reproduce the properties of transition metal complexes although we would have to take MO theory substantially further than we have taken it in this class For that reason we aren t going to go in that direction Instead we are going to utilize a different theory known as crystal eld theory This is that last thing that you will have to know for the nal by the way and you will just have to know it a little bit O C 00quot dX2y2 d Z2 8 Figure 4 CrCO6 The top image shows the central green Cr atom and the 6 carbonyl groups arranged in an octahedral bonding geometry The bottom image shows how these carbonyl groups point directly toward the two dorbitals Crystal eld theory differs from MO theory in that we don t exactly ever get as far as making a bond Instead we just consider the energies of the atomic dorbitals on the central transition metal atom under the in uence of the surrounding ligands Consider the molecule CrCO5 which we can call Chromiumhexacarbonyl Note that the Cr exists as Crquot and that the electronic con guration is 481 3D5 Further note that this structure is octahedral Each CO looks like CO where the lone pair on the carbon is what interacts with the Cr atom The Chromium atom has one electron in each of the available 5 3d orbitals The 6 carbonyl ligands are aligned along the x y and z cartesian axis as shown in gure 4 So now what is the nature of this chemical interaction It turns out that the electrons in the d orbitals are repelled by the two electrons in the carbonyl lone pairs However not all dorbitals are affected in the same way The dorbitals that are aligned along the cartesian axis the xzyz and the z2 d orbitals experience this repulsion more strongly than do the dxy dyz or dXZ orbitals The result is that the energies of the dorbitals on the Cr atom are split This is called crystal eld splitting The energy of all 5 dorbitals are raised relative to their energies in the isolated atombut the energies of the xzyz and the z2 dorbitals are raised more than the dxy dyz or dXZ orbitals The LOW f e d resulting splitting is shown T in Figure 1817 of your text Sphtnng 0396 A0 and I reproduce it here as 39 Figures 5a and 5b In these gures I have denoted Ao d d d as the total energy value of W W X2 the crystal eld splitting Notice that all the orbitals are raised in energy relative X to where they start In this I gure I show two cases High eld and these cases correspond Splitting 06 A0 to what are classi ed as 39 39 A0 weak crystal eld splittings 5 0394 A0 and strong crystal eld 5quot splittings What determines d d d whether a crystal eld xy zy xz splitting is weak or strong Figure 5 The two cases of crystal eld splitting for the The nature of the ligand is ocmhedral geometry the most important part Certain ligands will interact 39 39 9 43 DD with the dorbitals only a little bit and therefore lead to a weak or low crystal eld splitting while others will interact more strongly Also note the arrangement of the electrons in the various orbitals In the low eld splitting case at the top of Figure 5 the most energetically favorable way to a1range the electrons is to keep them all in separate orbitals with their spins aligned However if we split the orbitals by a large amount bottom of Figure 5 then it becomes more favorable to put all of the electrons into the lowest energy orbitals Although both of these electronic con gurations are paramagnetic the number of unpaired spins in the low eld case is much greater than in the high eld case Thus if one could measure the number of unpaired electrons and one can then this would serve as a measurement for whether the system was in the low or high eld case There is another way to get this information and that is to do spectroscopy In both of these systems the lowest energy electronic transitions correspond to taking an electron from one of the 3 lowest energy dorbitals and putting it into one of the 2 higher energy dorbitals A photon can be used to supply the energy necessary to move these electrons and a measurement of the photon energy that is required is a direct measurement of A0 or the crystal eld splitting energy Obviously in the high eld case A0 is greater than it is in the low eld case and so the energy that is required to Freshman Chemistry 20A Winter Quarter Lecture Week of Jan 11 Chemistry and the Periodic Table The point of this course is to learn how to think about chemical structure the periodic table and the relevance of these things to the world around us Consider the periodic table The elements at the far right are essentially inert The elements at the far left are amongst the most reactive of all elements Yet the diiTerence between the two elements is that the ones on the far right are characterized by 1 less electron and 1 less proton than the corresponding element 1 row down on the far le If you don t know what an electron or a proton is just wait a bit We ll get to that Let s take one of the simplest reactions possible 2Na Brz 9 2NaBr There are several things to note about this reaction First Na is a covalent solid and it is a metal 7 which means that it is very shiny like a silvered mirror In fact in terms of its electrical conductivity it is one of the best metals there is Recall also that by nature of its position at the far le of the periodic table it is a very reactive element Brz is a molecule and it is a liquid until 58 C above which it is a gas and it is red in color It turns out that Brz is only slightly reactive on a relative scale that would compare the reactivity of various elemental molecules NaBr is an ionic solid 7 which implies that it is not a metal but rather an insulator In fact it is one of the best insulators that there is In addition it is nearly completely inert or nonreactive and it is clear and colorless So we combine a highly reactive covalent metallic solid with a quasireactive red liquid or gas and we get an inert product that is an ionic insulating clear colorless solid Let s look at another reaction Na H20 9 NaOH 12 H2 In this reaction we are taking a reactive metallic shiny covalent solid and reacting it with a clear colorless molecular liquid how is the liquid held together and producing a colorless white reactive ionic solid and a slightly reactive colorless molecular gas We have only written down two apparently simple and common chemical reactions and we have already introduced a tremendous amount of complexity Other questions that we haven t brought up yet include How fast are these reactions Do they proceed to completion etc With this complexity already introduced how can we ever hope to understand truly complicated chemical systems such as living organisms or the earth s atmosphere or the chemical processes that are involved in fabricating electronics or plastics etc In this class we will only begin to shed light on these issues and we will do it slowly one step at atirne We will rst introduce the language of chemistry and we will follow this with a highly simpli ed description of what makes up an atom This description will necessarily involve some quantum mechanics starting with Chapter 15 We will discuss the types of experiments that have led to various descriptions of chemistry including spectroscopy which is the interaction of light with matter A er we discuss the atom we will move onto the structure of simple molecules and how simple models may be used with caution to predict such structures Finally we will increase the complexity of the systems we are trying to describe and include organic molecules polymers and transition metal complexes although we may not quite get to the transition metals For now let s back up a bit and just try to understand the language of the reactions we have already mentioned and discuss why they were written they way they were First let s return to the reaction 2Na Brz 9 2NaBr What we are saying when we write this reaction down is that 2 units of sodium atoms react with 1 unit of bromine gas to produce two units of sodiumbromide The numbers in front of the chemical species are intended to balance the chemical reaction We could have just as easily written the reaction as Na 12 Brz 9NaBr Notice that this works out just as well if we are trying to conserve mass In fact that is just what we are trying to do when we balance a chemical equation In all chemical reactions mass is conserved This is a principle that was rst stated by Lavoisier early in the last century Notice that we did something similar when we wrote down our second reaction of the day Na H20 9 NaOH 12 H2 However this reaction is a little more complicated 7 2 reactants to produce 2 products Balancing equations never involves more than algebra but nevertheless the algebra can get a little complicated Molecular Unitsa Molesa and the Periodic Table When we say that 1 unit of sodium plus 1 unit of water reacts what do we mean by units Chemists use all types of units to describe quantities of chemicals Below we have listed a few Mass Volume Grams g liters l 02 gallons roughly milligrams mg 0001 g milliliter ml 0001 I kilograms kg 1000 g 22 pounds microliter ul 10396 1 However when we are dealing with reactions such as those described above we need to work in units that are related to the number of molecules or atoms that are reacting or are produced We could certame just say that 1 atom of sodium reacts with 12 molecule bromine to produce 1 molecule of sodiumbromide However this is not very much material For example 1 atom of sodium only weighs 4 x 103923 grams The most sensitive mass scales measure only a few nanograms 10399 grams or so and so even those scales are 14 orders of magnitude away from being able to measure what happens in this reaction 103923 grams is so very small that it is incomprehensible to most people including me Thus we would like to work with amounts that are comprehensible The weight ofa mole ofmolecules is such an amount 1 mole ofthings is 6022x1023 of those same things so 1 mole of sodium atoms weights about 23 grams This is much better 6022x1023 is called Avogadro 3 number and is represented in shorthand by NA How did someone ever come up with the de nition of a mole as 6022x1023 It turns out that 1 mole of carbon atoms weighs 12 grams exactly It is good to remember that 1 mole of things is a tremendously large number of things 1 mole of seconds is longer than the age of the universe 1 mole of grams is approximately the mass of water on the earth We will nd that many things in chemistry are characterized by masses charges velocities etc that are of a scale that makes them very dif cult to comprehend Things like N A are used to make such physical quantities a little more manageable Scienti c notation Notice that above we have written that 1 ul 10396 Z We could have written 1 ul 0000001 1 however such notation becomes rather painful and unmanageable Imagine writing out Avagadro s number This type of preferred notation is called scienti c notation and if you have any problems with it you should review the supplement to the course text and the appendices in the text Now that we have a number the mole with which to deal with 1 1 chemical quantities how do we relate that to our metric system units the gram and the liter N a In the front inside cover to your textbook you can nd a periodic table Let s consider one of the elements listed in that table Its entry looks something like the box on the le 22990 There are three parts to this box At the top is the number 11 which is often shorthanded by Z and means the atomic number of the element In the middle is the abbreviation for the element This particular box is for sodium Why is the symbol Na Well sodium was one of the rst elements discovered and it was initially given the Latin name Natrium hence Na There are other elements like this For example Tungsten is shorthanded W standing for Wolfram However most of the abbreviations of the elements make a little more sense It is an excellent idea to memorize the symbols for the first 3 rows of the periodic table I won t ever ask you to regurgitate this information but it will make your life easier Finally at the bottom of the box is listed the number 22990 This is the atomic weight of sodium and means that 1 mole of sodium atoms weighs 22990 grams or that 1 atom of sodium weighs 22990 atomic mass units or amu s Note that it is always an atomic weight For example in the above equations we referred to Brz and H2 which don t normally exist as atoms but rather as diatomic molecules Their weight as given in the periodic table is for the individual atoms not the molecules So what are the molecular weights of Brz or Hz They are simply 2 times the atomic weights of those respective elements For Hz this is 2 x 1 g 2 g and for Brz it is 2 x 799 g 1598 g Molecular weights of more complicated molecules are calculated in the same way Let s use atomic weights and molecular weights to gure out how much material we would have to measure out to do our favorite reaction Na 12 Brz 9NaBr Assume that we want to end up with 10 gram of product The molecular weight ofNaBr is 229 799 1028 gramsmole Therefore 1 gram ofNaBr 1g1028g mole39l 97x10393 moles That means that we will need 97 millimoles 1 millimmole 10393 moles and we will abbreviate millimoles as mmoles of Na metal and 972mmoles of Brz gas So we need 97x10393 moles x 229 gmole weight Na 0222 g Na And 1297 x10393 moles Brz1598 gmol 0775 g Brz Let s check our calculations We add 022 g Na to 077 g Br to get 0997 g NaBr Si i cant Figgresa Calculatorsa and Wrong Answers It turns out that this number 099 g is the same as 10 g to the accuracy that we specified at the start of the problem As the matter of fact it should actually be reported as 10 g since we really don t know this number any more accurately than that One thing that calculators will give you is lots and lots of numbers to the right of the decimel place However we must constantly ask ourselves how well do we know the numbers that we are calculating This is not the same thing as how many gures that the calculator gives you At the start of this problem I specified that I wanted to have 10 grams of product I speci ed this number to the nearest tenth of a gram This means that any number between 095 and 104 grams would have been just ne 7 they are all the same to me since 095 rounds up to 10 and 104 rounds down to 10 Therefore even though a calculator will turn the 077 g Brz listed above into 077503 7 don t be fooled and lose points Your answer should be quoted only to as many signi cant gures as you know it Certain numbers that are used in algebraic equations like this are going to be absolutely known or known to a very high accuracy In this case we of course could look up the atomic weights of Na and Br to a tremendous accuracy and the 12 is known exactly as is the stoichiometry of the reactants and the products However since we asked for 10 gram of product we only need an accuracy of a tenth of a gram and we should only quote our answer to a tenth of a gram This is a way that many students lose credit on homework and test problems Signi cant gures Signi cant gures are so important that I think that we should do another example Assume that we have a compound of chemical formula C8H1102 and we want to burn it which means react it with oxygen However we only have 25 grams of oxygen How much of the compound can we burn The equation for this process is C8H1202 02 9 C0 H20 As a note whenever we burn a compound that contains only carbon hydrogen and oxygen we will need oxygen to burn it and we will produce only C02 and H20 This is the de nition of burning such a compound First let s balance this equation A good way to approach this problem would be to rst balance the carbons and the hydrogens on the le and the right sides of the equation A er doing that then you will only need to adjust the relative amount of 02 to make the reaction stoichiometrically balanced Since we have 8 C s on the le and 12 H s on the le then we can do the following C8H1202 on 9 8C0 6 H20 Note that now the carbons and the hydro gens are balanced on both sides The amount of oxygen is not correctly balanced yet At le we have 2 2X 0 s and on right we have 16 6 22 O s Therefore X 10 and so the balanced equation is C8H1202 1002 9 8 C0 6 H20 Now recall that we only had 25 grams ofOz to burn 25 g 32 gmol 078 moles of Oz That means we can only burn l100078 moles of CngzOz or we can burn 0078 moles Now the molecular weight ofthis compound is 81201g12l01g 21600g 1402 gmole So l4ll gmol X 0078 moles 10936 g compound 11 grams ofcompound The underlined number is the one that should be quoted as an answer Make certain that you understand why Unitsa Units Systems and Algebra Notice in the above calculation that the units work out correctly This always has to be the case If the units don t work out then you most certame have the wrong answer You can always use the units as a check to see if you have set up the algebraic equation correctly Although just having the units work out doesn t mean that the answer is correct it is certame a necessary condition for getting the correct answer Lets work out an example with just units Recall from high school physics that kinetic energy E 12 massvelocity2 l2mv2 an equation that is hopefully familiar to you According to our units above we can use kilograms for mass Velocity is in units of distance per unit time If we use kg s for mass then we should use meters for distance and seconds for time This is the MKS units system also called SI units MKS stands for Meters Kilograms Seconds Another unit system is the cgs centimeters grams seconds system So E 12 kgms2 Do these units describe an energy In the MKS units system the unit for energy is the Joule The definition of the Joule is in fact kg m2 s39z If we had used cms for velocity then we probably should have calculated out the energy in the cgs centimeters grams seconds unit system Lets convert l J to an energy inthe cgs system H 1 kg m2 s392 now multiply by the conversion factor for meters to cm s lJ 1 kg m2 s392100cmm2 104 kg cm2 s392 Now convert kg to g 104 kg cm2 s39z1000 gkg 107 g cm2 s392 107 cgs energy units There is a name for this cgs energy unit and it is ergs 7 ie the erg is the energy unit in the cgs system Therefore 1 erg 10397 J There are other units for energy including the electron volt eV the kiloWatt hour kW h the british thermal unit BTU etc The units that you use depend upon the problem that you are solving Densities Now let s return to our favorite chemical reaction Na 12 Brz 9NaBr Recall that in this reaction we are reacting a solid metallic element with a gas We have already discovered how to think about this reaction in terms of moles and how to convert from moles to grams and thus calculate the masses of equivalent amounts of reactant and product O en times chemists will not just consider the masses of the chemicals but their volumes also To do this we need to understand the concept of density Density is a massperunit volume and is usually described in the units of gcm3 Note that cm3 is a volume A cm is about as long as a trimmed fmgemail so a cm3 corresponds to the volume of a box with each side about the size of a ngernail Let s list the densities for a few common substances Here we are not going to be exact but just provide rough estimates Brz gas 006 gcm3 Water H20 1 gcm3 Organic materials 07 7 09 gcm3 Iron Fe 8 gcm3 Copper Cu 9 gcm3 Silver Ag 11 gcm3 Gold Au 19 gcm3 Let s say we had a set of 4 immiscible they don t mix but rather separate materials that we put into a closed container These materials were Brz benzene liquid formula C5H5 water and Iron A er we had prepared this rather bizarre mix we can use the densities to predict how they would appear in our container The iron has the highest density so it sinks to the bottom Although the densities of benzene and water are very close to BI one another the benzene has a slightly lower density so it oats 2 on to fth t F th 1 f 39 gas p o e wa er or ano er examp e 0 an organ1c aqueous mix consider cooking oil on water or an oil slick The oil is always on top of the water Finally the density of the Brz gas is much lower than any other component and so it lls the volume C6H6 above the liquid surface water Note that in the table listed above the densities of Cu Ag and Au are given and that their density increases in the direction of Au Now note their positions on the periodic table We will return to this later when we come to atomic structure So now let s convert densities of some substance into moles of that same substance Let s assume that benzene s density is in the middle of the range listed 08 gcm3 How many moles in 1 cm3 1 mole ofbenzene weighs 6 C atoms times 12 gmol 6 H atoms times 1 gmol 78 gmol lcm3 08 g benzene so 08 g cm3 78 gmol 10 mmol cm3 or 001 moles How about for Brz Although Brz is a liquid is becomes a vapor at around 58 C so let s just assume we are doing these measurements at 60 C 1 cm3 006g Brz so 0006 g cm393 l60 gmol 38 mol cm393 4 x 10395 moles Note that equivolumes of a liquid contain approximately 103 times more molecules or atoms that for a corresponding gas Typically the densities of a liquid are only slightly more 20 or so than for the corresponding solid of the same material water is a very notable exception However when a substance is converted from a liquid to a gas its volume increases 103 Avogadro s Hypothesis 0n gases This brings us to a lndamental point concerning gases It turns out that equi volumes of two gases contain just about the same number of moles of those gases This equality was rst noted by Avogadro Avogadro s hypothesis is Equal volumes of dz erent gases at the same temperature and pressure contain equal numbers of particles Let s de ne this a little lrther by the following example Assume that you were going to burn hydrogen with oxygen to make water and you were doing this reaction at a temperature and pressure so that both the reactants and water product were gases The reaction would be H2V2029H20 Assume that this reaction goes all the way to products It does This reaction by the way is a very famous one 7 it was the reaction responsible for ending the age of the Zeppelins ie this reaction was responsible for the Hindenburg disaster Anyway we have 15 units particles on the le and 1 unit of particles on the right According to Avogadro s hypothesis the volume of the product will then be 23 the volume of the reactants It turns out that this is indeed the case So we nd that Avogadro s de nition of a particle is either an atom or a molecule Does Avogadro s hypothesis hold for liquids and solids With the information thus far provided you should be able to prove that it does not Volume is not conserved when liquids or solids react The structure of the Atom Recall that we de ned 1 mole of C atoms as weighing 12 grams Let s check this de nition by looking at the periodic table under C Wait We see that the mass of carbon is listed as 12011 gmol not 12000 gmol Did we screw up Well let s consider the various components of an atom of carbon or an atom of any element for that matter It turns out that atoms are made up of protons positively charged particles neutrons no charge but roughly the same mass as protons and electrons negatively charged particles and much lighter than protons or neutrons Let s tabulate the masses and charges on these particles Particle mass charge Electron me 911 x 103931 kg 549x10394u 16 x 103919 Coulombs Proton mp 167 x 103927 kg 10087 u 16 x 1039 Coulombs Neutron n1n 168 x 103927 kg 10078 u no charge How do we know these quantities Let s imagine setting up an experiment to measure these separate components of an atom Recall that Kinetic Energy 12 mv2 Now imagine that we could somehow smash apart some atoms and then we could give each fragment an identical amount of kinetic energy then we could determine the masses of those fragments by measuring their velocities Consider the experiment described in the picture below In this experiment an atom A is hammer smaCkS apart disassembled by hitting it with an atom A into Protons imaginary hammer Remember 0 trons and demons we don t really have to do this experiment we can just think about it This is one of the A luxuries of theory Now ifwe smack atom A while atom A is in the presence of an electric eld then any charged components of the atom will be accelerated by Voltage the electric eld In fact the applied here electric eld imparts the same amount of kinetic energy to each neutrons Electrons travel charged component that is fast to right frotolns p remain produced This is just what we rave 39 Slowly p n need So all we have to do 1s put to le p n l e detectors some d1stance away from where we smashed the atom and we can measure how long it took for the fragments to get to the detector This is therefore a measurement of the fragment s velocity and thus its mass The first thing that we would notice is that some of the particles are accelerated in one direction by the electric eld and some fragments are accelerated in the other direction This tells us that we have two dilTerent charges present 7 a negative and a positive charge We would also nd that the positively charged particles travel much more slowly than the negatively charged particles and therefore they must be much heavier Finally we would add up our collected masses of the charged particles and we would compare the sum of those masses to the weight of our original atom We would nd that we are missing some weight This means that some fragments were produced that were not affected by the electric eld and thus those fragments are neutral By doing a series of care ll experiments on a series of atoms we would eventually obtain numbers for the charge and mass that are associated with the electron the proton and the neutron I think that one of the most amazing things to consider is that it has only been within the past century that the electron was discovered by J J Thompson So now let s return back to the question of why Carbon apparently weighs 12000 gramsmol from the de nition of Avogadro s number while it weighs 12011 grams according to most periodic tables Let s say that we could do the experiment discussed above on a whole bunch of carbon atoms looking at one atom a time What we would Figure 2 Experiment to determine the various components of an atom nd is that the vast majority of carbon atoms would contain 6 electrons 6 neutrons and 6 protons However occasionally about 1 out of every 100 measurements we would measure a carbon atom with 6 electrons 7 neutrons and 6 protons This is called an isotope of carbon Recall that we discussed what the various numbers and symbols mean for a given entry in the periodic table and that the number at the top of each entry was Z or the atomic number The atomic number refers to the number of protons or electrons in a charge neutral atom of the element described by the entry However many actually most elements have a number of dilTerent isotopes Hydrogen for example only has 1 proton and 1 electron If you consider the atomic weight of hydrogen you should be able to gure this out However there are two other isotopes of hydrogen 7 deuterium with one neutron and tritium with two neutrons Deuterium weighs 2 amu hence the deu in the name while tritium weights 3 amu and so the tri in the name Most isotopes found in nature are stable although some are not and will decay by the emission of neutron by ssioning in to new lighter elements or by other processes associated with radioactive decay Now let s return to carbon It has three isotopes 12C 13C and 14C When one is writing out elemental symbols and specifying particular isotopes then at the top le of the element symbol one writes the total number of neutrons protons contained by that element The molecular weight quoted in most periodic tables is the isotopically averaged molecular weight For carbon for example this means that it is going to be something like 99 X 120000 1 X 13000 001 or something X 14000 which turns out to be 12011 amu The isotope 14C is of particular interest because it is taken up by living organisms but it is not taken up by dead organisms It is also a radioactive isotope with a half life of I m guessing 6000 years This means than in 6000 years half ofthe element will have decayed into something else 7 usually something that is not radioactive Therefore the amount of 14C that one measures in a fossilized lifeform yields information about when that lifeform was alive The process of dating fossils by measuring the ratio 14C 12C is called radiocarbon dating and was invented by Bill Libby who was on the chemistry faculty at UCLA and who won the 1961 Nobel Prize in Chemistry for discovering the technique of radiocarbon dating Atomic Structure So now that we know the ingredients of an atom what is its structure The structure of an atom turns out to be a wonder Jl and surprising thing Knock on a piece of wood or your notebook or your desk Feels pretty solid huh That is eXactly how most scientists thought of the structure of atoms up to the turn of this century Atoms were solid billiard balls packed together in a tightly con gured lattice Fig 3 closestpacked billiard balls Consider Figure 3 Notice that there is not much dead space between the billiard balls and if one stacked the billiard balls to be many layers thick then there would be no way to shoot a bullet through the lattice and not hit a ball even if the bullet were infmitely small However Rutherford had a good idea He decided to take a very thin metal foil 39 and send very small bullets in the form of electrons through the foil In this way he could confum the atomic billiard ball model The electrons should be re ected back or should bounce off at random angles rather than passing through the lattice However what he found is that by far the vast majority of electrons passed through the metal lm as if nothing was there The very few that didn t pass through the metal lm were sharply de ected It was as if the electrons either hit nothing and this was most of the electrons or they hit a very solid billiard ball This indicated that matter was not made up of continuous stuff but was very unevenly Fig 4 i the T11101an mOdel distributed Most of the mass was concentrated into very tiny Of the atom regions much smaller than individual atoms Thus the model that was presented for the atom was the nuclear model given in Figure 4 In this model the neutrons and protons are all concentrated in the middle while the electrons ll the space between the nuclei This drawing greatly overexaggerates the size of the nucleus relative to the atomic radius The nucleus is really really small How small7 we don t really care It is just small Remember that Why is this important for chemistry What this means is that when atoms encounter one another it is really clouds of electrons encountering one another In fact protons and neutrons are relatively unimportant in chemistry and we will really not mention them very much the rest of this course All of chemistry is really determined by how dif lse clouds of electrons encounter one another For this reason we will spend quite a bit of time discussing just what is the nature of those dif lse couds of electrons Are they random or is there some order to those couds etc You will note that the book launches right into chemical structure and bonding and periodic trends relevant to the periodic table We aren t going to do that Rather than having you all memorize lots of different chemical structure and bonding rules we are going to try to probe the andamental issues that lead to these various rules Because of this approach we won t get to chemical structure and bonding for several weeks yet Ionic and Covalent Bonding Mostly ionic and part of Chapter 13 Before we move on to the details relating the electronic structure of atoms let s nish up our chemistry review by discussing the various types of chemical bonds Once again consider the reaction Na 12 Brz 9NaBr Na metal is a covalent solid Brz gas is a covalently bound diatomic molecule and NaBr is an ionic solid In a covalent bond electrons are shared between the atoms and each atom has the same number of electrons associated with it that it does as a neutral isolated atom In an ionic solid or molecule however one or more charges are donated by an atom or a part of the molecule to another atom or another part of the molecule The molecule remains charge neutral overall but charge within the molecule is separated For example NaBr might also be written NaBr39 Note that the number of positive charges in this molecule is equal to the number of negative charges 7 the molecule is charge balanced We are a little bit away from being able to rationalize this behavior However we can use the principle of charge balancing within a molecule to predict how various ions will combine with one another Take the following examples of ions At right is shown how the negative and positive ions will combine with each other to form a charge neutral ionic molecule Cation Anion Na 02 K Cl39 Call 003239 Fe3 Br39 Charge Balanced Molecule Na2O NaCl Na2C03 NaBr K20 KCl K2C03 KBr CaO CaCl2 CaC03 CaBr2 Fe203 FeClg Fe2CO33 FeBr3 In an ionic molecule the two connterions are held together by Coulombic forces 7 which means forces that are related to interacting charges If we consider bringing two oppositely charged ions together then one can imagine that they will attract each other opposites attract and conversely if we bring likecharged ions together they should repel one another It turns out that this energy of attraction or repulsion scales as 1r where r is the distance that separates the ions A more accurate equation looks like E Q1Q241180r where Q1 and Q2 are the charges on ions 1 and 2 and so is a fundamental physical constant called the permittivity of the vacuum 80 is equal to so 854X103912 CZJ391 1 m The charge on an ion labelled Q above can be further broken down so that Q Ze where Z is the number of charges Z is a positive integer for positive charges like protons and a negative integer for negative charges like electrons and e is the fundamental unit of charge To be exact e is the charge on a single electron e 160 X 10391 C We use meters for the separation between the ions denoted as r in the formula So now just plugging in units we have E CoulombsCoulombsCoulombszl391m391m Joules an energy Good Repulsive XQlerE Energy a Q1Qz lt 0 Attractive Fig 5 7 The Coulombic Potential Energy for ionic interactions Notice that ifQ1 and Q2 are of opposite sign as for an anion and a cation then E is negative This means that the energy of the system is reduced when the ions are brought together However if Q1 and Q2 are both of the same sign two cations or two anions then E increases as the ions are brought near each other Let s plot the energy for these two cases Figure 5 defining E0 as the case for when the ions are very far apart from each other Thus when oppositely charged ions are brought close to one another the energy of the system is decreased However if those ions are of like charge then the energy of the system is increased This energy is called the potential energy Imagine the following analogy Let s take a spring As we compress the spring we put energy into it and that energy is stored 7 ready to cause the spring to stretch back to its initial con guration The case is similar here If we used atweezer to compress likecharged ions together it is similar to compressing the spring As soon as we remove our tweezer or whatever it is we are using to press the ions together then the ions will quickly move away from one another 7 turning that potential energy into kinetic energy Why kinetic energy Recall that kinetic energy 12 mvz so that when the ions move away from one another their kinetic energy may be calculated from the masses of those ions and the velocity at which they move away Similarly if we have opposite ions that are far apart then they have a large amount of potential energy This is like an expanded spring As soon as we release the ions they will quickly move toward one another and the potential energy of the system is lowered Thus the two curves drawn here are called potential energy curves Now there is something interesting that we can do with this graph Let s say that we put a cation and an anion at a distance corresponding to the dot at r ri on the attractive potential energy curve We know that if we release them then they will begin moving toward each other and so as we hold them we Force need to exert a force to keep them apart How much Slope OfEnergy force is necessary It turns out that if we move the VS39 r curve ions just a little bit ie we change r by a small amount atrri Ar then the potential energy will change by a small amount AE Now if we calculate AEAr we are measuring the slope of the potential energy curve in the vicinity of the dot r ri Some of you may recognize this as evaluating dEdr at ri or taking the derivative of the potential at ri 71 times this slope or 71AEAr is a measurement of the force on the ions at ri You can Fig 6 The relationship between estimate this force by drawing a straight line tangent to The Coulombic potential and the the curve at ri such as is shown in Fig 6 The slope of interiOHiC fOI39CCS this line is the force on the ions Obviously as the slope gets steeper the force gets greater Notice that for the attractive potential the force is negative remember to multiply by 7l pulling the ions together For the repulsive potential however the force has the opposite sign The ions are pushed apart Let s do a units check on this Force is in the MKS units system measured in Newtons A Newton is a kg m s39z Energy is a Joule or a kg m2 s39z Thus a Joulem the units of AEAr the units of a Newton See 7 if the units work out then you are probably doing the problem correctly Energy O The important part of this diversion into ionic bonding is that we took an equation that represented a physical quantity we graphed it and we drew conclusions from the graph You should be able to do this routinely If I said that some physical quantity Q varied sinousoidally with distance what would that graph look like What would AQAr look like Note that it was not necessary to know the whole Coulombic equation to draw these conclusions 7 ie the values for so and e were not needed We just needed to know whether the ions were like or oppositely charged and that two charged systems interacted as lr All the constants and other bits of the equation simply scale our answers but they don t change them First set of assigned problems from syllabus optional problems highlighted in red Jan ll 13 15 Language ofChemistry App A818AppB 12 Jan 20 Basic Physical Quantities App C212l4cd Ch 110 1224 26 39 Ch 210 1316 193236 37 Optional problem really good one to work out One positively charged particle is headed right toward a second one which is xed in space ie it can t move For each of these particles the number of charges Z is equal to l and each has a mass of l amu what does this say about the nature of the particles If the two particles approach to within 1 Angstrom 103910 meters of each other what was the initial kinetic energy in Joules of the moving particle To do this problem rst calculate out the potential energy that the two particles have when they are l Angstrom away from each other Then think about relating this to the kinetic energy of the moving particle END CHAPTER 12 LECTURES Terms and Other Stuff from This series of Lectures Balancing Reactions Terms Ionic Covalent Metallic Stiochiometry Molecular formulas Balancing Chemical Reactions EX Burning an organic substance A limiting reagent chemical reaction Molecular Weights Atomic Weights Atomic Numbers Avagadro s Number Scienti c Notation Units Always make sure units work out Energy Densities Gram densities of gases liquids solids Avogadro s Hypothesis on Gases The Structure of the Atom Electrons Protons Neutrons Kinetic Energy Isotopes Radioactive Decay Rutherford s Experiments on the Structure of an Atom Ionic Bonding Cations Anions Counterions Coulombic Energy Coulombic Forces Lecture Set 2 The Paradoxes of Classical Mechanics and the Origins of Quantum Theory If we are to ever understanding chemical bonding then it is important that we rst understand the structure of the atom Unlike most of the principles involving conservation of mass volumes of gases etc the structure of the atom is a truly 20 century discovery As the matter of fact about 100 years ago many scientists felt that they pretty much understood the physical universe 7 there were only one or two nagging problems and those problems would soon work themselves out 7 no doubt about it Well the solutions to those nagging problems turned out to completely revolutionize both science and society Let s go over these problems as they existed 100 years ago and try to understand how they were solved In the latter half of the l9Lh century James Clarke Maxwell uni ed the study of electricity and the study of magnetism Although Oersted had shown in 1820 that electricity and magnetism were related it was Maxwell who really uni ed the two phenomena What Maxwell showed is that an oscillating electric magnetic eld must produce a complimentary and orthogonal magnetic electric eld The term orthogonal refers to a magnetic eld that is at right angles to an electric eld For example a zaxis is orthogonal to both the x and yaxis To understand his model let s rst describe what a wave is This Figure is a periodic function If we were watching some phenomena such as waves at an ocean beach or if we were listening some particular musical note then we might receive a signal such as this one The x axis here is time and the yaxis represents the strength of the signal There are three properties of this wave that we need to worry about First is the amplitude 7 or how strong is the Amplitude or strength of signal signal Second is the frequency of the wave in s39l denoted by the symbol v 7 ie how many crests pass by a xed point in a given period of time Finally we have the wavelength of the wave which we denote by the symbol 7 7 is de ned as the velocity of the wave in ms for example divided by the frequency of the wave and so its units are those of length For example assume that we have an acoustic wave travelling through some medium that is characterized by a speed of sound of l x 103 ms Let s further assume for the above graph the xaxis spans 01 second We have about 85 complete waves that pass us in 01 second v 85 waves01 second 85 s39l or 85 Hz 7t 1 x 103 ms85 s39l 115 m 7 a pretty long wavelength Let s try to develop a feeling for waves Waves at the beach Audible Sound 7 20 m v 01 s39l velocity 2 ms A in air 16 m to 016 m v 7 20 s391 to 20000 s39l velocity 331 ms in air at 25 C 1500 ms in H20 at 25 C Light 01 electromagnetic radiation 7 103916 to 108 m spanning yrays to radio waves v 1024 s391 to 100 s391 100 1 by the way velocity 3 x 108 ms in vacuum Velocity in air is very similar Visible Light A7750nmto400mn1nm10399m Note that electromagnetic radiation spans wavelengths covering 24 orders of magnitude This is a lot We can break this down into yrays 7t 103916 to 103911 m xrays 7t 103911 to 10398 m UV A 10398 to 4 x10quot8 m Vis A 7 4 x10quot8 to 75 x10quot8 m IR A 7 75 x10quot8 to 10393 m Microwave 7 10393 to 100 m Longer wavelengths are FM the AM radio waves and the Long radio waves A major difference between all these waves is that acoustic waves ocean waves and similar phenomena propagate through some definite medium but electromagnetic EM waves can propagate through vacuum This was a rather bizarre concept around the turn of the century 7 many people searched for the ether which was the supposed omnipresent medium that supported electromagnetic wave propagation The are a couple of other differences between EM and other waves as well Above we stated Magnftic eld 1 w Figure 2 A current loop generating a magnetic field The arrows indicate the direction of current flow and the direction of the generated field that J C Maxwell had found the connection between electricity and magnetism At left is shown this connection 7 a loop of current generates a magnetic field at right angles orthogonal to the current ow The direction of the magnetic field is given by a cross product Conversely a magnetic field such as that shown in Fig 2 will exert an electromotive force on the wire and thus generate a current One field does not exist without the other You will learn about this in physics We have referred to the electromagnetic properties of radiation Now we make the connection between these terms and Maxwell s discoveries Imagine an alternating electric field ie a sinusoidal field passing through vacuum The existence of this alternating field means that orthogonal to the field there will be a magnetic field alternating at the same frequency We could have made this statement the other way around This is shown in Figure 3 Note that the magnetic eld one of the two colors and the electric eld the other color are at right angles to one another just as in Figure 2 This is more or less what we mean by orthogonal So now we have a picture for light 7 it is an electromagnetic wave travelling at a velocity given by c where c represents the speed of light 30 X 108 ms and characterized by a wavelength or a frequency that can vary over many many orders of magnitude Paradox 1 Black Bod Radiation and Planck So this picture was understood at the turn of the century However there were glitches Let s assume that we can describe a wave by the following equation Wave A Sinvnt Now look at the expression VTEL v is a constant the frequency of the wave and t is a variable Plot this function 7 you will get a wave Why Now the energy carried by this wave of electromagnetic radiation is proportional to A2 7 or the square of the amplitude of the wave We haven t said much about the amplitude of the wave yet However more intense light equals a higher amplitude wave Fig 3 An electromagnetic wave Note that the blue and red NSWV f we apply heat to some waves are travelling at right angles to one another One mammal 1t W111 emu energy m the form represents an electric field and the other represents a magnetic 0f radiation Since we can get the field energy of this wave of light from the above wave equation there should be a very straightforward way to calculate the energy given off by some substance at some temperature T It turns out that this energy was calculated to be proportional to Tk4 Let s not worry too much about just how this calculation was done It is going to turn out to be an incorrect result so we don t really care too much The major point is that when the light coming off a heated object was characterized by measuring the intensity of the light as a function of wavelength very different behavior than Tk4 was observed This was the glitch that bothered people 100 years ago Heated objects are often called black body radiation emitters Let s go back to some of the implications that the above wave equation carries Wave A Sinvm One implication is that the energy of the wave is tunable continuously from 0 to infinity All we have to do is to turn the knob that controls A To see how the prefactor A control the energy of a wave consider ocean waves at Hawaii as compared to Lake Tahoe The frequency velocity and hence wavelength of those waves may be very similar However the waves in Hawaii are simply bigger Each wave packs a more powerful punch 7 it has more energy An equation describing Lake Tahoe waves would have a relatively small A compared to a similar equation describing waves in Hawaii The energy carried by the waves would be proportional to Planck in trying to understand the glitch in the theory of light made a tremendous leap He postulated that the energy of the wave was not inifinitely tunable Rather it was only tunable only in discrete amounts For an electromagnetic wave of a given frequency he postulated that the minimum energy that such a wave could carry was equal to hv This implies that there is a minimum unit of light called a photon One photon equals one quanta of light In MKS or SI units h 662608 x 103934 J s This is the famous waveparticle duality that is so prevalent in quantum mechanics So now we come to a rather confusing point If something is heated and it gives off energy in the form of light then which energy do we care about We now have two energies of relevance here The first is related to our original wave equation and is the prefactor A A more intense wave means more energy in the wave The second energy is related to our description of a photon and is E hv The answer is we care about both I won t hold you responsible for knowing the following equations but I will give them to you anyway Planck showed that using his assumption and what is known as a Boltzmann distribution he could reproduce the intensity and wavelength distribution of light coming from a thermal source His equation was introduced in October of 1900 and was Radiance A c1x4expczAT 1391 c1 2nc2h and c2 hck where c is the velocity of light and k is Boltzmann s constant You will encounter Boltzmann s constant throughout chemistry and physics k is the atomic equivalent of R in the famous ideal gas law PV nRT Divide R by N A and you have k So let s consider this a little bit and make sure that the units work out The units of Radiance are known to be Watts m392 J s391 m392 kg m2 s392 s391 m392 kg s393 Units of R0 in Planck s equation mzs39zJ s m394 Recall that a J kg m2 s2 SoRm2s392 zs39zsm39 kgs393 What does a plot of the radiation coming from a heated source look like We can make some connections here between energy Temperature and photon wavelength The temperature of this radiation is 3000 Kelvin We can convert this into an energy by using the equation EkT and show that 3000 K is equal to 414XlO3920 J Now E6m Intensity 4 136 m 01e6 m wavelength what is the energy of a Fig 4 A plot of emitted radiation from a heated mass This is the curve photon with a wavelength of that Planck successfully explained in 1900 18X106 m We can get this by E hv Converting wavelength into frequency we get a photon of energy E 6266XlO3934 J s3X108msl8X10396m l04El9 J These energies are not the same because the relationship between the Energy curve and the wavelength is not linear However they are only about a factor of 25 different If we were to heat up the mass to temperatures above 3000K the curve would shift to shorter wavelengths higher energy photons and if we were to cool it down then the curve would shift to longer wavelengths lower energy photons So what Planck did is to state that light came in discrete packets and that the energy of such a discrete packet was proportional to its frequency with a proportionality constant equal to h now known as Planck s constant One question that bothered me for a long time had to do with the equation that Planck used to describe black body radiation How did he come up with this The answer 7 he first recognized that the curve had the shape of what is known as a lognormal distribution and he fit that curve However the only way that he could rationalize a physical equation that described that curve was to propose the quantization of light into discrete packets of energy Paradox 2 Einstein and the Photoelectric Effect The was another glitch in the classical model of the universe and it was the following Shine light on a metal surface As we discussed above the more intense the light you shine onto the surface the more energy you put into the metal If you put energy into the metal then you can presumably activate some chemical or physical process 7 and one such process is that you can ionize the metal 7 ie emit electrons from the surface of the metal Let s assume that the energy required to eject l electron from the metal surface is 35 eV and we ll label this the ionization potential or the IP For bulk materials the IP is also called the work function The eV is an energy unit we haven t encountered yet We use it here because we want a manageable unit of energy For example if we listed this energy in Joules it would be 56x103919J Joules are useful when we discuss quantities such as moles For a single molecule or single electron processes the eV is much better lJ 62x1018eV So now if we shine 10 Watts of infrared light on the sample in one second we are putting lOJ or 62x1019eV s of energy into the sample However no electrons come off What s going on here Recall that we have two energies that we are concerned with here 7 the intensity of the light beam measured in Watts or J s39l and the energy of an individual photon hv We can increase the energy of a photon by increasing the frequency decreasing the wavelength of the light source What we find is that when hv gt 35 eV we suddenly get a shower of electrons from the surface of the metal As we increase the photon energy beyond 35 eV then the electrons coming from the surface become more and more energetic and it turns out that the energy carried by the photon was hv 7 IP 12 mev2 the kinetic energy of the ejected electron This is what Einstein explained He borrowed Planck s ideas for the energy of a photon and stipulated that if hv was above the ionization threshold for the metal then electrons would come off now matter how intense the light beam was More intense light beams produce more electrons but even a single photon can produce an electron For hvIP lt 0 no electrons are emitted from the metal regardless of how intense the light source is Einstein won the Nobel prize in physics for explaining the photoelectric effect Let s do an example here Say that we have a H atom and we want to ionize it Its IP is 136 eV What is the longest wavelength photon that will emit an electron If we have a photon with 7t50 nm then what is the velocity of the emitted electron 136 eV x 16x103919 JeV 7 hv 7 6626x103934 J s 3x108 ms7t391 91x10398 In or 91 nm light is necessary So if we have 50 nm light what is the velocity of the emitted electron 12 mev2 Kinetic Energy of electron hv 7t50 nm 7 hv 7t9l nm 7 66261039343X108n s150X10399191X1039918X103918 J So 12 mevz 7 059 1 1X103931kgv2 7 18x103918 J and v 7 198x109 ms Paradox 3 The structure of the Atom Recall from our previous lectures that we discussed that the structure of the atom appeared from Rntherford s experiments to consist of a very dense and tiny nucleus surrounded by a dilute and relatively large cloud of electrons Recall also that we discussed Coulombic potentials 7 ie the potential energy curves that describe the interactions between charges There was a nagging question here that we didn t answer If like charges repel each other as we know that they do and opposite charges attract then 0 Why are all the positive charges segregated from all the negative charges and 0 Why doesn t the atom collapse ie why don t electrons get sucked into the core of an atom Good thing that this doesn t happen by the way It turns out that the actual answer to this dilemma is related to what is known as the Heisenberg Uncertainty Principle 7 one of the most nonintuitive concepts in all of chemistry and physics However there are other more intuitive models that steered people toward the right pathways and the adoption of these models was aided not only by Rutherford s experiments but also by a set ofrecently collected emission spectra of various atoms To understand what an emission spectrum is just look at a neon sign As electric current is run through the neon gas the neon gas will convert that electric current into light 7 uorescent lamps work in the same way Intensity of Em itted Radiation Wavelength Figure 5 What the emission spectra of some arbitrary atom might look like People had collected emission spectra from a whole number of atoms and ions including H He and Li an isoelectronic set of atoms and ions Isoelectronic means that they each have the same number of electrons 7 exactly 1 each in this case What they had found was very surprising If emission intensity were plotted vs wavelength they would see something like what is shown in Fig 5 So there were two clues atomic structure The electrons existed as diffuse clouds surrounding tiny positively charged neuclei and the emission spectrum of the atoms consisted of discrete lines Niels Bohr rationalized these observations in the following way the electrons and the nuclei had a relationship with each other that was similar to that of the sun and the planets Notice that the gravitational attraction between the sun and the earth is tremendous However the earth doesn t collapse into the sun but rather it orbits around it 7 a planet orbiting the sun is sort of like a child spinning an airplane around on a string The child is pulling on the string but the airplane doesn t turn inward and collide with the child It has its on angular velocity that keeps in spinning around the child in balance with the child s pull In a similar way Niels Bohr imagined the electrons rotating around the nuclei The discrete lines in the emission spectrum resulted when an electron in one orbit switched to a different orbit 7 one that required a less energetic electron The change in energy between the two orbits was then given off as a photon This model turns out to be quite wrong Nevertheless it was the first time that discrete orbits of the electrons were proposed a correct notion and it was the first model that could correctly account for the electronnuclei model of the atom So how did Niels Bohr come up with discrete orbits about the nuclei After all planets circling the sun and electrons circling a nucleus are quite different things Obviously gravity plays no role in atomic structure and so the physics must be quite different Recall our equation for the Coulombic potential energy between two charges q1qz4118 r391 Potential energy If the electron is travelling about the nucleus with some velocity then the total energy of the electron is given by its kinetic energy 12 mvz the Coulombic potential Force E 12 mv2 q1qz41180r391 Slope ofEnergy vs r curve Note that in this equation we list the potential energy as negative implying oppositely charge particles an electron and a proton Now recall what we said about the force of attraction between two oppositely charged particles that are characterized by a Coulombic potential energy of interaction The force is related to the slope of the potential energy function The figure we generated previously is given here as Fig 6 It turns out that as we move along various points of the potential energy curve and draw lines atrri Energy O Fig 6 The Coulombic potential tangent to the curve at those points then the slope of those lines changes as r39z This is the same thing as saying that if E N r39l the AEAr N r39z Some of you may recognize this from calculus dEdrlr lr2 Thus if Potential Energy PE q1qz41180r391 then Force q1qz4118039 lr392 For a single electron attracted to a positively charged nucleus with a positive charge of Z q1 e the fundamental unit of charge the charge on 1 electron and q Ze Z times the charge on one proton Therefore PE Ze241t80r391 and F Ze24n80391r392 Now recall Newton s law relating force and acceleration F ma With m me the mass of an electron this gives us the following equality F Ze241180391r392 mea mevzr391 where we equate a with VZI the circular acceleration of an orbiting particle So far we have just set up some classical equations of motion and we haven t yet explained why the electrons don t collapse in to the nucleus or why they have discrete orbits This is where Bohr made his leap of faith He assumed that the angular momentum of the electrons could only assume discrete values and that those values were integral multiples of Planck s constant Mathematically this is the approximation he made An angular momentum is defined as is shown in Fig 7 A particle is orbiting around the the center of the box at a distance r from the center It has a momentum associated with it which is its mass times its velocity or mv The angular Figure 7 Angular momentum ofa particle ofmass m momentum 1 equals rmVSine Where 9 is and velocity v orbiting some Spot the center of the box de ned as Shown For 9 90 the angular at a radlus 1quot momentum l rmv The discrete values that Bohr assigned to the angular momentum were I nh21t So now we can use all of our equations coupled with Bohr s hypothesis to describe the orbits If I nh211 rmv then we can return to our equation for the Coulombic force F Ze241180391r392 mea mevzr391 Rearrange this equality to solve for v so that v Zez411somer39112 Now nh211 rmev rme Zez411smer39112 Rearrange once again to solve for the radius of one of the orbits and we get Lecture Series 6 Molecular Orbitals Introduction Now that we have some feeling for molecular structures let s begin talking about the actual molecular orbitals 7 that is the orbitals that contain the electrons that are involved in the chemical bonds Recall that for molecules that are characterized by ionic bonding we didn t even need to think about such things 7 we just had two oppositely charged atoms attracted to each other through a Coulombic potential However when we talk about covalent bonds we can no longer do this We need to think about electrons that are shared between the atoms that are bonded to each other How are they shared It turns out that the atomic orbital picture that we developed from Chapter 13 will be very helpful here Many of the structures that we have been covering 7 especially the organicmolecule structures should have raised some questions For example when we learned about the atomic orbitals that are used by a carbon atom we had a 2s orbital and 3 2p orbitals Each 2p orbital was aligned with an axis so we had 2px 2py and 2pz The 2s orbital was spherically symmetric If these are the atomic orbitals that we will need to form molecular orbitals then shouldn t the molecular shapes somehow resemble the spatial distribution of the atomic orbitals In other words since all of the porbitals are at right angles to each other shouldn t carbon be characterized by bond angles that are 90 instead of 1800 AB2 1200 AB3 and 10950 AB4 Think about this question It is important Although we won t actually answer this question until Lecture Series 7 think about it as you read through these notes Sh apes and Signs of Atomic Orbitals and Implications for Chemical Bonds Recall that when we discussed atomic orbitals we spoke of them in terms of wave functions Just like a sine or cosine wave the wave functions that described atomic orbitals could have either positive or negative sign For atomic orbitals that had n2 or greater the orbitals were characterized by wave functions that had nodes A node was defined as a surface in space such that when a wavefunction passes through that surface it changes sign For example the 2pxorbital has a node at the origin The sign of the orbital is negative to the left of the origin while it is positive to the right of the origin We saw a movie about this remember At the time however all we worried about was the number of nodes 7 a higher principle quantum number n means that there will be more nodes in the atomic orbital nl nodes to be exact Well the sign of a wavefunction is going to become important here when we begin talking about chemical bonding Let s take the simplest possible orbital the 18 orbital and lets take the simplest possible atom the H atom Using the 18 orbital and two H atoms we are going to construct the simplest possible molecule the H2 molecule In Fig l I have drawn two Hatoms and below those atoms I have ls on ls on indicated the 18 H1 H2 wavefunctions that are on each of the atoms In this I figure the atoms are far sI apart so that the D1 tance wavefunctions don t Figure 1 Two Hatoms and their 18 orbitals at far separation interact These two orbitals represent more or less the sign and the amplitude of the electron density on each of the atoms You may or may not recall that it is actually the square of the wavefunction that equals the amplitude of the electron density not the wavefunction itself However the square of the waveftmctians here are going to appear very similar to the actual wavefunaions themselves So the question is If we bring these two atoms together can we kl Angstrom Fig 2 Bringing two 18 orbitals together make a chemical bond It turns that a slightly more appropriate question might be to ask If we bring these two atomic orbitals together how can we make a molecular orbital that will give us a chemical bond Let s just bring these two orbitals together and see what happens We have done this in Figure 2 Notice that we now have a region of enhanced electron density between the two nuclei 7 we have shaded this region dark to emphasize it What we have done is simply added 1SH1 ISH2 Since now there is enhanced electron density in the space inbetween the Hatoms we might expect that this type of combination of the IS H1IS H2 there is actualy a node separating the two atoms Figure 3 A 18 orbital is subtracted from another one Note that there is a depleted region of electron density between the two Hatoms and that orbitals will lead to a chemical bond and hence we call it a bonding combination However we could have combined these two orbitals together in a different way There is no rule that says that you have to add the orbitals You can also subtract one of the orbitals from the other Let s subtract the IS on H2 from the IS on H1 and show the result as Figure 3 Now we see quite a different result as one would expect In between the two Hatoms there is now a region of depleted electron density and in the cartoon of the two spheres we white out this region to emphasize it Furthermore note that there is now a nodal plane between the two atoms We would not expect this combination of the atomic orbitals to lead to a stable molecule In fact this combination of the 1S orbitals actually leads to a higher energy system than simply having the two Hatoms isolated from one another We call this combination an antibonding combination Assume for the moment that the two lS orbitals have combined to form a new orbital 7 an orbital that we will call a molecular orbital In that case it turns out that we started out with two atomic orbitals and we made two molecular orbitals One molecular orbital looks a lot like 1SH11SH2 7 MO1 The other molecular orbital MOZ looks like lsHl1SH2 MO1 doesn t have any nodes and MOZ has one node Recall that the principle differences between a lS orbital and a ZS orbital are that l the 1S orbital has fewer nodes it has zero nodes to be exact while the ZS has one node and 2 the 1S orbital is lower in energy than the ZS We could make the same statements when we compare the ZS to the 3S ie the ZS is lower in energy and has on fewer nodes and so on It turns out that this will always be true Fewer nodes means lower energy more nodes means higher energy Thus we can already say with some certainty that MOZ is a higher energy molecular orbital than is MOl What we have done is to create molecular orbitals b b from a Linear gombination ie adding or subtracting of Atomic Qrbitals 7 a technique that is known as LCAO theory Assume for the moment that we have natoms that we are going to use to make an natom molecule Further assume that each atom has 1 atomic orbital it can be a lS orbital if you like The general rules for using LCAO to make molecular orbitals are the following 1 For n 7 atom systems each with l orbital take all possible unique linear 01OZO3 a a 00 I O 01 02 03 Figure 4 The various linear combinations of 3 sorbitals When there is a bonding interaction we place a b between the atoms When there is an antibonding interaction we place an a Ol 02 O3 rotate b o 0102 O3 1800 Fig 5 Two equivalent LCAO s combinations of the orbitals What is unique 01 39 O2 03 Assume that we have 3 1 2 3 atoms A with 3 orbitals a b O O O 0quot One linear combination is O1 O2 03 Another combination is O1 O2 03 and a final combination is O1 O2 03 Are there any more No Let s assume that the molecule A3 is linear and draw out these molecular orbitals in Figure 4 How about the combination 01 O2 03 We haven t listed this yet It turns out that the combination 01 O2 O3 is equivalent to O1 O2 03 Why is this Look at O1 O2 03 at the bottom of Fig 4 Count the number of nodes you should get 1 Now draw out the combination 01 O2 03 Notice that it also has just one node Now look at Figure 5 In this gure we show the equivalence of the two LCAO s O1 O2 O3 and O1 O2 03 Thus if we multiply O1 O2 03 by a negative 1 and then we rotate that LCAO by 180 then we produce 01 O2 03 This means that these two LCAO s are the same molecular orbitals and that they are not independent of each other Notice that we can t do similar tricks to inter convert any of the combinations shown in Figure 4 with each other This is an important point and you should reread this section until you really understand it Notice that we started out with 3 atomic orbitals and we ended up with three molecular orbitals The number of molecular orbitals that we generate has to equal the number of atomic orbitals that we start with Always N0 exceptions So to recap rule number 1 to do LCAO to generate molecular orbitals take all independent linear combinations of the atomic orbitals that you started with If you start with n atomic orbitals you will end up with n molecular orbitals 2 The second rule is Once you have all of the possible independent combinations order them with respect to energy by putting the molecular orbital that has the fewest nodes as the lowest energy orbital and the orbital that has the most nodes as the highest energy orbital Thus returning to the orbitals we drew in Figure 4 The combination 01 O2 03 does not have any nodes and so it is the lowest energy combination that is possible The next lowest energy is O O2 03 Note that there is one bonding interaction between 01 and Oz and one antibonding combination between 02 and 03 The highest energy LCAO is O1 O2 03 Here we have an antibonding interaction in both locations Let s return to the H2 molecule We know that 18 H1 lSH2 is a bonding molecular orbital and that lSH1lSH2 is an antibonding molecular orbital Let s save space and call molecular orbitals MO s from now on If we want to order them with respect to some energy scale then a good starting point is the energy of the isolated hydrogen atoms We know that the bonding MO is lower in energy than the isolated atoms and that the antibonding MO is higher in energy We can draw a diagram to show this and such a diagram is presented in Fig 6 If we set the potential energy of the isolated hydrogen atoms to equal 0 LP 2 1 2 then when we bring the atoms together we form two MO s Energy One MO has an energy that is H1 39 H2 less than 0 7 ie there is 0 O potential energy to be gained when these atoms are brought 00 together This is similar to the case for when we bring an electron and a proton together Fig 6 Ordering the MO s with respect to energy Our For these two H atoms we energy zero is the energy of two isolated H atoms As we also form a MO that has 1 bring the atoms together we form two MO s one we call energy that IS gt 3 and this Pb where the b stands for bonding and one we call 1 MO we label T Where where the stands for antibonding Note that energy is denotes that it is an conserved here in that 1 is raised by the same energy antlbondmg orbltal and T amount that I b is lowered denotes that it is a quot How would we order our 3 atoms that we used in Figures 4 and 5 and the related discussion We present those MO s in Figure 7 In this gure we de ne our energy 0 to be that of three separated atoms The bonding MO 1 is lowered in energy the 1 k is raised by an equivalent amount and Tub or 1 nonbonding is at the same energy as the isolated LP atoms Why Because we have one bonding interaction and one antibonding interaction They O 0 cancel each other out and so the O 39 I nb energy of the system is the same as that for nonbonded atoms m LIJb Question Use LCAO theory to Figure 7 The MO s for the 3 atom system of Fig 4 generate the M055 and the MO ordered with respect to energy energy leVel dlagram for a SyStem identical to that shown in Figure 7 except that there are 4 atoms not three What do the MO s look like if the atoms are arranged in a line What if they are arranged in a square Label the bonding nonbonding and antibonding MO s as is done in Fig 7 This is a hard question but treat it like a homework question and work it out I won t ask for you to hand it in but de nitely care that you know this stuff 0 energy Filling M 0 s with electrons MO s are very similar to atomic orbitals in that each molecular orbital can hold at most 2 electrons What this statement is saying is that MO s also obey the Pauli Exclusion Principle 7 no two electrons can have the same set of quantum numbers Thus if two electrons go into a 39 orbital then they are going to be of opposite spin It turns out that lling MO s is just Slgma bondmg like lling atomic orbitals Fill the lowest one rst 1 1 then the next one and then the next When you have two or more degenerate MO s ie two or more MO s that are at the same energy then ll them each rst with one electron apiece and then go back 0 and pair those electrons C Ban ding from porbitals 12 039quot I 1L2 01 2 2 Before we do talk much more about lling MO s let s rst discuss MO s that are generated from porbitals Consider the types of bonding interactions that can occur between porbitals We show them in Figures 8 and 9 Porbitals can interact with each other in two ways First the lobes of the I porbitals can point right at each other In this case we say the porbitals form a sigma bond or for the case of negative bond overlap a sigma antibond Fig 8 sigma bonding with p0rbitals The rules for bonding and antibonding are identical Note that 12 is an antibond while 1 2 is a bond to those that we just covered for sorbitals However there are some differences which if you graph the orbitals are readily apparent For example notice that for two 2px orbitals 1 12 is an antibond not a bond Make sure on understand wh as 6b and 6 It turns out that regardless of the orbital if the bond is generated from orbitals that are directed at one another then it is a sigma bond Thus sorbital bonding is always sigma bonding and we could have also labeled the bonding MO in Figure 6 as 6b and the antibonding MO as 6 In Figure 9 we show a different type of bonding 7 the pi bond or the nbond In this case the orbitals don t point at one another but rather they line up sidebyside For two px orbitals 1 12 is a bond and we label it 1Tb 1 12 is an antibond and we label it 11 Nonbonding Interactions Now let s look at a pi orbital interacting with a p J orbital as well as an interaction with an sorbital We draw these interactions out in Figure 10 Here i and j refer to either X y or z with i j In Figure 10 left we show a px orbital interacting with a py orbital The pZ orbital interaction would give us the same answer Note that the positive lobe of the pK orbital is pointing at both the positive and negative lobes of the py orbital so that the negative overlap is equal to the positive overlap The I These types ofpbonds are labeled result is that this is a non bonding interaction The s orbital 9 py orbital interaction shown at the right side of Fig 10 This is a similar situation Once bonding again there is no net bonding or antibonding 39 interaction antibonding If the lobe of the p orbital were pointing directly at the sorbital then there would be a net a bonding interaction Pi Bonding 1 b 2 12 1th I I 2 12 1t I Fig 9 Pibonding with porbitals Note that 12 is a nbond while 1 2 is a 11 antibond NonBonding Interactions I Fig 10 Note that two differently aligned porbitals or an s orbital and a porbital arranged as shown can never produce p orbi a1 bonding sorbital antibonding bonding or antibonding overlap possible and we would get an spc bond In Figure 11 we draw all of the possible bonding configurations and we order them with respect to energy We make a couple of assumptions when we generate such a drawing First we assume that all of the orbitals are characterized by the same principal quantum number n or at least by a similar quantum number n Why this stipulation If we have two orbitals of very different nvalue then their spatial extent is very different and the electron clouds don t effectively overlap with each Catalog of Bonding Interactions It 1 P39TE 2 173 a E g l O 329 I C 1 2 psc kA O 173 a g 86 g H I l 2 Figure 11 The various type of bonding interactions To generate the various types of antibonding interactions simply multiply one of the participating orbitals by a l The strongest bonds produce the most repulsive antibonds All of these interactions except for the psc bond are covered in previous gures other very well The result would be that the energy ordering of the bonding interactions could change a bit Note that we label the 56 bond as the strongest and the p11 bond as the weakest The energy ordering of these bonds is related to how much electron cloud overlap is possible given the orbitals and how they are oriented with respect to each other The 11 interaction has the orbitals oriented sideby side and not much overlap is possible However all of the sigma interactions are characterized by orbitals that are pointing directly at one another You migh be wondering why is the pc interaction weaker than a 56 interaction The p orbitals are more extended in space than the sorbitals and so when those electron clouds overlap there is a slighly smaller net total overlap than for the case of a 56 bond Thus the energy of the psc bond is somewhere between that of the pc and the 56 bond Can we order the antibonding interactions based on how the bonding interactions are ordered in Fig 11 Yes we can Return to Figures 6 and 7 Recall that if a bonding interaction lowers the energy of the system compared to two isolated atoms to produce a bonding MO with an energy AE then the antibonding MO is raised in energy by AE Thus the strongest bonds produce the most repulsive antibonds and the weakest bonds produce the most weakly repulsive antibonds From Figure 11 plus our knowledge of how antibonding and bonding are related to each P39G HP HP atom 1 atom 2 I sc5 ns ns atom 1 atom 2 Figure 12 The MO diagram for most diatomic molecules from the rst 3 rows of the periodic table other we can put together a molecular orbital diagram for homonuclear diatomic molecules This is similar to what we did in Figures 6 and 7 except that we will now be able to include all of the types of bonds that we have discussed so far We present this MO diagram in Figure 12 Let s consider Figure 12 for a moment Why are there twopn39MO s and only 1 pO39 Recall that the pc MO comes from two porbitals pointing right at each other If we bring two atoms together so that their px orbitals point right at each other then their py and pZ orbitals will be lined up sidebyside or in a p11 con guration Think about this point It turns out that we can use this MO diagram to describe the chemical bonding of nearly any diatomic that doesn t have dorbitals involved in the bonding This includes all homonuclear diatomics that can be constructed from the rst three rows of the periodic table as well as a bunch of heteronuclear diatomic molecules Let s work out the cases for H2 Hez C2 and NO The H 2 molecule 5395 For the H2 molecule all we need to concern ourselves with is the ls orbitals Thus similar to Fig 15 18 6 we can present the MO diagram as in Fig 13 We H l H 2 have one electron from each H atom and both electrons are spinpaired and placed in the lowest 56 energy M0 the 6bonding orbital We can look at Fig 13 and extract the bond order for H2 We have 1123011116 MO dlagram for the H2 one completely lled bonding MO and and empty antibonding MO Thus we have net 1 bond and the 545 bond order is l ls ls The Hez molecule He 1 He 2 The Hez molecule like the H2 molecule SG involves only lSorbitals so we can stay with the same MO diagram that we used in Fig 13 However Flgure 1439 The M0dlagram for Hez39 No stable molecule 1s formed now each atom has a completely lled ls shell p6 and so we have four total electrons Two are placed in the 6bonding MO and the other two pn are paired and placed in the 6 antibonding MO Since we have one bond and one 2 antibond we say that the bond order is 0 bond 2P pn C 2 order lled bonding MO S lled C 1 p6 antibonding MO S This diagram is shown in Fig 14 With a bond order of 0 there is no reason why He should form a stable molecule and in fact it doesn t This MO diagram explains why He does not form The C2 molecule C l C 2 The C atom has the electronic con guration of ls2 2s2 2p2 Since the ls shell is closed we will get 1 bond and l antibond 2amp6 Fig 14 MO Diagram for C2 from those electrons which means that they don t participate in bonding 7 ie they yield a net bond order of 0 Let s just worry about the 2s and 2p electrons We can take our MO diagram from Fig 12 and simply ll it according to the electronic con guration of the carbon atoms This is shown in Fig 14 Note that we could have also neglected the 2s electrons since they yield net 0 bonds The 2p electrons yield a total of 2 bonds and 0 antibonds We have a pc bond and two 12 p11 bonds that are lled Since each one ofthe p11 bonding MO s has only 1 electron we call those 12 bonds The pc bond is substantially stronger than the p11 bond Thus we predict that C2 has a bond order of 2 What would we have predicted from a Lewis dot structure The Lewis dot structure of C2 looks like CEC which means that Lewis dot theory predicts a triple bond or a bond order of 3 We will return to this a little later but for the moment I will just point out that the MO picture gives a more correct answer than does the Lewis dot picture There is one more thing that the MO picture predicts for C2 7 it predicts that there will be 2 unpaired electrons This means that C2 molecule is predicted to be paramagnetic If all electrons are paired such as for the case of H2 or even the He2 nonmolecule we call the electron structure diamagnetic The NO molecule This heteronuclear diatomic molecule has an odd number of electrons and so it will be paramagnetic since we simply can t completely pair an odd number of electrons The N atom has the 39T 391 electronic con guration 2s22p3 and the 0 2p atom has the electronic con guration of N 2s22p4 From our experience with C2 we can neglect the contribution from the 2s Figure 15 The MO diagram for the NO electrons since we know that they will yield a net bond order of 0 Let s just concentrate on the 2p electrons The MO diagram is shown in Figure 15 From the diagram we can see that there are 3 bonds 7 l p6 and 2p11 while there is 12 of a p11 antibond Thus the net bond order is 25 Are we allowed to have nonintegral bond orders Certainly Since each MO can contain either 0 l or 2 electrons there is a possibility of 0 12 or 1 total contribution to antibonding or bonding from each MO We would expect the NO molecule to be a free radical meaning that has an unpaired electron This implies that NO would be expected to be quite reactive and in fact it is NO is an important molecule in various reactions that occur in the atmosphere 7 both at low and at high elevations Since it is a free radical it can initiate a number of chemical processes What does the Lewis Dot structure predict for NO We show this prediction at right NO appears to have a double bond from the Lewis dot structure and it also appears to have an unpaired electron Thus the Lewis dot structure misses on the bond order but it gets the unpaired electron correct Lewis dot structures are certainly useful for many things However they do have their failings such as they can miss bond order remember that Lewis dot structures missed bond order for the C2 diatomic also MO theory tends to get bond order correct 7 at least it gets bond order correct more often than Lewis dot structure does molecule CHEM 20A1 winter 2001 LECTURE NOTES sixth set MATTER WAVES In the mid1890 s J J Thomson was able to create beams of electrons He showed t hat these beams were composed of charged particles electrons which when subjected to ele ctric and magnetic eld behaved exactly as expected for particles moving under the classical laws of mechanics And indeed our view of matter is of material composed of small particles electrons and nuclei within atoms atoms combined to form molecules and molecules aggre gated to form macroscopic material But in the mid 1920 s Thomson39s so and independentl y Davissonn and Germer showed that a beam of electrons behaved not like ballistically lik e bullets or particles but like waves Thus it seems that matter in this case electrons behave s like particle in some experiments and wavers in other experiments What is going on W e have already seen that this wave garticle dualng applies to electromagnetic radiation whic h forms interference pattems as expected for waves but behaves like particles photons wh en impinging on a surface and breaking electrons loose out of the material as in the photoe lectric experimentsAnd now it seems that the same waveparticle duality exists for electron s ie for matter Indeed that is so and it is very important First we want to learn a bit about matter waves also known as deBroglie waves If a beam of electrons is focused onto a crystalline target the electrons are scattered off the ato ms and the scattered electrons can go in all directions The scattered electrons can be collect ed on a screen at some distance from the target material and one nds that at some position s on the screen many scattered electrons strike while at others no electrons strike This is c haracteristic of an interference pattern formed by a wave for scattered materials the interfere nce pattem is called a diffraction pattern One can interpret this experiment much like the Young 2slit experiment with light only here we effectively have many slits We don39t actua lly have slits through which the beam passes but the scattered electrons originate on the ato ms much as the interfering light beam originates at the slits With light waves the interferen ce pattern depends upon both the separation of the slits and the wavelength of the light wit h the electrons the diffraction pattern depends upon the separation of the scatterers the ato ms in the crystal and a wavelength Problem For the two slit experiment the bright lines on a screen at a large distance R from the slits are positioned at an angle 9 where n d sine 1 and d is the separation between slits 7 is the wavelength and n is an integer d ltlt R the distance R is measured from the midpoint between the slits and the screen and th e angle 9 is that between the horizontal line connecting the midpoint between the slits and the line connecting the midpoint between the slits and a bright spot on the screen Be sure you understand this formula Try to derive this formula but don39t be dismay ed if you can t For electrons scattering off a crystal one might think of d as the distance between atoms an d 7 can be determined by the measured angles at which bright spots spots where many elec trons strike are located But what is this wavelength that has been determined for electrons other than something that makes the diffraction pattern intelligible One can measure the velocity u of the incident electron beam in fact we can contro l the velocity of the the electrons by subjecting them to known electric elds forces And one nds that the measured wavelengths as determined by the measured 9 s substituted into Eq 1 vary with the electron velocities in fact the 7 is proportional to u39l In fact one can do the same thing with a beam of neutrons and it is found that 7 is not only proportional to u39l but to m39l as well where m is the mass of the particle DeBroglie f ormulated these results as h Am m 2 where the measured constant of proportionally h is the same Planck39s constant as appears i n the Einstein expression 8 hv for the photon energy We have added a subscript m to 7 to indicate that it is a matter wave and not an electromagnetic wave This gives us some exp erimental insight into the meaning of Am but we still might ask quotwhat is wavingquot We addr ess this question below but first we summarize some important uses for diffraction DIFFRACT ION Equations such as 1 above relate the angle 9 for bright spots in the diffraction patte rn to distances d between scatterers If 7 gt d no pattem is observed and if 7 ltlt d the brig ht spots are so close to each other that they cannot be distinguished Why But if 7 is co mparable to but smaller that d ie 7 S d then one expects good diffraction patterns Why Since the wavelengths for photon electromagnetic radiation electron and neutron beams can be controlled by scattering these beams off crystalline material with scattering centers atoms spaced a distance d apart one can make use of such scattering experiments to determ ine the interatomic spacings d in materials 1m gortcmt Problem If the separation between atoms is d 01 nm the best diffraction patterns are obtained with wave lengths of slightly less than 01 nm Calculate the corresponding energies for photons electron and neutrons in eV39s Such experiments known as x ray diffraction electron diffraction and neutron diffrac tion experiments are the prima technique for determing the structure of matter Of course the crystalline material consists of many scatterers not just two and there are many releva nt interatomic distances not just d and the formulas relating interatomic distances to positi ons of bright spots is far more complicated than Eq 1 but the overall concept is that of the 2slit interference experiment Note that these expeiments depend upon the wavenature of photons electrons and neutrons QUANTUM MECHANICS WAVES amp PROBABILITY In our discussion of electromagnetic waves we discussed the relationship between w avefunctions and probability The average value of the square of the quotnormalizedquot wave fun ction I rtl2 was taken as proportional to the probability density the probability per unit v olume that a photon would be found at a point at position r A probability is a positive fracti on that ranges from 0 to l the probability of finding a photon in a small volume 5V about p oint r is the time averaged value of l I rtl5V and when summed over all small volumes at a 11 points r in space average over time Zl Prtl25V l 3 Problem Be sure you understand all this Look back at the discussion of probability density in the last set of notes Remember that for very small volume elements one can replace 5V by dxdydz and the sum in Eq 3 can be replaced by an integral 00 00 00 average over time fdx fdy f dz l I 1 tl2 l 4 oo oo oo Remember that a function I rt that has this property is said to be quotnormalizedquot The same interpretation of the wave function squared as a probability density holds f or mass particles such as electrons and neutrons The diffraction experiments with electron s and neutrons can be understood in terms of probability densities much as they were under stood in the last section for photons In quantum mechanics the appropriate theory for desc ribing matter on the atomic scale the wave functions I rt are the fundamental quantities that contain all possible nformation about the system Oddly the wave dnctions themselves have no direct physical significance but with knowledge of the wave dnctions one can obtain everything that can be known about a system In particular the quantity l I 1 tl2 is the probability density the probability per unit volume that the system will be near the point 1 at time t In turn this means that the wave dnction I rt must be normalized i e that 00 00 00 fax fdy f dz pump 1 5 OO 00 00 These are very important statements about quantum mechanics don t just memorize them be sure you understand them Not only is I rt not physically meaningful it can be negative or even complex W hat is a complex number Note that there is no average over time required in Eq 5 As a matter of fact most of the wave functions we encounter will not be dependent upon time i e time t will not enter into the problem Remember in our discussion of diffraction of electromagnetic waves we poited out t hat the diffraction patter could be built up one photon at a time where one could not predict the exact spot that the photon would strike only the probability of it striking near any spot The same is true for electron and neutron diffraction Whereas classical motion of particles t raveling ballistically like a missile ar deterministic with point of impact well speci ed qua ntum mechanically one can only predict probabilistically where the particle will strike This probabilistic picture is not a shortcoming of quantum mechanics because if you believe qu antum mechanics and there is ample reason to do so states that the probabilistic information is all that can be obtained QUANTUM MECHANICS DISCRETE ENERGIES A property of many quantum mechanical systems is that they have discrete energies said differently their energies are quantized This is strange If we throw a ball gently it has little energy but if we throw it hard it has a g reat deal of energy We can throw it so that it moves with all intermediate speeds i e with a 11 intermediate energies Thus the energy that the ball can have is continuous But that is not true for the electron in an H atom it can only have certain discrete quantized values and all intermediate values are forbidden A given quantum mechanical system for example an H atom can exist in many different states The state is llly described by a particular wave function I r even though as indicated ab ove this wave function has not direct physical signi cance One state differs from another if there is physical property that is different equivalently each state has its own distinctive wav e function One often indicates the quantum state by an index k thus each value of k indica tes a different state The kth state is then specified by its wave function I kr The system in a given state described by Pkr has a given energy Ek If the energies are quantized i e if many intermediate energies are forbidden the states are said to be discrete or quantized The state with the lowest allowed energy is known as the ground state whereas the states with higher energies are known as excited states Several states may have the same energy in which case this group of states with the same energy is said to be degenerate The index k is often denoted a quantum number In some cases the index k as represented here may represent several quantum numbers DISCRETE STATES amp SPECTROSCOPY If a quantum system such as an H atom is struck by a photon it may absorb the p hoton in which case the photon disappears and the system incorporates its energy But the quantum system can only absorb discrete amounts of energy This can be understood by no ting that the system is initially in a state LI nr with energy En and it can only absorb just th e right amount of energy to quotexcitequot it to a quothigher statequot with some quotallowed energyquot Em T hus the system can only absorb energy EmEn from the photon where En and Em are the in itial and final allowed energies respectively Furthermore it is highly unlikely that the quottransitionquot from the nth to the mth state can occur except by full absorption of a single photon Therefore since the energy of the photon is hv only photons with frequency In n V T 6 can be absorbed by the system Problem Be sure you understand Eq 6 As a consequence of Eq 6 it can be seen that if electromagnetic radiation of all fre quencies is incident upon a sample only certain frequencies those that satisfy Eq 6 will b e absorbed Thus if we plot the number of photons absorbed against the frequency of the p hotons we get a series of lines at the discrete quotabsorption frequenciesquot such a plot is know n as the absorption spectrum and because it is a series of lines it is known as a line spec trum The existence of line spectra which is a clear indication of quantization of energy w as one of the principal reasons that scientists developed quantum mechanics Note that excit ation for small energy gaps between quantum states ie small values of EmEn require low frequency photons whereas excitation over large energy gaps require high frequency photo ns Many systems are found primarily in their ground states so that absorption carries them from the ground state to various excited states but this is not always the case In some cases the allowed energy levels En form a continuum 139e the energy gaps are und etectably small Usually the energies of a free particle be it a free electron in contrast to one bound to a nucleus or a free atom in contrast to one bound within a molecule exist in co ntinuum states One thus finds that the absorption spectrum of ground state H atoms consi sts of a series of absorption lines but that above some very high frequency v0 all frequen cies are absorbed 139e the line spectrum becomes a continuum spectrum The discrete sp ectral lines are indicative of the quantized energy gaps for the electron bound to the nucleus but for frequencies above vO the electron has absorbed so much energy that it breaks free i e it is ionized Problem Why is the ionization energy equal to hvo The process above is called photo ionization Why Problem If the absorbed frequency is 2hv0 what is the kinetic energy of the ionized electron How does this relate to the explanation of the photoelectric effect as described by Einstein Problem The photo dissociation of a molecule occurs when a photon is absorbed with sufficient energy to break the chemical bond holding the molecule together Discuss some of the qualitative features of the spectrum of a molecule Quite generally chemical reactions can be initated by radiation Each chemical system has its specific wave functions Pkr and corresponding allo wed energies Ek Thus each chemical system has its own specific allowed energy gaps Em En and its own characteristic absorption line spectrum By identifying the spectral lines 139 e the absorption frequencies one can identify the substance this is one of the best tools fo r identifying chemical substances Systems that are in excited quantum states o en emit a photon so as to retum to a lo wer state usually the ground state Such a process is called emission The photo emitted ca n only have frequencies that correspond to the allowed energy gaps in the system Why Thus the emission spectrum the plot of number of photons emitted versus frequency mus t also be a line spectrum and it too is characteristic of the particular substance generating the spectrum The compositions of the sun and stars have been studied by way of their emissi on spectra Problem Modify Eq 6 for emission SCHROEDINGER EQUATION The allowed energies or at least the allowed energy gaps for each system can be det ermined experimentally by means of the spectra But we are interested in more than this we are interested in knowing why a particular system has a particular set of discrete energies w ith specifc energy gaps To do this we must first obtain the potential energy of the system we have already shown how to write down the exact potential energy for atoms and molecul es One then has to assign the appropriate masses to each particle 139e to the electrons and nuclei this is easily done One then has to quotsolvequot the equations of motion in this case the CHEM 20A1 winter 2001 LECTURE NOTES twelfth set MOLECULAR ORBITALS Recall that our model for the electronic structure of atoms involves treating an electron as an identi able particle with spin that obeys the Pauli exclusion principle The state of each electron except for spin is specified completely by a one electron wave function wnylyml known as an orbital or more completely because of the inclusion of spin as a spin orbital The state of the individual electrons is not assumed to be independent of the presence of all the other electrons but the effect of the other electrons is incorporated through shielding and the exclusion principle For obvious reasons the orbitals in atoms are known as atomic orbitals or AO s Remember that we also denoted the oneelectron wave functions in a box as orbitals perhaps we could call them quotbox orbitalsquot although I know ofno one who does so A similar description can be developed for the electrons in a molecule The one electron functions are known as molecular orbitals or MO s Here too we must take account of the exclusion principle and shielding Thus we find many parallels between the electronic structure of atoms and molecules However it is more difficult to identify and describe the relevant MO39s than the corresponding AO s the AO s are all concentrated about a single nucleus whereas the MOS are delocalized over two or more nuclei We address the problem of MOS below We stress that this description in terms of MOS is a model and that alternative descriptions have also been successfully used The H2 Molecule Ion Let us first consider the Hf ion This molecular ion really does exist or at least itcan be formed and kept around for a short time Note that this molecular ion has only one electron so that in some ways it is the molecular analogue of the hydrogen atom The wave function or orbital for this molecule must give rise to a probability density that is equally distributed on both sides of the midpoint between the H nuclei Why When the electron is near one of the nuclei nucleus A it should look pretty much like a ls electron around that nucleus ie it is well described by the atomic orbital W1 SA1 when it is near the other nucleus nucleus B it should look pretty much like a ls electron around that nucleus and it should be welldescribed by the atomic orbital W1 5131 Because the electron binds the two nuclei in the region between the two nuclei we expect the electron density to be somewhat higher than if the electron were entirely associated with nucleus A or entirely associated with nucleus B This expected behavior can be more or less approximated by the wave function 1 1 e NI11sA1 WIsB1 1 Note that indeed in the vicinity of A quite far from B the function W1 5131 is close to zero so that I l m NW1SA1 and that in the vicinity ofB quite far from A w1SAl is close to zero so that I l m Nw1sBl In the region about half way between A and B the wave function I l is somewhat larger than would have been expected from NwsAl or NwsBl alone this then results in a larger electron density at the halfway point than would have been expected from the individual atomic orbitals alone and this is what is expected if the electron is to promote bonding This can be understood by recognizing that electron density is proportional to l Pll2 l 1 1l2 N2lWlsA1 t WlsB1l2 2 NzllWlSA1l2 lWlsB1l2 2WlsA1WlsB1l and that the cross term 2W1 s A1411 5131 represents the increased electron density at the midpoint The I 1 in Eql is an approximation to the molecular orbital for the electron in Hf we have not shown that it is exact only that it seems to have many of the expected properties ofthe exact MO The I l in Eq 1 is an approximation ofthe MO expressed as a sum or linear combination of atomic orbitals AO s such an approximate MO is known as an LCAO or linear combination of atomic orbitals Note that the functional dependence represented by 1 indicates that the wave functions are determined by the position of electron 1 since there is only one electron in this case one should note that the description in the equation refers only to a single electron in the vicinity of M nuclei Problem Place the molecule so that its internuclear axis lies along the xaxis Place the midpoint at x 0 Sketch w1sAl W1 5131 and I l as functions of x on the same graph Then sketch the l I ll2 given in Eq 2 as a function of x on the same graph Since orbitals are quotwave functionsquot they can be superposed added or subtracted and this is exactly what has been done in Eq 1 The probability density in Eq 2 clearly illustrates the constructive interference of the two atomic orbitals constituting the LCAO It is this constructive interference that accounts for the enhanced electron density at the midpoint the enhanced electron density that shields the positively charged nuclei from each other anal thereby promotes bonding If one plots the probability density of the MO as required in the problem above one sees that the electron is highly delocalized as compared to the electron on a single H atom Why this delocalization also promotes bonding The constructive interference that results in the bondpromoting densitybuildup between the nuclei requires that the AO s overlap each other Why If the nuclei are far apart there is not much overlap and 2x111 s A1411 5131 is small if the nuclei are quite close the overlap is large the constructive interference is large and the bondpromoting energy density between the nuclei is great Consequently large overlap gives rise to strong bonding One must keep in mind however that maximum overlap is not the only property controlling bonding although overlap clearly increases with decreasing internuclear separation if the nuclei are pushed too close together the energy begins to increase sharply and the nuclei are pushed apart Why Problem Refer back to the graphs in the last problem in order to sort out the concepts of overlap interference electron density delocalization and bonding We shall not dwell on the constant N that appears in Eq 1 but we note that it is a normalization constant which must be adjusted so that the probability of finding the electron someplace between oo and 00 must be one Explain A diiTerent MO could be formed such that 1 1 N39V1sA1 WlsB1l 3 This MO differs from the one above in that it exhibits destructive interference at the midpoint between the nuclei and the probability density around the midpoint is decreased below what it would be for isolated atoms In fact there is a node at the midpoint Consequently the unshielded nuclei repel each other vigorously The electrons are delocalized but to the outside regions where they cannot be of any use in binding the two nuclei together And indeed if an electron is in this MO it does not form a bond in fact the two nuclei repel each other Such an orbital is called an antibonding MO whereas that in Eq 1 is called a bonding MO An antibonding MO always has an extra node in addition to those expected for comparable bonding orbitals in the region of the midpoint If an orbital leaves the electron pretty much as is was on one atom or the other it is called a non bonding MO There is a mathematical rule that the number of independent LCAO MO39s is equal to the the number of component AO39s Thus in the problem above we have two AO39s W1SA W1SA and correspondingly the two LCAOMO s P T39 that appear in Eqs 1 and 3 We next examine the orbital energies and the corresponding energy level diagrams We can represent the AO s as ISA and 1sB and the MOS as 16 and 16 for the bonding and antibonding functions in Eqs 1 and 3 respectively We can draw an energy level diagram with the ground state atomic orbital energy corresponding to Is A on the left that of lsB on the right and the two molecular energies in between The ISA and lsB atomic orbital energies are of course equal the 16 bonding MO energy level is lower than that of the ls AO by an amount that is approximately the same as that by which the 16 antibonding orbital energy is higher The 16 orbital energy represents the ground state of the molecule the 16 orbital energy represents its rst excited state and because this latter orbital is antibonding H is not stable in the first excited state Problem Draw this energy level diagram The H2 Molecule Next we tum to the H2 molecule with its two electrons In the ground state we can put both electrons into the molecular orbital 1 given above 139 e LI161 NlWlsA1 t WisB1l 4a Tm 2 NlWIsA2 t WlsB2 4b Note that we have affixed the 16 quantum number introduced above to the MO Note also that the number in parenthesis indicates the particular electron of interest If one electron in this bonding orbital lowers the energy of the system one might expect two electrons to do so even more And indeed the two electron bond in H2 is stronger than the one electron bond in H21 Note that although the presence of increased probability density around the midpoint shields the two nuclei more effectively from each other thereby promoting bonding this quotstabilizingquot effect is partially cancelled by the destabilizing effect due to the electron electron shielding resulting from the crowding of two electrons into the same small region between the nuclei Explain We can excite the H2 molecule into an excited state in which one electron is in a bonding and the other in an antibonding orbital T150 NW1sA1 W1sB1 5a LI1642 N39w1sA2 W1SB2 5b Note the 16 notation for the antibonding MO In this state the H2 molecule falls apart into two H atoms because the bonding due to the I orbital is pretty much undone by the antibonding due to the I orbital thereby leaving the system in a nonbonding state As for the case of atoms we can refer to the electronic con gurations for molecules i e we can denote the MOS in which the electrons are found The ground state con guration for the H molecular ion is 16 The ground state con guration for the H2 molecule is 16239 the rst excited state con guration is 1616 We refer back to the energy level diagram for H and for the two H atoms A and B We recognize that this also represents the orbital energies for H2 and for the A and B hydrogen atoms Explain The total energy of the H2 molecule would then be the sum of the orbital energies of the two electrons In the ground state for the atoms one electron is in the lsA A0 and one electron is in the lsB AO whereas in the ground state for the H2 molecule both electrons are in the lowerenergy 16 orbital It can then readily be seen that the energy of the H2 molecule is lower than that of the two separated H atoms ie that the two H atoms bond together Problem Illustrate this on the energy level diagram This picture is somewhat too oversimplified Because of the electronelectron repulsions ie the shielding the orbital energy for the 16 MO in H2 is higher than that for the 16 orbital in Hzl Problem Explain this last sentence and discuss its implications for bonding Problem Discuss the first excited configuration 1616 of the H2 molecule in terms of the above energy level diagram The Non Existant Hz Molecular Ion Suppose a third electron were to be added so as to form the H239 molecular ion would it be stable Before worrying about the additional electronelectron repulsion which would tend to destabilize the molecule by increasing the energy let us focus on the consequences of the exclusion principle In the ground state one has the bonding MO given in Eq 1 For H we put two electrons into that one orbital see Eqs 4ab But the exclusion principle will permit only two electrons in a given orbital and so the third electron must go into the antibonding orbital of Eq 3 and this cancels the bonding effect of one of the bonding MO39s This suggests that Hz39 is less stable than H2 and less stable than H because of the increased electronelectron repulsions Explain this last point And indeed HZ39 is not stable Problem Discuss this last paragraph in terms of the energy level diagram developed above for H and H2 The configuration of the unstable ground state of HZ39 is 16216 because the 163 configuration is forbidden by the exclusion principle The two electrons in the 16 orbital must have opposite spins whereas the one in the 16 orbital can have either spin up mS 12 or spin down ms l2 The two electrons in the 16 orbital are therefore paired and have a net spin of zero However the unstable HZ39 molecule has a net spin of either ms 12 or 12 Why Consequently the H239 molecule interacts strongly with an applied magnetic eld If we look back at the H2 molecule we see that it has no net spin and therefore does not interact strongly with a magnetic eld But H does have a net spin and so does interact strongly with an applied magnetic eld Problem Explain this last paragraph Make use of energy level diagrams and indicate electrons occupying orbitals by arrows up and down to represent the electron spins Free Radicals Almost all molecules have an even number of electrons and in their ground states have no net spin 139e all the electron spins are paired This can be understood by analysis of the three molecules Hf H2 and HZ39 The rst and third have an odd number of electrons and hence a net spin the second has an even number of electrons and their spins are paired H2 is the most stable of the three so that an H ion is eager to add another electron and become an H2 molecule while an HZ39 ion is eager to eject an electron and become an H2 molecule The 16 orbital is bonding so if it is occupied by only one electron the energy of the system is lowered by placing another electron in that orbital However since because of the exclusion principle a third electron cannot enter this orbital and must go into the higher energy antibonding 16 orbital the system does not readily accept a third electron Molecules with an odd number of electrons are called free radicals They are usually very reactive because as explained above they usually like to complete the occupancy of bonding MO39s therefore they either add an electron if the bonding orbital is half occupied ot lose an electron if the odd electron is in an antibonding orbital It is for this reason that most molecules have even numbers of electrons and do not interact strongly with an applied magnetic eld the free radicals are reactive and so don39t stick around long On the other hand because they are so reactive the free radicals play a major role in chemistry a role as rapidly reacting intermediates in chemical reactions Note that contrary to what some say chemical bonds are not formed by the pairing of electron spins one electron bonds exist as seen in Hf However as we have seen paired electrons o en give rise to strong two electron bonds The statements above are not absolute There are molecules with even numbers of electrons that also interact strongly with magnetic elds There are free radicals that are quite stable We will look into these matters later on The Li2 Molecule The Li atom has the con guration ls22s The two ls electrons form a tight core so that from the chemical point of view Li looks like a single 2s electron held by a nucleus with an effective charge of not much over le Thus it looks much like a hydrogen atom in the 2s state but a hydrogen atom for which the ls orbitals are not available If two Li atoms are brought close to each other their 2s electrons overlap and form molecular orbitals The bonding in the Li2 molecule is thus much like that in H2 except that the valence electrons in the atoms are 2s rather than ls and the MOS can be envisaged as due to overlap of 2s rather than ls AO39s Be sure you understand all this The relevant bonding orbital in analogy to that in Eq 1 is then T250 NquotV2sA1 f W2sB1l 6 The discussion of electrons in the 26 bonding MO is quite analogous to that presented in the last section for the 16 bonding MO in H2 However there are some differences due to the differences between the ls and 2s atomic orbitals Both ls and 2s orbitals have maxima near the nucleus and decrease as the distance r from the nucleus increases but the 2s wave function decreases more rapidly and passes through a node then through a minimum after which its negative value approaches zero as r gets very large In the bound Li2 molecule the 12510 and 125130 functions overlap quite effectively more so than the ls functions ofH in H2 because the 2s functions are more spread out from the nucleus than are the ls functions Explain And overlap as we have learned is good for bonding Problem Draw some graphs to illustrate the differences between the overlap of ls functions and 2s functions Problem Discuss the antibonding 26 MO corresponding to the 26 bonding MO in Eq 6 Comment on the extra node Later we shall improve on this picture of covalent bonding in Li2 by introducing the concept of hybridization The LiH Molecule We next study the bond in lithium hydride LiH This molecule is formed when an H atom with configuration ls reacts with a Li atom with configuration ls22s Again the ls electrons on Li form a core which does not participate appreciably in the chemical reaction these core electrons reduce the effective nuclear charge on the valence 2s electron to slightly over le Thus the overlapping electrons might be thought to occupy bonding molecular orbitals of the form T50 N39quotV1sH1 W2sL11l 7a T50 N39quotII1sH2 WZSL12l 7b Note that xylsH is a ls AO on H while 125m is a 2s orbital on Li The discussion follows much like that for the examples above but the one surprise might be the minus signs in Eqs 7ab unexpected for a bonding MO That is readily explained by noting that the 2s function is negative near the LiH midpoint and so if one seeks constructive interference to promote high midpoint energy density and consequently bonding one needs the minus sign Be sure you understand this We have to do better The ionization energies and affinities of the ls electron on H and of the 2s electron on Li are not the same both because of the shielding in Li and the the fact that the n quantum numbers are different In fact H has the greater electronegativity so that the 2s electron on Li migrates somewhat more towards the H than the ls electron on H migrates toward the Li We thus expect ionic character to the bond and consequently a dipole moment In terms of the molecular orbitals we expect the one electron probabilities to be weighted more towards the H than towards the Li We can express these ideas by reformulating Eqs 7ab as T60 N 3921fII1sH1 fW2sL11l 8a T60 N 3921fII1sH2 fW2sL12l 8b where fis afraction 139e fis less than 1 and greater than zero We write this as 1 2 f 2 0 For LiH the fraction f is less than 05 which suggests that the ionic character of LiH can be expressed as LilH39 Why The ionic character of the bond is sometimes expressed as l4f052 Problem Note that if f 05 Eqs 8ab reduce to Eqs 7ab and the ionic character is zero 139e the bond is perfectly covalent This is the case for H2 and Li2 but not for LiH Problem Note that iff 0 both electrons would reside completely on the H and the ionic character would be unity i e the bond would be completely ionic This is an exaggeration for LiH it would imply an enormous difference in electronegativities Why Problem Note that if f 1 both electrons would reside completely on the Li and the ionic character would once again be unity and the bond completely ionic Not only is the ionic character of LiH less than one but its actual ionic character is such that it places the excess electronic charge on the H whereas this formulation places it on the Li Our discussion has not indicated how to determine the fraction f for any given bond AB That can be a difficult task But one can sometimes make rough estimates of f of the ionic character and of the dipole moment by knowing the electronegativities of the atoms A and B Explain Back to the F2 Molecule Let us return to the covalent bond in F2 as described by the bonding MO in Eq 6 rewritten here as T250 Nquot112sA1 W2sB1l 9 Why not try an M0 ofthe form wgpA1p2pBl One reason for not doing so is that as a consequence of shielding the 2p atomic orbitals have higher energy than the 2s orbitals It is not obvious that even if the wzpl wzpB bonding M0 were to lie further below the 41sz1 and wzpBl AO s than the WZSAWZSB bonding MO lies below the WZSA and 41sz AO39s whether the stabler bond would be represented by the 412S AWZSB or the WZpAW2pBl MO Problem Sort this out In almost all cases it is indeed the w2p5A WZSB MO that lies lowest 139e that is stabler than the 41sz WZPB MO One would therefore expect the 41sz wzpB M0 to be an excited orbital for Lil And an excited state for Li2 might be one with a configuration described as VZSA WZSB WZpA WZpB 139e one electron in the wZSAw25B MO and the other in the wgpAw2pB MO But not all wsz12pB molecular orbitals are the same And this is where chemistry begins to be interesting Let us define the line between the nuclei as the zaxis then we see that although everything of physical interest in the F2 molecule is the same in the y and x directions they are different in the zdirection We then might expect very different science to be associated CHEM 20A winter 2001 LECTURE NOTES ninth set MULTI ELECTRON SYSTEMS EXCLUSION We are now ready to address the problem of atoms other than hydrogenlike atoms The basic difference between hydrogenlike atoms and other atoms and atomic ions is that the former contain a single electron and the latter more than one electron There are three concepts to keep in mind in studying the more complicated atoms and atomic ions 1 One must include the energies of all the electrons 2 One must take account of the fact that electrons are spin 12 particles that are subject to the Pauli exclusion principle 3 One must recognize that electrons repel each other ie that there are repulsive inter electron interactions that give rise to shielding raise the energy of the atom and therefore make it less stable than if there were no interelectron interactions We address these issues in turn Multi non interacting particle systems l dimensional box Let us return to the easiest of all problems a particle of mass me in a ldimensional box of length L Remember that the allowed energies are given by the formula Sn thZSmeL2 r1 l234 1 Suppose we add a second particle and that there are no interactions between the particles ie the particles are totally oblivious of each other s existence what then is the energy of the system It is simply the sum of the energies of the two particles We could then write the energy as EM hiSmesznfmf n1 1234 2 n2 1234 where III and n2 are the quantum numbers associated with particles 1 and 2 respectively For 3 particles in the box we have En1n2n3 hiSmeL2n12n22n32 n1 12535 3 n2 l23 n3 l23 Problem Draw an energy level diagram for this 3particle system Label the levels n1n2n3 and indicate the degeneracies It can also be shown that the wave functions for the system are simply the products of the wave functions of the individual particles So for the 2particle system Tn1n2X15XZ Wn1X1Wn2X2 3 where X1 and X2 are the position of particle l and particle 2 respectively and wn1xl and wnzxz are the wave functions for particle l and 2 respectively The state of each electron is described by the oneparticle functions 41quotX also denoted an orbital and the energy corresponding to this is the orbital or oneelectron energy The state of a 2particle system is denoted by the orbital states of each of the particles 139e by the product function in Eq 3 and by the notation n1n2 Atoms If one neglects all interactions among electrons in an atom then the energy of the He atom would be EM 4RHn1392n2392 n1 123 4 n2 123 and the energy of the Li atom would be En1n2n3 9RHn12n22n32 n1 12535 5 n2 l23 n3 l23 Be sure you understand these Similarly the wave functions for these system are simple products of the oneelectron wave functions for the He atom the wave function would be TnlJLmanZJZwlZ rlarl Wm11m11r1Wn212m12r2 6 where 1 1 and 1 2 represent the vectorial positions of particles 1 and 2 respectively By vectorial position we mean specification not only of the distance from the nucleus but specification also of the angles xing the direction in which the electron may be situated These oneelectron functions are the orbitals discussed in the last section For the Li atom the wave function would be Tnlllmlln212m12n313m13 T112913 Wnlllmllr1Wnl12m12r2Wn313m13r3 7 In this discussion we have neglected spin we cannot neglect it Furthermore we have neglected interelectron interactions we cannot neglect them Problem Draw the appropriate energy level diagrams corresponding to Eqs 4 and 5 Label the states with these energies and indicate the degeneracies Note that specification of an orbital is nlm1 and specification of the state con guration of a twoelectron atom would be n1llm11 n2lzmlz Pauli exclusion principle An electron is not just any old particle and therefore the problem of multiple electrons in a box is somewhat more complicated than that above There is a fundamental postulate of quantum mechanics known as the Pauli exclusion principle which states that in a given system atom molecule or box each electron must be in a different oneelectron quantum state By state one means the fully speci ed state including spin This concept is denoted the exclusion principle because it excludes all but a single electron from occupying a given state We summarize the Pauli exclusion principle by stating that in a given system only two electrons can occupy a given orbital or only one electron can occupy a given spin orbital We turn back and reconsider the problems discussed above but with the inclusion of spin and subject to the exclusion principle Pauli exclusion principle for electrons in a box Consider a 1dimensional box with three noninteracting electrons The allowed energies can be given by the expression Enlngmn3 hiSmeL2n12n22n32 8 n1n2n3 123 where n1 n2 n3 are the quantum numbers for electron 1 2 and 3 respectively Now ask for the ground state If one could put all three electrons in the n 1 oneelectron states ie all three in the same orbital then the ground state energy would be 3h28meL2 Here orbitals mean the one electron wave functions that describe each of the electrons independently But only two electrons can be placed in n 1 states and so the 111 state is forbidden by the exclusion principle At least one of the electrons must be placed in an n 2 state Thus the ground state of the threeelectron system is 112 or 121 or 211 and the ground state energy is 6h28meL2 twice that calculated without regard to the exclusion principle This is a very significant effect ie doubling the ground state energy and totally changing the way in which the electrons occupy the levels Furthermore we see that for this particular problem the As we shall see the very structure of the periodic table of elements is determined primarily by the exclusion principle Above we discussed the filling of orbitals two electrons per orbital but a more complete description is one in which it is the spinorbitals that are filled Where n1n2n3 describes orbital states nymsl n2mszn3ms3 describes spinorbital states with the spin quantum number ms taking on the value 12 or l2 Thus if we place three electrons in the ground state of a ldimensional box two of the electrons quotoccupyquot the ll2 and 112 state and one must occupy either the 212 or 2l2 state Problem Be sure you understand all of the above Indistinguishability of electrons Although we analyze atoms and molecules in terms of numbered electrons there is a fundamental postulate of quantum mechanics that all electrons in a given system are indistinguishable Thus in the twoelectron ground state problem above we cannot distinguish which electron is in the ll2 and which in the 112 Thus the ground state can be represented as ll2ll2 or as ll2ll2 since the two are indistinguishable Problem Explain why the two states are indistinguishable and why this ground state is not degenerate For the threeelectron problem the ground state is ll2ll22l2 or ll2ll22l2 consequently this state is doubly degenerate because of spin Be sure you understand this Spectroscopy for electrons in a box Let us next turn to the spectroscopy of electrons in a box Impinging photons can be absorbed with consequent increase in energy of the electrons in the box Of course this is a quantum system and allowed transitions can take place only between allowed energy levels Furthermore in most cases all cases of interest to us transitions induced by electromagnetic radiation excite only a single electron Although all absorption lines must correspond to transitions from an occupied state but in addition for systems with many electrons absorption must involve transitions to an unoccupied state To understand this consider the ground state of a ldimensional box in which four electrons have been placed The electrons occupy the degenerate ll2 and 112 orbitals as well as the degenerate 212 and 2l2 orbitals The system can be excited by the absorption of photons that carry it to an allowed excited state by exciting one electron For example one of the higher energy 2ms electrons can be excited to a 3ms 4ms or higher state But the lower energy lms electrons cannot be excited to a 2ms state because both of these are already occupied although it can be excited to a 3ms 4mg or higher state Problem Depict the allowed transitions discussed in the last paragraph along with the occupation of states on an energy level diagram Arrange the Ns of the various absorption lines mentioned in order of decreasing wave length Problem Do the same for a box with only three electrons Pauli exclusion principle for electrons in an atom We can think of an atom as a species of box to which the electrons are con ned Perhaps thinking of it as a potential well is more useful For the moment we neglect all interactions among the electrons other than those ascribed to the exclusion principle In this case the only forces on the electrons are those due to the attraction of the nucleus Since the electrons are independent of each other they each behave as though they were the single electron in a hydrogenlike atom Or at least they would behave that way were it not for the exclusion principle The energy of each electron is known and the total energy of the atom is the sum of the individual electron energies Thus for the Li atom the energy is given by Eq 5 and the wave function by Eq 7 One might imagine the ground state of the Li atom to have n1 n2 n3 ie the ground state energy to be 9RH3 but that cannot be because of the exclusion principle The n l orbitals are doubly degenerate because of spin degeneracy but because of the exclusion principle only two electrons can be placed in the n l orbitals the third electron must go into an n 2 orbital Thus the ground state energy of Li is 9RH94 This then means that the ground state energy as well as the ground state occupation of orbitals is greatly affected by the exclusion principle Problem Don39t move on until you understand this Summary concepts for of atoms with no electron electron interactions Before leaving this topic let39s give a summary of terms and concepts Oneelectron wave functions that describe the behavior of single independent electrons are called orbitals if they do not specify spin and are called spin orbitals if they do The wave function for a collection of independent electrons is the product of the oneelectron functions or orbitals The energy corresponding to each individual electron ie the energy of each orbital is called an orbital energy The energy of an atom consisting of independent electrons is the sum of the orbital energies An orbital is occupied if it contains an electron it is fully occupied if it contains two electrons The nl specification of each and every electron i e the specification of the quantum numbers n and l for each of the occupied orbitals is known as the con guration Although we refer to the electrons as noninteracting we take into account the interaction due to the exclusion principle By non interacting we mean that there are no perceptible electrostatic interactions among the electrons Periodic Table With knowledge of the energylevel pattern for hydrogenlike atoms of the degeneracies and of the exclusion principle we can make use of the model of noninteraction electrons to formulate or at least outline the periodic table of elements Note that a model is an approximation that simpli es the problem a good model both simpli es and gives almost exact results The model of non interactions electrons in an atom is a terrible one but sufficiently good to yield the main outlines of the periodic table As we stated earlier each element has a nucleus with a palticular charge Ze surrounded by Z electrons Z is the atomic number The energy of each non interacting electron is Z2RHn392 where n l23 In the ground state the electrons go into the lowest available energy orbitals but only two electrons can be accommodated in each orbital So the ground states evolve as follows H atom Z 1 one electron in the nll0 orbital We indicate the con guration as ls1 He atom Z 2 two electrons in the nll0 orbital We indicate the configuration as ls2 This completes the first row of the periodic table Li atom Z 3 two electrons in the nll0 orbital and one electron in the n2l0 orbital We indicate the configuration as ls22s1 We could just as well make use of the ls22p1 orbital because the 2s and 2p orbitals are degenerate however as we shall see later this is true only for noninteracting electrons anal for interacting electrons the ns electrons have lower energy than the np electrons which have lower energy than the nal electrons Be atom Z 4 The ground state configuration is ls22s2 B atom Z 5 The ground state configuration is ls22s22p1 C atom Z 6 The ground state configuration is ls22s22p2 N atom Z 7 The ground state configuration is ls22s22p3 We have taken advantage of the fact that the 2p orbitals are 6fold degenerate so that there is no dif culty in placing three electrons into 2p orbitals O atom Z 8 The ground state con guration is ls22s22p4 F atom Z 9 The ground state con guration is ls22s22p5 Ar atom Z 10 The ground state con guration is ls22s22p6 This completes the second row of the periodic table Note that at the end of each row the orbitals with the n value characteristic of that row is saturated and the next electron has to go into a level with higher n and consequently much higher energy Na atom Z 11 The ground state con guration is ls22s22p63s1 Ca atom Z 12 The ground state con guration is ls22s22p63s2 Cl atom Z 17 The ground state con guration is ls22s22p63s23p5 Ne atom Z 18 The ground state con guration is ls22s22p63s23p6 We stop at this point because things get a bit complicated and we must rst take account of the electronelectron interactions that we have neglected Note that H Li and Na all have their highest energy electrons in an ns1 con guration We therefore divide the table into rows with a given n and columns with H above Li above Na The elements Ar and Ne have all their highest energy electrons in an np6 con guration therefore we place Ne in a column below Ar Similar reasoning leads to Ca being in a column below Be and Cl below F Elements in a given column have many properties in common Ionization Energies We will come back to the periodic table after considering the electron electron interactions But even at this stage we can get some insight into the properties of different atoms For example we can get some estimates of the ionization energies 10 the energy necessary to extract an electron from the atom in its ground state For H the ionization energy is RH Why For the Hel ion I0 4RH Why For the He atom we would also get I0 4RH since in our model the electrons are independent of each other This is wrong the elecytrons are not independent of each other and the ionization energy is less that 4RH For Li the ionization energy I0 9RH Why

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