Principles of Chemistry I
Principles of Chemistry I CHEM 115
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Chapter 11 States of Matter We constantly deal with three states of matter 0 Gases are compressible fluids Their molecules are widely separated o Liquids are relatively incompressible fluids Their molecules are more tightly packed o Solids are nearly incompressible and rigid Their molecules or ions are in close contact and do not move A change of state or phase transition is a change of a substance from one state to another n limation Gas Liquid Solid crystal behavior fluid fluid rigid Viscous viscous elastic compressible nearly nearly incompressible incompressible structure random random ordered diIUte dense dense molecules in motion in motion fixed in place on weak strong lattice interactions interactions strong interactions The phase transitions take place at different temperatures for different substances The variation is driven by interactions between molecules or ions comprising the substance At equilibrium liquids are continuously vaporizing and vapor lique es lfa liquid is in a closed vessel with space above it a partial pressure ofthe vapor state builds up in this space The vapor pressure ofa liquid is the partial pressure of the vapor over the liquid measured at equilibrium at a given temperature The vapor pressure ofa liquid depends on its temperature As the temperature increases the kinetic energy ofthe molecular motion becomes greater and vapor pressure increases Liquids and solids with relatively high vapor pressures at normal temperatures are said to be volatile The temperature at which the vapor pressure ofa liquid equals the pressure exerted on the liquid is called the boiling point or freezes is called the freezing point 0 As the temperature ofa liquid increases the vapor pressure increases until it reaches atmospheric pressure At this point stable bubbles of Vapor pressure mmHg Chlorofon39n Diethyl ether Tup ur mercury Column before the experiment Wnlcl39 molecule Wulcr 1 surlucl Temperature C vapor form within the liquid This is called boiling o The normal boiling point is the boiling point at 1 atm The temperature at which a pure liquid changes to a crystalline solid o The melting point is identical to the freezing point and is de ned asthe temperature at which a solid becomes a liquid 0 Unlike boiling points melting points are affected signi cantly by only large pressure changes Heat of Phase Transition To melt a pure substance at its melting point requires an extra boost of energy to overcome lattice energies o The heat needed to melt 1 mol ofa pure substance is called the heat of fusion and denoted AHfus o For ice the heat of fusion is 601 kJmol H20s 9 H20I AHfuS 601 kJ To boil a pure substance at its melting point requires an extra boost of energy to overcome intermolecular forces 0 The heat needed to boil 1 mol ofa pure substance is called the heat ofvaporization and denoted AHvap o For water the heat of vaporization is 4066 kJmol H200 9 H20g AHvap 4066 kJ Example of calculations The heat ofvaporization of ammonia is 234 kJmol How much heat is required to vaporize 100 kg of ammonia AH 1 OOOOg170 gmol 234 kJmol 134gtlt1O4 kJ Another example What is the temperature of water originally 100 g at 30 C to which100 g of ice at 0 C is added Two steps a the amount of heat consumed by 10g of ice 140 upon melting into the water 120 AH100 g180 gmol 601 kJmol 334 kJ b the heat conservation cpm1T1Tf cpm2T2 39Tf AH 0 40 Iccand 4184 Jg 0C 100 g30o CT 20 3 100 g0o C T 334 kJ 0 3000 110T 3339 4184 gt 110T 7983000 20 gt T 200 C Time4gt Heat added at constant rate Water and steam Tum pcralurc C ClausiusClapeyron Equation It has been demonstrated that the logarithm of the vapor pressure of a liquid varies linearly with absolute temperature Consequently the vapor pressure of a liquid at two different temperatures is described by lnP2Pl AHvapRmrn 1m Example 1 Carbon disul de 082 has a normal boiling point of 46 C and a heat ofvaporization of 268 kJmol What is the vapor pressure of carbon disul de at C Since boiling T1 is when vapor pressure Pl 1atm 76 orr nP21atm 268 kJmol l0008314 kJmol K 1319 1273 1703 gt P2 1atm exp1703 0182 atm 138 torr Phase Diagrams When considering phase transitions not just at different pressures the peratures of e h se under which the different states of a pure substance are stable For example the phase diagram for water illustrates that at 1 atm only solid exists b l 0 Cwhich transformsto liquid above DEC At DUO two phases ice and water coexist At 100 C water and water TBmPemmfe vapor coexist Pressure atm o A typical phase diagram consists of three curves that divide the diagram into regions labeled solid liquid and gasquot like the one for water or that for C02 o Curve AB dividing the solid region from the liquid region represents the conditions under which the solid and liquid are in equilibrium ie coexist o Usually the melting point is only slightly affected by pressure For this reason the melting point curve AB is nearly vertical o Typically the melting curve has a small positive slope ie solids are typically heavierthan corresponding liquids If a liquid is more dense than its solid the curve leans slightly to the left causing the melting point to decrease with pressure That is a case of water and a reason for ice floating and almost frictionless skating o Curve AC which divides the liquid region from the gaseous region represents the boiling points of the liquid for various pressures o Curve AD which divides the solid region from the gaseous region represents the vapor pressures of the solid at various temperatures o The curves intersect at A the triple point which is the temperature and pressure where three phases of a substance exist in equilibrium o The temperature above which the liquid state of a substance no longer exists regardless of pressure is called the critical temperature o The vapor pressure at the critical temperature is called the critical pressure Note that curve AC ends at the critical point C Pressure atm Pressure atm Temperature 4 Temperature 4 A B Properties of Liquids Surface Tension and Viscosity The molecular structure ofa substance de nes the intermolecular forces holding it together The intermolecular forces define physical properties 0 Vapor pressure and boiling point The stronger are the intermolecular forces the higher is boiling point and the lower is vapor pressure 0 Surface tension the energy required to increase the surface area ofa liquid by a unit amount 0 A molecule within a liquid is pulled in all directions whereas a molecule on the surface is only pulled to the interior 0 As a result there is a tendency for the surface area of the liquid to be minimized raindrops are nearly spherical In comparisons of substances as intermolecular forces between molecules increase the apparent surface tension also increases 0 Capillary action is related to surface tension but involves a A a liquid and a surface of a solid or liquid The surface tension counteracts the wetting of the glass surface by water or the lack ofwetting by Hg 0 Viscosity is the resistance to ow exhibited by all liquids and gases 0 Viscosity can be illustrated by measuring the time required for an object to fall through a column ofthe liquid as well for liquids leaking through a hole 0 Viscosity is usually greater at lower temperature liquids are usually more viscous than gases 0 Fluidity is opposite to viscosity 0 In comparisons of substances as intermolecular forces increase viscosity usually increases 0 Intermolecular Forces Many of the physical properties of liquids and certain solids can be explained in terms of intermolecular forces the forces of attraction between molecules Three types of forces are distinguished for interaction between neutral molecules 0 Dipoledipole forces Together are termed as 0 London or dispersion forces LDF Van derWaals forces VdW 0 Hydrogen bonding Van der Waals forces are the weak attractive forces in a large number of substances Hydrogen bonding occurs in substances containing hydrogen atoms bonded to certain very electronegative atoms London forces are the weak attractive forces resulting from instantaneous dipoles that occur due to the distortion of the electron cloud surrounding a molecule 0 London forces increase with molecular weight The larger a molecule the more easily it can be distorted to give an instantaneous dipole o If everything is equal the shape of a molecule becomes important Elongated molecules have greater polarizability and thus greater LDF o All covalent molecules exhibit some London force The dipoledipole force is an overall attractive intermolecular force resulting from the tendency of polar molecules to align themselves positive end to negative end 0 Dipdip force increases with density of dipoles ie with increasing molecular dipole moment and with decreasing volume for the same dipole moment Hydrogen bonding is a force that exists between a hydrogen atom covalently bonded to a very electronegative atom X and a lone pair of electrons on a very electronegative atom Y c To exhibit hydrogen bonding one of the following three structures must be present NH 0 H FH Only N O and F are electronegative enough to leave the hydrogen nucleus exposed o Molecules exhibiting hydrogen bonding have abnormally high boiling points compared to molecules with similar van der Waals forces 0 For example water has the highest boiling point ofthe Group Vl hydrides 0 Similar trends are seen in the Group V and VII hydrides loo 0 80 720 O 60 O 40 E 40 E 40 o c a m 5 20 9 781 393 HqTe ED 0 m 00 ll 2 40 440 HZSe 760 460 HIS 20 40 60 80 100 120 20 40 60 80 quot30 l lO Mnlccular wcigln Molecular weight A B o A hydrogen atom bonded to an electronegative atom appears to be special 0 The electrons in the 0H bond are drawn to the O atom leaving the dense positive charge of the hydrogen nucleus exposed 0 It s the strong attraction ofthis exposed nucleus for the lone pair on an adjacent molecule that accounts for the strong attraction o A similar mechanism explains the attractions in HF and NH3 Electrical Character of Molecules and lons m permanent or charge example Na SO42 8 ipole polar molecule neutral gt 2539 species total charge zero but there 5 is permanent charge separation example H20 Induced dipoe nonpolar molecule no permanent charge separation but can be induced by proximity to ion or dipole The molecule is polarizable example 02 charge density becomes asymmetric in 00 the presence of an external dipole Interactions are based on electrical charges stronger i0 diPOIe occurs between ions and polar 1050 kJmOI 399 Q solvents eg Na dissolved in H20 dipole dipole occurs between polar molecules in 310 kJmo Q pure substance or solution eg HCI 1040 kJmol molecules with 0 H N H or H F bond have superstrong dipoledipole forces called diPOIe induced hydrogen bonding eg NH3 H20 dipole London dispersion forces in induced induced solutions or pure substances eg CO2 dissolved in H20 or dipole dipole 110 kJmol liquid methane CH4 weaker Solid State A solid is a nearly incompressible state of matter with a welldefined shape The units making up the solid are in close contact and in fixed positions 0 Solids are characterized by the type of force holding the structural units together o In some cases these forces are intermolecular but in others they are chemical bonds metallic ionic or covalent From this point of view there are four types of solids Molecular Van der Waals forces Metallic Metallic bond Ionic Ionic bond Covalent Covalent bond A molecular solid is a solid that consists of atoms or molecules held together by intermolecular forces Many solids are of this type Examples include solid neon solid water ice solid carbon dioxide dry ice A metallic solid is a solid that consists of positive cores of atoms held together by a surrounding sea of electrons metallic bonding In this kind of bonding positively charged atomic cores are surrounded by delocalized electrons Examples include iron copper and silver An ionic solid is a solid that consists of cations and anions held together by electrical attraction of opposite charges ionic bond Examples include cesium chloride sodium chloride and zinc sulfide but ZnS has considerable covalent character A covalent network solid is a solid that consists of atoms held together in large networks or chains by covalent bonds Opposite to molecular substances the covalent bonds in network solids are not limited to molecules but Examples include carbon in its forms as diamond or graphite asbestos silicon carbide silicon oxide Structureproperty relation Tmemnq Hardness EIectrConduct Molecular Lowlt300 C only Soft and brittle Nonconductors intermolecular isolators interactions must be broken Metallic Vary Malleable Conductors low for 1 amp 2 groups highest in the middle of trans elements lonic High Hard and Nonconductors increase with brittle but become charge density conductors in high charge mag their molten nitude small size state Covalent High Very hard Usually network chemical bonds nonconductors must be broken except for graphite Solids can be crystalline or amorphous A crystalline solid is composed of one or more crystals each crystal has a welldefined ordered structure in three dimensions 0 Examples include sodium chloride and sucrose An amorphous solid has a disordered structure It lacks the well defined arrangement of basic units found in a crystal 0 Glass is an amorphous solid A crystal lattice is the geometric arrangement of lattice points in a crystal A unit cell is the smallest boxlike unit from which you can construct a crystal by stacking the units in three dimensions There are seven basic shapes possible for unit cells which give rise to seven crystal systems used to classify crystals Xray diffraction is a method for determining the structure and dimensions ofa unit cell in a crystalline compound Table 1 1 7 The Seven Crystal Systems Crystal System Edge Length Angles Examples Cubic a b c 90 NaCl Cu Tetragonal a 1 5 90 TiOz rutile Sn white tin Onhorhombic a at b c 90 CaCO aragonite B21804 Monoclinic a 17 L 90 90 PbCrO4 Hexagonal a b 0 90 120 C graphite ZnO Rhombohedral a b 2 c i 90 CaC03 calcite HgS Cinnabar Triclinic 1 b c at a 90 KZCr207 CuSOA SHZO 90 3 90 7 e K C W 90 C c a a 90 1 90 0 90 b 90 bd 390 742 90 a K I r a Cubic Tetragmml Orthothth lt4 7 1whi KP rquot W s a 6 a 90 E 7 a f a 90 900 a 900 K b L r b L r r m a a d a Monocljnic Hexagonal Rhombohedral Tridinic Crystal structure and density A simple cubic unit cell sc is a cubic cell in which the lattice points are situated only at the corners 1 atom per unit cell A bodycentered cubic bcc unit cell is one in which there is a lattice point in the center ofthe cell as well as at the corners 2 atoms per unit cell A facecentered cubic fcc unit cell also cubic close packed cop is one in which there are lattice points at the center of each face of the cell as well as at the corners 4 atoms per unit cell Simple cubic Body centered cubic Faceicenlered cubic Calculations with unit cell a volumeside3 b density mass volume MW atomsvolume Examples Platinum crystallizes in an fcc lattice and has density of 2145 gcm3 What is the length ofthe unit cell edge of Pt crystall MW of Pt 19508 amu 1950860221023 3239103922 v 3239103922 g 4 21 45 gcm3 6042 k103923 cm3 gt 392410398 cm Defects There are principally two kinds of defects that occur in crystalline substances Chemical impurities such as in rubies where the crystal is mainly aluminum oxide with an occasional A3 ion replaced with Cr which gives a red color Defects in the formation of the lattice Crystal planes may be misaligned or sites in the crystal lattice may remain vacant Chapter 8 Multielectron Atoms An electron configuration ofan atom is a particular distribution of electrons among available subshells The notation for a con guration lists the subshell symbols with subscripts identifying number of electrons 1s22s 2p1 Every atom has an in nite number of possible electron con gurations The configuration associated with the lowest energy level of the atom is called the ground state Other configurations correspond to excited states An orbital diagram is used to show how the orbitals ofa subshell are occupied by electrons The two spin projections are given by arrows pointing up ms 12 and down m5 12 Thus electronic con guration 1s22s22p1 corresponds to the orbital diagram The number of electrons allowed to be in one orbital cannot exceed two More precisely it is formulated by the Pauli exclusion principle no two electrons can have the same u set of quantum numbers in other words no more than two electrons can occupy the same I Direction of orbital two electrons ca be place external eld into the same orbital only with opposite spin projections ms quotn m N V N External magnet External magnet Wavefunctions or one more look at orbitals The wave we used to describe electron motion is called the wave function and usually is labeled 14 which is a function of coordinates In general the wave function is obtained by solving the Shrodinger equation The probability offinding electron in a volume dVis given by Lp2dV It is fairly 15 easy to tell apart different orbitals by their contours in 25 accordance with the value of l but their internal structure such 39 nodal a mammal on C r m 0 m 3 U o 3 Q 3 m U E U 0 called radial nodes has to be characterized in a different n do it by introducing radial probability w24nr2 Three pictures on the 2p left illustrate such radial probabilities for hydrogen atom 25 orbitals The radius is given in units of Bohr radius au053 A Few important observations 1 something we anticipated U 1 the number of radial nodes equals 11 l s m 3 15 In 25 2 the size of orbitals increase 7 3d with n both average distance m ltIgt and the maximum of the distribution rmax 35 3 for a fixed 11 and increasing 1 the distribution becomes 3 broader and the probability of W nding electron near nucleus decreases So called penetration decreases with l Parlml l Prnhahlllrv Padlal Prnhahlllrv Consequences of different radial probabilies Shielding When trying to find the lowest energy electron configuration ground state one has to put each consecutive electron into the lowest energy available orbital to the extend permitted by the Pauli exclusion principle If electrons did not feel each other that would be easy Since the charge of the nucleus is now Ze the energy of orbitals in a shell should be En R22n2 where R is the Rydberg constant R 218gtlt103918J still increasing with the value of n But in a multielectron atom each electron interacts not only with the nucleus but also with other electrons That interaction makes energies of subshells differ from each other sometimes even orbitals from different shells can disobey that simple order Screenin shieldin Effects The predominant screening of outer valence electrons is by inner shell or core electrons ie valence electrons are predominantly screened by electrons of lower principle quantum number n Electrons of the same principle quantum number also screen each other but the effect is small and we will treat it as a secondary effect in our discussions s p d f electrons of a given principal quantum number penetrate to the nucleus to different extents ns gt np gt nd gt nf thus ns electrons are screened by inner electrons less than np electrons which are screened less than nd electrons etc As a result Ens lt Enp lt End lt Enf lt Let s consider Li atom with its 3 protons in the nucleus When constructing its ground state we shall place consecutively its 3 electrons so that to make the total minimal ie by choosing every time the lowest energy available orbital lfwe apply En R22n2 to Li the predicted energy of the 2s and 2p electrons should be the same and 9 times the energy of the 1s electron in hydrogen This is not observed experimentally The shielding effects make it fall to an effective 29 13 The Aufbau principle Combination ofthe above trends with the Pauli exclusion principle result in the so called Aulbau principle for buildingup electron con gurations ofth e ground state of a multielectron atom The basic order in which atomic subshells are lled in a manyelectron atom are given by the diagram Examples for ground state electronic configurations B Z5 1s22s22p1 or He2s22p1 where electrons from inner subshells having the same configuration as a smaller noble gas are represented as a noblegas core He The remaining 3 in this case electrons are called valence electrons and are given explicitly Cl Z17 1s22s22p61 s23p5 or Ne3s23p5 7 valence electrons Ca Z20 Ar4s2 2 valence electrons Sc Z21 Ar 3d14s2 3 valence electrons Zn 230 Ar 3d1 4s2 2 valence electrons When n1d subshell is filledie n1d1 it is added to noble gas configuration that is then called pseudo noblegas core Hund s rule and Exceptions from the Aufbau principle Pauli exclusion principle can be interpreted as electrons disliking each other to the extend that they would rather occupy different orbitals if possible Thus for carbon the ground state configuration 1s22s22p2 should be represented by the orbital diagram which is not but rather one of these m m T u I Hund s rule clarifies it by stating that the ground state has the highest spin when electrons are in not fully filled subshell ie orbital diagram of the ground state of carbon is The preference for electron spins to spread out over the whole subshell causes some exceptions to the Aufbau principle in which stability of halffilled and fully filled subshells is realized Cr 224 is not Ar 3d44s2 but Ar1 3d54s1 halffilled 3d subshell Cu 229 is not Ar 3d94s2 Ar 3d 04s1 filled 3d subshell Substances that have nonzero magnetic moments nonzero spins are called paramagnetics Paramagnetics are attracted to magnetic field while diamagnetics are repelled by it Diamagnetic substances have zero total spin If all electrons are paired then the atom is diamagnetic in all other cases it is paramagnetic The former happens only for the totally filled subshells Thus for the ground state of an atom The shape of an atom is defined by its outmost electrons and since orbital radius increases with the principle number n valence electrons are responsible for atom s shape Obviously sorbitals are spherical independent on whether there are two or one electron in it The situation for nonspherical orbitals depends on the number of electrons in the subshell There is a mathematical theorem called Unsold39s Theorem which mathematically describes a condition when electrons in the shell make up a spherical shape That condition realizes for the fully filled and halffilled subshells Periodic Properties The periodic law states that when the elements are arranged by atomic number their physical and chemical properties vary periodically We will look at three periodic properties in the main group groups 121318 Atomic radius Ionization energy Electron affinity Atomic Radius Two factors determining the size of an atom the principal quantum number n the larger is n the largerthe size ofthe orbital the effective nuclear charge which is the positive charge an outer electron experiences from the nucleus reduced due to shielding effects from intervening electrons The trend Within each period horizontal row the atomic radius tends to decrease with increasing atomic number nuclear charge Within each group vertical column the atomic radius tends to increase with the period number new shell Since electron density is given as distribution 2r the convention for defining atomic radius is the covalent radius the average radius of an atom in molecular compouns such as halfthe distance between the atoms in the diatomic molecule 02 IA HA HA VA VA VIA VIIA VHIA Atomic radius Periodl dquot 4 quote a d3 6 do 6 do 6F 6 6 eeeeo Weeeeeeee Periods 60 eeoe weewww Ionization energy ionization potential IE The first ionization energy ofan atom is the minimal energy needed to remove the highest energy outermost electron from the neutral atom equals minus the energy of the highest occupied orbital It is always positive value Li1s2251 a Li1s2 e39 E 520 kJmol V I b 5 5 aquot p 2100 p eth Peliw 2mm lSOO ionization Entrgy IkJmol 5m 2 ll Is 35 i4 so Alunlil number Z Trends in ionization potentials increase with atomic number within a given period followsthe trend in size as it is more dif cult to remove an electron that is closer to the nucleus decrease down the group for the same reason The second ionization energy is the lowest energy needed to remove electron from the singly charged ion and so forth Electron affinity EA The electron affinity EA is the energy released upon adding an electron to a neutral atom in the gaseous state to form a negative ion NB Your book defines it differently as the enthalpy energy required to add an electron ie it has an opposite sign to my definition which is usually used The book s definition is usually CNe3s23p5 e C39Ne3s23p6 AH 349 kJmol EA349 kJmol The electron affinity for various elements varies in value and even in sign but in general it follows the same trend as the ionization potential Electron affinities EA lemol H 73 Helt0 Li 60 Be 18 B 27 C 122 N 7 O 141 F 328 Nelt0 Na 53 Mg 21 AI 43 Si 134 P 44 S 200 CI 349 Ar lt0 K 48 Ca 186 Ga 29 Ge 116 As 78 Se 195 Br 325 Kr lt0 Rb 47 Sr 146 In 29 Sn 116 Sb101 Te 190 295 Xelt0 Cs 46 Ba 46 TI 19 Pb 35 Bi 91 P0 183 At 270 Rnlt0 It is noticeable that the additional stability of halffilled and especially the totally filled subshells has its toll on electron affinity Alkaliearth metals and noble gases filledup subshells in both cases have negative or very small EA Electron affinity of N as well as other group 15 elements is smaller that that of neighboring atoms C 0 More than one electron can be added In the gas phase the second step is endothermic 09 e39 gt Oz39g AH 703 kJmol EA 703 kJmol 99 e39 gt 8239g AH 332 kJmol EA 332 kJmol Summary of trends As atomic size decreases along the diagonal I First ionization energy increases electrons are harder to remove I Adding more electrons is easier less straight forward Bottom line Metals are larger so they tend to lose electrons Nonmetals are smaller so they tend to gain electrons Chapter 10 Molecular Geometry and Chemical Bonding MOLECULAR SHAPES Molecular geometry is the general shape of a molecule as determined by the relative positions of the atomic nuclei and can be related to electronic structure To determine the shape of a molecule or molecular ion one must identify correct Lewis Dot Structure apply the ValenceShell ElectronPair Repulsion Theory also known as the VSEPR Theory VSEPR predicts the shapes of molecules and ions by considering arrangement of electron pairs around a given atom a central atom o Valence shell electron pairs repel each other assuming orientations to minimize repulsion and establishing certain groups of molecular shapes All bonds single double or triple are treated the same o The shapes can be classified according to how many bonds surround the central atom gttwo bonds linear or angular bent gtthree trigonal planar trigonal pyramidal or Tshaped gtfour tetrahedral seesaw or square planar gtfive trigonal bipyramidal or square pyramidal gtsix octahedral gtseven pentagonal bipyramidal o Not all electron pair repulsions are equal Lone pair electrons occupy larger volume and have a stronger repulsion than bonding pair electrons The result of this is that lone pair electrons squeeze bonding pair electrons together This causes the bond angle between atoms to become smaller than the ideal angle VSEPR NOTATION First the central atom is called quotAquot Second all the outer atoms are designated with an quotXquot For now it is not important whetherthe atoms are different Also a double bond is treated similarto a single bond Finally any lone pair electrons are designated with an quotEquot For instance water molecule has 2 hydrogen atoms and 2 lone pair electrons surrounding the central oxygen atom Therefore its VSEPR notation is AX2E2 Now we can distinguish the shapes furtherfor various number of bonds two linear is AXZ or AX2E3 and angular is either AXZE or AXZEZ three trigonal planar is AXg trigonal pyramidal is AXgE and Tshaped is AX3E2 four tetrahedral is AXA seesaw is AXAE and square planar is AX4E2 five trigonal bipyramidal is AX5 and square pyramidal is AX5E six octahedral is AXE seven pentagonal bipyramidal is AX7 109347 180 120 Linear Trigonal Tetrahedral planar 90 quot a 120 39 2 Trigonal Octahedral bipyramidal 16 Electron pair Arrangement vs Molecular geometry a Up to four pairs ELECTRON PAIRS ARRANGEMENT MOLECULAR Total Bonding 0 OF PAIRS GEOMETRY EXAMPLE X 2 2 0 Linear Lg 0 00 BER F Be F 2 F Trigonal I 3 0 planar BF B Ax1 3 F F 3 Trigonal planar I Bent or Lone pair 17 2 l amular so S Ail 0 quot 0 o l Telrahedral 4 0 AX H4 H 1CH H Trigonal 4 3 l Termlledral pyramidal NH AXZE Bent or 2 2 angular 1120 A b More than four paws ule ILW a mu Lmnm39 m m mm H l hwnmml 1 K mm 7 7 mmm B pvamm Aml mhmmm mm Ax m m my 45mm 39 u wde ax W m 7W um mm M m m run mmm awmm AX mum m m Electron Arrangement and Angles If all electron pairs are presumed totally identical the angles between bonds can be calculated according o t 109 47 150 120 CL 5 Linear Trigonal Tetrahedral planar an 9 120 9 90 72quot Trigonal Octa edral Pentagonal bipyramidal 1 6 ipyramidal Since it is not the case the molecular geometry around the central atom slightly varies even for the same arrangement of pairs one pairs Strength of repulsive forces decreases in the order LonePair LonePair gt LP BondingPair gt BP BP Multiple bonds require more space because ofa greater number of electrons ie also cause lowering the angle between neighboring single bonds DIPOLES and POLARITY of Molecules In a diatomic molecule knowing the dipole moment the magnitude of the charge displacement can be calculated physical gt Dipole moment charge displacement x bond length H q X r chemical the dipole moment is usually given in units of Debyes 48 D 1 electron charge x 100 pm molecules with polar bonds can be polar or nonpolar depending on their shapes Example if2 of the same polar bonds point in exactly opposite directions their polarity cancels the bonds are still polar but the molecule can become nonpolar Ber is this way There are several steps involved in estimating molecular polarity write a correct Lewis Structure determine the AEN electronegativity Examp39e 0f dipo39e cance39ing difference for each bond 4 O gt determine the molecular shape determine whether the bond dipoles cancel or not This concept is important as molecular polarity influences the physical properties like solubility of molecules Valence Bond Theory ORBITALS and BONDING Valence bond theory is an approximate theory to explain the covalent bond from a quantum mechanical view According to this theory a bond forms between two atoms when the following conditions are met Two atomic orbitals overlap The total number of electrons in both orbitals is no more than two First very simple approach Since covalent bonds involve sharing of electrons why notjust share from half lled ground state atomic orbita s In this method unpaired electrons from the valence orbital of each atom are assumed to interact pair and form the bond The molecular geometry is thus determined by the nature of the orbitals that interact Orbitals bond in the direction they protrude in order to maximize the overlap In some cases this approximation works well and agrees with the real molecular geometry see H2 HF F2 owever many shortcomings exist with this mo e a maximum of 3 bonds are expected with the representative elements the shape of the ground state orbitals in isolated atoms is determined by E Hydmgenlluoride HF and cannot describe all geometries Since the orbitals are occupied by 5 shared or bonding electrons there is no reason to expect that the shape of these orbitals should stay the same as in isolated atoms Le atomic orbitals c Fluorine F2 A better description of bonding involves replacing pure orbitals with mathematically reformulated orbitals this is called hybridization In the hybridization process electrons in the ground state orbitals can be imagined as being quotexcitedquot to higher energy hybrid orbitals which have different shapes and properties Through binding with other atoms this excitation energy for the atom is canceled with a I result of lower overall energy of a molecule Examples of this x are shown below The number of hybrid quot1 orbitals always equals the 13115251332 number of atomic orbitals that hybridize Hybrid orbitals are named by 39gt indicating the number and kind 39 of atomic orbitals hybridized Thus when 4 valence electrons bond sp3 orbitals are formed In this case t e ground state electrons can be imagined to be excited to the hybrid configuration The figure above shows the process for the formation of 2 5 5 methane In this case the Wde WWW central carbon is s hybridized and a tetrahedral shape is obtained The above hybridization also 2 describes orbital shape in E E 20 2 L55 a other representative elements umoam me moalnm where nonbonding pairs exist in NH3 one lone pair is also excited to an sp3 orbital during the bonding process The overall shape between the central N and the three bonded hydrogens becomes trigonal pyramidal in H2O two lone and two bonding pairs are sp3 hybrids bent The hybridization process can be conveniently represented by a valence shell orbital diagram this is the energy diagram shown at the left of the figures m Bundng In analogous fashion other hybridization processes can be systemically considered A For instance in the formation of BeClz sp orbitals form as is shown below Note that for completeness the two empty 2p orbitals are not shaded in the diagram The BeClz molecule is then linear as shown on the left and below In the formation of BF3 the diagram below shows that B is hybridized to three sp2 hybrid orbitals which bond to F and the remaining unfilled p orbital is empty Important Property of multiple bonds If only a single bond is formed the overlapping is of a 6 character It can be between two 5 orbitals s and p p and p or between any hybridized orbitals In a double bond one bond is o and the other one is 1 The latter is between two p orbitals In a triple bond one bond is o and two are 1 The latter are between two pairs of p orbitals a bond Single bond a band it bond endwand Sideiodide overlap overlapl Double hand Double and triple bonds are stronger than a single one but not by a factor of two and three respectively They are also shorter but more importantly double bonds lock molecular configuration which allows variety of isomers Cis and trans lsomers of dichloroethylene How to find out the hybridization for an atom count the number of c bonds or number of bonded atoms and the number of lone pairs starting from 5 add p orbitals rst and then d orbitals to make an appropriate number of orbitals equal the number of c bonds above Example AX2E2 need four orbitals ie it has to be sp3 hybridization there are only three p orbitals available and only one s LINEAR 180 Linear Examples 82 HEN BeF2 A x sp hybridization sp hybridization T leNAL EIPVRNNDAL Estuaarl Eximulei claw W413 Ll p x g WI Li 3L sp3 hybridization dsp3 hybridization dzsp3 hybridization Example XCF42 SdQQQQQ SPQQQ Xe atom ground state saQQQ Q Q Q MD W J KJ lone pairs Xe F Xe atom in XeF4 5s Shortcomings of the Valence bond theory Bonds are localized cannot represent resonances W W HCCCH HCCCH lt gt HCCCH HCC3CH H H Some cases even in diatomic molecules cannot be explained such as paramagnetism of 02 and 82 The Molecular Orbital Theory Molecular orbital theory is a theory of the electronic structure of molecules in terms of molecular orbitals which may spread over several atoms or the entire molecule 0 As atoms approach each other and their atomic orbitals overlap molecular orbitals are formed 0 In the quantum mechanical view both a bonding and an antibonding molecular orbital are formed ED 615 For example when two hydrogen atoms bond a 018 bonding molecular orbital is formed as well as a 018 antibonding molecular orbital with energies of the molecular orbitals MO different from original atomic orbitals are different Because the energy of the two electrons in the lower bonding MO is lower than the energy of the individual atoms the molecule is stable E Each MO is different from the atomic orbitals but often is represented as a linear combination of atomic orbitals a hybrid made from A0 of notjust one but many atoms In this case it is abbreviated as MOLCAO Molecular Orbitals as Linear Combinations of Atomic Orbitals These linear combinations can be either from A0 inphase ie constructively interfering bonding MO or outof phase destructiver interfering antibonding MO If an atomic orbital does not overlap with other AO ie is not involved in bonding then it is called nonbonding MO Hf ion At large separations the overlap is small Emmy 39 7472 u 2 4 2 4 5 a Inbemuclea sepamuon H2 Ubonding molecular in 4 z 72 u 2 4 At short distances due to overlap the bonding orbital 045 is of a lower energy Emmy 2 4 5 a Inbemuclea sepm on Hf ion At large separations the overlap is small At short distances due to overlap the nonbonding orbital 015 is of a higher energy Bond Order The term bond order refers to the number of bonds that exist between two atoms 0 Ifthere are no resonances required for representing the molecule s Lewis structure and there are no unpaired electrons then the bond order is an integer For cases with resonances it might become a noninteger o The bond order of a diatomic molecule is defined as onehalf the difference between the number of electrons in bonding orbitals nb and the number of electrons in antibonding orbitals na Electronic Configurations of Diatomic Molecules Again in accordance with the Pauli exclusion principle each molecular orbital MO can hold 2e with opposite spins l Homonuclear diatomic molecules H2 again E For simplicity the MOs are Q 6 constructed from only two A0 of two 15 hydrogen atoms f1 theAOon HAis1sAandon HBis1sB 8A 8B 61s 8 In the molecule only the molecular orbitals are available The 1s AC on H no longer exist on its own 018 is lower in energy than the 1s orbitals of free H atoms while M18 is higher in energy Thus if the electrons occupy c518 ratherthan the 1s orbitals of free H they will have lower energy 018 has the e density between the nuclei but M02 has the e density on opposite sides of the nuclei This corresponds to our intuitive sense of the bonding and antibonding situation Molecular electron configurations are written in the same way as atomic electron configurations c5192 Orbitals are conserved the number of MOs will always be the same as the number of AO used to construct them Consider Hz 2 e39 are bonding and 1 e39 is antibonding We expect H2 to be twice as stable as Hf Now electron con guration is orsfww We are also able to handle odd e39 cases easily Bond Order H2 Bond order 202 1 H27 Bond order 212 12 H 52 Bond order 222 0 H52 Is not stable Period 2 Liz Li has 1s2 2s1 con guration The core orbitals are smaller and E 0 35 have lesser overlap with orbitals ow U25B on other atoms But even if we include them into the picture there H 62 is no net effect from them because the bene t from the placing 2e39 H ms into 015 is fully compensated by WSA H H W55 having 2e39 in the antibonding 09 Thus the bond order is de ned 015 exclusively by the valence ele rons Bond order 202 1 B82 has 1s2 2s2 con guration The bond order 222 0 so the Bez molecule is not stable However Be metal is stable 32 B He2s2 2p1 configuration Both the 2s and three 2p orbitals can overlap when the B nuclei form a E 2 E5 l B2rgt 05g ilt5 62p 625 82s H HBZs ND G25 molecule One set of p orbitals will overlap end on csbonding and 2 will overlap in a parallel fashion 7 bonding The 2p orbitals that overlap end on form 02 bonds and 02p while those that overlap in parallel form 72 and 7cgtxlt2p MOs The TC overlap is smaller and thus we draw a smaller gap between 72 and 7cgtxlt2p MOs as compared to 02p and 02p BO 422 1 This MO energy diagram assumes no interaction between the 2s and 2p orbitals and predicts that 82 is diamagnetic In fact 82 is paramagnetic A Paramagnetic substance is attracted into a magnetic field A Diamagnetic substance is repelled from a magnetic field What is wrong with the model The 2s and 2p orbitals actually do interact cs28 and 02p and this changes the MO diagram This interaction inverts the order of the 72 and 02p orbitals and the 028 and 028 orbitals are no longer equally spaced relative to the 2s atomic orbitals This MO diagram predicts that 82 ll 62p E 2p Ear U M Bzp Li ll 82s H 28 H 82s Z will be paramagnetic 0282G282 7292 CZ N25 025 F2 The importance of2s2p mixing decreases across the period and for N2 803 02 802 and F2 BO1 the unmixed diagram is appropriate because of increasing 2s 2p gap in A0 greater differentiation in s and p electron shielding 62p N2 E 2 Z ii lt 4 75 H m H paramagnetic The VB model predicts that Oz has only paired electrons and should be diamagnetic The MO model predicts that 02 has 2 unpaired electrons and should be paramagnetic Thus the MO model better reflects reality 2s H Another gain of the MO theory is ability to predict optical absorption and luminescence spectra that involve transitions from occupied M0 to empty MO ltulrp 625 2s 625 l There is a correlation between bond order bond energy and bond length But comparison of 82 and F2 shows that a particular bond energy cannot be associated with just a particular bond order F2 has unusually large ee repulsions because it has 14 valence electrons and is a very small molecule The 02 molecule is 6 2P 02 gt J E i2 atm Splgp H2 ll Heteronuclear diatomic molecules In heteronuclear diatomic molecules such as HCI or CO The energies of AOs are different which affects the energies of resulting MOs HCI 62p 6 628 Cgtlt3 28 lt H 72 28H The overlap of2p orbital of Cl and 1s orbital of H result in a set of G orbitals one bonding and one antibonding and the remaining 4 electrons in 2p of chlorine do not overlap with that of hydrogen ie stay in nonbonding orbitals often labeled as n The picture on the left illustrates that the net effect of overlap is zero The MO energy diagram also makes it clear why HCI molecule is more stable than Clz is despite that they have the same bond order lll Polyatomic molecules Delocalized bonding Advantages of the MO theory are especially clear In describing delocalized bonding in polyatomic molecules Ozone OASN m M M bonding 11 nonbonding 11 antibonding 11 Chapter 6 Thermochemistry Thermodynamics is the science of the relationship between heat and other forms of energy Thermochemistry is the study of the quantity of heat absorbed or evolved by chemical reactions Energy can be in various forms Radiant Energy Electromagnetic radiation Thermal Energy Associated with random motion of a molecule or atom Chemical Energy Energy stored within the structural limits of a molecule or atom Nuclear Energy Energy stored within the nuclei of atoms There are three types of energy Kinetic Energy is the energy of motion Ek mv22 depends only on mass and velocity of an object Potential Energy is the energy of position in a field of force eg Ep mgh is the potential energy of a mass m at the height h in the gravitational field characterized by the acceleration constant 9 980 msz Internal Energy is the sum of the kinetic and potential energies of the particles making up a substance the inside of an object The total energy of a system is the sum of its kinetic energy potential energy and internal energy E101 EkEp The SI unit of energy is Joule J H kg mzs39z Other unit often used is calorie cal 1 cal 4184 J Consider a 10 kg bucket of water lifted 50 m up above the ground The potential energy ofthe bucket is equal Ep mgh 10 kg x 980 ms392 x 50m 490 kg m2s392 490 J lfthe bucket is let to a free fall according to the law of energy conservation at the bottom its potential energy is fully transformed into the kinetic energy which allows us to calculate the bucket velocity at that point 5 mv22 490 J gt v 2Epm 2 2 x 490 J lO kg12 99 ms When the bucket eventually stops at the bottom the energy is transformed into internal energy U some heat q released to surroundings of the bucket and likely some work done W as a dent on the ground Transformation of energy is possible in either direction and one specific side of these phenomena that we will deal with is associated with chemical reactions We will exclude potential and kinetic energy of a vessel with reactants and products as a whole unit Instead we concentrate on heat exchange in the course of reactions and work done onby the system The substance or mixture of substances under study in which a change occurs will be called the thermodynamic system or simply system The surroundings are everything in the vicinity of the thermodynamic system Heat is defined as the energy that flows into or out of a system because of a difference in temperature between the system and its surroundings Heat flows from a region of highertemperature to one of lowertemperature once the temperatures become equal heat flow stops That situation is called thermal equilibrium Heat is denoted by the symbol q The sign convention considers heat flow from the point of view of the system The sign of q is positive if heat is absorbed by the system The sign of q is negative if heat is evolved by the system Heat of Reaction Heat of Reaction is the value of q required to return a system to the given temperature at the completion ofthe reaction An exothermic process is a chemical reaction or physical change in which heat is lost by the system q lt 0 An endothermic process is a chemical reaction or physical change in which heat is absorbed by the system q gt 0 Surroundings Surroundings qlt0 The heat absorbed or released by a reaction depends on the conditions under which it occurs For example when the heat is supplied is a closed reservoir ie heat at constant volume qV it is all consumed by the system to increase its internal energy U When volume increases system does some work The equivalent to internal energy in this case is enthalpy Enthalpy Usually a reaction takes place in an open vessel and therefore at the constant pressure of the atmosphere The heat ofthis type of reaction is denoted qp the heat at constant pressure Enthalpy denoted H is an extensive property of a substance ie depends on the quantity of substance that can be used to obtain the heat absorbed or evolved in a chemical reaction Enthalpy is a state function a property of a system that depends only on its present state and is independent of any previous history of the system The change in enthalpy for a reaction at a given temperature and pressure called the enthalpy of reaction is obtained by subtracting the enthalpy of the reactants from the enthalpy ofthe products AH H products H reactants The change in enthalpy is equal to the heat of reaction at constant pressure This re resents the entire change in internal energy ALI minus any expansion work done by the system AH 01p Enthalpy and internal energy The enthalpy of a system H is de ned as We also said that enthalpy change AH equals the heat transfer at constant pressure while the internal energy change AU Energy equals the heat transferred at constant volume As we will see they differ by the work done by the W lt 0 WW system The work is energy exchange that results when a force F moves an object through a distance d W gt 5 k w F X d In chemical systems doing expansion work it is equal the change in volume at a given pressure q lt 0 he times the pressure wpAV FAgtltAgtlt d The sign convention is the same as for q q gt 0 Since AH represents the heat at constant pressure 0 then AH HrHy Ur er U pW Ur U pVf V AU pAV or AU AHpAV qw the rst law ofthermodynamics conservation of energy Thermochemical Equations A thermochemical equation is the chemical equation for a reaction including phase labels in which the equation is given a molar interpretation and the enthalpy of reaction for these molar amounts is written directly after the equation N2g 3H2g 9 2NH3Q AH 918 kJ Stoichiometric coefficients numbers of moles In a thermochemical equation it is important to note phase labels because the enthalpy change AH depends on the phase of the substances 2H2g 029 gt 2H20g AH 4837 kJ 2H2g 02g gt 2H20 AH 5717 kJ The reason 2H20g e 2H20l AH 880 kJ Manipulation with enthalpy of a balanced chemical reaction should be exactly the same as with the quantities of reactants and products in the reaction When a thermochemical equation is multiplied by any factor the value of AH forthe new equation is obtained by multiplying the AH in the original equation by that same factor When a chemical equation is reversed the value of AH is reversed in sign Consider the reaction of methane CH4 burning in the presence of oxygen at constant pressure Given the following equation how much heat could be obtained by the combustion of 100 grams CH4 CH4g 2029 9002g 2H20 AH 8903 kJ 100 g CH4160 gmol x 8903 kJ 556 kJ lost by the system ie 556 kJ obtained from it Calorimetry The Measurement of Heat Exchange To see how heats of reactions are measured we must look at the heat required to raise the temperature of a substance because a thermochemical measurement is based on the relationship between heat and temperature change The heat required to raise the temperature of a substance is its heat capacity The heat capacity C of a sample of substance is the quantity of heat required to raise the temperature of the sample of substance one degree Celsius Kelvin C qAT Changing the temperature of the sample requires heat equal to q CAT Suppose a piece of iron requires 670 J of heat to raise its temperature by one degree Celsius The quantity of heat required to raise the temperature ofthe piece of iron from 250 C to 350 C is C67OJ C gt q Cgtlt AT670J Specific heat 5 specific heat capacity is the quantity of heat required to raise the temperature of 1 g of substance by one degree Celsius Kelvin sqATm gtqmxsxAT Calculate the heat absorbed when the temperature of 150 grams of water is raised from 200 C to 500 C The specific heat of water is 4184 Jg C q m x s x AT150 g x 418 Jg C x 500 C 200 0C 188 kJ A calorimeter is a device used to measure the heat absorbed or evolved during a physical or chemical change The measurements can be at constant pressure or constant volume Constant volume calorimeter bomb calorimeter measurements of AU Constant pressure calorimeter measurements of AH Classifications of Systems Open system can exchange mass and energy with surroundings Closed system exchange energy but no mass Am 0 Isolated system no energymass exchange q 0 Am 0 Heat Exchange in isolated systems q 0 Hot object cools down and the cold one warms up so that the heat released by the former is fully consumed by the latter qsystem qcold qhot 0 1 For example if 300 g of benzene at 25 C were mixed with 500 g the same solvent at 50 C what the resultant temperature would be 0 S X m1gtlt T1 Tlinal S X mzgtlt T2 Tlinal 0 w m1gtlt T1 Tlinal mzgtlt Tlinal T2 or Egt Tlinal m1T1 r39f122m1 m Egt T na 300X25 500X50300 500 405 C Similarly in a calorimeter qsystem qcalorimeter qreaction 0 mo 1quot Example Calculate the molar heat of combustion for naphthalene if 1433 g of naphthalene C10Hg was burned in a constant pressure calorimeter cause the temperature of water raise from 2067 C to 2634 C The calorimeter has 200 kg of water and the heat capacity ofthe bomb is 180 kJ C qrxn 39 qwater qbomb 39 mwater X 8water Cbomb AT gt qrxn 2000 g X 4184 JQOC 180 kJOC 26340C720570C 1017 kJOC 557 C 5765 kJ Since the molar mass of naphthalene is 12816 gmol the molar heat of combustion is AHc 5765 X 12817 1 433 kJmo 5156 kJmol Instead of measuring enthalpy of reaction can be calculated using known enthalpies of other reactions Hess s law of heat summation states that for a chemical equation that can be written as the sum of two or more steps the enthalpy change forthe overall equation is the sum of the enthalpy changes forthe individual steps Note that it does not require for the reaction to actually proceed through those steps For example suppose you are given the following data Ss 02g 9 SOZg AH1 297 kJ rxn 1 2SOsg 9 SOzg 02g AH2 198 kJ rxn 2 Could we use these data to obtain the enthalpy change for the reaction 2Ss 302g 9 2SOsg rxn 3 Since rxn 3 2 rxn l rxn2 gt AHquot 2AH1 AH2 2297 198 792 kJ Another example Get enthalpy of the rxn CH4g 2 029 9 0029 2 H200 From the following information CH4g 9 C S graphite 2 H2g AHO AfH CH4 2 02g 9 2 02g AH 0 Csgraphite 02g 9 COZg AH AfH COZ 2H29 029 9 2 H2O AHO AfHo H2O ArHo Afr0 Ci i4 AH 002 AilP H20 We managed to express the enthalpy of rxn through some standardized ones Standard Enthalpy of Formation The standard enthalpy of formation of a substance denoted AHfo is the enthalpy change for the formation of one mole of a substance in its standard state from its component elements in their standard state and reference form The term standard state refers to the standard thermodynamic conditions chosen for substances when listing or comparing thermodynamic data 1 atmosphere pressure and the specified temperature usually 25 C The enthalpy change for a reaction in which reactants are in their standard states is denoted AH delta H zero or delta H naught Note that the standard enthalpy of formation for a pure element in its standard state is zero Most importantly the enthalpy of reaction can be now expressed in a rigorous manner using enthalpies of the products and reactants Allotrope is one of two or more distinct forms of an element in the same physical state graphite diamond and 060 The reference form of an element is usually the stablest allotrope of the element under standard thermodynamic conditions Examples Calculate AriP for 200 g 02 g 9 20029 Ca 8 0029 12 02 g 9 CaCOsS 4NH3Q 5029 9 4N0 g 6H20 9 Fuels A fuel is any substance that is burned to provide heat or other forms of energy Foods as fuels Fossil fuels Coal gasification and liquefaction Food fills three needs of the body olt supplies substances for the growth and repair of tissue olt supplies substances forthe synthesis of compounds used in the regulation of body processes olt supplies energy About 80 of the energy we need is for heat The rest is used for muscular action and other body processes A typical carbohydrate food glucose 06H1206 undergoes combustion according to the following equation CeH1206s 127202 g 9 45002g 43H20 I AHquot 2803kJ One gram ofglucose yields 156 kJ 373 kcal when burned A representative fat is glyceryl trimyristate C45H8606 The equation for its combustion is 04548606 s so2 g e coo2 g 6H20 I AHquot 27820kJ One gram of fat yields 385 kJ 920 kcal when burned Note that fat contains more than twice the fuel per gram than carbohydrates contain Fossil fuels account for nearly 90 ofthe energy usage in the United States Anthracite or hard coal the oldest variety of coal contains about 80 carbon Bituminous coal a younger variety of coal contains 45 to 65 carbon Fuel values of coal are measured in BTUs British Thermal Units A typical value for coal is 13200 BTUlb 1 BTU 1054 kJ Natural gas and petroleum account for nearly threequarters of the fossil fuels consumed per year Purified natural gas is primarily methane CH4 but also contains small quantities of ethane C2H6 propane C3H8 and butane C4H10 The fuel value of natural gas is close to that for the combustion of methane CH4 g 2 02 g 9 C02 g 2 H20 I AHquot 802kJ Petroleum is a very complicated mixture of compounds Gasoline obtained from petroleum contains many different hydrocarbons one of which is octane C8H18 C8H18l 252 02 g e 8002 g 9H20 I AHquot 5074kJ This value of AHquot is equivalent to 444 kJg Chem 115 Final exam study guide Time Monday December 8 8 10 am Location CB 111 regular lecture room Format The nal will be worth 100 points twice as many as the midterm exams It will consist of three parts Part I worth z40 points will cover new material that has not been tested previously Part II worth z4550 points will be comprehensive Most problems in this section will be similar but not identical to problems on the previous midterms Part III worth z1015 points will be designed to test how you can integrate or extend your knowledge This should be the most difficult portion of the exam All portions of the final will be handed out simultaneously You may use a calculator on the entire exam Summary for Part I Chapter 9 o Sections 9193 very important VSEPR model electronic and molecular geometries molecular polarity Note drawing Lewis structures was covered on Exam 3 but is a skill essential for application of the VSEPR model This topic may therefore appear in either PartI or Part II of the exam o Sections 9496 very important Valence bond theory hybridization c and 11 bonding o Sections 9798 moderately important Be able to draw MO diagrams for homonuclear or heteronuclear diatomics second row elements only and determine bond order and paramagnetism Chapter 10 o Sections 101103 moderately important This section will not be directly covered in class o Sections 104106 very important Ideal gas law applications of the ideal gas equation and partial pressures of gas mixtures o Sections 107109 moderately important Kinetic molecular theory molecular effusion and diffusion and deviations from ideal gas behavior The material covered in parts II and III will be the same as in the three midterm examinations The study guides for these exams are available on the class webpage My solutions for all worksheet and homework assignments are also available on the class webpage You will also nd the keys to the two class quizzes and the three midterm examinations
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