Chemistry 142 Lecture Notes Week 3 and 4
Chemistry 142 Lecture Notes Week 3 and 4 Chem 142
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This 16 page Class Notes was uploaded by Chantellioh on Saturday October 24, 2015. The Class Notes belongs to Chem 142 at University of Washington taught by Xiaosong Li in Summer 2015. Since its upload, it has received 161 views. For similar materials see General Chemistry in Chemistry at University of Washington.
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Date Created: 10/24/15
Monday October 12 2015 Chemistry Notes Week 3 Day 1 Chapter 12 Quantum Mechanics and Atomic Theory continued 3 IMPORTANT EQUATIONS TO KNOW Bohr s equation E 2178J22n2 just an approximation de Broglie s equation A hmv any particle w mass has wave calculate wavelength for particle using mass must use ONLY for particles WITH mass Schrodinger s Equation Hoperator x W Eoperator x W Hoperator KEoperator PEoperator if you put an H hydrogen wave in you get energy out ALL particles have KE equal to 12mv2 particles PE s differ by source SOLUTION TO SCHRODINGER S EQUATION FOR HYDROGEN E 2178J22n2 exact to Bohr s shows wave is quantized by n n distance between nucleus and electron initiation state no interaction between nucleus and electron o n infinity EO ONLY THE SOLUTION FOR HYDROGEN Monday October 12 2015 THE SHAPE OF WAVES Solution to Schrodinger s Equation the solution possible wave shapes you have in a 3D space stype orbital ptype orbital dtype orbital ftype orbital we are required to know the shapes in the s p and d orbitals http4bpblogspotcomxjcwwAnj8kTOWMXHh52HlAAAAAAAAA8kulBOhug8sZMs1600 orbsgit good idea of what they look like DEFINING WAVE FUNCTION WITH ONE ELECTRON 3 NOTATIONS 1 Distance n1 only stypm n2 s and p type waves n3 s p and d type waves 2 Shape G see above n2 3 Orientation n 1 The same link provides a good example of different orientations there s 1 orientation for stype 3 for ptype and 5 for dtype Monday October 12 2015 ELECTRON ORBITAL ENERGIES IN THE H ATOM Energy Level of H E 2178 x 1018J 1n2 ENERGIES ARE IDENTICAL whether in 33 3p or 3d degenerate different shapes of waves have same energy Electron Orbital Shapes all stypes look the same from the outside but very different on inside Outside 3s orbital 1s orbital 28 orb39tal lnside crosssection 1s orbital 2s orbital 3s orbital black ring where electrons are present white ring nodes where electrons are absent Monday October 12 2015 Atomic Emission Experiment light solution on fire observe the color the flame gives off n3 n4 n burning heatcolor burning burning the solution excitesenergizes electrons to move it to a higher energy level heatcoor this extra energy is released in the form of heat and color explains how you feel see light of flame 39 AEho release of energy in terms of light Wednesday October 14 2015 Chemistry Lecture Notes Week 3 Day 2 Chapter 12 Quantum Mechanics and the Atomic Theory The class had a substitute so I m not sure how much of this is relevant to Professor Li s exam I d like to assume that it s what Prof Li reviewed in Day 3 but I m including everything I have just in case electron itself has a magnetic moment imagine as a charged sphere spinning to create a magnetic field Note that electron is NOT a sphere but a point 2 types of electron spinning spin up and spin down Angular Momentums n1s O n2s O n2p 1 n 3d 2 if the atom has 1 electron the energy across an orbital is the same regardless of shape ELECTRON CONFIGURATION electrons fill lowest orbitals first Each n Z m gets 2 electrons n 2p 1 m 1 has two electrons n2p 1 m1 has two electrons and n2p 1 mO has two electrons 2p has a total of 6 electrons s p and d are no longer degenerate no longer have the same energy Wednesday October 14 2015 EVERY ELECTRON in an atom has a unique set of quantum numbers n Km ms POLYELECTRONIC ATOMS for A direct solution to sohrodinger equation is not possible uses orbital approximation these have attraction to nuoleus BUT ALSO repulsion to other eleotrons COMPARE ts 2s 2p 3s 3p 3d 4s lt one electron ts 2s 2p 3s 3p 4s 3d lt polyelectronio FILLING ORBITALS fill starting from lowest energy 151 the two electrons in one orbital MUST have opposite spins How to Draw Hydrogen 21 131 f spinup electron ts Helium 22 132 N ts Beryllium 132232 N N ts 2s Boron 1322322p1 titlt 1s 2s 2p Wednesday October 14 2015 HUND S RULE 1 there is a favoring of unpaired electrons for lowest energy configuration 2 there is a favoring of parallel spins for lowest energy configuration This means Carbon s Electron configuration looks like flflff 1s 2s 2p NOTthis N N N I I 1s 2s 2p orthis n N f i if I 1s 2s 2p The further an electron is the easier it is to give up the electron Na becomes Na because it s ready to lose it s extra electron THE AUF BAV PRINCIPLE start w lowest and build up from lowest to highest filled in order of increasing energy not in order of increasing n means that in multi electronic atoms electrons fill order of electron configuration 4s comes before 3d etc Friday October 16 2015 Chemistry Lecture Notes Week 3 Day 3 Chapter 12 Quantum Mechanics and the Atomic Theory MOST IMPORTANT THINGS TO KNOW IN THIS SECTION solution to sohrodinger s equation in terms of diagram only for H atom 33 3p 3d 23 2p Energy 1s No Change in E from 2p gt2 DEGENERATE En 2178 x1o18 J 22n2 energy released is in terms of photon so must use AE hv Polyelectronic atom energy diagram must be memorized A 3d 4s gt 9 2 3p UJ 3s 2p 23 1s OR 1s gt 28 gt 2p gt 3s gt 3p gt 4s gt 4d Friday October 16 2015 of electrons in highest p and s orbital valence electrons each orbital has 2 electrons s can have a max of 2 electrons 1 orbital p can have a max of 6 electrons 3 orbitals d can have a max of 10 electrons 5 orbitals Memorize Hund s Rules there s a favoring of unpaired electrons there s a favoring of parallel spins Exceptions to normal electron configuration in transition metals orbital filling favors more stable configurations having a half filled or completely filled orbital is more stable than having the normal configuration EXAMPLE for chromium normal rules would make the configuration Ar 4323d4 but because a half filled d shell is much more stable one of the sshell electrons occupy the dshell gt Ar 4s13d5 EXAMPLE for copper normal rules would make the configuration Ar4323d9 but because a fully filled d shell is much more stable one of the sshell electrons occupy the dshell gt Ar4s13d1o Ionization Energy energy required to take one electron out of an atom to form a cation HEnergy gt H e First Ionization Energy Trend increases towards the top and right of periodic table increases from left to right across a period Friday October 16 2015 increases from bottom to top up a group Electron Affinity energy change associated w addition of electron Halogen elements have great electron affinity Atomic Radius distance from nucleus to outermost electron orbital Decreases across the row due to the increase in Zeff decreases up a column also due to that Monday October 19 2015 Chemistry Notes Week 4 Day 1 Review of Ch 12 important points taken off of clicker questions 1Know relationship between energy freq wavelength Einstein s equation E energy is proportional to freq energy is proportional to inverse of wavelength 2Speed of light is constant in general chemistry c blue is on the left of visible spectrum shorter wavelength and higher frequency absorption is opposite of emission in terms of spectrum range frequency should be the same in terms of photons Workfunction 12mv2hv Workfunction higher work function lower velocity when hv is kept constant SAMPLE PROBLEM Wavelength of energy ejected when given photon wavelength and workfunction find nu from the photon s wavelength nu cwavelength plug nu into Eh v work function use 12 mv2hv work function to find velocity plug velocity into wavelength hmv MUST UNDERSTAND WAVELENGTH OF PHOTON vs WAVELENGTH OF PARTICLES W MASS wavelength of photon lambda cnu wavelength of part w mass lambda hmv i Calculate energy level spacing of Hydrogen using En2178 x 1018J1n2 Q AE4gt1gt AE4gt2 and therefore lambda4gt1 lt lambda4 gt2 Monday October 19 2015 L Hydrogen has 1 electron 13 ONLY IF IT IS AT GROUND STATE therefore hydrogen has a 33 orbital even if the electron is not currently occupying that orbital the electron jumps from 13 to 33 if given the appropriate amnt of energy 8 Energy configuration of N can be 1322322p2331 if one energy jump from 2p gt 3s 1322322p3 ground state 1322322p2 N 1322322p4 ground state of N 9 Lowest ionization energy one occupying highest orbital bc electron furthest from nuclear has easiest time leaving 1 Electron affinity of F add one electron to fill orbital level A ionization of F 1322322p5 gt 1322322p4 F 39 He s making this question harder by making us draw the electrons in the boxes We must know how to arrange electrons fill boxes w one before adding a second diagram 12 electron configuration for element given the periodic table Notes about Final given by professor Most of these questions will be made harder to keep grade average 60 Statistically we should not get a score lower than 20 15 questions will be on this chapter and 5 will be on chapter 13 Wednesday October 21 2015 Title Ch 13 Types of Bonds bonding gt change molecules we ll only focus on covalent pure and polar and ionic bonds there are other bonds such as metallic bond bond between metals covalent bonds gtweak gt very easy to change there are no pure ionic compounds Ch 13 focuses mostly on covalent bonds IONIC BONDS formed by a metal and a nonmetal network of ions very high melting point bc very hard to break an ionic bond conducts electricity when melted or dissolved in water liquidaqueous an insulator when it s a solid but once it s dissolved NaCl becomes Na and Cl which conducts electricity COVALENT BONDS sharing of electrons both electrons orbit both nuclei electronegativity Fe gt F energy EN is not physical it s a measurable quantity Polarity electrons are shared unequally H F forms a covalent bond 2 nonmetals F has a strong electron affinity extra negative charges around fluorine and fewer around hydrogen so hydrogen has a slightly positive charge Wednesday October 21 2015 we MUST know how to identify polar vs non polar but we don t need to know how to calculate a dipole moment The net charge of a molecule is always zero LEWIS STRUCTURE most important thing in ch 13 Valence Electrons only define valence electrons w main group A elements only count the outermost s and p orbitals Ne3sZ3p5 has 7 VE Lewis Dot Structure no absolutely right way to draw for single atoms 00 or C or C 39 H39 gtH only way to use covalent interactions not good to use for ionic interactions don t really draw LDS for transition metals bc the have many active electrons Friday October 23 2015 Chemistry Lecture Notes Week 4 Day 3 Chapter 13 Chemical Concept LEWIS DOT STRUCTURE Valence Electrons electrons in the highest s and p type orbitals often times they ll form an ionic bond indicated through lewis dot symbols each dot one valence electron 8 dots makes a full octet used to understand bonding Lewis Dot Structures atoms have the most stable electron configuration when they have the noble as configuration bonding pair two dots put inbetween two atoms to signify bonding usually shown through a single line the line represents two dots lone pairs each couplet of two electrons that is not bonded to another atom Count of valence electrons it takes to create complete octets for all atoms divided by two bonding pairs total valence electrons divided by two minus bonding pair pairs lone pairs M first find the amount of bonding pairs each atom needs to have a full octet the amount of boding electrons the amount of lone electrons should equal to the total number of valence electrons Charge to calculate the formal charge identify valence electrons chlorine has compare to electrons owned by chlorine in LDS in lewis dot structure you only count half of the shared electrons one electron per bond Friday October 23 2015 Multiple Bonds the one that needs more bonds should be in the middle carbon needs 4 bonds and each oxygen needs 2 easy to see that double bond needed just by knowing this using book procedure first draw out standard LDS with one bond then count number of full octets for each atom add bonds to satisfy needs Resonance Structures you can move structures around to form isomers as long as octet structure is okay 03 is a classic example two structures are degenerate they are both correct but they have different energy One problem on exam asks you to identify the correct structure identify valence electrons identify valence electrons represented in LDS for each atom formal charge charge of each atom you want to have the smallest amt of charge distribution Structural lsomers whole idea of molecular structure is to minimize amnt of electron repulsion don t really know structure of molecule from the LDS
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