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by: Carmela Kilback


Carmela Kilback
GPA 3.92

Daniel Chiu

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Daniel Chiu
Class Notes
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This 13 page Class Notes was uploaded by Carmela Kilback on Wednesday September 9, 2015. The Class Notes belongs to CHEM 152 at University of Washington taught by Daniel Chiu in Fall. Since its upload, it has received 13 views. For similar materials see /class/192558/chem-152-university-of-washington in Chemistry at University of Washington.




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Date Created: 09/09/15
Lecture 20 VESPR VESPR Background The Lewis Dot Structure approach provided some 0 Reading Zumdahl 1313 insight into molecular structure in terms of bonding but what about geometry 39 OUtline Recall from last lecture that we had two types of Concept behind VESPR electron pairs bonding and lone Molecular geometries Valence Electron Shell Pair Repulsion VESPR 3D structure is determined by minimizing repulsion of electron pairs VESPR Background cont VESPR Background cont Must consider both bonding and lone pairs in minimizing repulsion Example NH3 both bonding and lone pairs Example CH4 H H I u o i H N gt I I H c H gt H p c H a b C 0 Lewis Structure VESPR Structure LeWIS Structure VESPR Structure VESPR Applications VESPR Applications Linear Structures angle between bonds is 180 The previous examples illustrate the stratgey for applying VESPR to predict molecular structure Example BeF2 1 Construct the Lewis Dot Structure 13 Be39F39 2 Arranging bondinglone electron pairs 39 39 in space such that repultions are minimized DOC F Be F V 180 VESPR Applications VESPR Background cont 39 Trigonal Planar StTUCtur653 angle between 0 Pyramidal Bond angles are lt120 and structure bonds is 120 is nonplanar Example BF3 Example NH3 1 20 0 i 2F H 4quotno I HP F B P I I H H a b VESPR Applications VESPR Applications 0 Tetrahedral angle between bonds is lO95 Tetrahedral angle may vary from 1095 exactly due to size differences between bonding and lone pair electron densities x H bonding pair a X 0 Example CH4 lone pair b VESPR Applications VESPR Applications 0 Classic example of tetrahedral angle shift from 0 Comparison of CH4 NH3 and H20 1095 is water Methane Ammonia Water Bonding pair Bonding l a b C VESPR Applications Trigonal Bipyramidal 120 in plane and two orbitals at 90 to plane Example PC15 VESPR Applications Octahedral all angles are 90 Example PCl6 Advanced VESPR Applications Square Planar versus See Saw 163th 9 See Saw No dipole moment F Square Planar Advanced VESPR Applications Driving force for last structure was to maximize the angular separation of the lone pairs Advanced VESPR Applications VESPR and resonance structures Must look at VESPR structures for all resonance species to predict molecular properties It on no intI1 Inq VESPR Applications Provide the Lewis dot and VESPR structures for CFZClz Does it have a dipole moment 39F39 F I I 32 e39 c1 C F I I II F 0 1 III Cl Tetrahedral Lecture 14 Intro to Quantum Mechanics Reading Zumdah1125126 Outline 7 Basic concepts 7 A model system particle in abox 7 Other contining potentials Quantum Concepts hrmodel was capable of descn The Bo 39hing the discrete or quantized emission spectrum ofH But the failure ofthe model tormultielectron systems combined with other issues the ultraviolet catastmphe workfunctions otmetals 39c etc suggested that anew description ofatoml matter was needed Quantum Concepts cont This new description was known as wave mechanics or quantum mechanics Recall photons and electrons readily demonstrate waveparticle duality The idea behind wave mechanics was that the existence ofthe electron in xed energy levels could be though of as a standing wave Quantum Concepts cont Whatis a standing wave 7 A standing waveis amuuun in whleh txanslauun ufthe wave dues not occur ln the guitar string analogy g g lllustxated note that standing waves involyenodes in which no on o the string oeeurs ote also thatinteger and has 39 integeryalues otthewayelength cunespund to standing waves Quantum Concepts cont Louis de Bruglle suggeststhat furthe e orloits enyisioned by Buhr only certain orbits are allowed sineethey sausly the standing wave eondiuon not allowed Quantum Concepts cont Erwin Schrodinger develops a mathematical formalism that incorporates the wave nature of matter WEw ineti e Energy PE The Hamiltonian H x TheWavefuneuun 11 E energy Quantum Concepts cont I What is a wavefunction 11 aprobability amplituole I Consider a wave y AeIWWJIW yr AMMXAWwe A I Probability of nding a particle in space i Probability 11 11 I With the wavefunction we can describe spatial distributions Quantum Concepts cont I Another limitation of the Bohr model was that i assumed we could know both the position and momentum of an electron exactly I Werner Heisenberg development of quantum mechanics leads him to the observation that there is a fundamental limit to how well one can know both the position and momentum ofa particle h Ax I Ap 2 47r Uncertainty in position Uncertainty in momentum Quantum Concepts cont Example What is the uncertainty in velocity for an electron in a 1A radius orbital in which the positional uncertainty is 1 ofthe radius Ax 1 A0 01 1 x 103912 m h 6626x103934Jr u 527x1039 kgmr 47m 4lx10 um AP Ap 527x10 23kgm r 7mj39m 7 huge Quantum Concepts cont Example you re quantum as well What is the uncertainty in position for a 80 kg student walking across campus at 1 3 ms with an uncertainty in velocity of 1 Ap m Av 80kg0 013 ms 1 04 kg ms h 7 6626x103934Jr7 Ax e 507x10 3 m 4rAp 41 04kgms Very small we know Where you are Potentials and Quantization I Consider a particle free to move in l dimension P The Free Particle x Potential E 0 I The Schrodinger Eq becomes W4WWWWZW Mr I Energy ranges from 0 to in nity not quantized Potentials and Quantization cont I What ifthe position ofthe particle is constrained by a potential int Particle in a Box Potential E 0 for 0 S X S L no all other x I Now position of particle is limited to the dimension of the box Potentials and Quantization cont What do the wavefunctions look like hiz Like a standing wave Potentials and Quantization cont What does the ehehgy look like Potentials and Quantization cont Cunsiderthe fulluwing dye muleeule thelength ufwhieh can be eunsidered the length ufthe hm an eleetxun is limited tn L a A Whatwavelength uflight emespuhds tn AE 39um hi tn hzv hz hz M my quot 2 7 quot3quot 8mm2 22 71 2 8x10 quot its ohm shuuld he m hm Potentials and Quantization cont One effect ofa constraining potential is that the ehehgy ofthe system becomes quantized Back to the hydrogen atom n 3 1 V0 eunstxaming putential Potentials and Quantization cont Also in the case ofthe hydrogen atom ehehgy becomes quantized due to the phesehee ofa constraining potential I E Schrudinger Equauuh V0 397 Reeuvers the Buhiquot hehaviur Lecture 16A Polyelectronic Atoms I Reading Zurndahl12101213 I Outline 7 Spin 7 The Aufbau Principle 7 Filling up orbitals and the Periodic Table Spin I Further experiments demonstrated the need for one more quantum number I Speci cally some particles electrons in partic ar demonstrated inherent angular momentum Spin cont I The new quantum n 5 m5 12 analagous to m I For the electron ms has two Values 12 and 12 ms12 The Aufbau Principal I For polyelectronic atoms a direct solution of the Schrodinger Eq is not possible I When we construct polyelectronic atoms we use the hy ogenatom orbital nomenclature to discuss in which orbitals the electrons reside I This is an approximation and it is surprising how well it actually works The Aufbau Principal cont I When placing electrons into orbitals in the construction of polyelectronic atoms we use the Aufbau Principle I This principle states that in addition to adding protons and neutrons to the nucleus one simply adds electrons to the hydrogenlike atomic orbitals I Pauli exclusion principle No two electrons may have the same quantum numbers Therefore only 0 electrons can reside in an orbital differentiated by ms The Aufbau Principal cont I Finally orbitals are lled starting from the lowest energy I Example Hydrogen D EDD Is 2 2p I Example Helium Z 2 EDD The Aufbau Principal cont Limaum 73 152251 29 152252 29 15 25 Ben39llium z 4 BED The Aufbau Principal cont Carbon 25 mm Zp Hund39s Rule Luwest energy eun guranunis e ne in which the maximum numb er ufunp aired eleetmns are distributed amungst aset ufdegenerate mums 15 25 Nitrogen z 7 1s22szzp3 The Aufbau Principal cont Oxygen 28 1s12szzp 1s12s22p5 1s22szzp6 mu The Aufbau Principal cont Sodium 211 1szzszzp zs Igt Ne35 Ne 3523p The Aufbau Principal cont We nuW have the urbital eun guratmns fur the rst 18 elements L m u l w I F m m m w y 2p a up z N My A x v s m A a 1 In an aw i In is Elements in same Eulumn have the same ufvalenee eleetmnsl The Aufbau Principal cont Slmllartu Sudlum Webegm menexmwufmepenme tableby adding eleetxuns m Lhe45 urbital Why nut 3a befure 457 3d 15 elusertu Lhenucleus Minimum mm 45 311W fur Elmer appmach Lherefure is energetically preferred The Aufbau Principal cont 39 Elements 219 and Z 20 219 Futassium isZZszzp zszszs A445 zznCa1emm isIZszzp zsupMszArm2 Elements Ztho 230 have occupied d orbitals 221 Scandium isZZszzp zszszszzd Ax 452m 224 chmmmm Ax45 3d5 exceptmn 2 3n Zinc ISZZSZZp Kszlp Aszld Ar 452de The Aufbau Principal cont m m M m n a M E This urbltal 511mg scheme gives nse m themudem penudlc table The Aufbau Principal cont We m n M m M MEI V 2 3 5 A er Lanthanum Xe6575d we start 511mg A er Acumum Rn The Aufbau Principal cont m H M M u H mm 175mm we Stan 511mg 5f The Aufbau Principal cont ylt m l l 39 39 39 47v 7 H 5 e l e l l Heading un culumn given tutal number ufvalence e1 ectmns The Aufbau Principal cont Summary I Electrons go into quot orbitals to construct polyelectronic atoms Aufbau Principle I Rem ember the Lecture 16B Periodic Trends Reading Zumdahl 12141216 Outline 7 Periodic Trends I Ionization Energy Electron Af nity and Radii 7 A Case Example Periodic Trends I The Valence electron structure ofatoms can be used to explain Various properties of atoms I In general properties correlate down a group of elemenw Warnin such discussions are by nature Very generalized exceptions do occur Periodic Trends Ionization I IfWe put in enough energy We can remove an electron from an atom Zr an Energy gt er I The electron is completely removed from the atom potential energy 0 Periodic Trends Ionization I Generally done using photons With energy measured in eV 1 eV 16 x103919 J I The greater the propensity for an atom to hold on to is electrons the higher the ionization potential Will be I Koopmans Theorem The ionization energy of an electron is equal to the energy of the orbital from Where the electron came Periodic Trends Ionization I One can perform multiple ionizations Aug Alg e39 11 580 kJmol rst Alg gt A12g e39 12 1815 kJmol second Alzg I A13g e39 13 2740 kJmol third A13g I A14g er 1 11500 kJmol fourth Periodic Trends Ionization First Ionization Potentials i D it rllumuK ul l Periodic Trends Ionization First Ionization Potentials nereases as ne gues hum tn n 123 git Reasuu increased 2 Deerease as one gues uuwri mup Reasuu increased distanee umuueleus Periodic Trends Ionization Removal ofvalence Versus eore eleetrous Nag Na g e l495 kJmul Ne35 Ne remumng valmcequot eleetruri Na g Naz g e l2 456m kJmul Ne lsZZSZZpS remuvlng curequotelectmn Takes signi cantly more energy to remove a eore eleetrouteudauey for eore con gurations to be energetically stab e Periodic Trends Electron Affinity Eleetror Af nity the energy change associated with the addition of an eleetror to a gaseous atom aoj gt Ems Periodic Trends Electron Affinity We will stick with our thermodynamic de nition with energy released being a negative quantity J o Hl l w more number l Wuwl Periodic Trends Electron Affinity Elements that have high eleetror af nity Gmup 7 thehalugens and Gmup o o and s speulieally Periodic Trends Electron Affinity Some elements will not form ions t2 I4 Mum uumm Orbital con gurations can explain both observations Periodic Trends Electron Affinity 39 Why is EA so great for the halogens Fg e39 gt F39g EA 327 8 kJmol 1s22s22p5 1sZZsZZp 1 Ne 39 Why is EA so poor for nitrogen Ng e39 gt N39g EA gt 0 unstable 1s22s22p3 1s22s22p4 e39 must go into occupied orbital Periodic Trends Electron Affinity 39 How do these arguments do for O Og e39 O39g EA 140 kJmol 1s22s22p4 1s22s22p5 Bigger Zt overcomes e39 repu 39 39 What about the second EA for O O39g e39 OZ39g EAgt 0 unstable 1s22s22p5 1sZZsZZp Ne con guration but electron repulsion isjust too great Atomic Radii Atomic Radii are de ned as the covalent radii and are obtained by taking 12 the distance of a bond 2r r atomic radius Atomic Radii 39 Decrease to right due 3 g c a due increase in Z k 9 0 a a o a 1 a a a 39 Increase down column a due to population of 0 0 O 0 0 a orbitals ofgreatern 09 9 Looking Ahead We can partition the periodic table into general types of elements Metals tend to give up e39 nonMetals tend to gain e39 Metalloids can do either


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