Physical Chemistry (II) CHEM Lecture 12 Molecular Structure. Lecturer: Hanning Chen, Ph.D. 02/28/2018
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1 Physical Chemistry (II) CHEM Lecture 12 Molecular Structure Lecturer: Hanning Chen, Ph.D. 02/28/2018
2 Quiz 11 5 minutes Please stop writing when the timer stops!
3 Three Types of Chemical Bonds What is a chemical bond? an electronic force of attraction holding atoms together in a molecule or crystal, resulting from sharing or transferring of electrons A e e B conduction electrons e homogeneous nonmetallic nonmetallic metallic nonmetallic heterogeneous covalent bond ionic bond delocalized metallic bond
4 Molecular Potential Energy Surface repulsion : nucleus-nucleus interaction Energy attraction : electron-nucleus interaction at short bond length, r < R 0 repulsion > attraction R 0 R e at long bond length, r > R 0 D e repulsion < attraction R 0 : repulsive separation R e : equilibrium bond length dissociation energy when r D e = E min ( ) R e D 0 D e D 0
5 Born-Oppenheimer Approximation atom pair Na Cl separated ions Na + Cl which trajectory is for the ultra-slow dissociation of ground state NaCl? t d Answer: lower black curve Energy Na Cl which trajectory is for the ultra-fast dissociation of ground state NaCl? separated atoms t d 0 Na + Cl ion pair Answer: green curve charge distribution is conserved, Na-Cl distance because no time for electron relaxation Born-Oppenheimer approximation: tion in the diabatic and adiabatic representations. The ionic ( green) and covalent (blue) diabatic n the same character across the potential energy surface, whereas the adiabatic states (black) The electronic motion and the nuclei motion are entirely separated. Electrons run much faster than nuclei.
6 Valence Bond Theory A chemical bond stems from the sharing of two anti-parallel electrons A B Homonuclear diatomic molecules: A = B H H 1 H 1 H 1 H 2 H (chemically identical) two-electron-two-center two possible electronic configurations electron overlap A B A B Red + Blue = Purple A(1)B(2) combination rule? A(2)B(1)
7 Linear Combination of Atomic Orbitals LCAO A B + A B A(1)B(2) addictive superposition A(2)B(1) deeper red Ψ( 1,2 ) = A(1)B(2) + A(2)B(1) bonding orbital deeper blue electron accumulation A B in the bonding region A B wavefunction of the first electron wavefunction of the second electron
8 Anti-bonding Orbital A B A B subtractive superposition A(1)B(2) A(2)B(1) nodal points Ψ( 1,2 ) = A(1)B(2) A(2)B(1) anti-bonding orbital electron depletion nodal points A B in the bonding region A B E anti bonding > E bonding wavefunction of the first electron wavefunction of the second electron
9 Where Are the Spins? bonding orbital: Ψ( 1,2 ) = A(1)B(2) + A(2)B(1) Does it satisfy the anti-symmetric requirement of multi-electron wavefunction? Answer: No! Ψ( 2,1) = A(2)B(1) + A(1)B(2) = Ψ( 1,2 ) Ψ( 1,2 ) = 1 2 ( A(1)B(2) + A(2)B(1) )( α(1)β(2) α(2)β(1) ) singlet state spatial component spin component bonding orbital is a mixture of four spin-orbital configurations Aα(1)Bβ(2) Aβ(1)Bα(2) Bα(1)Aβ(2) Bβ(1)Aα(2) Two anti-parallel electrons are needed to form a chemical bond
10 Triplet State anti-bonding orbital: Ψ( 1,2 ) = A(1)B(2) A(2)B(1) spatial component is already anti-symmetric α(1)α(2) Ψ( 1,2 ) = 1 2 ( A(1)B(2) A(2)B(1) ) ( α(1)β(2) + α(2)β(1) ) triplet state β(1)β(2) spin component For example: Aα(1)Bα(2) Bα(1)Aα(2) The anti-bonding orbital must be a triplet state with a higher energy.
11 Classification of Covalent Bonds number of nodal planes σ bond: head-to-head overlap s s p s π bond: side-by-side overlap p + p = 1 number of nodal planes containing the molecular axis
12 Bond Strength number of nodal planes δ bond: (very rare) + = y 2 d d Please sort σ, π and δ bonds in order of increasing bond strength weakest bond δ < π < strongest bond σ Why? The formation of nodal planes are energetically expensive. The orbital overlap is substantially diminished by those nodal planes.
13 O O Polyatomic Molecules electronic configurations : O :1s 2 2s 2 2 p 4 1s 2s 2 p H :1s 1s p x p y p z p y p z H 2 O O O 1s 2 p σ σ 2 p 1s Why is the water molecule bond angle not 90 degree? HOH 104 because of the orbital hybridization E 2s E 2 p shielding constant σ σ 2s 2 p
14 Orbital Hybridization 2s 2 p p x p y p z sp 3 E 2s E 2 p energetically degenerate Mathematical Justification 109! h n = a n ϕ s + b nx ϕ px + b ny ϕ py + b nz ϕ pz h n :hybrid orbitals { } n = 1,2,3,4 ϕ s,ϕ px,ϕ py,ϕ pz :orthonormal atomic orbitals tetrahedral geometry dτϕ * ϕ = δ = i j ij 1 i=j 0 i j maximally alleviate electron repulsion
15 Mathematical Justification h n = a n ϕ s + b nx ϕ px + b ny ϕ py + b nz ϕ pz h m = a m ϕ s + b mx ϕ px + b my ϕ py + b mz ϕ pz Overlap between two hybrid orbitals: δ mn = ( ) * ( ) a ϕ + b ϕ + b ϕ + b ϕ n s nx px ny py nz pz dτh * h = dτ a m n ϕ + b ϕ + b ϕ + b ϕ m s mx px my py mz pz Hybrid orbitals are also orthonormal! { } m = { 1,2,3,4 } n = 1,2,3,4 4 4 = 16 a total of 16 coupled linear equations to determine 16 orbital coefficients h = ϕ +ϕ +ϕ +ϕ h = ϕ ϕ ϕ +ϕ 1 s px py pz sp 3 2 s px py pz h = ϕ ϕ +ϕ +ϕ h = ϕ +ϕ ϕ ϕ 3 s px py pz 4 s px py pz E h1 = E h2 = E h3 = E h4
16 π Orbital Hybridization for -bonds Ethene electronic configurations : C :1s 2 2s 2 2 p 2 1s 2s H :1s 1s p x p y p z For each carbon atom 3 σ bonds can be formed but only two p electrons are available orbital promotion: C C 2s C C 1s 2 p σ p x p y p z Now, we have one excess p electron C C
17 sp 2 Hybridization 2s p p x y p z sp 2 hybrid orbitals p z 120! trigonal planar geometry formation of a π bond HCH 117 E C=C = 146 < 2E C C = 2 83 = 166 (kcal / mol) a π bond is usually weaker than a σ bond
18 sp Hybridization Ethyne electronic configurations : 1s 2s Why is ethyne linear? 1s p x p y p z H :1s C :1s 2 2s 2 2 p 2 In order to maximize the number of chemical bonds, orbital promotion is needed 2s (kcal / mol) E C C = 200 E C=C + E C C = 229 p p x y p z sp hybrid orbitals p p y z two bonds σ two π bonds For each carbon atom 3E C C = 249 The linear sp hybrid orbitals ensure maximum electron separation
19 Review of Homework An emission line from K atoms is found to have two closely spaced components, one at nm and the other at nm. Account for this observation, and deduce what information you can. ground state K: [Ar]4s 1 S= 1 2 L=0 J= 1 2 first excited state K: [Ar]4p 1 S= 1 2 L=1 J= 1 2 or S P 1 2 or 2 P 3 2 two transitions: 2 S P 1 2 and 2 S P 3 2 Δ!v = nm nm = 57.7cm 1 Δ!v = 1 2 ( ( ) j ( j +1) ) = !A j 2 j 2 +1!A = 3 2!A!A = 38.50cm 1
20 Homework 12 Reading assignment: Chapters 10.1 and 10.2 Homework assignment: Exercises 10.4(a) Problems Homework assignments must be turned in by 5:00 PM, March 1st, Thursday to my mailbox in the Department Main Office located at Room 4000, Science and Engineering Hall
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