Physical Chemistry (II)

Lecture 12

CHEM 3172-80

Lecturer: Hanning Chen, Ph.D.03/01/2017

Molecular Structure

Quiz 11

5 minutes

Please stop writing when the timer stops !

Three Types of Chemical BondsWhat 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 Be−

e−

covalent bond

homogeneous

heterogeneousionic bond

e−

metallic↔ nonmetallic

nonmetallic↔ nonmetallic

metallic bond

conduction electrons

delocalized

Molecular Potential Energy SurfceEnergy

Re

Re : equilibrium bond length

De = Emin Re( )

repulsion : nucleus-nucleus interactionattraction : electron-nucleus interaction

De

dissociation energy D0 ≈ De

when r→∞

D ≈ 0

at short bond length, r < R0

repulsion > attraction

at long bond length, r > R0

repulsion < attraction

R0

R0 : repulsive separation

Born-Oppenheimer Approximation

ANRV408-PC61-08 ARI 27 February 2010 15:27

Adiabatic state: aneigenstate of theBorn-OppenheimerelectronicHamiltonian

Diabatic state: anelectronic state thatdoes not changecharacter as a functionof molecular geometry

ET: electron transfer

CDFT: constraineddensity functionaltheory

1. INTRODUCTIONQualitatively, a diabatic electronic state is one that does not change its physical character as onemoves along a reaction coordinate. This is in contrast to the adiabatic, or Born-Oppenheimer, elec-tronic states, which change constantly so as to remain eigenstates of the electronic Hamiltonian. Aclassic example of the interplay between diabatic and adiabatic pictures is sodium-chloride dissoci-ation (Figure 1). Here, the ground adiabatic state is thought of as arising from the avoided crossingbetween an ionic and a covalent state. The adiabatic state thus changes character—transformingfrom Na-Cl to Na+-Cl− as the bond gets shorter—whereas the ionic and covalent configurationsplay the role of diabatic states.

Diabatic electronic states play a role in a variety of chemical phenomena but are, at the sametime, underappreciated by many chemists. For example, diabats are often used in the constructionof potential energy surfaces because they are smooth functions of the nuclear coordinates (1–3). In spectroscopy, diabatic states are invoked to assign vibronic transitions and rationalize therates of interstate transitions (4–6). In the general description of electronically excited dynamics,diabatic states are advantageous because they typically have a small derivative coupling, simplifyingthe description of electronic transitions (7–11). In scattering theory, diabatic states connect toclearly defined product channels (12–14). Finally, diabatic states play a qualitative role in ourunderstanding of molecular bonding (15–17) (as illustrated by the NaCl example above), electrontransfer (ET) (18, 19), and proton tunneling (20–22).

This review article is intended as an introduction to the basic concepts behind the constructionof diabatic states and their use in describing chemical phenomena. After summarizing differ-ent definitions of diabatic states—and in particular discussing why so many competing definitionsexist—we focus on a particular definition based on constrained density functional theory (CDFT).We show how CDFT-derived diabatic states are computed in practice and discuss several illustra-tive chemical applications.

Na-Cl distance

Ener

gy

Na

Na Cl

Cl

Cl–

Cl–

Na+

Na+

Figure 1NaCl dissociation in the diabatic and adiabatic representations. The ionic ( green) and covalent (blue) diabaticstates maintain the same character across the potential energy surface, whereas the adiabatic states (black)change.

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which trajectory is for the ultra-slow dissociation of ground state NaCl ?

td →∞lower black curveAnswer:

which trajectory is for the ultra-fast dissociation of ground state NaCl ?

td → 0Answer: green curve

charge distribution is conserved, because no time for electron relaxation

Born-Oppenheimer approximation: The electronic motion and the nuclei motion are entirely separated.

Electrons run much faster than nuclei.

ion pair

atom pair

separated atoms

separated ions

Valence Bond TheoryA chemical bond stems from the sharing of two anti-parallel electrons

A B↓

↑

Homonuclear diatomic molecules: A = B H − H(chemically identical)

1H − 1H1H − 2H

two possible electronic configurations

A

two-electron-two-center

B

A(1)B(2)

A B

A(2)B(1)

electron overlap

Red + Blue = Purple

combination rule ?

Linear Combination of Atomic Orbitals

A B

A(1)B(2)

A B

A(2)B(1)

+addictive superposition

Ψ 1,2( ) = A(1)B(2)+ A(2)B(1)

A B

wavefunction of the first electron

A B

wavefunction of the second electron

electron accumulationin the bonding region

bonding orbital

deeper reddeeper blue

LCAO

Anti-bonding Orbital

A B

A(1)B(2)

A B

A(2)B(1)

−subtractive superposition

Ψ 1,2( ) = A(1)B(2)− A(2)B(1)

A B A Belectron depletion

in the bonding region

anti-bonding orbital

wavefunction of the first electron wavefunction of the second electron

Eanti−bonding > Ebonding

nodal points nodal points

Where Are the Spins ?

Ψ 1,2( ) = A(1)B(2)+ A(2)B(1)bonding orbital:

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( ) = 12A(1)B(2)+ A(2)B(1)( ) α (1)β(2)−α (2)β(1)( )

spatial component spin component

bonding orbital is a mixture of four spin-orbital configurations

singlet state

Two anti-parallel electrons are needed to form a chemical bond

Aα (1)Bβ(2) Aβ(1)Bα (2) Bα (1)Aβ(2) Bβ(1)Aα (2)

Triplet StateΨ 1,2( ) = A(1)B(2)− A(2)B(1)anti-bonding orbital:

Ψ 1,2( ) = 12A(1)B(2)− A(2)B(1)( )

α (1)α (2)α (1)β(2)+α (2)β(1)( )

β(1)β(2)

⎧

⎨⎪

⎩⎪⎪

spin component

triplet state

The anti-bonding orbital must be a triplet state with a higher energy.

spatial component is already anti-symmetric

For example:

Aα (1)Bα (2) Bα (1)Aα (2)

Classification of Covalent Bonds

σ bond:

number of nodal planes containing the molecular axis

s s

number of nodal planes

0

s phead-to-headoverlap

0

0

π bond:side-by-side

overlapp

+p

= 1

Bond Strength

δ bond:

number of nodal planes

+ =

d d

2(very rare)

Please sort σ , π and δ bonds in order of increasing bond strength weakest bond strongest bond

δ σ< π <

Why ?

The formation of nodal planes are energetically expensive. The orbital overlap is substantially diminished by those nodal planes.

y

Polyatomic Molecules

H2O

electronic configurations :

O :1s2 2s2 2p4

H :1s

O O

1s 2p σO O

1s2pσ

1s

1s

2s 2p

Why is the water molecule bond angle not 90 degree ? ∠HOH ≈104°because of the orbital hybridization

px py pzpy ⊥ pz

E2s ≈ E2 p σ 2s ≈σ 2 p

shielding constant

Orbital Hybridization

2s 2p

px py pz

E2s ≈ E2 p energetically degeneratesp3

tetrahedral geometry

109!

maximally alleviate electron repulsion

Mathematical Justification

hn = anϕs + bnxϕ px+ bnyϕ py

+ bnzϕ pz

n = 1,2,3,4{ }hn :hybrid orbitalsϕs ,ϕ px

,ϕ py,ϕ pz

:orthonormal atomic orbitals

dτ∫ ϕi*ϕ j = δ ij =

1 i=j0 i ≠ j

⎧⎨⎪

⎩⎪

Mathematical Justification

hn = anϕs + bnxϕ px+ bnyϕ py

+ bnzϕ pzhm = amϕs + bmxϕ px

+ bmyϕ py+ bmzϕ pz

Overlap between two hybrid orbitals:

δmn = dτhm*hn∫ = dτ amϕs + bmxϕ px

+ bmyϕ py+ bmzϕ pz( )* anϕs + bnxϕ px

+ bnyϕ py+ bnzϕ pz( )∫

Hybrid orbitals are also orthonormal !

n = 1,2,3,4{ } m = 1,2,3,4{ }a total of 16 coupled linear equations to determine 16 orbital coefficients

h1 =ϕs +ϕ px+ϕ py

+ϕ pz h2 =ϕs −ϕ px−ϕ py

+ϕ pz

h3 =ϕs −ϕ px+ϕ py

+ϕ pzh4 =ϕs +ϕ px

−ϕ py−ϕ pz

Eh1= Eh2

= Eh3= Eh4

sp3

4 × 4 = 16

Orbital Hybridization for -bondsπEthene electronic configurations :

C :1s2 2s2 2p2

H :1s

1s

1s

2s

px py pz

C C

C C

1s 2p σC C

For each carbon atom3 σ bonds can be formed but only two electrons are availablep

orbital promotion:

2s

px py pz

Now, we have one excess p electron

sp2 Hybridization2s

px py pz sp2 pzhybrid orbitals

trigonal planar geometry

120!

∠HCH ≈117°

formation of a bondπEC=C = 146 < 2EC−C = 2 × 83= 166 (kcal /mol)

a bond is usually weaker than a bondπ σ

sp Hybridization

Ethyne

Why is ethyne linear ?

electronic configurations :

C :1s2 2s2 2p2H :1s

1s

1s

2s

px py pz

In order to maximize the number of chemical bonds, orbital promotion is needed2s

px py pz sp pzhybrid orbitals pytwo bondsσ two bondsπ

For each carbon atom

The linear sp hybrid orbitals ensure maximum electron separation

EC≡C = 200(kcal /mol)

EC=C + EC−C = 2293EC−C = 249

Review of Homework 6Review of Homework 119.26. An emission line from K atoms is found to have two closely spaced components, one at 766.70 nm and the other at 771.11 nm. Account for this observation, and deduce what information you can.

ground state K: [Ar]4s1 S= 12

L=0 J= 12

first excited state K: [Ar]4p1 S= 12

L=1 J= 12

or 32

2S12

2P12

or 2P32

2S12

→2 P12

and 2S12

→2 P32

two transitions:

Δ!v = 12!A j2 j2 +1( )− j1 j1 +1( )( ) = 12 !A

3232+1⎛

⎝⎜⎞⎠⎟ −

1212+1⎛

⎝⎜⎞⎠⎟

⎛⎝⎜

⎞⎠⎟= 32!A

Δ!v = 1766.70nm

− 1771.11nm

= 57.7cm−1

!A = 38.50cm−1

Homework 12

Reading assignment: Chapters 10.1 and 10.2

Homework assignment: Exercises 10.4(a) Problems 10.17

Homework assignments must be turned in by 5:00 PM, March 2nd, Thursday

to my mailbox in the Department Main Office located at Room 4000, Science and Engineering Hall