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Lecture 14 February 5, 2010PowerPoint Presentation

Lecture 14 February 5, 2010

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Nature of the Chemical Bond with applications to catalysis, materials science, nanotechnology, surface science, bioinorganic chemistry, and energy

William A. Goddard, III, [email protected]

316 Beckman Institute, x3093

Charles and Mary Ferkel Professor of Chemistry, Materials Science, and Applied Physics, California Institute of Technology

Teaching Assistants: Wei-Guang Liu <[email protected]>

Ted Yu <[email protected]>

Course schedule

Friday Feb. 5, 2pm L14 TODAY(caught up)

Midterm given out on Friday. Feb. 5, due on Wed. Feb. 10

It will be five hour take home with 30 min. break, open notes

Separated atoms limit

Note that in each case we get one bonding combination (no new nodal plane) and one antibonding combination (new nodal plane, red lines)

At large R 2ps better bonding than 2pp

In earlier lectures we considered the strength of one-electron bonds where we found that

Since the overlap of ps orbitals is obviously higher than pp

We expect that

bonding

antibonding

Correlation diagram for Carbon row homonuclear diatomics

C2

N2

O2

F2

United atom limit

O2+

separated atom limit

N2+

Homonuclear Diatomics Molecules – the valence bond view

Consider bonding two Ne atoms together

Clearly there will be repulsive interactions as the doubly occupied orbitals on the left and right overlap, leading to repulsive interactions and no bonding. In fact as we will consider later, there is a weak attractive interaction scaling as -C/R6, that leads to a bond of 0.05 kcal/mol, but we ignore such weak interactions here

The symmetry of this state is 1Sg+

Halogen dimers

Next consider bonding of two F atoms. Each F has 3 possible configurations (It is a 2P state) leading to 9 possible configurations for F2. Of these only one leads to strong chemical binding

This also leads to a 1Sg+ state.

Spectroscopic properties are listed below .

Note that the bond energy decreases for Cl2 to Br2 to I2, but increases from F2 to Cl2. we will get back to this later.

Di-oxygen or O2 molecule

Next consider bonding of two O atoms. Each O has 3 possible configurations (It is a 3P state) leading to 9 possible configurations for O2. Of these one leads to directly to a double bond

This suggests that the ground state of O2 is a singlet state.

At first this seemed plausible, but by the late 1920’s Mulliken established experimentally that the ground state of O2 is actually a triplet state, which he had predicted on the basis of molecular orbitial (MO) theory.

This was a fatal blow to VB theory, bringing MO theory to the fore, so we will consider next how Mulliken was able to figure this out in the 1920’s without the aid of computers.

O2 MO configuration

(1pg)2

2

For O2 the ordering of the MOs

Is unambiguous

4

2

2

2

Next consider states of (1pg)2

2

2

States based on (p)2

D-

S-

S+

D+

Have 4 spatial combinations

Which we combine as

where x and y denote px and py

φ1, φ2 denote the angle about the axis

and F is independent of φ1, φ2

Rotating about the axis by an angle g, these states transform as

States arising from (p)2

Adding spin we get

MO theory explains the triplet ground state and low lying singlets

O2

Energy (eV)

1.636

(p)2

0.982

0.0

Ground state

Using the correleation diagram

Choices for N2

In order to use the correlation diagram to predict the states of diatomic molecules, we need to have some idea of what effective R to use (actually it is the effective overlap with large R small S and small R large S).

Mulliken’s original analysis [Rev. Mod. Phys. 4, 48 (1932)] was roughly as follows.

1. N2 was known to be nondegenerate and very strongly bound with no low-lying excited states

2

4

2

2

4

2

4

2

2

2

2

N2 MO configurations

This is compatible with several orderings of the MOs

Largest R

2

4

2

4

2

2

2

4

2

2

Smallest R

2

N2+

But the 13 electron molecules BeF, BO, CO+, CN, N2+

Have a ground state with 2S symmetry and a low lying 2S sate.

In between these two 2S states is a 2P state with spin orbital splitting that implies a p3 configuration

This implies that

Is the ground configuration for N2 and that the low lying states of N2+ are

This agrees with the observed spectra

Correlation diagram for Carbon row homonuclear diatomics

C2

N2

O2

F2

United atom limit

O2+

separated atom limit

N2+

First excited configuration

(1pu)3

(1pu)3

(1pg)3

(1pg)3

(1pg)2

Ground configuration

excited configuration

1Su+

1Du

Only dipole allowed transition from 3Sg-

3Su-

3Su+

1Su-

3Du

Strong transitions (dipole allowed) DS=0 (spin)

Sg Su or Pu but S- S-

Role of O2 in atmosphere

Moss and Goddard JCP 63, 3623 (1975)

Strong

Get 3P + 1D O atom

Weak

Get 3P + 3P O atom

Implications

UV light > 6 eV (l < 1240/6 = 207 nm) can dissociate O2 by excitation of 3Su+ which dissociates to two O atom in 3P state

UV light > ~7.2 eV can dissociate O2 by excitation of 3Su- which dissociates to one O atom in 3P state and one in 1D (maximum is at ~8.6 eV, Schumann-Runge bands)

Net result is dissociation of O2 into O atoms

Regions of the atmosphere

mesosphere

O + hn O+ + e-

Heats from light

O2 + hn O + O

100

stratopause

altitude (km)

O + O2 O3

stratosphere

50

O3 + hn O + O2

Heats from light

30

20

tropopause

10

troposphere

Heated from earth

200

300

Temperature (K)

ionosphere

night

Heaviside-Kennelly layer

Reflects radio waves to allow long distance communications

D layer day

nightglow

At night the O atoms created during the day can recombine to form O2

The fastest rates are into the Herzberg states,

3Su+

1Su-

3Du

Get emission at ~2.4 eV, 500 nm

Called the nightglow (~ 90 km)

Problem with MO description: dissociation

3Sg- state: [(pgx)(pgy)+ (pgy) (pgx)]

As R∞ (pgx) (xL – xR) and (pgy) (yL – yR)

Get equal amounts of {xL yL and xR yR} and {xLyR and xR yL}

Ionic: [(O-)(O+)+ (O+)(O-)]

covalent: (O)(O)

But actually it should dissociate to neutral atoms

Back to valence bond (and GVB)

Four ways to combine two 3P states of O to form a s bond

bad

Closed shell

Open shell

Looks good because make p bond as in ethene, BUT have overlapping doubly occupied orbitals antibonding

Each doubly occupied orbital overlaps a singly occupied orbital, not so repulsive

Analysis of open shell configurations

Each can be used to form a singlet state or a triplet state, e.g.

Singlet: A{(xL)2(yR)2[(yL)(xR) + (xR)(yL)](ab-ba)}

Triplet: A{(xL)2(yR)2[(yL)(xR) - (xR)(yL)](ab+ba)} and aa, bb

Since (yL) and (xR) are orthogonal, high spin is best (no chance of two electrons at same point) as usual

Bond H to O2

Bring H toward px on Left O

Overlap doubly occupied (pxL)2 thus repulsive

Overlap singly occupied (pxL)2 thus bonding

Get HOO bond angle ~ 90º

S=1/2 (doublet)

Antisymmetric with respect to plane: A” irreducible representation (Cs group)

Bond weakened by ~ 51 kcal/mol due to loss in O2 resonance

2A” state

Bond 2nd H to HO2 to form hydrogen peroxide

Bring H toward py on right O

Expect new HOO bond angle ~ 90º

Expect HOOH dihedral ~90º

Indeed H-S-S-H:

HSS = 91.3º and HSSH= 90.6º

But H-H overlap leads to steric effects for HOOH, net result:

HOO opens up to ~94.8º

HOOH angle 111.5º

trans structure, 180º only 1.2 kcal/mol higher

Rotational barriers

7.6 kcal/mol Cis barrier

HOOH

1.19 kcal/mol Trans barrier

HSSH:

5.02 kcal/mol trans barrier

7.54 kcal/mol cis barrier

Compare bond energies (kcal/mol)

O23Sg-

119.0

50.8

67.9

HO-O

68.2

H-O2

51.5

17.1

HO-OH

51.1

HOO-H

85.2

Interpretation:

OO s bond = 51.1 kcal/mol

OO p bond = 119.0-51.1=67.9 kcal/mol (resonance)

Bonding H to O2 loses 50.8 kcal/mol of resonance

Bonding H to HO2 loses the other 17.1 kcal/mol of resonance

Intrinsic H-O bond is 85.2 + 17.1 =102.3

compare CH3O-H: HO bond is 105.1

Bond O2 to O to form ozone

Require two OO s bonds get

States with 4, 5, and 6 pp electrons

Ground state is 4p case

Get S=0,1 but 0 better

Goddard et al Acc. Chem. Res. 6, 368 (1973)

Pi GVB orbitals ozone

Some delocalization of central Opp pair

Increased overlap between L and R Opp due to central pair

Bond O2 to O to form ozone

lose O-O p resonance, 51 kcal/mol

New O-O s bond, 51 kcal/mol

Gain O-Op resonance,<17 kcal/mol,assume 2/3

New singlet coupling of pL and pR orbitals

Total splitting ~ 1 eV = 23 kcal/mol, assume ½ stabilizes singlet and ½ destabilizes triplet

Expect bond for singlet of 11 + 12 = 23 kcal/mol, exper = 25

Expect triplet state to be bound by 11-12 = -1 kcal/mol, probably between +2 and -2

Alternative view of bonding in ozone

Start here with 1-3 diradical

Transfer electron from central doubly occupied pp pair to the R singly occupied pp.

Now can form a p bond the L singly occupied pp.

Hard to estimate strength of bond

Ring ozone

Form 3 OO sigma bonds, but pp pairs overlap

Analog: cis HOOH bond is 51.1-7.6=43.5 kcal/mol. Get total bond of 3*43.5=130.5 which is 11.5 more stable than O2.

Correct for strain due to 60º bond angles = 26 kcal/mol from cyclopropane. Expect ring O3 to be unstable with respect to O2 + O by ~14 kcal/mol,

But if formed it might be rather stable with respect various chemical reactions.

Ab Initio Theoretical Results on the Stability of Cyclic Ozone L. B. Harding and W. A. Goddard III J. Chem. Phys. 67, 2377 (1977) CN 5599

Photochemical smog

High temperature combustion: N2 + O2 2NO

Thus Auto exhaust NO

2 NO + O2 2 NO2

NO2 + hn NO + O

O + O2 + M O3 + M

O3 + NO NO2 + O2

Get equilibrium

Add in hydrocarbons

NO2 + O2 + HC + hn Me(C=O)-OO-NO2

peroxyacetylnitrate

More on N2

The elements N, P, As, Sb, and Bi all have an (ns)2(np)3 configuration, leading to a triple bond

Adding in the (ns) pairs, we show the wavefunction as

This is the VB description of N2, P2, etc. The optimum orbitals of N2 are shown on the next slide.

The MO description of N2 is

Which we can draw as

The configuration for C2

Si2 has this configuration

2

1

1

2

1

2

4

3

4

4

2

2

From 1930-1962 the 3Pu was thought to be the ground state

Now 1Sg+ is ground state

2

2

Ground state of C2

MO configuration

Have two strong p bonds,

but sigma system looks just like Be2 which leads to a bond of ~ 1 kcal/mol

The lobe pair on each Be is activated to form the sigma bond. The net result is no net contribution to bond from sigma electrons. It is as if we started with HCCH and cut off the Hs

C2, Si2,

Re-examine the energy for H2+

For H2+ the VB wavefunctions were

Φg= (хL + хR) and

Φu= (хL - хR) (ignoring normalization)

where H = h + 1/R. This leads to the energy for the bonding state

eg = <L+R|H|L+R>/ <L+R|L+R> = 2 <L|H|L+R>/ 2<L|L+R>

= (hLL + hLR)/(1+S) + 1/R

And for the antibonding state

eu = (hLL - hLR)/(1-S) + 1/R

We find it convenient to rewrite as

eg = (hLL + 1/R) + t/(1+S)

eu = (hLL + 1/R) - t/(1-S)

where t = (hLR - ShLL) includes the terms that dominate the bonding and antibonding character of these 2 states

The VB interference or resonance energy for H2+

The VB wavefunctions for H2+

Φg= (хL + хR) and Φu= (хL - хR) lead to

eg = (hLL + 1/R) + t/(1+S) ≡ ecl + Egx

eu = (hLL + 1/R) - t/(1-S) ≡ ecl + Eux

where t = (hLR - ShLL) is the VB interference or resonance energy and

ecl = (hLL + 1/R) is the classical energy

As shown here the t dominates the bonding and antibonding of these states

Analysis of classical and interference energies

The classical energy, ecl = (hLL + 1/R), is the total energy of the system if the wavefunction is forced to remain an atomic orbital as R is decreased.

The exchange part of the energy is the change in the energy due to QM interference of хL and хR, that is the exchange of electrons between orbitals on the L and R nuclei

The figure shows that ecl is weakly antibonding with little change down to 3 bohr whereas the exchange terms start splitting the g and u states starting at ~ 7 bohr.

Here the bonding of the g state arises solely from the exchange term, egx = t/(1+S) where t is strongly negative, while the exchange term makes the u state hugely repulsive, eux = -t/(1-S)

Analysis of classical and interference energies

egx = t/(1+S) while eux = -t/(1-S)

Consider first very long R, where S~0

Then egx = t while eux = -t

so that the bonding and antibonding effects are similar.

Now consider a distance R=2.5 bohr = 1.32 A near equilbrium

Here S= 0.4583

- = -0.0542 hartree leading to
- egx = -0.0372 hartree while
- eux = + 0.10470 hartree
- ecl = 0.00472 hartree
Where the 1-S term in the denominator makes the u state 3 times as antibonding as the g state is bonding.

Analytic results - details

- Explicit calculations (see appendix A of chapter 2) leads to
- S = [1+R+ R2/3] exp(-R)
- ecl = - ½ + (1 + 1/R) exp(-2R)
- = -[2R/3 – 1/R] exp(-R) – S(1+1/R) exp(-2R)
t ~ -[2R/3 – 1/R] exp(-R) neglecting terms of order exp(-3R)

Thus for long R, t ~ -2S/R

That is, the quantity in t dominating the bond in H2+ is proportional to the overlap between the atomic orbitals.

At long R this leads to a bond energy of the form

- ~ -(2/3) R exp(-R)
That is the bond strength decreases exponentially with R.

- has a minimum at ~ R=2 bohr, which is the optimum R.
But S continues to increase until S=1 at R=0.

Contragradience

хL

хR

The above discussions show that the interference or exchange part of KE dominates the bonding, tKE=KELR –S KELL

This decrease in the KE due to overlapping orbitals is dominated by

tx = ½ [< (хL). ((хR)> - S[< (хL)2>

Dot product is large and negative in the shaded region between atoms, where the L and R orbitals have opposite slope (congragradience)

The VB exchange energies for H2

For H2, the classical energy is slightly attractive, but again the difference between bonding (g) and anti bonding (u) is essentially all due to the exchange term.

1Eg = Ecl + Egx

3Eu = Ecl + Eux

-Ex/(1 - S2)

Each energy is referenced to the value at R=∞, which is

-1 for Ecl, Eu, Eg 0 for Exu and Exg

+Ex/(1 + S2)

Analysis of classical and exchange energies for H2

For H2 the VB energies for the bonding state (g, singlet) and antibonding (u, triplet) states are

1Eg = Ecl + Egx

3Eu = Ecl + Eux

Where Ecl = <ab|H|ab>/<ab|ab> = haa + hbb + Jab + 1/R

Egx = Ex/(1 + S2)

Eux = - Ex/(1 - S2)

where Ex = {(hab + hba) S + Kab –EclS2} = T1 + T2

Here T1 = {(hab + hba) S –(haa + hbb)S2} = 2St contains the 1e part

T2 ={Kab –S2Jab} contains the 2e part

The one electron exchange for H2 leads to

Eg1x ~ +2St /(1 + S2)

Eu1x ~ -2St /(1 - S2)

compared to the H2+ case

egx ~ +t/(1 + S)

eux ~ -t/(1 - S)

Analysis of the VB exchange energy, Ex

where Ex = {(hab + hba) S + Kab –EclS2} = T1 + T2

Here T1 = {(hab + hba) S –(haa + hbb)S2} = 2St

Where t = (hab – Shaa) contains the 1e part

T2 ={Kab –S2Jab} contains the 2e part

Clearly the Ex is dominated by T1 and clearly T1 is dominated by the kinetic part, TKE.

T2

T1

Ex

Thus we can understand bonding by analyzing just the KE part if Ex

TKE

Analysis of the exchange energies

E(hartree)

Eu1x

Eg1x

R(bohr)

The one electron exchange for H2 leads to

Eg1x ~ +2St /(1 + S2)

Eu1x ~ -2St /(1 - S2)

which can be compared to the H2+ case

egx ~ +t/(1 + S)

eux ~ -t/(1 - S)

For R=1.6bohr (near Re), S=0.7 Eg1x ~ 0.94t vs. egx ~ 0.67t

Eu1x ~ -2.75t vs. eux ~ -3.33t

For R=4 bohr, S=0.1

Eg1x ~ 0.20t vs. egx ~ 0.91t

Eu1x ~ -0.20t vs. eux ~ -1.11t

Consider a very small R with S=1. Then

Eg1x ~ 2t vs. egx ~ t/2

so that the 2e bond is twice as strong as the 1e bond but at long R, the 1e bond is stronger than the 2e bond

Van der Waals interactions

For an ideal gas the equation of state is given by pV =nRT

where p = pressure; V = volume of the container

n = number of moles; R = gas constant = NAkB

NA = Avogadro constant; kB = Boltzmann constant

Van der Waals equation of state (1873)

[p + n2a/V2)[V - nb] = nRT

Where a is related to attractions between the particles, (reducing the pressure)

And b is related to a reduced available volume (due to finite size of particles)

London Dispersion

The universal attractive term postulated by van der Waals was explained in terms of QM by Fritz London in 1930

The idea is that even for spherically symmetric atoms such as He, Ne, Ar, Kr, Xe, Rn the QM description will have instantaneous fluctuations in the electron positions that will lead to fluctuating dipole moments that average out to zero. The field due to a dipole falls off as 1/R3 , but since the average dipole is zero the first nonzero contribution is from 2nd order perturbation theory, which scales like

-C/R6 (with higher order terms like 1/R8 and 1/R10)

Consequently it is common to fit the interaction potentials to functional froms with a long range 1/R6 attraction to account for London dispersion (usually refered to as van der Waals attraction) plus a short range repulsive term to acount for short Range Pauli Repulsion)

Noble gas dimers

- LJ 12-6
- E=A/R12 –B/R6
- = De[r-12 – 2r-6]
- = 4 De[t-12 – t-6]
- = R/Re
- = R/s
where s = Re(1/2)1/6

=0.89 Re

Ar2

s

Re

De

Remove an electron from He2

Ψ(He2) = A[(sga)(sgb)(sua)(sub)]= (sg)2(su)2

Two bonding and two antibonding BO= 0

Ψ(He2+) = A[(sga)(sgb)(sua)]= (sg)2(su) BO = ½

Get 2Su+ symmetry.

Bond energy and bond distance similar to H2+, also BO = ½

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