Lecture 14 February 5, 2010

© copyright 2010 William A. Goddard III, all rights reservedCh120a-Goddard-L14 1

Nature of the Chemical Bond with applications to catalysis, materials

science, nanotechnology, surface science, bioinorganic chemistry, and energy

Lecture 14 February 5, 2010

William A. Goddard, III, [email protected] 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 scheduleFriday Feb. 5, 2pm L14 TODAY(caught up)Midterm given out on Friday. Feb. 5, due on Wed. Feb. 10It will be five hour take home with 30 min. break, open notes


Last time


Separated atom limitSeparated atoms notationMO notation


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 2p better bonding than 2p

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

Since the overlap of p orbitals is obviously higher than pWe expect that

bonding

antibonding


Summarizing united atom limit

Note for 3d, the splitting is

3d < 3d < 3d

Same argument as for 2p


Correlation diagram for Carbon row homonuclear diatomics

United atom limit

separated atom limit

F2

O2

O2+

N2C2

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 1g+


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 1g+ 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

2

4

2

2

2

2

For O2 the ordering of the MOs

Is unambiguous

2

(1g)2

Next consider states of (1g)2


States based on ()2

Have 4 spatial combinations

Which we combine aswhere x and y denote

x and y

φ1, φ2 denote the angle about the axis

and F is independent of φ1, φ2 Rotating about the axis by an angle , these states transform as


States arising from ()2

Adding spin we get

()2

Ground state 0.0

0.982

1.636

O2

Energy (eV)

MO theory explains the triplet ground state and low lying singlets


Using the correleation diagram

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

22

2

2

4

2

4

2

2

Choices for N2


N2 MO configurations

2

4

22

2

2

4

2

4

2

2This is compatible with several orderings of the MOs

Largest R

Smallest R


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 2 states is a 2 state with spin orbital splitting that implies a 3 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

United atom limit

separated atom limit

F2

O2

O2+

N2C2

N2+


1s and 2s cases

BA

BA


AntiBond

BO1

2

2.5

3

2.5

0

1

2


More about O2


First excited configuration

(1g)2 Ground configuration

(1g)3(1u)3 excited configuration

(1g)3(1u)3

u

u-

u+

u

u+

u-

Strong transitions (dipole allowed) S=0 (spin)g u or u but - -

Only dipole allowed transition from 3g

-


The states of O2 molecule

Moss and Goddard JCP 63, 3623 (1975)

(u)4(g)2

(u)3(g)3


Role of O2 in atmosphere

Moss and Goddard JCP 63, 3623 (1975)

StrongGet 3P + 1D O atom

WeakGet 3P + 3P O atom


Implications

UV light > 6 eV (< 1240/6 = 207 nm) can dissociate O2 by excitation of 3u

+ which dissociates to two O atom in 3P state

UV light > ~7.2 eV can dissociate O2 by excitation of 3u-

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

Temperature (K)

altit

ude

(km

)

tropopause

stratopause

102030

100

50

troposphere

stratosphere

mesosphere

O2 + h O + O

O + h O+ + e-

O + O2 O3

O3 + h O + O2

300200

Heated from earth

Heats from light

Heats from light


ionosphere

night

D layer day

Heaviside-Kennelly layerReflects radio waves to allow long distance communications


nightglow

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

The fastest rates are into the Herzberg states, u

u-u

+

Get emission at ~2.4 eV, 500 nm

Called the nightglow (~ 90 km)


Problem with MO description: dissociation3g

- state: [(gx)(gy)+ (gy) (gx)]

As R∞ (gx) (xL – xR) and (gy) (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 bond

Open shell Closed shell

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

bad

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)]()}

Triplet: A{(xL)2(yR)2[(yL)(xR) - (xR)(yL)]()} and

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


VB description of O2

+

+

+

-

Must have resonance of two VB configurations


Bond H to O2

Bring H toward px on Left O

Overlap doubly occupied (xL)2

thus repulsive

Overlap singly occupied (xL)2

thus bonding

2A” state

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


Bond 2nd H to HO2 to form hydrogen peroxide

Bring H toward py on right O

Expect new HOO bond angle ~ 90ºExpect HOOH dihedral ~90ºIndeed 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

HOOH

1.19 kcal/mol Trans barrier

7.6 kcal/mol Cis barrier

HSSH:

5.02 kcal/mol trans barrier

7.54 kcal/mol cis barrier


Compare bond energies (kcal/mol)

O2 3g- 119.0

HO-O

HO-OH 51.1

68.2 H-O2 51.5

HOO-H 85.217.1

67.9 50.8

Interpretation: OO bond = 51.1 kcal/molOO bond = 119.0-51.1=67.9 kcal/mol (resonance)Bonding H to O2 loses 50.8 kcal/mol of resonanceBonding H to HO2 loses the other 17.1 kcal/mol of resonanceIntrinsic 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

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

Require two OO bonds get

States with 4, 5, and 6 pelectrons

Ground state is 4 case

Get S=0,1 but 0 better


sigma GVB orbitals ozone


Pi GVB orbitals ozone

Some delocalization of central Op pair

Increased overlap between L and R Op due to central pair


Bond O2 to O to form ozone

New O-O bond, 51 kcal/mol

lose O-O resonance, 51 kcal/mol

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

New singlet coupling of L and R orbitalsTotal 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 ozoneStart here with 1-3 diradical

Transfer electron from central doubly occupied ppair to the R singly occupied p.

Now can form a bond the L singly occupied p.

Hard to estimate strength of bond


New material


Ring ozone

Form 3 OO sigma bonds, but p 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 + h NO + O

O + O2 + M O3 + M

O3 + NO NO2 + O2

Get equilibrium

Add in hydrocarbons

NO2 + O2 + HC + h 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


GVB orbitals of N2

Re=1.10A

R=1.50A

R=2.10A


Hartree Fock Orbitals N2


2

2

2

2

2

2

4

4

2

The configuration for C2

4

1

1

3

1


2

2

2

2

2

2

4

4

2

The configuration for C2

4

1

1

3

1

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

Now 1g+ is ground state

Si2 has this configuration


Ground state of C2

MO configuration

Have two strong 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,


Low-lying states of C2


Include B2, Be2, Li2, Li2+


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

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

where = (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 tog = (hLL + 1/R) + /(1+S) ≡ ecl + Eg

x

u = (hLL + 1/R) - /(1-S) ≡ ecl + Eux

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

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

As shown here the 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, eg

x = /(1+S) where is strongly negative, while the exchange term makes the u state hugely repulsive, eu

x = -/(1-S)


Analysis of classical and interference energies

egx = /(1+S) while eu

x = -/(1-S)

Consider first very long R, where S~0Then eg

x = while eux = -

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 = 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)

~ -[2R/3 – 1/R] exp(-R) neglecting terms of order exp(-3R)

Thus for long R, ~ -2S/R

That is, the quantity in 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

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

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

Dot product is large and negative

in the shaded region between

atoms, where the L and R orbitals have

opposite slope (congragradience)

x = ½ [< (хL). ((хR)> - S [< (хL)2>

хL хR


The VB exchange energies for H2

1Eg = Ecl + Egx

3Eu = Ecl + Eux

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.

Each energy is referenced to the value at

R=∞, which is

-1 for Ecl, Eu, Eg 0 for Ex

u and Exg

+Ex/(1 + S2)

-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 are1Eg = Ecl + Eg

x

3Eu = Ecl + Eux

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

x = Ex/(1 + S2)

Eux = - Ex/(1 - S2)

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

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

T{Kab –S2Jab} contains the 2e partThe one electron exchange for H2 leads to

Eg1x ~ +2S /(1 + S2)

Eu1x ~ -2S /(1 - S2)

compared to the H2+ case

egx ~ +/(1 + S)

eux ~ -/(1 - S)


Analysis of the VB exchange energy, Ex

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

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

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

T{Kab –S2Jab} contains the 2e part

Clearly the Ex is dominated by T and clearly T is dominated by the kinetic part, T. E

x

T2

T1

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


Analysis of the exchange energies

The one electron exchange for H2 leads to

Eg1x ~ +2S /(1 + S2)

Eu1x ~ -2S /(1 - S2)

which can be compared to the H2

+ case

egx ~ +/(1 + S)

eux ~ -/(1 - S) Eu

1x

Eg1x

R(bohr)

E(hartree)

Consider a very small R with S=1. Then

Eg1x ~ 2vs. eg

x ~

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

For R=1.6bohr (near Re), S=0.7 Eg

1x ~ 0.94vs. egx ~

Eu1x ~ -2.75vs. eu

x ~ For R=4 bohr, S=0.1

Eg1x ~ 0.20vs. eg

x ~

Eu1x ~ -0.20vs. eu

x ~


Van der Waals interactions

For an ideal gas the equation of state is given by pV =nRTwhere p = pressure; V = volume of the containern = 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

Ar2

Re

De

LJ 12-6E=A/R12 –B/R6

= De[12 – 26]De[12 – 6]= R/Re= R/where = Re(1/2)1/6

=0.89 Re


Remove an electron from He2

Ψ(He2) = A[(g)(g)(u)(u)]= (g)2(u)2

Two bonding and two antibonding BO= 0

Ψ(He2+) = A[(g)(g)(u)]= (g)2(u) BO = ½

Get 2u+ symmetry.

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


stop

Lecture 14 February 5, 2010

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