© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L11 Ch120a- Goddard- L01...

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L11 Ch120a-Goddard-

L01

1

Nature of the Chemical Bond with applications to catalysis, materials

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

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

Lecture 13 February 5, 2014

Homonuclear diatomics

Course number: Ch120aHours: 2-3pm Monday, Wednesday, Friday

Teaching Assistants:Sijia Dong <[email protected]>Samantha Johnson <[email protected]>

mailto:[email protected]

mailto:[email protected]

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L12 2

Homonuclear Diatomics


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 orbital (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.


The homonuclear diatomic correlation diagram

Mulliken knew the ordering of the atomic orbitals and considered how combinations of the atomic orbitals would change as the nuclei were pushed together to eventually form a united atom.

First consider the separate atoms limit where there is a large but finite distance R separating the atoms.

The next slide shows the combinations formed from 1s, 2s, and 2p orbitals.


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)


Splitting of levels

General nodal arguments allow us to predict that

But which is lower of

and which is lower of

Here the nodal plane arguments do not help


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 ppWe expect that

bonding

antibonding


Separated atom limitSeparated atoms notationMO notation


Homonuclear Diatomics Molecules – Ne2

compare the VB and MO views


MO view 3 antibonds cancel 3 bonds, thus no net bonding

Generally bond ~ 1/(1+S)

Antibond ~ 1/(1-S)

Where S= overlap. Thus net is antibonding left and right overlap, leading to repulsive interactions and no bonding. 2

2

2

2

2

2

4

4

1s22s22p6=10 e

VB

MO

1s22s22p6=10 e20 e

3 bonds

3 anti bonds


Homonuclear Diatomics Molecules – F2



MO view 2 antibonds cancel 2 bonds, but still net single bond

2

2

2

2

2

0

4

4

1s22s22p5=9 e

VB

MO

1s22s22p5=9 e20 e

3 bonds

2 anti bonds


Homonuclear Diatomics Molecules – O2



MO view 1 antibonds cancel 1 bonds, but still net double bond

But now we have 2 electrons to put into the two pgx and pgy orbitals, can get either singlet or triplet states

Lets examine this in more detail

2

2

2

2

2

0

4

2

1s22s22p5=9 e

VB

MO

1s22s22p5=9 e20 e

3 bonds

1 anti bonds


States based on (p)2

D-

S-

D+

S+

Have 4 spatial combinations

Which we combine aswhere 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

Ground state

MO theory explains the triplet ground state and low lying singlets

xy-yx better than xy+yxBecause no chance for both electrons at same spot (like Hund’s rule on spin of atoms)Triplet lower by Kxy, singlet higher by Kxy

xx-yy better than xx+yyBecause decrease chance for both electrons at same spot (like Hund’s L rule atoms)Triplet lower by Kxy, singlet higher by Kxy



Adding spin we get

(p)2

Ground state 0.0

0.982

1.636

O2

Energy (eV)

MO theory explains the triplet ground state and low lying singlets

MO exact



(p)2

Ground state 0.0

0.982

1.636

MO exact

Spin and angular

momentum forbidden

Spin and parity forbidden

Mulliken knew that these splittings would be ~ 1 eV in infrared region, so he looks near dusk to get the longest path of the sunlight through the atmosphere and saw the transitionImmediately MO big winner and VB big loser


First excited configuration of O2 (pu)3(pg)3

(1pg)2 Ground

(1pg)3(1pu)3 excited

(1pg)3(1pu)3

3Du

1Su-

3Su+

1Du

1Su+

3Su-

Strong transitions (dipole allowed) DS=0 (spin)Sg Su or Pu but S- S-

Only dipole allowed transition from 3Sg

-

18

2

2

2

2

2

0

4

2

MO


The states of O2 molecule

Moss and Goddard JCP 63, 3623 (1975)

(pu)4(pg)2

(pu)3(pg)3


Homonuclear Diatomics Molecules – N2



MO view 0 antibonds

get triple bond

2

2

2

2

2

0

4

0

1s22s22p3=7 e

VB

MO

1s22s22p3=7 e20 e

3 bonds

0 anti bonds

Get singlet state, 1Sg+


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


Homonuclear Diatomics Molecules – N2+


This suggests that N2+

2

2

2

2

2

0

3

0

1s22s22p3=7 e

MO

1s22s22p3=6 e20 e

2.5 bonds

0 anti bonds

VB

2Pu ground state

1

1

3

2Sg+

1st exc. state

2Su+

2nd exc. state

But really ground state is 2Sg+

, 1st exc. State is 2Pu

Something wrong with order of orbitals


United atom limit

Next consider the limit in which the two nuclei are fused together to form a united atom

For N2 this would lead to a Si atom.

Here we get just the normal atomic aufbau states

1s < 2s < 2p < 3s < 3p < 4s,3d < 4p etc

But now we consider an itty bity elongation of the Si nucleus toward two N nuclei and how the atomic states get perturbed

For the 1s orbital all that happens is that the energy goes up (less electron density on the nuclei) and the symmetry becomes sg


2s and 2p united atom orbitals

Similarly 2s just goes to 2sg (and a lower binding)

But the 2p case is more interesting

For the 2ps state the splitting of the nuclei lead to increased density on the nuclei and hence increased binding while for 2pp there is no change in density

Thus 2psu < 2ppu


Summarizing united atom limit

Note for 3d, the splitting is

3ds < 3dp < 3dd

Same argument as for 2p


Summary more united atom levels


Correlation diagram for Carbon row homonuclear diatomics

United atom limit

separated atom limit

F2

O2

O2+

N2C2

N2+


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 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

United atom limit

separated atom limit

F2

O2

O2+

N2C2

N2+


O2 MO configuration

2

4

2

2

2

2

For O2 the ordering of the MOs

Is unambiguous

2

(1pg)2

Next consider states of (1pg)2


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 3Pu was thought to be the ground state. Now exper. Ground state is 1Sg

+ is (0.09ev lower)


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,


Low-lying states of C2


2

2

2

2

2

2

4

4

2

The configuration for Si2

4

1

1

3

1

Si2 has this configuration

C2 has this configuration

Effectively Si2 has lower overlap than C2 (toward Sep. atom limit)

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L12

Do B2, Be2, Li2, Li2+

45


MO and VB view of He dimer, He2

VB view

MO view

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

ΨVB(He2) = A[(La)(Lb)(Ra)(Rb)]= (L)2(R)2

Substitute sg = R + L and sg = R - L

Get ΨMO(He2) ≡ ΨMO(He2)

Net BO=0

Pauli orthog of R to L repulsive


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)


Noble gas dimers

Ar2

Re

De

s

LJ 12-6 Force FieldE=A/R12 –B/R6

= De[r-12 – 2r-6] = 4 De[t-12 – t-6]r = R/Ret = R/swhere s = Re(1/2)1/6

=0.89 Re

No bonding at the VB or MO level

Only simultaneous electron correlation (London attraction) or van der Waals attraction, -C/R6


London Dispersion

The weak binding in He2 and other noble gas dimers was explained in terms of QM by Fritz London in 1930

The idea is that even for a spherically symmetric atoms such as He 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)


London Dispersion

The weak binding in He2 and other nobel gas dimers was explained in terms of QM by Fritz London in 1930



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


Remove an electron from He2 to get 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

MO view


Remove an electron from He2 to get 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

VB view

MO view

Substitute sg = R + L and sg = L - R

Get ΨVB(He2) ≡ A[(La)(Lb)(Ra)] - A[(La)(Rb)(Ra)]

= (L)2(R) - (R)2(L)

-


He2+

He2 Re=3.03A De=0.02 kcal/molNo bond

2Su+

2Sg+

(sg)2(su)

(sg)1(su)2

-

+

BO=0.5

H2 Re=0.74xA De=110.x kcal/molBO = 1.0H2

+ Re=1.06x A De=60.x kcal/molBO = 0.5

Check H2 and H2+ numbers

MO good for discuss spectroscopy,

VB good for discuss chemistry


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) + 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


1s and 2s cases

BA

BA


More on O2

57


Exitation energies (eV) to O2 excited states

vertical


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 ( 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

Temperature (K)

altit

ude

(km

)

tropopause

stratopause

102030

100

50

troposphere

stratosphere

mesosphere

O2 + hn O + O

O + hn O+ + e-

O + O2 O3

O3 + hn 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, 3D

u

1Su-3Su

+

Get emission at ~2.4 eV, 500 nm

Called the nightglow (~ 90 km)


Problem with MO description: dissociation3Sg

- 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

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)]( -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


GVB wavefunction of triplet O2: sigma orbitals

R=4 bohr

(O2sL)2 O2pzL O2pzRbond

R=3 bohr

Re=2.28 bohr

Moss, Bobrowicz, Goddard JCP 63, 4632 (1975)

Get orthogonal to O2s on other center

Causes some (2s-lpz) to stay orthogonal to bond pair

(O2sR)2


GVB wavefunction of triplet O2: pi orbitals

R=4 bohr

(OpxL)2 O2pxR O2pyL

Spin paired

R=3 bohr

Re=2.28 bohr

Moss, Bobrowicz, Goddard JCP 63, 4632 (1975)

Get orthogonal to O2pp on other center

Doubly occupied orbtial delocalizes (bonding)

(OpyR)2


GVB orbitals at Re

Problem: one VB configuration not enough

+


VB description of O2

+

+

+

-

Must have resonance of two VB configurations


Back to valence bond (and GVB)

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


Bond energies

5.2 eV


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

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


Stopped Feb. 5, 2014

75


Add material for O2 + C2H4 (sing and trip)


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 3Sg- 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 s bond = 51.1 kcal/molOO p 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


Add material for O2 + C2H4 (sing and trip)


Bond O2 to O to form ozone

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

Require two OO s bonds get

States with 4, 5, and 6 p pelectrons

Ground state is 4p case

Get S=0,1 but 0 better


sigma GVB orbitals ozone


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

New O-O s bond, 51 kcal/mol

lose O-O p resonance, 51 kcal/mol

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

New singlet coupling of pL and pR 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 p p 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


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)


Noble gas dimers

Ar2

Re

De

s

LJ 12-6 Force FieldE=A/R12 –B/R6

= De[r-12 – 2r-6] = 4 De[t-12 – t-6]r = R/Ret = R/swhere s = Re(1/2)1/6

=0.89 Re

No bonding at the VB or MO level

Only simultaneous electron correlation (London attraction) or van der Waals attraction, -C/R6


London Dispersion

The weak binding in He2 and other noble gas dimers was explained in terms of QM by Fritz London in 1930




London Dispersion

The weak binding in He2 and other nobel gas dimers was explained in terms of QM by Fritz London in 1930



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

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L11 Ch120a- Goddard- L01...

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Transcript of © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L11 Ch120a- Goddard- L01...