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Lecture 13 February 3, 2010PowerPoint Presentation

Lecture 13 February 3, 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

Wednesday Feb. 3, 2pm L13 TODAY(caught up)

Friday Feb. 5, 2pm L14

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

It will be four hour take home, open notes

Reconstruction of (110) surface, side view along [-1,1,0]

Surface As has 3 covalent bonds to Ga, with 2 e in 3s lone pair, relaxes upward until average bond angle is 95º Surface Ga has 3 covalent bonds leaving 0 e in 4th orbital, relaxes downward until average bond angle is 119º. GaAs angle 0º 26º

Si has dangling bond electron at each surface atom

54.7º

54.7º

As

Ga

54.7º

[110]

Si (110)

GaAs (110)

[001]

Reconstruction of GaAs(110) surface, discussion

We consider that bulk GaAs has an average of 3 covalent bonds and one donor acceptor (DA) bond. But at the surface can only make 3 bonds so the weaker DA bond is the one broken to form the surface.

The result is that GaAs cleaves very easily compared to Si. No covalent bonds to break.

As has 3 covalent bonds, leaving 2 electrons in 3s lone pair. AsH3 has average bond angle of 92º. At the GaAs surface As relaxes upward until has average bond angle of 95º

Ga has 3 covalent bonds leaving 0 eletrons in 4th orbital. GaH3 has average bond angle of 120º. At the GaAs surface Ga relaxes downward until has average bond angle of 119º.

This changes the surface Ga-As bond from 0º (parallel to surface to 26º. Observed in LEED experiments and QM calculations

Analysis of charges

Ga

As

Ga

As

Ga

As

Bulk structure: each As has 3 covalent bonds and one Donor-accepter bond(Lewis base – Lewis acid). This requires 3+2=5 electrons from As and 3+0=3 electrons from Ga.

We consider that each bulk GaAs bond has 5/4 e from As and ¾ e form Ga.

Each surface As has 5/4+1+1+2 = 5.25e for a net charge of -0.25 each surface Ga has ¾+1+1+0= 2.75 e for a net charge of +0.25 Thus considering both surface Ga and As, the (110) is neutral

5.25e

2.75e

Net Q =0

0

2

0

2

0

2

1

1

1

1

1

1

1

1

1

1

5/4

3/4

3/4

3/4

5/4

5/4

5/4

3/4

3/4

5/4

5/4

3/4

5/4

3/4

3/4

5/4

5/4

3/4

a

g

a

g

a

g

5/4

3/4

3/4

5/4

5/4

3/4

5/4

3/4

3/4

5/4

5/4

3/4

The GaAs (100) surface, unreconstructed

Every red surface atom is As bonded to two green 2nd layer Ga atoms, but the other two bonds were to two Ga that are now removed. This leaves three non bonding electrons to distribute among the two dangling bond orbitals sticking out of plane (like AsH2)

1st Layer RED

2nd Layer GREEN

3rd Layer ORANGE

4th Layer WHITE

GaAs(100) surface reconstructed (side view)

For the perfect surface, As in top layer, Ga in 2nd layer, As in 3rd layer, Ga in 4th layer etc.

For the unreconstructed surface each As has two bonds and hence three electrons in two nonbonding orbitals.

Expect As atoms to dimerize to form a 3rd bond leaving 2 electrons in nonbonding orbitals.

Surface As-As bonds

As

Ga

As

Ga

As

Ga

As

Ga

Charges for 2x1 GaAs(100)

2

2

2

2

2nd layer ga has 3 e

1

1

Top layer, As

2nd layer, ga

5/4

5/4

3/4

3/4

3/4

3/4

3/4

3/4

3/4

3/4

5/4

2e As-ga bond

5/4

3rd layer, as

1

1

2e As LP

5/4

5/4

Each surface As has extra 0.5 e dimer has extra 1e Not stable

3/4

3/4

3/4

3/4

3/4

1st layer As has 5.5 e

3/4

3/4

2e As-As bond

3/4

Now consider a missing row of As for GaAs(100)

0

0

0

0

1

1

Top layer, As

2nd layer, ga

5/4

5/4

3/4

3/4

3/4

3/4

3/4

3/4

ga empty LP

3rd layer, as

2nd layer ga has 2.25e

Each 2nd layer ga next to missing As is deficient by 0.75e extra 0.5 e 4 ga are missing 3e

3/4

3/4

3/4

1st layer As has 5.5 e

3/4

3/4

3/4

Consider 1 missing As row out of 4

Extra 1e

missing 3e

-1-1-1+3=0 net charge

Extra 1e

Thus based on electron counting expect simplest surface reconstruction to be 4x2. This is observed

Extra 1e

Extra 1e

missing 3e

Different views of GaAs(100)4x2 reconstruction

-1.0e

+1.5e

Two missing As row plus missing Ga row

Exposes 3rd row As

Agrees with experiment

Previous page, 3 As dimer rows then one missing

Hashizume et al Phys Rev B 51, 4200 (1995)

summary

- Postulate of surface electro-neutrality
- Terminating the bulk charges onto the surface layer and considering the lone pairs and broken bonds on the surface should lead to:
- the atomic valence configuration on each surface atom. For example As with 3 covalent bonds and a lone pair and Ga with 3 covalent bonds and an empty fourth orbital
- A neutral surface
- This leads to the permissible surface reconstructions

To be added – band states

IP(P)=4.05 eV

0.054 eV

Remove e from P, add to conduction band = 4.045-4.0 = 0.045 eV

Thus P leads to donor state just 0.045eV below LUMO or CBM

To be added – band states

EA(Al)=5.033 eV

0.045 eV

Add e to Al, from valence band = 5.1 -5.033 = 0.067 eV

Al leads to acceptor state just 0.067eV above HOMO or VBM

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.

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 togtether 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 pp

We expect that

bonding

antibonding

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

Correlation diagram for Carbon row homonuclear diatomics

C2

N2

O2

F2

United atom limit

O2+

separated atom limit

N2+

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+

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

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-

Exitation energies (eV) to O2 excited states

vertical

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

GVB wavefunction of triplet O2: sigma orbitals

(O2sL)2

O2pzL

O2pzR

(O2sR)2

bond

R=4 bohr

R=3 bohr

Get orthogonal to O2s on other center

Re=2.28 bohr

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

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

GVB wavefunction of triplet O2: pi orbitals

Spin paired

(OpxL)2

O2pxR

O2pyL

(OpyR)2

R=4 bohr

R=3 bohr

Get orthogonal to O2pp on other center

Re=2.28 bohr

Doubly occupied orbtial delocalizes (bonding)

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

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

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

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