lecture 5,6, january 19, 2011 bonding in nefoncbbe hydrides

1© copyright 2011 William A. Goddard III, all rights reserved Ch120a-Goddard-L5-6

Nature of the Chemical Bond with applications to catalysis, materials

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

Lecture 5,6, January 19, 2011

Bonding in NeFONCBBe hydrides

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]>Caitlin Scott <[email protected]>


Course schedule

Friday January 14: L3 and L4

Monday January 17: Caltech holiday (MLKing)

Wednesday January 19: wag L5 and L6

Friday January 21: wag L7 and L8, caught up

Monday January 24: wag L9

Wednesday January 26: wag L10 and L11

Friday January 28: wag participates in a retreat for our nanotechnology project with UCLA

Back on schedule

Monday January 31: wag L12


Last time


He, 2

Ne, 10

Ar, 18

Zn, 30Kr, 36

Aufbau principle for atoms

Particularly stable atoms, closed shells

Xe, 54

Rn, 86

Uuo, 118


More detailed description of first row atoms

Li: (2s)

Be: (2s)2

B: [Be](2p)1

C: [Be](2p)2

N: [Be](2p)3

O: [Be](2p)4

F: [Be](2p)5

Ne: [Be](2p)6


Summary of ground states of Li-Ne

N 4S (2p)3

O 3P (2p)4

F 2P (2p)5

Ne 1S (2p)6

Li 2S (2s)1

Be 1S (2s)2

B 2P (2p)1

C 3P (2p)2

Ignore (2s)2


Bonding H atom to He

Start with the ground state of He, (1s)2 = A(He1s)(He1s) and bring up an H atom (H1s), leads to

R

HeH: A[(He1s)(He1s)(H1s)]

But properties of A (Pauli Principle) tell us that the H1s must get orthogonal to the He 1s since both have an spin.

A[(He1s)(He1s)(θ)]

Where θ = H1s – S He1s

Consequently θ has a nodal plane, increasing its KE. Smaller R larger S larger increase in KE. Get a repulsive interaction, no bond


Bonding H atom to Ne

Start with the ground state of Ne, (1s)2(2s)2(2p)6

Ψ(Ne)= A{(2px)(2px)(2py)(2py)(2pz)(2pz)} (omitting the Be)

and bring up an H atom (H1s) along the z axis, leads to

A{(2px)2(2py)2(Ne2pz)(Ne2pz)(H1s}

Where we focus on the Ne2pz orbital that overlaps the H atom

R

The properties of A (Pauli Principle) tell us that the H1s must get orthogonal to the Ne 2pz since both have an spin.

θ = H1s – S Ne2pz

θ has a nodal plane, increasing its KE. Smaller R larger S larger increase in KE. Get a repulsive interaction, no bond


Now consider Bonding H atom to all 3 states of F

R

Bring H1s along z axis to F and consider all 3 spatial states.

z

x

F 2pz doubly occupied, thus H1s must get orthogonal repulsive

F 2pz doubly occupied, thus H1s must get orthogonal repulsive

F 2pz singly occupied, Now H1s need not get orthogonal if it has opposite spin, can get bonding

A{(2px)1(2py)2(F2pz)(F2pz)(H1s}

A{(2px)2(2py)1(F2pz)(F2pz)(H1s}


Now consider Bonding H atom to x2y2z1 state of F

z

xFocus on 2pz and H1s singly occupied orbitals

Antibonding state (S=1, triplet)

[φpz(1) φH(2) - φH(1) φpz(2)]()

Just like H2.

Bonding state (S=0, singlet)

[φpz(1) φH(2) + φH(1) φpz(2)]()

(Pz H + H Pz)

(Pz H - H Pz)

energy

R

Full wavefunction for bond becomes

A{(F2px)2(F2py)2[(Fpz)(H+(H)(Fpz

1+

3+

Full wavefunction for antibond becomes

A{(F2px)2(F2py)2[(Fpz)(H-(H)(Fpz


Schematic depiction of HF

Denote the ground state of HF as

Where the line connecting the two singly occupied orbitals covalent bonding

We will not generally be interested in the antibonding state, but if we were it would be denoted as


Bond a 2nd H atom to the ground state of HF?

Starting with the ground state of HF as

can a 2nd H can be bonded covalently, say along the x axis?

This leads to repulsive interactions just as for NeH.

Since all valence orbitals are paired, there are no other possible covalent bonds and H2F is not stable.

A{(F2px)(F2px)(Hx) (F2py)2[(Fpz)(Hz+(Hz)(Fpz

antibond bond


Now consider Bonding H atom to all 3 states of O

R

Bring H1s along z axis to O and consider all 3 spatial states.

O 2pz doubly occupied, thus H1s must get orthogonal repulsive

O 2pz singly occupied.

Now H1s need not get orthogonal if it has opposite spin, can get bonding

Get S= ½ state,

Two degenerate states, denote as 2

z

x


Bonding H atom to x1y2z1 and x2y1z1 states of O

(Pz H + H Pz)

R

z

x

A{(O2px)2(O2py)1[(Opz)(H+(H)(Opz

A{(O2px)1(O2py)2[(Opz)(H+(H)(Opz

The full wavefunction for the bonding state2y

2x

2

bond

bond


Bond a 2nd H atom to the ground state of OH

Starting with the ground state of OH, we can ask whether a 2nd H can be bonded covalently, say along the x axis.

Bonding a 2nd H along the x axis to the 2y state leads to repulsive interactions just as for NeH. No bond.

2y

2xBonding a 2nd H along the x axis to the 2x state leads to a covalent bond

A{(O2py)2[(Opx )(Hx)+(Hx)(Opx[(Opz)(Hz+(Hz)(Opz

z

x

bondbond

A {(O2px)(O2px)(Hx) (O2py)1[(Opz)(H+(H)(Opz

antibondbond


Analize Bond in the ground state of H2O


z

x

bondbond

This state of H2O is a spin singlet state, which we denote as 1A1.

For optimum bonding, the pz orbital should point at the Hz while the px orbital should point at the Hx

Thus the bond angle should be 90º.

In fact the bond angle is far from 90º for H2O, but it does approach 90º for S Se Te

θe Re


What is origin of large distorsion in bond angle of H2O


Bonding Hz to pz leaves the Hz orbital orthogonal to px and py while bonding Hx to px leaves the Hx orbital orthogonal to py and pz, so that there should be little interference in the bonds, except that the Hz orbital can overlap the Hx orbital.

OHx bond OHz bond

Since the spin on Hx is half the time and the other half while the same is true for Hz, then ¼ the time both are and ¼ the time both are . Thus the Pauli Principle (the antisymmetrizer) forces these orbitals to become orthogonal. This increases the energy as the overlap of the 1s orbitals increases. Increasing the bond angle reduces this repulsive interaction

z

x


Testing the origin of bond angle distorsion in H2O

If the increase in bond angle is a response to the overlap of the Hz orbital with the Hx orbital, then it should increase as the H---H distance decreases.

In fact:

Thus the distortion increases as Re decreases, becoming very large for R=1A (H—H of 1.4A, which leads to large overlap ~0.5).To test this interpretation, Emily Carter and wag (1983) carried out calculations of the optimum θe as a function of R for H2O and found that increasing R from 0.96A to 1.34A, decreases in θe by 11.5º (from 106.5º to 95º).

z

x

θe Re


Validation of concept that the bond angle increase is due to

H---H overlap

Although the driving force for distorting the bond angle from 90º to 104.5º is H—H overlap, the increase in the bond angle causes many changes in the wavefunction that can obscure the origin. Thus for the H’s to overlap the O orbitals best, the pz and px orbtials mix in some 2s character, that opens up the angle between them. This causes the O2s orbital to build in p character to remain orthogonal to the bonding orbitals.


Bond a 3rd H atom to the ground state of H2O?

Starting with the ground state of H2O

We can bring a 3rd H along the y axis. This leads to

This leads to repulsive interactions just as for NeH.

Since all valence orbitals are paired, there are no other possible covalent bonds and H3O is not stable.

OHy antibond

A{(O2py)(O2py)(Hy) [(Opx )(Hx)+(Hx)(Opx[(Opz)(Hz+(Hz)(Opz

OHx bond OHz bond


Now consider Bonding H atom to the ground state of N

R

Bring H1s along z axis to O and

N 2pz singly occupied, forms bond to Hz

z

x

A{(N2px)(N2py)[(Npz)(Hz+(Hz)(Npz

NHz bond

Two unpaired spins, thus get S=1, triplet state

Denote at 3-


Bond a 2nd H atom to the ground state of NH

Starting with the ground state of NH, bring a 2nd H along the x axis. This leads to a 2nd covalent bond.A{(N2py)[(Npx )(Hx)+(Hx)(Npx[(Npz)(Hz+(Hz)(Npz

bondbond

z

x

Denote this as 2B1 state.

Again expect 90º bond angle.

θeRe

Indeed as for H2O we find big deviations for the 1st row, but because N is bigger than O, the deviations are smaller.


Bond a 3rd H atom to the ground state of H2N

Starting with the ground state of H2N

We can a 3rd H along the y axis. This leads to

NHy bond

A{[(Npy )(Hy)+(Hy)(Npy[(Npx )(Hx)+(Hx)(Npx[(Npz)(Hz+(Hz)(Npz

NHx bond NHz bond

Denote this as 1A1 state.

Again expect 90º bond angle.

θeRe

Indeed as for H2N we find big deviations for the 1st row, but because there are now 3 bad H---H interactions the deviations are larger.


The ground state of H3N has all valence orbitals are paired, there are no other possible covalent bonds and H4N is not stable.

Bond a 4th H atom to the ground state of H3N?


New material


Now consider Bonding H atom to all 3 states of C

No singly occupied orbital for H to bond with

(2px)(2py)

(2py)(2pz)

Bring H1s along z axis to C and consider all 3 spatial states.

(2px)(2pz)


H1s can get bonding

Get S= ½ state,


z

x


Ground state of CH (2)

z

x

A{(2s)2(OH bond(O2px)1}

The full wavefunction for the bonding state2x

2y

A{(2s)2(OH bond(O2py)1}


Bond a 2nd H atom to the ground state of CH

z

x

Starting with the ground state of CH, we bring a 2nd H along the x axis.

Get a second covalent bond

This leads to a 1A1 state.

No unpaired orbtial for a second covalent bond.


Analyze Bond in the ground state of CH2

z

x

Ground state has 1A1 symmetry. For optimum bonding, the pz orbital should point at the Hz while the px orbital should point at the Hx. Thus the bond angle should be 90º.

As NH2 (103.2º) and OH2 (104.5º), we expect CH2 to have bond angle of ~ 102º

θe Re


But, the Bending potential surface for CH2

3B1

1A1

1B1

3g-

1g

9.3 kcal/mol

The ground state of CH2 is the 3B1 state not 1A1.

Thus something is terribly wrong in our analysis of CH2


Re-examine bonding to Be, B, and C.

Ground state of Be atom: A[(1s)2(2s)2] = A[(1s)(1s)(2s)(2s)]

1.06A2s

R~0.07A0.14A

1s

For the 1s orbital of Be2+, Zeff = 4- 5/16 =3.6875=1/R1s

So R1s = 1/3.69 = 0.27 bohr = 0.14 A For the 2s orbital of Be+, Zeff = 4 – 1.72 = 2.28; hence R2s = 4/2.28=1.75 bohrE[Be+ (2s)1] = - ½ Zeff/R2s = 0.65 h0

For the 2s orbital of Be, Zeff = 4 – 1.72 – 0.29 = 1.99 Thus R2s = n2/Zeff = 2.01 bohr = 1.06AE[Be (2s)2] = 2 (- ½ Zeff/R2s )= -0.99 h

Thus IP(Be) = 0.99 – 0.65 = .34 h0 = 9.25 eV (exper 9.32 eV)


consider the e-e interactions within the (2s)2 pair

Each electron has its maximum amplitude in a spherical shell centered at R2s ~ 1.06 A = 2.01 bohr

Thus the two electrons will on the average be separated by 2*sqrt(2) = 2.8 bohr leading to an ee repulsion of ~1/2.8 hartree= 9.5 eV. This is BIG, comparable to the predicted IP2 a0

2 a0

2 a0

e1

e2

2 a02.8 a0


hybridization of the atomic 2s orbitals of Be.

2s

pz

φ2s + φpz

φ2s - φpz

So far we have assumed the Be wavefunction to be A[(1s)2(2s)2] = A[(1s)(1s)(2s)(2s)]

In fact this is wrong. Writing the wavefunction as A[(1s)(1s)(φa)(φb)] and solving self-consistently (unrestricted Hartree Fock or UHF calculation) for φa and φbleads to

φa = φ2s + φpz and φb = φ2s - φpz

where φpz is like the 2pz orbital of Be+, but with a size like that of 2s rather (smaller than a normal 2p orbital) This pooching or hybridization of the 2s orbitals in opposite directions leads to a much increasedaverage ee distance, dramatically reducing the ee repulsion.


analyze the pooched or hybridized orbitals

z z

zx

zx

φ2s + φpzφ2s - φpzPooching of the

2s orbitals in opposite

directions leads to a dramatic

increase in the ee distance, reducing ee repulsion.

1-D

2-D

Schematic. The line shows symmetric pairing.

Notation: sz and sz bar or ℓ and ℓ bar. Cannot type bars. use zs to show the bar case


Problem with UHF wavefunction for Be

A[(φa)(φb)] = [φa(1)][(φb(2)(2)] – [φb(1)(1)] [φa(2)]

Now φahas spin rather than .

Does not have proper spin or space permutation symmetry.

Combine to form proper singlet and triplet states.1Ψ(1,2) = [φa(1)φb(2)+φb(1)φa(2)][(2)–(1)]3Ψ(1,2) = [φa(1)φb(2)-φb(1)φa(2)][(2)+(1)] and The Generalized Valence Bond (GVB) method was developed to optimize wavefunctions of this form. The result is qualitatively the same as UHF, but now the wavefunction is a proper singlet.

I do not have handy a plot of these GVB orbitals for Be but there are similar to the analogous orbtials for Si, which are shown next

Interchange s1 and s2: get

A[(φa)(φb)] = [φa(1)][(φb(2)(2)] – [φb(1)(1)] [φa(2)]


The GVB orbitals for the (3s)2 pair of Si atom

Long dashes indicate zero amplitude, solid lines indicate positive amplitude while short dashes indicate negative amplitude. The spacing between contours is 0.05 in atomic units


Analysis of the GVB singlet wavefunction1Ψ(1,2) = [φa(1)φb(2)+φb(1)φa(2)][(2)–(1)]

Substituting φa = φ2s + φpz and φb = φ2s - φpz into the spatial factor leads to

(ab+ba) = (s+z)(s-z)+(s-z)(s+z) = [s(1)s(2) - 2 z(1)z(2)]

(ignoring normalization), which we will refer to as the CI form (for configuration interaction).


Analysis of the GVB singlet wavefunction1Ψ(1,2) = [φa(1)φb(2)+φb(1)φa(2)][(2)–(1)]


(ab+ba) = (s+z)(s-z)+(s-z)(s+z) = [s(1)s(2) - 2 z(1)z(2)]

(ignoring normalization), which we will refer to as the CI form (for configuration interaction).

In the GVB wavefunction it is clear from the shape of the sz and zs wavefunctions that the average distance between the electrons is dramatically increased. This is a little more complicated to see in the CI form.

Consider two electrons a distance R from the nucleus. Then the probability for the two electrons to be on the same side is s(R)s(R)-2 z(R)z(R) which is smaller than s(R)s(R) while the probability of being on opposite sides is s(R)s(-R)- 2 z(R)z(-R) = s(R)s(R)+2 z(R)z(R) which is increased.


Analysis of the GVB triplet wavefunction3Ψ(1,2) = [φa(1)φb(2)-φb(1)φa(2)][(2)+(1)] and


(ab-ba) = (s+z)(s-z)-(s-z)(s+z) = [s(1)z(2)-z(1)s(2)]

(ignoring normalization).

This is just the wavefunction for the triplet state formed by exciting the 2s electron to 2pz, which is very high (xx eV).

Thus we are interested only in the singlet pairing of the two lobe or hybridized orbitals. This is indicated by the line pairing the two lobe functionsBottom line: 2s orbitals of (1s)2(2s)2 state of Be hybridize in ±z direction in order to reduce electron repulsion (as the cost of promotion to the 2p orbital 16% of the time)


Role of pooched or hybridized atomic lobe orbitals in bonding of BeH+

Consider the bonding of H to Be+

The simple VB combination of H1s with the 2s orbital of Be+ leads to a very small overlap and contragradience

In fact optimizing the wavefunction for BeH+ leads to pooching of the 2s toward the H1s with much improved overlap and contragradience.


Thus the wave function is A{[(sz)(zs)+(zs)(sz)]()(H

where sz≡(s+z) and zs ≡(s-z)

Here the H overlaps slightly more with sz than with zs, but the spin on sz is half the time

Thus at large R we obtain a slightly repulsive interaction.

Role of pooched or hybridized atomic lobe orbitals in bonding of BeH neutral

At large R the the orbitals of Be are already hybridized

Hszzs


At small R the H can overlap significantly more with sz than with zH, so that we can form a bond pair just like in BeH+. This leads to the wavefunction







Hszzs

Hszzs

A{[(sz)(H)+(H)(sz)]()(zsIn which the zs hybrid must now get orthogonal to the sz and H bond pair. This weakens the bond from that of BeH+ by ~ 1 eV


More analysis of the GVB singlet wavefunction1Ψ(1,2) = [φa(1)φb(2)+φb(1)φa(2)][(2)-(1)]

where φa = φ2s + φpz and φb = φ2s - φpz

The optimum value of it is 0.376 for Si) which leads to a significant increase in the average ee distance, but from the CI expansion

[s(1)s(2) - 2 z(1)z(2)]/sqrt(1+4)

We see that the wave function is still 86% (2s)2 character. This is expected since promotion of 2s to 2p costs a significant amount in the one electron energy. This promotion energy limits the size of

Normalizing the GVB orbitals leads to φa = (φ2s + φpz)/sqrt(1+2)

Thus the overlap of the GVB pair is

< φa | φa > = (1-2)/(1+2)=0.752,

similar to a bond pair


Problem with the GVB wavefunction

A problem with this simple GVB wavefunction is that it does not have the spherical symmetry of the 1S ground state of Be.

This problem is easily fixed in the configuration interaction (CI) form by generalizing to

{s(1)s(2) - 2 [z(1)z(2)+x(1)x(2)+y(1)y(2)]}

which does have 1S symmetry. Here the value of 2 ~ 2/3

This CI wavefunction can be solved for self-consistently. It is referred to as MC-SCF for multiconfiguration self-consistent field. But the simple GVB description of independent electrons in each orbital is obscured.


Contragradience

We saw that the bonding in H2+ and H2 was dominated by

the decrease in KE as the atoms were brought together.

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

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

½ [< (хL). ((хR)>

хL хR

skip









Hszzs

Hszzs

A{[(sz)(H)+(H)(sz)]()(zsIn which the zs hybrid must now get orthogonal to the sz and H bond pair. This weakens the bond from that of BeH+ by ~ 1 eV


Compare bonding in BeH+ and BeH

BeH+

BeH

3 eV

2 eV

1 eV

Long range Repulsive interaction

with H

Short range Attractive interaction sz with H

1eV Repulsive orthogonalization of

zs with sz H

BeH+ has long range attraction no short

range repulsion

TA’s check numbers, all from memory


Now bond 2nd H to BeH

Expect linear bond in H-Be-H and much stronger than the 1st bond

Expect bond energy similar to BeH+, maybe stronger, because the zs orbital is already decoupled from the sz.

1+

Cannot bind 3rd H because no singly occupied orbitals left.

~3.1 eV


Compare bonding in BeH and BeH2

BeH+

TA’s check numbers, all from memory

MgH+

3.1 eVR=1.31A

1.34 eVR=1.73A

Expect linear bond in H-Be-H and much stronger than the 1st bond

Expect bond energy similar to BeH+, maybe stronger, because the zs orbital is already decoupled from the sz.

1+

2+

1+

linear

Cannot bind 3rd H because no singly occupied orbitals left.

2.1 eVR=1.65 A

2.03 eVR=1.34A

~3.1 eV ~2.1 e


The ground state for C atom

Ψyz=A[(sx)(xs)+(xs)(sx)]()(y)(z)] which we visualize as

z

x

Based on our study of Be, we expect that the ground state of C is

2s pair pooched +x and –x

yz open shell pz

sx

xs

py

2s pair pooched +z and –z

xy open shell

px

szzs

py

Ψyx=A[(sz)(zs)+(zs)(sz)]()(y)(x)] which we visualize as

Ψxz=A[(sy)(ys)+(ys)(sy)]()(x)(z)] which we visualize aspx

pz

sy,ys

2s pair pooched +y and –y

xz open shell


The GVB orbitals of Silicon atom

Long dashes indicate zero amplitude, solid lines indicate positive amplitude while short dashes indicate negative amplitude. The spacing between contours is 0.05 in atomic units


Now reconsider Bonding H atom to all 3 states of C

Now we can get a bond to the lobe orbital just as for BeH


(2px)(2pz)

C 2pz singly occupied.

H1s can get bonding

Get S= ½ state,


(2px)(2py)

(2py)(2pz)


Is the 2 state actually 2

The presence of the lobe orbitals might seem to complicate the symmetryΨyz=A[(sx)(xs)+(xs)(sx)]()(y) (zbond)2]

Ψxz=A[(sy)(ys)+(ys)(sy)]()(x)(zbond)2)]

To see that there is no problem, rewrite in the CI form (and ignore the zH bond)Ψyz=A[(s2 – x2)]()(y)]

Now form a new wavefunction by adding - y2 to Ψyz

Φyz ≡ A[s2 – x2 – y2]()(y)]

But the 3rd term is A[y2]()(y)]= – A[(yy(y)]=0

Thus Φyz = Ψyz and similarly

Φxz = A[s2 – x2 – y2]()(x)] = Ψxz

Thus the 2s term [s2 – x2 – y2] is clearly symmetric about the z axis, so that these wavefunctions have 2symmetry

skip


Bonding of H to lobe orbital of C, Long R

Thus the wave function is A{(px)(py)[(sz)(zs)+(zs)(sz)]()(H

At large R the lobe orbitals of C are already hybridized

2s pair pooched +z and –z

xy open shell

Unpaired H




Bonding of H to lobe orbital of C, small R


A{[(sz)(H)+(H)(sz)]()(zspxpy

But now the zs hybrid must now get orthogonal to the sz and H bond pair. This destabilizes the bond by ~ 1 eV

The symmetries of the nonbond orbitals are: zs=, px=x, py=y

Since the nonbond orbitals, , x, y are orthogonal to each other the high spin is lowest (S=3/2 or quartet state)

We saw for NH that (xy –yx)() has 3- symmetry. CH has one additional high spin nonbond orbital, leading to 4-

Hszzs

pxpy

Sz-H bond pair nonbond orbitals


GVB orbitals of SiH 4- state


GVB orbitals of SiH 2 state

H

sx

xs

pz

py


The bonding states of CH and SiH

The low-lying state of SiH are shown at the right. Similar results are obtained for CH.

The bond to the p orbital to form the 2 state is best

CH SiH

De(2) 80.0 70.1

Kcal/mol

(24-) De(4-) 62.9 33.9

17.1 36.2

p bond

sz bond

The bond to the lobe orbital is weaker than the p, but it is certainly significant


Analysis of bonding in CH and SiH

180º

The reason is that as the pH bond is formed, the incoming H orbital overlaps the 2s part of the lobe orbital. To remain orthogonal, the 2s orbital builds in some –z character along with the x character already there. This rotates the lobe orbital away from the incoming H. This destabilizes the lobe pair making it easier for the 2nd H to bond to the lobe pair.

104º

Bond to p orbital is still the best for C and Si but the lobe bond is also quite strong, especially for CH

Thus hydridization in the atom due to electron correlation leads naturally to the new 4- bonding state.

Note that although the (sx)(xs) lobe pair for the atom are at 180º in the atom, they bend to ~104º for CH and SiH


Bonding the 2nd H to CH and SiH

As usual, we start with both components of the ground state of CH or SiH, 2x and 2y and bond bring an H along some axis, say x.

H

sx

xs

pz

py

2y

2x


Bonding the 2nd H to CH and SiH

Now we get credible bound states from both components

H

sx

xs

pz

py

2y

2x

A bond to the sx lobe orbital of CH (2y)

A bond to the p orbital of CH (2x)

This leads to the 1A1 state of CH2 and SiH2 that has already been discussed.

This leads to the 3B1 state that is the ground state of

CH2


The p bond leads to the 1A1 state

GVB orbitals for SiH(1A1)

Ψ=A{[(sy)(ys)+(ys)(sy)(SiHLbond)2(SiHRbond)2}

The wavefunction is

Applying C2z or in the plane interchanges (sy) and (ys) but the (sy)(ys) pair is symmetric under this interchange. Thus the total symmetry is 1A1.


Bonding the 2nd H to the lobe orbital

At large distances the bond to the lobe orbital will be slightly repulsive and at an angle of 128 to the already formed p bond.

However at short distances, we form a strong bond. After forming the bond, each bond pair readjusts to have equivalent character (but an average of lobe bond and p bond).

The wavefunction becomes

Ψ=A{(SiHLbond)2(SiHRbond)2 [(ℓx}

ℓx

Here the two bond pairs and the ℓ orbital have A1 while x has b1 symmetry so that the total spatial symmetry is B1.

This leads to both 3B1 and 1B1 states, but triplet is lower (since ℓ and x are orthogonal).


Analyze Bond angles of CH2 and SiH2θe Re

1A1 state. Optimum bonding, pz orbital points at Hz while px orbital points at Hx, leading to a bond angle of 90º. We expect that HH repulsion increases this by slightly less than the 13.2º for NH2 and 14.5º for OH2 and indeed it increases by 12.4º. But for Si the increase from 90º is only 2º as for P and As.

θe Re3B1 state. Optimum bonding, the two bonds at ~128º. Here HH orthogonalization should increase this a bit but Cmuch less than 12º since H’s are much farther apart. However now the ℓ orbital must get orthogonal to the two bond pairs a bond angle decrease. The lone pair affect dominates for SiH2 decreasing the bond angle by 10º to 118º while the HH affect dominates for CH2, increasing the bond angle by 5º to 133º

ℓ


The GVB orbitals for SiH2 (3B1)


Analysis of bond energies of 1A1 state

Consider the first two p bonds. Ignoring the affect of the bonds on the lobe orbitals, the main difference arises from the exchange terms.

For C or Si A[(2s)2(pz)(px)] leads to a term in the energy of the form (Jxz –Kxz) since the x and z orbitals have the same spin.

But upon bonding the first H to pz, the wavefunction becomes A{(2s)2[(pzH+Hpx)()(px)}. Now the pz and px orbitals have the same spin only have the time, so that this exchange term is decreased to - ½ Kxz.

However in forming the second bond, there is no additional correction.

Since Kxz ~ 14 kcal/mol for C and ~9 kcal/mol for Si. This means that the 2nd bond should be stronger than the first by 7 kcal/mol for C and by 4.5 kcal/mol for Si. E(kcal/mol)1st bond 2nd bond

C 80 90Si 70 76.2

This is close to the observed differences.


Analysis of bond energies of CH2 and SiH2 state

2

4-

CH

62.9ℓ

SiH

3P

80.0p

2

4-33.9ℓ

70.1p

3B1

1A1 90.0p

99.1ℓ

3B1

1A1

56.9ℓ

76.1p

CH2

SiH2

CH Lobe bonds: 63 and 9950% increase

SiH Lobe bonds: 35 and 5750% increase

We find that forming the first bond to the p orbital, destabilizes the lobe pair so that bonding to the lobe orbital of XH is 50% stronger than bonding to lobe orbital of X


The Bending potential surface for CH2

3B1

1A1

1B1

3g-

1g

9.3 kcal/mol



Analysis of bond energies of 3B1 state

For CH the lobe bond is 17 kcal/mol weaker than the p bond while for SiH it is 37 kcal/mol weaker.

Forming the lobe bond requires unpairing the lobe pair which is ~1 eV for the C row and ~1.5 eV for the Si row. This accounts for the main differences suggesting that p bonds and lobe bonds are otherwise similar in energy.

Forming a lobe bond to CH or SiH should be easier than to C or Si, because the first p bond has already partially destabilized the

lobe pair. Since the SiH2(3B1) state is 19 kcal/mol higher than SiH2(1A1) but SiH(4-) is 35 kcal/mol higher than Si(2), we conclude that lobe bond has increased in strength by ~16 kcal/mol

Indeed for CH the 3B1 state is 9.3 kcal/mol lower than 1A1 implying that the lobe bond has increased in strength relative to the p bond by 26 kcal/mol.


CH2 GVB orbitals


Add 3rd H to form CH3

120º

133º

For CH2 we start with the 3B1 state and add H. Clearly the best place is in the plane, leading to planar CH3, as observed.

As this 3rd bond is formed, each bond pair readjusts to a mixture of p and lobe character to become equivalent ~sp2 orbitals

We could also make a bond to the out-of-plane p orbital but this would lead to large HH repulsions.

However the possibility of favorable out-of-plane bonding leads to an extremely flat potential curve for CH3.

Since the lobe orbital is already unpaired, we get a very strong bond energy of 109 kcal/mol.


Add 3rd H to 1A1 SiH2 to form SiH3.

For SiH2 we start with the 1A1 state and add H to the lobe pair. Clearly this leads to a pyramidal SiH3. Indeed the optimum bond angle is 111º. Since we must unpair the lobe orbital this 3rd bond is relatively weak, 72 kcal/mol.


Add 4th H to form CH4 and SiH4.For CH3 we start with the planar molecule and bring the H up to the out of plane p orbital. As the new bond forms all four bonds readjust to become equivalent leading to a tetrahedral CH4 molecule. This bond is 105 kcal/mol, slightly weaker than the 3rd 109 kcal/mol) since it is to a p orbital and must interact with the other H’s.


Add 4th H to form CH4 and SiH4.For CH3 we start with the planar molecule and bring the H up to the out of plane p orbital. As the new bond forms all four bonds readjust to become equivalent leading to a tetrahedral CH4 molecule. This bond is 105 kcal/mol, slightly weaker than the 3rd 109 kcal/mol) since it is to a p orbital and must interact with the other H’s.For SiH2 we start with the pyramidal geometry (111º bond angle) and add to the remaining lobe orbital. As the new bond forms, all fourbonds readjust to become equivalent, leading to a tetrahedral SiH4 molecule. No unpairing is required a strong bond, 92 kcal/mol (the 3rd bond was 72)


GVB orbitals of CH3 and CH4


GVB orbitals of SiH3 SiH bond pair

Dangling bond orbital


GVB orbitals of SiH4


Hybridization of GVB Orbitals

Idealized case.Tetrahedral: sp3

Orthogonal and point to vertices of a tetrahedron.

Rationalizes bonding in CH4.

Assumes 75% p character

Tepe

ratu

re

x

CubicI-43m

TetragonalI4cm

RhombohedralR3m

OrthorhombicPmn21

y

z

o

FE AFE/

FE AFE/

FE AFE/

Px Py Pz

+ +

+ +

+ +

+ +

=

=

=

=

MacroscopicPolarization

Ti atom distortions

=

=

=

=

Microscopic Polarization

(s-x+y-z)/2

(s+x+y+z)/2(s-x-y+z)/2

(s+x-y-z)/2

GVB: CH4 is 70% p

Atom: lobe is 13%p: total =226/4=57% p


Comparisons of successive bond energies SiHn and CHn

p lobe

p

lobe

p

p

lobe

lobe

The numbers for Si changed slightly between 1980 and now. These are the older numbers, I think


Now do the bonding to B atom

Ψyz=A[(sx)(xs)+(xs)(sx)]()(z)] which we visualize as

z

x

Based on our study of C, we expect that the ground state of B is

2s pair pooched ±x

(or ±y)z open shell pz

sx

xs

2s pair pooched ± z x open shell

px

szzs

Ψyx=A[(sz)(zs)+(zs)(sz)]()(x)] which we visualize as

Ψxz=A[(sz)(zs)+(zs)(sz)]()(y)] which we visualize as

2s pair pooched ± z y open shell szpy

zs


form BH and AlH by bonding along the z axis

Ψyz=A[(sx)(xs)pair](zz]Bonding to the pz state of B we obtain

2s pair pooched ±x

(or ±y)

H-z covalent bond

pz

sx

xs

H-sz covalent bond open shell

pxszzs

Ψyx=A[(sz)(H)+(H)(sz)]()(x)(zs]

Ψxz yx=A[(sz)(H)+(H)(sz)]()(y)(zs]

open shell szpy

zs

128º

H

H

HH-sz covalent bond

1+

3x

3y


The bonding states of BH and AlH

The ground state of BH and AlH is obtained by bonding to the p orbital (leading to the 1+ state.

However the bond to the lobe orbital (leading to the 3stateis also quite strong.

The bond to the lobe orbital is weaker than the p, but it is certainly significant

1+

3

1

3+

2P+ H


BH2 and AlH2

128º

Starting from the ground 1+ state of BH or AlH, the second bond is to a lobe orbital, to form the 2A1 state.

Just as for the 3B1 state of CH2 and SiH2 the bond for BH2 opens by several degrees to 131º while the bond to the AlH2 closes down by ~9º.

θe Re


BH3 and AlH3

Bonding the 3rd H to the 2A1 state of BH2, leads to planar BH3 or AlH3.

Now all bond become equivalent, ~sp2.

But there is no 4th bond since there remain no additional unpaired orbitals to bond to.


Re-examine bonding in NH, OH, and FH

Why did we ignore hybridization of the (2s) pair for NH, OH, and FH?

The reason is that the ground state of N atom

A[(2s)(2s)(x)(y)(z)]

Already has occupied px,py,pz orbitals. Thus

Pauli annihilates any hybridization in the 2s orbital.


Re-examine bonding in NH, OH, and FH

However, the doublet excited state of N can have hybridization, eg

A(2s)2(y)2(z) A[(sx)(xs)+(xs)(sx)]()(y)2(z) which leads to the 2A1 excited states of NH2 of the form

θe Re


Bonding to halides (AXn, for X=F,Cl,…

The ground state of F has just one singly-occuped orbital and hence bonding to C is in many ways similar to H, leading to CF, CF2, CF3, and CF4 species. However there are significant differences.

Thus CF leads to two type so bonds, p and lobe just like CH

2

4

p bond

lobe bond

Covalent bond expect (CH)

80 kcal/mol

63 kcal/mol

actual bond (CF)

120 kcal/mol

63 kcal/mol


How can CF lead to such a strong bond, 120 vs 80 kcal/mol?

Consider the possible role of ionic character in the bonding

In the extreme limit

+

C F C+ F-

E (R=∞) = 0 E (R=∞) =IP (C) – EA (F)

=11.3 – 3.4 =7.9 eV

IP (C) = 11.3 eV = 260 kcal/mol EA (F) = 3.40 eV = 78.4 kcal/molCan Coulomb attraction make up this difference?


Estimate energy of pure ionic bond for 2 CF

Covalent limit (C + F)

Ionic limit (C+ + F-)

7.9 eV14.4/R

Re=1.27A

14.4/1.27 = 11.3 eV

Net bond = 11.3-7.9 = 3.4 eV= 78 kcal/mol

Units for electrostatic interactions E=Q1*Q2/(0*R)where 0 converts units (called permittivity of free space)E(eV) = 14.4 Q1(e units)*Q2(e units)/R(angstrom)E(kcal/mol) = 332.06 Q1(e units)*Q2(e units)/R(angstrom)

Ionic estimate ignores shielding and pauli repulsion for small R. Thus too large


CF has strong mixture of covalent and ionic character

+

Pure covalent bond ~ 80 kcal/mol (based on CH)

Pure ionic bond ~ 78 kcal/mol (ignore Pauli and shielding)

Net bond = 120 kcal/mol is plausible for 2 state

But why is the bond for the 4- state only 63, same as for covalent bond?


Consider ionic contribution to 4state

+

C F C+ F-

E (R=∞) = 0 E (R=∞) =IP (C) – EA (F)

=16.6 – 3.4 =13.2 eV

To mix ionic character into the 4state the electron must be pulled from the sz lobe orbital.

This leads to the (2s)1(2p)2 state of C+ rather than the ground state (2s)2(2p)1 which is 123 kcal/mol = 5.3 eV higher

Thus ionic bond is NEGATIVE (78-123=-45 kcal/mol)

Thus F only likes to bond to p orbitals, because only in this case can it gain ionic character


Bonding the 2nd F to CFWith the 4- state at 57 kcal/mol higher than , we need only consider bonding to 2, leading to the 1A1 state.

Bad Pauli repulsion increases FCF angle to 105º (CH2 was 102º)

1A1

3B1

57 kcal/mol


Now bond 3rd F to form CF3

Get pyramidal CF3 (FCF angle ~ 112º)

In sharp contrast to planar CH3

The 3rd CF bond should be much weaker than 1st two.

This strong preference for CF to use p character makes conjugated flourocarbons much less stable than corresponding saturated compounds.

Thus C4H6 prefers butadiene but C4F6 prefers cylcobutene

Of course the 4th bond to form CF4 leads to a tetrahedral structure


Summary, bonding to form hydrides

General principle: start with ground state of AHn and add H to form the ground state of AHn+1

Thus use 1A1 AH2 for SiH2 and CF2 get pyramidal AH3

Use 3B1 for CH2 get planar AH3.

For less than half filled p shell, the presence of empty p orbitals allows the atom to reduce electron correlation of the (ns) pair by hybridizing into this empty orbital.

This has remarkable consequences on the states of the Be, B, and C columns.


stopped


The role of symmetry in QM

In this course we are concerned with the solutions of the Schrodinger equation, HΨ=EΨ, but we do actually want to solve this equation.

Instead we want to extract the maximum information about the solutions without solving it.

Symmetry provides a powerful tool for doing this.

Some transformation R1 is called a symmetry transformation if it has the property that R1 (HΨ)=H(R1Ψ)

The set of all possible symmetries transformations of H are collected into what is called a Group.


The definition of a Group

1). Closure: If R1,R2 G (both are symmetry transformations) then R2 R1 is also a symmetry transformation, R2 R1 G

2. Identity. The do-nothing operator or identity, R1 = e G is clearly is a symmetry transformation

3. Associativity. If (R1R2)R3 =R1(R2R3).

4. Inverse. If R1 G then the inverse, (R1)-1 G ,where the inverse is defined as (R1)-1R1

= e.


The degenerate eigenfunctions of H form a representation

If HΨ=EΨ then H(R1Ψ)= E(R1Ψ) for all symmetry transformations of H.

Thus the transformations amount the n denegerate functions, {S=(RiΨ), where Ri Ψi G} lead to a set of matrices that multiply in the same way at the group operators. The Mathematicians say that these functions form a basis for a representation of G. Of course the functions in S may not all be different, so that this representation can be reduced. The mathematicians went on to show that one could derive a set of irreducible representations that give all possible symmetries for the H. reorientations from which one can construct any possible.


Example, an atom.

For an atom any rotations about any axis passing through the nucleus is a symmetry transformation. This leads to the group denoted as SO(3) by the mathematicians [O(3) indicates 3 three-dimensional real space, S because the inversion is not included).

The irreducible representations of O(3) are labeled as

S (non degenerate) and referred to as L=0

P (3 fold degenerate) and referred to as L=1

D (5 fold degenerate) and referred to as L=2

F (7 fold degenerate) and referred to as L=3

G (9 fold degenerate) and referred to as L=4


H2O, an example of C2v

consider the nonlinear H2A molecule, with equal bond lengths, e.g. H2O, CH2, NH2

The symmetry transformations are

1. e for einheit (unity) xx, yy, zz

2. C2z, rotation about the z axis by 2/2=180º, x-x, y-y, zz

3 xz, reflection in the xz plane, xx, y-y, zz

yz, reflection in the yz plane, x-x, yy, zz

Which is denoted as the C2v group.


Stereographic projections

Consider the stereographic projection of the points on the surface of a sphere onto a plane, where positive x are circles and negative x are squares.

Start with a general point, denoted as e and follow where it goes on various symmetry operations.

This make relations between the symmetry elements transparent.

e.g. C2zxz= yz

Combine these as below to show the relationships

xy e

C2z

xz

yz

C2z

yz

xz

xz

C2z

yz

xy e

xz

C2z

yz

C2v


The character table for C2v

Since C2zxz= yz the symmetries for yz are already implied by C2zxz. Thus there are only 4 possible symmetries.

Since (C2z)2 = e, (xz)2 = e, (yz)2 = eWe expect wavefunctions to be ±1 under each operationname

1-e N-ea1

a2

b1

b2


Symmetries for NH2

Ψ1 = A{(N2py)[(Npx )(Hx)+(Hx)(Npx[(Npz)(Hz+(Hz)(Npz

NHz bondNHx bond

Applying xz to the wavefunction leads to

Ψ2 = A{(N2py)0[(NpR)(HR)+(HR)(NpR[(NpL)(HL+(HL)(NpL

Each term involves transposing two pairs of electrons, e.g., 13 and 24 as interchanging electrons.

Since each interchange leads to a sign change we find that xzΨ1 = Ψ2 = Ψ1

Thus interchanging a bond pair leaves Ψ invariant

Also the N(1s)2 and (2s)2 pairs are invariant under all operations


NH2 symmetry continued

Consider now the singly occupied Npx orbitalC2z changes the sign but, xz does not. Thus Npx transforms as b1 andThe total wavefunction transforms as B1.Include the S= 1/2 , leads to The 2B1 state

1-e N-ea1

a2

b1

b2


Symmetries for H2O and CH2

H2O NH2


OHx bond OHz bond

H2O

A{(C2py)0[(Cpx )(Hx)+(Hx)(Cpx[(Cpz)(Hz+(Hz)(CpzCH2

CH2

Since have enen number electrons in 2py, wavefunction is invarient under all symmetry transformations, thus must be 1A1.


Now do triplet state of CH2

Soon we will consider the triplet state of CH2 in which one of the 2s nonbonding electrons (denoted as to indicate symmetric with respect to the plane of the molecule) is excited to the 2px orbital (denoted as to indicate antisymmetric with respect to the plane)A{(C2s2px)1[(CpL )(HL)+(HL)(CpL[(CpR)(HR+(HR)(CpR

CHR bondCHL bond

Thus the symmetry of triplet CH2 is 3B1

Since we know that the two CH bonds are invariant under all symmetry operations, from now on we will write the wavefunction as

A{[(CHL(CHRCC)1

Here is invariant (a1) while transforms as b1. Since both s and p are unpaired the ground state is triplet or S=1

2s

zy

2px


Second example, C3v, with NH3 as the prototype

We will consider a system such as NH3, with three equal bond lengths. Here we will take the z axis as the symmetry axis and will have one H in the xz plane.

The other two NH bonds will be denoted as b and c.

NHb bond

A{[(Npy )(Hy)+(Hy)(Npy[(Npx )(Hx)+(Hx)(Npx[(Npz)(Hz+(Hz)(Npz

NHx bond NHc bond

zx


x

b

c

x

b

c

Symmetry elements for C3v

Take the z axis out of the plane. The six symmetry operations are:

eC3

x

b

c C32= C3

-1

x

b

c

x

b

c

xz xzC3

x

b

c

xzC32

zx


The C3v symmetry group

x

b

c

e

C3

C32= C3

-1

xz

xzC3

xzC32

The xzC3 transformation corresponds to a reflection in the bz plane (which is rotated by C3 from the xz plane) and the xzC3

2

transformation corresponds to a reflection in the cz plane (which is rotated by C3

2 from the xz plane). Thus these 3 reflections are said to belong to the same class.

Since {C3 and C3-1 do similar

things and are converted into each other by xz we say that they are in the same class.


The E symmetry (irreducible representation) is of degree 2, which means that if φpx is an eigenfunction of the Hamiltonian, the so is φpy and they are degenerate.

This set of degenerate functions would be denoted as {ex,ey} and said to belong to the E irreducible representation.

The characters in this table are used to analyze the symmetries, but we will not make use of this until much later in the course.

The character table for C3v

Thus an atom in a P state, say C(3P) at a site with C3v symmetry, would generally split into 2 levels, {3Px and 3Py} of 3E symmetry and 3Pz of 3A1 symmetry.


Application for C3v, NH3

=A{(LP)2[(NHb bond(NHc bond(NHx bond

zx

Consider first the effect of xz. This leaves the NHx bond pair invariant but it interchanges the NHb and NHc bond pairs. Since the interchange two pairs of electrons the wavefunction does not change sign. Also the LP orbital is invariant.

We will write the wavefunction for NH3 as

where we combined the 3 Valence bond wavefunctions in 3 pair functions and we denote what started as the 2s pair as LP

x

cb

LP


Next consider the C3 symmetry operator. It does not change LP. It moves the NHx bond pair into the NHb pair, moves the NHb pair into the NHc pair and moves the NHc pair into the NHx pair. A cyclic permutation on three electrons can be written as (135) = (13)(35). For example.

φ(1)φ(3)φ(5) φ(3)φ(5)φ(1) (say this as e1 is replaced by e3 is replaced by e5 is replace by 1)

This is the same as

φ(1)φ(3)φ(5) φ(1)φ(5)φ(3) φ(3)φ(5)φ(1)

The point is that this is equivalent to two transpostions. Hence by the PP, the wavefunction will not change sign. Since C3 does this cyclic permutation on 6 electrons,eg (135)(246)= (13)(35)(24)(46), we still get no sign change.

Consider the effect of C3

(135)

(35) (13)

Thus the wavefunction for NH3 has 1A1 symmetry


New material


Linear molecules, C∞v symmetry

Consider the wavefunction for one electron in a linear molecule.

Here we use polar coordinates, = sqrt(x2+y2), z (axis along z) Since the wavefunction has period of 2 in , the dependence of any wavefunction can be expanded as a Fourier series,

φ(,,z)=f(,z){a0 + m=1m=∞ [am cos m + bm sin m]

O=C=O z

x






Clearly the kinetic energy will increase with m, so that for the same f(,z), we expect m=0 lowest, then m=±1, then m=±2, m=±3, etc

O=C=O z

x






Clearly the kinetic energy will increase with m, so that for the same f(,z), we expect m=0 lowest, then m=±1, then m=±2, m=±3, etc

Also if we rotate the molecule about the z axis by some angle , the states with the same m get recombined

[cos m()] = (cos m)(cosm) – (sin m)(sin m)

[sin m()] = (sin m)(cosm) + (cos m)(sin m)

O=C=O z

x

Which means that the wavefunctions with the same m are degenerate


The symmetry operators are:

Rz(): counterclockwise rotation by an angle about the z axis

xz: reflection in the xz plane (this takes + into –)

C∞v symmetry groupO=C=O z

x





’ = Rz() xz Rz(-); reflection in a plane rotated by an angle from the xz plane (there are an infinite number of these)

e: einheit (unity)

This group is denoted as C∞v,The character table (symmetries) are

C∞v symmetry group

name

1-e N-e

H--C=N z

x


Demonstrate that ’ = Rz() xz Rz(-) is a reflection in the plane rotated by an angle from the xz plane

Take the z axis out of the plane. The six symmetry operations are:

x

y

e

xzC-

x

y

e

x

y

e

C

x

y

e

x

y

e x

y

e

’

’







x






e: einheit (unity)

This group is denoted as C∞v,The character table (symmetries) are


x

name

1-e N-e


The symmetry functions for C∞v

Lower case letters are used to denote one-electron orbitals


Application to FH

The ground state wavefunction of HF is


In C∞v symmetry, the bond pair is (m=0),while the px

and py form a set of orbitals (m=+1 and m=-1).

Consider the case of up spin for both x and y

Ψ(1,2) = A{φxφyφxφyφyφx

Rotating by an angle about the z axis leads to

φa = cos φx + sin φy and φb = cos φy - sin φx

This leads to

φaφbφbφacos)2 +sin)2] φxφyφyφx

Thus φxφyφyφxtransforms as .


Continuing with FH

Thus the (px)2(py)2 part of the HF wavefunction


transforms as .

The symmetry table, demands that we also consider the symmetry with respect to reflection in the xz plane.

Here px is unchanged while py changes sign. Since there are two electrons in py the wavefunction is invariant.

Thus the ground state of FH has 1symmetry


Next consider the ground state of OH

Ψx=A{(OHbond(x)(y)(x)]}

Ψy=A{(OHbond(x)(y)(y)]}

We write the two wavefunctions for OH as

2y

2x

z

x

We saw above that

A{(x)(y)} transforms like thus we need examine only the transformations of the downspin orbital. But this transforms like .

Another way of describing this is to note that A{(x)2(y)2} transforms like and hence one hole in a (shell, (3 transforms the same way as a single electron, (

Thus the total wavefunction is 2.


Now consider the ground state of NH

z

x

A{(NH bond)2(N2px)(N2py

We saw earlier that up-spin in both x and y leads to symmetry.

Considering now the reflection, xz, we see that with just one electron in y, we now get -.Thus the ground state of NH is 3-.


Now consider Bonding H atom to all 3 states of C

No singly occupied orbital for H to bond with

(2px)(2py)

(2py)(2pz)


(2px)(2pz)


H1s can get bonding

Get S= ½ state,


z

x


Ground state of CH (2)

z

x

A{(2s)2(OH bond(O2px)1}

The full wavefunction for the bonding state2x

2y

A{(2s)2(OH bond(O2py)1}


Bond a 2nd H atom to the ground state of CH

z

x

Starting with the ground state of CH, we bring a 2nd H along the x axis.

Get a second covalent bond

This leads to a 1A1 state.

No unpaired orbtial for a second covalent bond.


Analyze Bond in the ground state of CH2

z

x

Ground state has 1A1 symmetry. For optimum bonding, the pz orbital should point at the Hz while the px orbital should point at the Hx. Thus the bond angle should be 90º.

As NH2 (103.2º) and OH2 (104.5º), we expect CH2 to have bond angle of ~ 102º

θe Re


But, the Bending potential surface for CH2

3B1

1A1

1B1

3g-

1g

9.3 kcal/mol


Thus something is terribly wrong in our analysis of CH2


stopped

lecture 5,6, january 19, 2011 bonding in nefoncbbe hydrides

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