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Lecture 9, September 30, 2009

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

Course number: KAIST EEWS 80.502 Room E11-101

Hours: 0900-1030 Tuesday and Thursday

William A. Goddard, III, [email protected]

WCU Professor at EEWS-KAIST and Charles and Mary Ferkel Professor of Chemistry, Materials Science, and Applied Physics, California Institute of Technology

Senior Assistant: Dr. Hyungjun Kim: [email protected]

Manager of Center for Materials Simulation and Design (CMSD)

Teaching Assistant: Ms. Ga In Lee: [email protected]

Special assistant: Tod Pascal:[email protected]

EEWS-90.502-Goddard-L09

There was no lecture on Sept. 22 because of the EEWS conference

Goddard will be traveling Oct 2-11 and will not give the lectures scheduled for Oct. 6 and 8

Consequently an extra lecture will be added at 2pm on Wednesday Sept. 30 and another at 2pm Wednesday Oct 14.

This will be in the same room, 101 E11

L8: Sept. 29, as scheduled, 9am

L9: Sept. 30, new lecture 2pm replaces Oct 6

L10: Oct. 1, as scheduled, 9am

L11: Oct. 13, as scheduled, 9am

L12: Oct. 14, new lecture 2pm, replaces Oct 8

L13: Oct. 15, as scheduled, 9am

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For those that will be bored with no lectures, we have a course project for extra credit.

The numbers in these notes were mostly put together in 1972 to 1986, when the QM calcualtions were less accurate and the experiments less complete.

I would like to update all the numbers using modern QM methods (B3LYP DFT with 6-311G** basis set) of moderate accuracy.

Dr. Hyungjun Kim has a number of systems we would like to update at this level and will work with interested students

Satisfactory completion of such a project, with a report summarizing the results can be used in lieu of the class midterm.

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EEWS-90.502-Goddard-L09

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 R1is 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.

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1). Closure: If R1,R2 e G (both are symmetry transformations) then R2 R1is also a symmetry transformation, R2 R1e G

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

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

4. Inverse. If R1e G then the inverse, (R1)-1e G,where the inverse is defined as (R1)-1R1 = e.

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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 RiΨieG} 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.

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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

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

- The symmetry transformations are
- e for einheit (unity) xx, yy, zz
- C2z, rotation about the z axis by 2p/2=180º, x-x, y-y, zz
- sxz, reflection in the xz plane, xx, y-y, zz
- syz, reflection in the yz plane, x-x, yy, zz
- Which is denoted as the C2v group.

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e

e

y

y

x

x

C2z

C2z

syz

sxz

sxz

syz

syz

C2z

C2z

syz

sxz

sxz

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. C2zsxz= syz

Combine these as below to show the relationships

C2v

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The basic symmetries (usually called irreducible representations) for C2v are given in a table, called the character table

My choice of coordinate system follows (Mulliken in JCP 1955). This choice removes confusion about B1 vs B2 symmetry (x is the axis for which sxz moves the maximum number of atoms

In the previous slide we saw that C2zsxz= syz which means that the symmetries for syz are already implied by C2zsxz. Thus we consider C2z andsxz as the generators of the group.

This group is denoted as C2v, which denotes that the generators are C2z and a vertical mirror plane (containing the C2 axis)

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CH2

NH2

H2O

A{(O2py)2[(Opx )(Hx)+(Hx)(Opx)](ab-ba)[(Opz)(Hz)+(Hz)(Opz)](ab-ba)}

H2O

OHx bond

OHz bond

1A1

2B1

1A1

NH2

A{(N2pya)[(Npx )(Hx)+(Hx)(Npx)](ab-ba)[(Npz)(Hz)+(Hz)(Npz)](ab-ba)}

NHx bond

NHz bond

CH2

A{(C2py)0[(Cpx )(Hx)+(Hx)(Cpx)](ab-ba)[(Cpz)(Hz)+(Hz)(Cpz)](ab-ba)}

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A{(C2sa)1(2pxa)1[(CpL )(HL)+(HL)(CpL)](ab-ba)[(CpR)(HR)+(HR)(CpR)](ab-ba)}

CHL bond

CHR bond

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

y

z

A{[(CHL)2(CHR)2](Csa)1(Cpa)1}

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

p=2px

s=2s

Thus the symmetry of triplet CH2 is 3B1

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z

x

A{[(Npy )(Hy)+(Hy)(Npy)](ab-ba)[(Npx )(Hx)+(Hx)(Npx)](ab-ba)[(Npz)(Hz)+(Hz)(Npz)] (ab-ba)}

NHx bond

NHc bond

NHb bond

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.

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z

x

The symmetry transformations of C3v are:

Hx

- e for einheit (unity) xx, yy, zz
- Cz3, rotation about the z axis by 2p/3=120º
- (Cz3)2 rotation about the z axis by 2(2p/3)=240º
- sxz, reflection in the xz plane, xx, y-y, zz
- syz, reflection in the plane through Hb and the z axis
- syz, reflection in the plane through Hc and the z axis

Hb

Hc

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b

sxzC32

C3

e

x

sxz

sxzC3

C32= C3-1

c

Clearly this set of symmetries is closed, which you can see by the symmetry of the diagram.

Also we see that the generators are sxz andC3

Note that the operations do not all commute.

Thus sxzC3 ≠ sxzC3

Instead we get

sxzC3 = sxzC32

Such groups are called nonabelian and lead to irreducible representations with degree (size) > 1

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Here the E irreducible representation is of degree 2.

This 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.

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.

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z

x

sLP

We will write the wavefunction for NH3 as

x

A{(sLP)2[(NHb bond)2(NHc bond)2(NHx bond)2]}

c

b

Where each of the 3 VB bond pairs is written as a pair function and we denote what started as the 2s pair as slp

The rotations and relections interchange pairs of electrons leaving the wavefunction invariant.

Thus we get 1A1 symmetry

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For a linear molecule (axis along z) the symmetry operators are:

e: einheit (unity)

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

sxz: reflection in the xz plane

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

This group is denoted as C∞v

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Lower case letters are used to denote one-electron orbitals

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The ground state wavefunction of HF is

A{(F2px)2(F2py)2[(Fpz)(H)+(H)(Fpz)](ab-ba)}

In C∞v symmetry, the bond pair is s (m=0),while the px and py form a set of p orbitals (m=+1 and m=-1).

Consider the case of up spin for both px and py

Ψ(1,2) = A{φxaφya}=(φxφy- φyφx) aa

Rotating by an angle g about the z axis leads to

(φaφb- φbφa) = [(cosg)2 +(sing)2] }=(φxφy- φyφx)

Thus (φxφy- φyφx) transforms as S.

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

A{(F2px)2(F2py)2[(Fpz)(H)+(H)(Fpz)](ab-ba)}

transforms as S.

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 1S+ symmetry

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x

z

Next consider the ground state of OH

We write the wavefunctions for OH as

2Px

Ψx=A{(sOHbond)2[(pxa)(pya)(pxb)]}

Ψy=A{(sOHbond)2[(pxa)(pya)(pyb)]}

We saw above that

A{(pxa)(pya)} transforms like S. thus we need examine only the transformations of the downspin orbital. But this transforms like p.

2Py

Thus the total wavefunction is 2P.

Another way of describing this is to note that A{(px)2(py)2} transforms like S and hence one hole in a (p)4shell, (p)3 transforms the same way as a single electron, (p)1

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x

z

Now consider the ground state of NH

A{(NH bond)2(N2pxa)(N2pya)}

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

With just one electron in py, we now get S-.

Thus the ground state of NH is 3S-.

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x

z

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

(2px)(2pz)

O 2pz singly occupied.

H1s can get bonding

Get S= ½ state,

Two degenerate states, denote as 2P

(2py)(2pz)

(2px)(2py)

No singly occupied orbital for H to bond with

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x

z

Ground state of CH (2P)

The full wavefunction for the bonding state

2Px

A{(2s)2(OHs bond)2(O2pxa)1}

2Py

A{(2s)2(OHs bond)2(O2pya)1}

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x

z

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

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.

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x

z

θe

Re

Analyze Bond in the ground state of CH2

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

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1B1

1Dg

1A1

3B1

3Sg-

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

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2 a0

1s

2s

1.06A

0.14A

R~0.14A

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

Each electron has its maximum amplitude in a spherical torus 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

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So assumed that the Be wavefunction is A[(1s)2(2s)2] = A[(1sa)(1sb)(2sa)(2sb)]

2s

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

φa = φ2s + lφpz and φb = φ2s - lφpz

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 increased average ee distance, dramatically reducing the ee repulsion.

pz

φ2s + lφpz

φ2s - lφpz

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x

x

z

z

Pooching of the 2s orbitals in opposite directions leads to a dramatic increase in ee distance, reducing ee repulsion.

φ2s + lφpz

φ2s - lφpz

z

z

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

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A[(φaa)(φbb)] = [φa(1)a(1)][(φb(2)b(2)] – [φb(1)b(1)] [φa(2)a(2)]

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)][a(1)b(2)–b(1)a(2)]

3Ψ(1,2) = [φa(1)φb(2)-φb(1)φa(2)][a(1)b(2)+b(1)a(2)] and aa, bb

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

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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

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1Ψ(1,2) = [φa(1)φb(2)+φb(1)φa(2)][a(1)b(2)–b(1)a(2)]

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

(ab+ba) = (s+lz)(s-lz)+(s-lz)(s+lz) = [s(1)s(2) - l2 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)-l2 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)- l2 z(R)z(-R) = s(R)s(R)+l2 z(R)z(R) which is increased.

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1Ψ(1,2) = [φa(1)φb(2)+φb(1)φa(2)][a(1)b(2)-b(1)a(2)]

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

The optimum value of l~0.4 (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) - l2 z(1)z(2)]/sqrt(1+l4)

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 l.

Normalizing the GVB orbitals leads to φa = (φ2s + lφpz)/sqrt(1+l2)

Thus the overlap of the GVB pair is

< φa | φa > = (1-l2)/(1+l2)=0.752,

similar to a bond pair

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3Ψ(1,2) = [φa(1)φb(2)-φb(1)φa(2)][a(1)b(2)+b(1)a(2)] and aa, bb

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

(ab-ba) = (s+lz)(s-lz)-(s-lz)(s+lz) = [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 functions

EEWS-90.502-Goddard-L09

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 CI form by generalizing to

{s(1)s(2) - m2 [z(1)z(2)+x(1)x(2)+y(1)y(2)]}

which does have 1S symmetry. Here the value of m2 ~ l2/3

This form of CI wavefunction can be solved for self-consistently, referred to as MC-SCF for multiconfiguration self-consistent field. But the simple GVB description is obscured.

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

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

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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

At large R the orbitals of Be are already hybridized

A{[(sz)(H)+(H)(sz)](ab-ba)(zsa)}

zs

sz

H

In 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

Thus the wave function is A{[(sz)(zs)+(zs)(sz)](ab-ba)(Ha)}

where sz≡(s+lz) and zs ≡(s-lz)

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

Thus at large R we obtain a slightly repulsive interaction.

zs

sz

H

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Short range Attractive interaction sz with H

Long range Repulsive interaction with H

BeH

TA’s check numbers, all from memory

2 eV

BeH+ has long range attraction no short range repulsion

3 eV

1 eV

BeH+

1eV Repulsive orthogonalization of zs with sz H

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BeH+

MgH+

1S+

3.1 eV

R=1.31A

2.1 eV

R=1.65 A

1.34 eV

R=1.73A

2.03 eV

R=1.34A

2S+

~3.1 eV

1S+

linear

~2.1 e

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.

TA’s check numbers, all from memory

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

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

Modified from the simple (1s)2(2s)2(2p)2 form

Ψyz=A[(2sa)(2sb)(ya)(za)]

Consider first pooching the 2s orbitals in the z direction

Ψyz=A[(2s+lz)(2s-lz)+(2s-lz)(2s+lz)](ab-ba)(ya)(za)]

2s pair pooched +z and –z

yz open shell

Expanding in the CI form leads to

Ψyz=A[(2s)(2s)-l2(z)(z)](ab-ba)(ya)(za)]

=A[(2sa)(2sb)](ya)(za)] -l2A[(za)(zb)(ya)(za)]

But the 2nd term is zero since the za is already occupied

Thus the 2s can only pooch in the x direction

sx

py

pz

Ψyz=A[(sx)(xs)+(xs)(sx)](ab-ba)(ya)(za)]

xs

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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

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x

z

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

Ψyz=A[(sx)(xs)+(xs)(sx)](ab-ba)(ya)(za)] which we visualize as

sx

py

2s pair pooched +x and –x

yz open shell

pz

xs

Ψyx=A[(sz)(zs)+(zs)(sz)](ab-ba)(ya)(xa)] which we visualize as

px

py

2s pair pooched +z and –z

xy open shell

zs

sz

Ψxz=A[(sy)(ys)+(ys)(sy)](ab-ba)(xa)(za)] which we visualize as

px

xz open shell

2s pair pooched +y and –y

pz

sy,ys

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(2px)(2pz)

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

C 2pz singly occupied.

H1s can get bonding

Get S= ½ state,

Two degenerate states, denote as 2P

(2py)(2pz)

(2px)(2py)

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

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The presence of the lobe orbitals might seem to complicate the symmetry

Ψyz=A[(sx)(xs)+(xs)(sx)](ab-ba)(ya) (zH bond)2]

Ψxz=A[(sy)(ys)+(ys)(sy)](ab-ba)(xa)(zH bond)2)]

To see that there is no problem, rewrite in the CI form (and ignore the zH bond)

Ψyz=A[(s2 – l x2)](ab-ba)(ya)]

Now form a new wavefunction by adding - l y2 to Ψyz

Φyz ≡A[s2 – l x2 – l y2](ab-ba)(ya)]

But the 3rd term is A[y2](ab-ba)(ya)]= – l A[(ya)(yb)(ya)]=0

Thus Φyz = Ψyz and similarly

Φxz = A[s2 – l x2 – l y2](ab-ba)(xa)] = Ψxz

But the 2s terms [s2 – l x2 – l y2] are clearly symmetric about the z axis. Thus these wavefunctions have 2P symmetry

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At large R the lobe orbitals of C are already hybridized

Thus the wave function is A{(pxa)(pya)[(sz)(zs)+(zs)(sz)](ab-ba)(Ha)}

Unpaired H

2s pair pooched

+z and –z

xy open shell

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

Thus at large R we obtain a slightly repulsive interaction.

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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

px

py

zs

sz

H

A{[(sz)(H)+(H)(sz)](ab-ba)(zsa)(pxa)(pya)}

Sz-H bond pair

nonbond orbitals

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=s, px=px, py=py

Since the nonbond orbitals, s, px, py are orthogonal to each other the high spin is lowest, get S=3/2 or quartet state

We saw for NH that (pxpy –pypx)(aa) has 3S- symmetry. CH has one additional high spin nonbond s orbital, leading to 4S-

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sx

pz

H

py

xs

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CH

SiH

Kcal/mol

p bond

De(2P)

80.070.1

D(2P-4S-)

17.136.2

De(4S-)

62.933.9

sz bond

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 2P state is best

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

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

Thus hydridization in the atom due to electron correlation leads naturally to the new 4S- 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

104º

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.

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sx

pz

H

py

xs

As usual, we start with the ground states of CH or SiH, 2Px and 2Py and bond bring an H along some axis, say x.

2Py

2Px

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Now we get credible bound states from both components

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

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

sx

2Px

pz

2Py

H

py

xs

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

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GVB orbitals for SiH(1A1)

The wavefunction is

Ψ=A{[(sy)(ys)+(ys)(sy)(ab-ba)](SiHLbond)2(SiHRbond)2}

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

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sℓ

px

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 [(sℓa)(pxa)]}

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

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

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θe

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

Re

sℓ

3B1 state. Optimum bonding, the two bonds at ~128º. Here HH orthogonalization should increase this a bit but C

much less than 12º since H’s are much farther apart. However now the sℓ 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

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

For C or Si A[(2s)2(pza)(pxa)] 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)(ab-ba)(pxa)}. 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

C8090

Si7076.2

This is close to the observed differences.

TA’s check numbers

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3P

4S-

33.9ℓ

4S-

62.9ℓ

2P

2P

70.1p

80.0p

SiH

CH

3B1

56.9ℓ

1A1

76.1p

1A1

90.0p

SiH2

3B1

99.1ℓ

CH2

CH Lobe bonds: 63 and 99

50% increase

SiH Lobe bonds: 35 and 57

50% increase

Assume 50% in lobe bond is from the first p bond destabilizing the lone pair

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We saw for CH that the lobe bond is 17 kcal/mol weaker than the p bond while it is 37 kcal/mol weaker for Si. 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(4S-) is 35 kcal/mol higher than Si(2P), 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.

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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

We could also make a bond to the out-of-plane pp 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 no longer paired, we get a very strong bond energy of 109 kcal/mol.

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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.

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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 four

bonds readjust to become equivalent leading to a tetrahedral SiH4 molecule. No unpairing is required so that we get a strong bond, 92 kcal/mol (the 3rd bond was 72)

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Idealized case.

Tetrahedral: sp3

(s+x+y+z)/2

(s-x-y+z)/2

(s-x+y-z)/2

(s+x-y-z)/2

Orthogonal and point to vertices of a tetrahedron.

Rationalizes bonding in CH4.

Assumes 75% p character (atom is 57% p) GVB 70% p

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Now bonding H along the z axis we get a 2S+ state and two components of 3P

2S wins

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