Lecture 4, January 14, 2011 aufbau principle atoms

1© copyright 2010 William A. Goddard III, all rights reserved Ch120a-Goddard-L04 © copyright 2011 William A. Goddard III, all rights reserved

Nature of the Chemical Bond with applications to catalysis, materials

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

Lecture 4, January 14, 2011

aufbau principle atoms

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]>


The excited states of H atom

hφk = ekφk for all excited states k

We will use spherical polar coordinates, r, θ, φ

where z=r cosθ, x=r sinθ cosφ, y=r sinθ sinφ

h = - ½ – Z/r is independent of θ and φ which can be seen by noting that d2/dx2 + d2/dy2 + d2/dy2 is independent of θ and φ (it transforms like r2 = x2 + y2 + z2).Consequently the eigenfunctions of h can be written as a factor depending only on r and a factor depending only on θ and φ φnlm = Rnl(r) Zlm(θ,φ) where the reason for the numbers nlm will become apparent later

xy

z

θ

φ


The ground state of H atom

h = - ½ – Z/r

φnlm = Rnl(r) Zlm(θ,φ)

For the ground state, R1s = exp(-Zr) and Z1s = Z00 = 1 (a constant), where l=0 and m=0again ignoring normalization. The radial wavefunction is nodeless, as expected The angular function is a constant, which is clearly the best for KE.

xy

z

θ

φ


The p excited angular states of H atomφnlm = Rnl(r) Zlm(θ,φ) Now lets consider excited angular functions, Zlm.They must have nodal planes to be orthogonal to Z00

x

z

+

-

pz

The simplest would be Z10=z = r cosθ, which is zero in the xy plane.Exactly equivalent are Z11=x = rsinθcosφ which is zero in the yz plane, and Z1-1=y = rsinθsinφ, which is zero in the xz planeAlso it is clear that these 3 angular functions with one angular nodal plane are orthogonal to each other. Thus the integrand of ∫yz has nodes in both the xy and xz planes, leading to a zero integral

x

z

+-

px

x

z

+

-

pxpz-

+

pz


More p functions?

So far we have the s angular function Z00 = 1 with no angular nodal planes

And three p angular functions: Z10 =pz, Z11 =px, Z1-1 =py, each with one angular nodal plane

Can we form any more with one nodal plane orthogonal to the above 4 functions?

x

z

+

-

px’

x

z

+-

pzpx’

+ -

For example we might rotate px by an angle about the y axis to form px’. However multiplying, say by pz, leads to the integrand pzpx’ which clearly does not integrate to zero

. Thus there are exactly three pi functions, Z1m, with m=0,+1,-1, all of which have the same KE.

Since the p functions have nodes, they lead to a higher KE than the s function


More angular functions?

So far we have the s angular function Z00 = 1 with no angular nodal planes

And three p angular functions: Z10 =pz, Z11 =px, Z1-1 =py, each with one angular nodal plane

Next in energy will be the d functions with two angular nodal planes. We can easily construct three functions

x

z

+

-

dxz-

+

dxy = xy =r2 (sinθ)2 cosφ sinφ

dyz = yz =r2 (sinθ)(cosθ) sinφ

dzx = zx =r2 (sinθ)(cosθ) cosφ

where dxz is shown here

Each of these is orthogonal to each other (and to the s and the three p functions)


In addition we can construct three other functions with two nodal planes

dx2-y2 = x2 – y2 = r2 (sinθ)2 [(cosφ)2 – (sinφ)2]

dy2-z2 = y2 – z2 = r2 [(sinθ)2(sinφ)2 – (cosθ)2]

dz2-x2 = z2 – x2 = r2 [(cosθ)2 -(sinθ)2(cosφ)2]

where dz2-x2 is shown here,

Each of these three is orthogonal to the previous three d functions (and to the s and the three p functions)

This leads to six functions with 2 nodal planes

More d angular functions?

x

z+

-

dz2-x2

-

+


In addition we can construct three other functions with two nodal planes

dx2-y2 = x2 – y2 = r2 (sinθ)2 [(cosφ)2 – (sinφ)2]

dy2-z2 = y2 – z2 = r2 [(sinθ)2(sinφ)2 – (cosθ)2]

dz2-x2 = z2 – x2 = r2 [(cosθ)2 -(sinθ)2(cosφ)2]

where dz2-x2 is shown here,

Each of these three is orthogonal to the previous three d functions (and to the s and the three p functions)

This leads to six functions with 2 nodal planes

However (x2 – y2) + (y2 – z2) + (z2 – x2) = 0

Which indicates that there are only two independent such functions. We combine the 2nd two as

(z2 – x2) - (y2 – z2) = [2 z2 – x2 - y2 ] = [3 z2 – x2 - y2 –z2] =

= [3 z2 – r2 ] which we denote as dz2

More d angular functions?

x

z+

-

dz2-x2

-

+


Summarizing we get 5 d angular functions

x

z+

-

dz2

-

+

57º

Z20 = dz2 = [3 z2 – r2 ] m=0, d

Z21 = dzx = zx =r2 (sinθ)(cosθ) cosφ

Z2-1 = dyz = yz =r2 (sinθ)(cosθ) sinφ

Z22 = dx2-y2 = x2 – y2 = r2 (sinθ)2 [(cosφ)2 – (sinφ)2]

Z22 = dxy = xy =r2 (sinθ)2 cosφ sinφ

We find it useful to keep track of how often the wavefunction changes sign as the φ coordinate is increased by 2 = 360º

When this number, m=0 we call it a function

When m=1 we call it a function



m = 1, d

m = 2, d


Summarizing the angular functions

So far we have

•one s angular function (no angular nodes) called ℓ=0

•three p angular functions (one angular node) called ℓ=1

•five d angular functions (two angular nodes) called ℓ=2

Continuing we can form

•seven f angular functions (three angular nodes) called ℓ=3

•nine g angular functions (four angular nodes) called ℓ=4

where ℓ is referred to as the angular momentum quantum number

And there are (2ℓ+1) m values for each ℓ


The real (Zlm) and complex (Ylm) momentum functions

Here the bar over m negative


Excited radial functions

Clearly the KE increases with the number of angular nodes so that s < p < d < f < g

Now we must consider radial functions, Rnl(r)The lowest is R10 = 1s = exp(-Zr)All other radial functions must be orthogonal and hence must have one or more radial nodes, as shown here

Note that we are plotting the cross section along the z axis, but it would look exactly the same along any other axis. Here

R20 = 2s = [Zr/2 – 1]exp(-Zr/2) and R30 = 3s = [2(Zr)2/27 – 2(Zr)/3 + 1]exp(-Zr/3)

Zr = 2 Zr = 1.9

Zr = 7.1


Combination of radial and angular nodal planes

Combining radial and angular functions gives the various excited states of the H atom. They are named as shown where the n quantum number is the total number of nodal planes plus 1

The nodal theorem does not specify how 2s and 2p are related, but it turns out that the total energy depends only on n.

Enlm = - Z2/2n2

The potential energy is given by

PE = - Z2/2n2 = -Z/ , where =n2/Z

Thus Enlm = - Z/2

1s 0 0 0 1.02s 1 1 0 4.02p 1 0 1 4.03s 2 2 0 9.03p 2 1 1 9.03d 2 0 2 9.04s 3 3 0 16.04p 3 2 1 16.04d 3 1 2 16.04f 3 0 3 16.0

nam

e

tota

l nod

al p

lane

sra

dial

nod

al p

lane

san

gula

r no

dal p

lane

s

R̄ R̄

R̄

Siz

e (a

0)


Hydrogen orbitals 1s, 2s, 2p, 3s, 3p, 3d, 4s, 4p, 4d, 4f

Angstrom (0.1nm) 0.53, 2.12, 4.77, 8.48

Sizes hydrogen orbitals

H--H C

0.74

H

H

H

H

1.7

H H

H H

H H

4.8

R̄ =(n2/Z) a0 = 0.53 n2 A


Hydrogen atom excited states

1s-0.5 h0 = -13.6 eV

2s-0.125 h0 = -3.4 eV

2p

3s-0.056 h0 = -1.5 eV

3p 3d

4s-0.033 h0 = -0.9 eV

4p 4d 4f

Energy zero

Enlm = - Z2/2n2 = -(1/2n2)h0 = 13.6 eV/n2


Plotting of orbitals: line cross-section vs. contour


Contour plots of 1s, 2s, 2p hydrogenic orbitals

1s, no nodes

2s, one radial node 2p, one angular node


Contour plots of 3s, 3p, 3d hydrogenic orbitals

3s, two radial nodes

3p, one angular node one radial node

3d, two angular nodes


Contour plots of 4s, 4p, 4d hydrogenic orbtitals

4s, 3 radial nodes

4p, one angular node two radial nodes

4d, two angular nodes and one radial node


Contour plots of hydrogenic 4f orbitals

All seven 4f have three angular nodes and no radial nodes


He atom – describe using He+ orbitalsWith 2 electrons, we can form the ground state of He by putting both electrons in the He+ 1s orbital, just like the MO state of H2

ΨHe(1,2) = A[(Φ1s)(Φ1sΦ1s(1)Φ1s(2) (

He<1s|h|1s> + J1s,1s

two one-electron energies one Coulomb repulsion



ΨHe(1,2) = A[(Φ1s)(Φ1sΦ1s(1)Φ1s(2) (

He<1s|h|1s> + J1s,1sFirst lets review the energy for He+. Writing Φ1sexp(-r) we determine the optimum for He+ as follows

<1s|KE|1s> = + ½ 2 (goes as the square of 1/size)

<1s|PE|1s> = - Z(linear in 1/size)



ΨHe(1,2) = A[(Φ1s)(Φ1sΦ1s(1)Φ1s(2) (

He<1s|h|1s> + J1s,1sFirst lets review the energy for He+. Writing Φ1sexp(-r) we determine the optimum for He+ as follows

<1s|KE|1s> = + ½ 2 (goes as the square of 1/size)

<1s|PE|1s> = - Z(linear in 1/size)

Applying the variational principle, the optimum must satisfy dE/d = -Z = 0 leading to =Z,

This value of leads to KE = ½ Z2, PE = -Z2, E=-Z2/2 = -2 h0.

write PE=-Z/R0, so that the average radius is R0=1/


J1s,1s = e-e energy of He atom – He+ orbitals

How can we estimate J1s,1s

Assume that each electron moves on a sphere

With the average radius R0 = 1/

And assume that e1 at along the z axis (θ=0)

Neglecting correlation in the electron motions, e2 will on the average have θ=90º so that the average e1-e2 distance is ~sqrt(2)*R0

Thus J1s,1s ~ 1/[sqrt(2)*R0] = 0.71

A rigorous calculation (notes chapter 3, appendix 3-C page 6)

Gives J1s,1s = (5/8)

R0

e1

e2

Now consider He atom: EHe = 2(½ 2) – 2ZJ1s,1s


The optimum atomic orbital for He atom

He atom: EHe = 2(½ 2) – 2Z(5/8)

Applying the variational principle, the optimum must satisfy dE/d = 0 leading to

2- 2Z + (5/8) = 0

Thus = (Z – 5/16) = 1.6875

KE = 2(½ 2) = 2

PE = - 2Z(5/8) = -2 2

E= - 2 = -2.8477 h0

Ignoring e-e interactions the energy would have been E = -4 h0

The exact energy is E = -2.9037 h0 (from memory, TA please check). Thus this simple approximation accounts for 98.1% of the exact result.


Interpretation: The optimum atomic orbital for He atom

ΨHe(1,2) = Φ1s(1)Φ1s(2) with Φ1sexp(-r)

We find that the optimum = (Z – 5/16) = 1.6875

With this value of , the total energy is E= - 2 = -2.8477 h0

This wavefunction can be interpreted as two electrons moving independently in the orbital Φ1sexp(-r) which has been adjusted to account for the average shielding due to the other electron in this orbital.

On the average this other electron is closer to the nucleus about 31% of the time so that the effective charge seen by each electron is 2.00-0.31=1.69

The total energy is just the sum of the individual energies.

Ionizing the 2nd electron, the 1st electron readjusts to = Z = 2

With E(He+) = -Z2/2 = - 2 h0. thus the ionization potential (IP) is 0.8477 h0 = 23.1 eV (exact value = 24.6 eV)


Now lets add a 3rd electron to form Li

ΨLi(1,2,3) = A[(Φ1s)(Φ1s(Φ1s

Problem with either or , we get ΨLi(1,2,3) = 0

This is an essential result of the Pauli principle

Thus the 3rd electron must go into an excited orbital

ΨLi(1,2,3) = A[(Φ1s)(Φ1s)(Φ2s

or

ΨLi(1,2,3) = A[(Φ1s)(Φ1s)(Φ2pz(or 2px or 2py)

First consider Li+ with ΨLi(1,2,3) = A[(Φ1s)(Φ1s

Here Φ1sexp(-r) with = Z-0.3125 = 2.69 and

E = -2 = -7.2226 h0. Since the E (Li2+) = -9/2 = -4.5 h0 the IP = 2.7226 h0 = 74.1 eV

The size of the 1s orbital is R0 = 1/ = 0.372 a0 = 0.2A


Consider adding the 3rd electron to the 2p orbital


Since the 2p orbital goes to zero at z=0, there is very little shielding so that it sees an effective charge of

Zeff = 3 – 2 = 1, leading to

a size of R2p = n2/Zeff = 4 a0 = 2.12A

And an energy of e = -(Zeff)2/2n2 = -1/8 h0 = -3.40 eV

0.2A

1s

2.12A

2p


Consider adding the 3rd electron to the 2s orbital


The 2s orbital must be orthogonal to the 1s, which means that it must have a spherical nodal surface at ~ 0.2A, the size of the 1s orbital. Thus the 2s has a nonzero amplitude at z=0 so that it is not completely shielded by the 1s orbitals.

The result is Zeff2s = 3 – 1.72 = 1.28

This leads to a size of R2s = n2/Zeff = 3.1 a0 = 1.65A

And an energy of e = -(Zeff)2/2n2 = -0.205 h0 = -5.57 eV

0.2A

1s

2.12A2s

R~0.2A


Li atom excited states

1s

2s

-0.125 h0 = -3.4 eV2p

Energy

zero

-0.205 h0 = -5.6 eV

-2.723 h0 = 74.1 eV

MO picture State picture

(1s)2(2s)

(1s)2(2p)

E = 2.2 eV17700 cm-1

564 nm

Ground state

1st excited state

Exper671 nmE = 1.9 eV


Aufbau principle for atoms

1s

2s2p

3s3p

3d4s

4p 4d 4f

Energy

2

2

6

2

62

6

10

10 14

He, 2

Ne, 10

Ar, 18Zn, 30

Kr, 36

Including shielding, 2s<2p3s<3p<3d, 4s<4p<4d<4f Get generalized energy spectrum for filling in the electrons to explain the periodic table.Full shells at 2, 10, 18, 30, 36 electrons


He, 2

Ne, 10

Ar, 18

Zn, 30

Kr, 36


General trends along a row of the periodic table

As we fill a shell, thus

B(2s)2(2p)1 to Ne (2s)2(2p)6

For each atom we add one more proton to the nucleus and one more electron to the valence shell

But the valence electrons only partially shield each other.

Thus Zeff increases leading to a decrease in the radius ~ n2/Zeff

And an increase in the IP ~ (Zeff)2/2n2

Example Zeff2s=

1.28 Li, 1.92 Be, 2.28 B, 2.64 C, 3.00 N, 3.36 O, 4.00 F, 4.64 Ne

Thus (2s Li)/(2s Ne) ~ 4.64/1.28 = 3.6


Many-electron configurations

General aufbau

ordering

Particularly stable


General trends along a column of the periodic table

As we go down a column

Li [He}(2s) to Na [Ne]3s to K [Ar]4s to Rb [Kr]5s to Cs[Xe]6s

Things get more complicated

The radius ~ n2/Zeff

And the IP ~ (Zeff)2/2n2

But the Zeff tends to increase, partially compensating for the change in n so that the atomic sizes increase only slowly as we go down the periodic table and

The IP decrease only slowly (in eV):

5.39 Li, 5.14 Na, 4.34 K, 4.18 Rb, 3.89 Cs

(13.6 H), 17.4 F, 13.0 Cl, 11.8 Br, 10.5 I, 9.5 At

24.5 He, 21.6 Ne, 15.8 Ar, 14.0 Kr,12.1 Xe, 10.7 Rn


Transition metals; consider [Ar] plus one electron

[IP4s = (Zeff4s )2/2n2 = 4.34 eV Zeff

4s = 2.26; 4s<4p<3d

IP4p = (Zeff4p )2/2n2 = 2.73 eV Zeff

4p = 1.79;

IP3d = (Zeff3d )2/2n2 = 1.67 eV Zeff

3d = 1.05;

IP4s = (Zeff4s )2/2n2 = 11.87 eV Zeff

4s = 3.74; 4s<3d<4p


3d = 2.59;


4p = 3.20;


3d = 4.05; 3d<4s<4p

IP4s = (Zeff4s )2/2n2 = 21.58 eV Zeff

4s = 5.04;


4p = 4.47;

K

Ca+

Sc++

As the net charge increases the differential shielding for 4s vs 3d is less important than the difference in n quantum number 3 vs 4Thus charged system prefers 3d vs 4s


Transition metals; consider Sc0, Sc+, Sc2+

3d: IP3d = (Zeff3d )2/2n2 = 24.75 eV Zeff

3d = 4.05;

4s: IP4s = (Zeff4s )2/2n2 = 21.58 eV Zeff

4s = 5.04;

4p: IP4p = (Zeff4p )2/2n2 = 17.01 eV Zeff

4p = 4.47;

Sc++

As increase net charge the increases in the differential shielding for 4s vs 3d is less important than the difference in n quantum number 3 vs 4.

(3d)(4s): IP4s = (Zeff4s )2/2n2 = 12.89 eV Zeff

4s = 3.89;

(3d)2: IP3d = (Zeff3d )2/2n2 = 12.28 eV Zeff

3d = 2.85;

(3d)(4p): IP4p = (Zeff4p )2/2n2 = 9.66 eV Zeff

4p = 3.37;

Sc+

(3d)(4s)2: IP4s = (Zeff4s )2/2n2 = 6.56 eV Zeff

4s = 2.78;

(4s)(3d)2: IP3d = (Zeff3d )2/2n2 = 5.12 eV Zeff

3d = 1.84;

(3d)(4s)(4p): IP4p = (Zeff4p )2/2n2 = 4.59 eV Zeff

4p = 2.32;

Sc


Implications on transition metals

The simple Aufbau principle puts 4s below 3dBut increasing the charge tends to prefers 3d vs 4s. Thus Ground state of Sc 2+ , Ti 2+ …..Zn 2+ are all (3d)n

For all neutral elements K through Zn the 4s orbital is easiest to ionize.

This is because of increase in relative stability of 3d for higher ions


Transtion metal orbitals


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


Consider the ground state of B: [Be](2p)1

Ignore the [Be] core then

Can put 1 electron in 2px, 2py, or 2pz each with either up or down spin. Thus get 6 states.

We will depict these states by simplified contour diagrams in the xz plane, as at the right.

Of course 2py is zero on this plane. Instead we show it as a circle as if you can see just the front part of the lobe sticking out of the paper.

2px

2pz

2py

z

x

Because there are 3 degenerate states we denote this as a P state. Because the spin can be +½ or –½, we call it a spin doublet and we denote the overall state as 2P


Consider the ground state of C: [Be](2p)2


Can put 2 electrons in 2px, 2py, or 2pz each with both up and down spin.

Or can put one electron in each of two orbitals: (2px)(2py), (2px)(2px), (2py)(2pz),

We will depict these states by simplified contour diagrams in the xz plane, as at the right.

Which state is better? The difference is in the electron-electron repulsion: 1/r12

z

x

z

x

z

x

(2px)2

(2pz)2

(2px)(2pz)

Clearly two electrons in the same orbital have a much smaller average r12 and hence a much higher e-e repusion. Thus the ground state has each electron in a different 2p orbital


Consider the states of C: formed from (x)(y), (x)(z), (y)(z)

Consider first (x)(y): can form two spatial products: Φx(1)Φy(2) and Φy(1)Φx(2)

These are not spatially symmetric, thus combine

Φ(1,2)s=φx(1) φy(2) + φy(1) φx(2)

Φ(1,2)a= φx(1) φy(2) - φy(1) φx(2)

y

x

(2px)(2py)

Which state is better? The difference is in the electron-electron repulsion: 1/r12

To analyze this, expand the orbitals in terms of the angular coordinates, r,θ,φ

Φ(1,2)s= f(r1)f(r2)(sinθ1)(sinθ2)[(cosφ1)(sinφ2)+(sinφ1) (cosφ2)]

=f(r1)f(r2)(sinθ1)(sinθ2)[sin(φ1+φ2)]

Φ(1,2)a= f(r1)f(r2)(sinθ1)(sinθ2)[(cosφ1)(sinφ2)-(sinφ1) (cosφ2)]

=f(r1)f(r2)(sinθ1)(sinθ2)[sin(φ2 -φ1)]


Consider the symmetric and antisymmetric combinations of (x)(y)

y

x

(2px)(2py)Φ(1,2)s= f(r1)f(r2)(sinθ1)(sinθ2)[sin(φ1+φ2)]

Φ(1,2)a= f(r1)f(r2)(sinθ1)(sinθ2)[sin(φ2 -φ1)]

The big difference is that Φ(1,2)a = 0 when φ2 = φ1 and is a maximum for φ2 and φ1 out of phase by /2.

But for Φ(1,2)s the probability of φ2 = φ1 is comparable to that of being out of phase by /2.

Thus the best combination is Φ(1,2)a

Thus for 2 electrons in orthogonal orbitals, high spin is best because the electrons can never be at same spot at the same time

Combining with the spin parts we get

[φx(1) φy(2) + φy(1) φx(2or spin = S = 0

[φx(1) φy(2) - φy(1) φx(2also andor spin = S = 1


Summarizing the states for C atom

Ground state: three triplet states=2L=1. Thus L=1, denote as 3P (xy-yx) ≡ [x(1)y(2)-y(1)x(2)] (xz-zx)(yz-zy)Next state: five singlet states=2L+1. Thus L=2, denote as 1D (xy+yx)(xz+zx)(yz+zy)(xx-yy)(2zz-xx-yy)Highest state: one singlet=2L+1. thus L=0. Denote as 1S(zz+xx+yy)

y

x

(2px)(2py)

Hund’s rule. Given n electrons distributed among m equivalent orgthogonal orbitals, the ground state is the one with the highest possible spin. Given more than one state with the highest spin, the highest orbital angular momentum is the GS


Calculating energies for C atom

The energy of xy is

Exy = hxx + hyy + Jxy = 2hpp +Jxy

Thus the energy of the 3P state is

E(3P) = Exy – Kxy = 2hpp +Jxy - Kxy

For the (xy+yx) component of the 1D state, we get

E(1D) = Exy + Kxy = 2hpp +Jxy + Kxy

Whereas for the (xx-yy) component of the 1D state, we get

E(1D) = Exx - Kxy = 2hpp +Jxx - Kxy

This means that Jxx - Kxy = Jxy + Kxy so that

Jxx = Jxy + 2Kxy

Also for (2zz-xx-yy) we obtain E = 2hpp +Jxx - Kxy

For (zz+xx+yy) we obtain

E(1S) = 2hpp + Jxx + 2 Kxy

y

x

(2px)(2py)


Summarizing the energies for C atom

E(1S) = 2hpp + Jxx + 2Kxy

E(1D) = 2hpp +Jxx - Kxy = 2hpp +Jxy + Kxy

E(3P) = 2hpp + Exy – Kxy = 2hpp +Jxy - Kxy

2Kxy

3Kxy


Comparison with experiment

E(1S)

E(1D)

E(3P)

2Kxy

3Kxy

C Si Ge Sn Pb

TA’s look up data and list excitation energies in eV and Kxy in eV.

Get data from Moore’s tables or paper by Harding and Goddard Ann Rev Phys Chem ~1985)


Summary ground state for C atom

z

x

(2px)(2pz)

Ψ(1,2,3,4,5,6)xz= A[(1s)(1s)(2s)(2s)(2px)(2pz)] =

= A[(1s(2s)2(2px)(2pz)] =A[(Be)(2px)(2pz)] =

= A[(x)(z)] = which we visualize as

Ψ(1,2,3,4,5,6)xy = A[(x)(y)]

which we visualize as

Ψ(1,2,3,4,5,6)yz = A[(y)(z)]


z

x

Note that we choose to use the xz plane for all 3 wavefunctions, so that the py orbitals look like circles (seeing only the + lobe out of the plane

z

x


Consider the ground state of N: [Be](2p)3


Can put one electron in each of three orbitals: (2px)(2py)(2pz)

Or can put 2 in 1 and 1 in another:

(x)2(y), (x)2(z), (y)2(x), (y)2(z), (z)2(x), (z)2(y)

As we saw for C, the best state is (x)(y)(z) because of the lowest ee repulsion.

xyz can be combined with various spin functions, but from Hund’s rule we expect A[(x)(y)(z)] = [Axyz] to be the ground state. (here Axyz is the antisymmetric combination of x(1)y(2)x(3)]The four symmetric spin functions areWith Mswhich refer to as S=3/2 or quartetSince there is only one xyz state = 2L+1 with L=0, we denote it as L=0, leading to the 4S state.


Energy of the ground state of N: A[(x)(y)(z)] = [Axyz]

Simple product xyz leads to

Exyz = 3hpp + Jxy + Jxz + Jyz

E(4S) = <xyz|H|A[xyz] <xyz|A[xyz]

Denominator = <xyz|A[xyz] = 1

Numerator = Exyz - Kxy - Kxz – Kyz =

E(4S) = 3hpp + (Jxy - Kxy) + (Jxz - Kxz) + (Jyz – Kyz)

E(2P)

E(2D)

E(4S)

4Kxy

2Kxy

=3hpp + 3Jxy - 3Kxy

=3hpp + 3Jxy + 1Kxy = 3hpp + 2Jxy + Jxx - 1Kxy

=3hpp + 2Jxy + Jxx + 1Kxy

TA’s check this

z

x

Pictorial representation of the N ground

state

Since Jxy = Jxz = Jyz

and Kxy= Kxz = Kyz


Comparison with experiment

E(1S)

E(1D)

E(3P)

4Kxy

2Kxy

N P As Sb Bi




Consider the ground state of O: [Be](2p)4

Only have 3 orbitals, x, y, and z. thus must have a least one doubly occupied

Choices: (z)2(x)(y), (y)2(x)(z), (x)2(z)(y)

and (z)2(x)2, (y)2(x)2, (y)2(z)2

Clearly it is better to have two singly occupied orbitals.

Just as for C atom, two singly occupied orbitals lead to both a triplet state and a singlet state, but the high spin triplet with the same spin for the two singly occupied orbitals is best

z

x

z

x

(2px)2(2py)(2pz)

(2pz)2(2px)2


Ψ(1,2,3,4,5,6)xz= A[(1s)(1s)(2s)(2s)(2py)(2py)(2px)(2pz)]

= A[(1s(2s)2(2py)(2py)(2px)(2pz)] =

= A[(y)(y)(x)(z)] = which we visualize as

Summary ground state for O atom

z

x

Ψ(1,2,3,4,5,6)xy = A[(z)(z)(((x)(y)]


Ψ(1,2,3,4,5,6)yz = A[(x)(x)((y)(z)]

which we visualize as (2px)2(2py)(2pz)

z

x

z

x

We have 3 = 2L+1 equivalent spin triplet (S=1) states that we denote as L=1 orbital angular momentum, leading to the 3P state

(2py)2(2px)(2pz)

(2pz)2(2px)(2py)


Calculating energies for O atom

is

Exz = E(Be) + 2hyy + hxx + hzz + Jyy+ 2Jxy+ 2Jyx+ Jxz – Kxy – Kyz – Kxz

Check: 4 electrons, therefore 4x3/2 = 6 coulomb interactions

3 up-spin electrons, therefor 3x2/2 = 3 exchange interactions

Other ways to group energy terms

Exz = 4hpp + Jyy + (2Jxy – Kxy) + (2Jyx – Kyz) + (Jxz– Kxz)

Same energy for other two components of 3P state

(2py)2(2px)(2pz)The energy of Ψ(1,2,3,4,5,6)xz= A[(1s)(1s)(2s)(2s)(2py)(2py)(2px)(2pz)]

= A[[Be](y)(y)(x)(z)]


Comparison of O states with C statesO (2p)4 3P C (2p)2 3PNe (2p)6 1S

Compared to Ne, we have

Hole in x and z

Hole in y and z

Hole in x and y

Compared to Be, we have

Electron in x and z

Electron in y and z

Electron in x and y

z

x

Thus holes in O map to electrons in C

z

x


Summarizing the energies for O atom

E(1S)

E(1D)

E(3P)

2Kxy

3Kxy

O S Se Te Po




Consider the ground state of F: [Be](2p)5

Only have 3 orbitals, x, y, and z. thus must have two doubly occupied

Choices: (x)2(y)2 (z), (x)2(y)(z)2, (x)(y)2(z)2

Clearly all three equivalent give rise to spin doublet. Since 3 = 2L+1 denote as L=1 or 2P

Ψ(1-9)z= A[(1s)(1s)(2s)(2s)(2py)(2py)(2px)(2pz)]

= A[(1s(2s)2(2py)(2py)(2px)(2pz)] =

= A{[Be](x)(x)(y)(y)(z)}


z

x


Calculating energies for F atom

is

Exz = 5hpp + Jxx+ Jyy+ (4Jxy – 2Kxy) + (2Jxz – Kxz) + (2Jyx– Kyz)


3 up-spin electrons, therefore 3x2/2 = 3 exchange interactions

2 down-spin electrons, therefore 2x1/2 = 1 exchange interaction



(2px)2(2py)2(2pz)The energy of

Ψ(1-9)z= A{[Be](2px)(2px)(2py)(2py)(2pz)]


Comparison of F states with B statesF (2p)5 2P B (2p)1 2PNe (2p)6 1S

Compared to Ne, we have

Hole in z

Hole in x

Hole in y

Compared to Be, we have

Electron in z

Electron in x

Electron in y

Thus holes in F map to electrons in B

z

x

z

x


Calculating energies for F atom

is

Exz = 5hpp + Jxx+ Jyy+ (4Jxy – 2Kxy) + (2Jxz – Kxz) + (2Jyx– Kyz)






(2px)2(2py)2(2pz)The energy of

Ψ(1-9)z= A{[Be](2px)(2px)(2py)(2py)(2pz)]


Consider the ground state of Ne: [Be](2p)6

Only have 3 orbitals, x, y, and z. thus must have all three doubly occupied

Choices: (x)2(y)2(z)2

Thus get spin singlet, S=0

Since just one spatial state, 1=2L+1 L=0.

denote as 1S

Ψ(1-10)z = A[[Be](x)(x)(y)(y)(z)(z)]

which we visualize asNe (2p)6 1S

z

x


Calculating energy for Ne atom

is

Exz = 6hpp +Jxx+Jyy+ Jzz + (4Jxy – 2Kxy) + (4Jxz – 2Kxz) + (4Jyx– 2Kyz)

Check: 6 p electrons, therefore 6x5/2 = 15 coulomb interactions



Since Jxx = Kxx we can rewrite this as

Exz = 6hpp +(2Jxx-Kxx ) +(2Jyy-Kyy ) +(2Jzz-Kzz ) + 2(2Jxy – Kxy) +

2(2Jxz – Kxz) + 2(2Jyx– Kyz)

Which we will find later to be more convenient for calculating the wavefunctions using the variational principle

(2px)2(2py)2(2pz)2

The energy of

Ψ(1-9)z= A{[Be](2px)(2px)(2py)(2py)(2pz)(2pz)}


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

Starting with the ground state of He, (1s)2 = A(He1s)(He1s) and bringing 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

Starting 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 bringing 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 reponse 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 orthonal to the bonding orbitals.


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

Starting with the ground state of H2O

We can 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. θe Re

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. Indeed as for H2N we find big deviations for the 1st row, but because there are not 3 bad H---H interactions the deviations are larger. θe Re


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

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


More on symmetry

We saw in previous lectures that symmetry often has a profound effect on the solutions of the Schrödinger Equation

For example inversion symmetry g or u states

Permutational symmetry in electrons symmetric and antisymmetric wavefunctions for transposition of space, spin, and space-spin coordinates.

More general discussions make use of group theory or more correctly group representation theory.

Here we will use outline some of the essential elements relevant to this course


The symmetry operations for the Schrödinger equation form a group

Consider some symmetry operator, R1, of the Hamiltonian, for example R1 = I (inversion) for H2 or the transposition, , for any two electron system. Then

If HΨ=EΨ if follows that

R1 (HΨ)=H(R1Ψ)= E(R1Ψ) so that R1Ψ is an eigenfunction with the same E. The set of symmetry operators {R1 , R2 , … Rn} = G forms a Group. This follows since:

1). Closure: If R1,R2 G (that is both are symmetry elements) then R2 R1 is also a symmetry element, R2 R1 G (we say that the set of symmetry operations is closed). This follows since

(R2 R1) (HΨ)= R2 H(R1Ψ)= R2 E(R1Ψ)= E(R2 R1Ψ)}, that is

(R2 R1Ψ) is also an eigenstate of H with the same energy E

2. Identity. Also R1 = e (identity) G. Clearly eΨ is a symmetry element


The symmetry operations for the schrodinger equation form a group, continued

3. Associativity. If {R1,R2,R3 } G then (R1R2)R3 =R1(R2R3). This follows since (R1R2)R3HΨ = (R1R2)ER3Ψ = E (R1R2)R3Ψ and R1(R2R3)HΨ = R1H(R2R3)Ψ = R1E (R2R3)Ψ= ER1(R2R3)Ψ

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

=e. This follows for any finite set that is closed since, there must be some integer p, such that (R)p = e. Thus [(R)p-1]R = e = R[(R)p-1] where (R)p-1 ≡ (R1)-1 The above four conditions are necessary and sufficient to define what the Mathematicians call a group. The theory of the properties of such a group is called Group Theory. This is a vast field, but the only part important to QM and materials is group representation theory.


Group representation Theory

If G is a group with n elements, consider the set of functions S={Ψ1= R1Ψ, Ψ2= R2Ψ, …Ψn= RnΨ}. From the properties of a group, any operation R G on any Ψi S leads to a linear combination of functions in S. This leads to a set of nxn matrices that multiple in exactly the same way as the elements of G , so the Mathematicians say that S is a basis for the group G and that these matrices form a representation of G. The mathematicians went on to show that one could derive a set of irreducible reorientations from which one can construct any possible representation. This theory was worked out ~1905 mostly in Germany but it had few real practical applications until QM.

In QM these irreducible representations are important because they constitute the possible symmetries of all possible eigenfunctions of the Hamiltonian H.


An example, C2v

The point of going through these definitions is to make sure that the students understand that the use of Group Theory in QM involves very simple concepts. No need to get afraid of the complex notations and nomenclature used sometimes. For those that want to see my views on Group Theory with some simple applications. Some notes are available for this course, denoted as Volume V, chapter 9. This dates from ~1972 when I used to teach a full year course on QM and from 1976 when I included such materials in my then full year course on chemical bonding

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


C2v

Czn denotes a rotation of 2/n about the z axis (C for cyclisch, German for cyclic)

xz denotes a reflection or mirror in the xz plane (for spiegel, German for mirror)

The symmetry operators are

And the group {e, C2z, xz, yz} is denoted as C2z


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

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

In the previous slide we saw that C2zxz= yz which means that the symmetries for yz are already implied by C2zxz. Thus we consider C2z and xz as the generators of the group.

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 xz moves the maximum number of atoms

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


More on C2v

Each of the symmetry elements are of order two,

That is Rp = e where p=2.

We saw earlier, with R = I (Inversion) and (transposition)

That with p=2 the eigenfunctions of the H must transform as symmetric (+) or antisymmetric (-) under the operation.

The character table reflects this.

Also since C2zxz= yz , we see that the symmetry of yz is determined by those of C2z and xz

The general naming convention is A and B are symmetric and antisymmetric with respect to the principal rotation axis.

Subscripts 1 and 2 are symmetric and antisymmetric with respect to the mirror plane generator.


Applications for C2v

Consider that an atom with a single electron in a p orbital (say B or Al) is placed at a site in a crystal with C2v symmetry. The character table tells us that in general, the px, py, and pz states will all have different energies. On the other hand if the symmetry were that of a square (D4h), px and py would be degenerate, but pz might be different, and in the symmetry of an octahedron (Oh) or tetrahedron (Td), the three p states will be degenerate.

We will get to such issues later in the course but for now we will use the symmetry only to provide names for the states

name1-e N-ea1

a2

b1

b2

N electron wavefunction, use A1,A2 etc one electron orbitals: use a1,a2 etc


Symmetries for H2O, NH2, and CH2

Previously we discussed the wavefunctions for H2O, NH2, and CH2 from bonding H to two p orbitals

H2O NH2


OHx bond OHz bond

H2O

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

NHz bondNHx bond

NH2

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

CH2


First do CH2 singlet

CHR bondCHL bond

Ψ1=A{(C2py)0[(CpL )(HL)+(HL)(CpL[(CpR)(HR+(HR)(CpR

We used x and z for the bond directions before but now we want new x,y,z related to the symmetry group.Thus we denote bonds as L and R for left and right. Note that the C (1s) and (2s) pairs are invariant under all operationsApplying xz to the wavefunction leads to

Ψ2=A{(C2py)0[(CpR)(HR)+(HR)(CpR[(CpL)(HL+(HL)(CpL

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 get that xzΨ1 = Ψ2 = Ψ1

Thus interchanging a bond pair leaves Ψ invariant

pR2s

zy HL

HR

pL


Finish CH2 singlet

CHR bondCHL bond

Ψ1=A{(C2py)0[(CpL )(HL)+(HL)(CpL[(CpR)(HR+(HR)(CpR

Applying Cz to the wavefunction also interchanges two bond pairs so that

C2zΨ1 = Ψ2 = Ψ1 A1 or A2 symmetryAlso xzΨ1 = Ψ2 = Ψ1 A1 or B1 symmetry

Thus we conclude that

the symmetry of this state of CH2 is 1A1,

where the superscript 1 spin singlet or S=0

pR2s

zy HL

HR

pL


Next consider NH2

The wavefunction for NH2 differs from that of CH2 only in having a singly occupied px orbital

A{(N2px)1[(NpL )(HL)+(HL)(NpL[(NpR)(HR+(HR)(NpR

NHR bondNHL bond

pR2s

zy HL

HR

pL

Since the symmetry operation simultaneously operates on all electrons, we can consider the effects on Npx separately.

Here we see that C2z changes the sign but, xz does not. Thus Npx transforms as b1. (note it is not necessary to examine xz since it is not a generator).

With one unpaired spin this state is S= ½ or doublet Thus the symmetry of NH2 is 2B1


Now do OH2

The wavefunction for OH2 differs from that of NH2 only in having px orbital doubly occupied

A{(O2px)2[(OpL )(HL)+(HL)(OpL[(OpR)(HR+(HR)(OpR

OHR bondOHL bond

2s

zy HL

HR

pL

Since the symmetry operation simultaneously operates on all electrons, we can consider the effects on Opx separately.

Since C2z changes the sign of Opx twice, the wavefunction is invariant. Thus (Opx)2 transforms as A1.

With no unpaired spin this state is S= 0 or singlet

Thus the symmetry of OH2 is 1A1

pR


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


C3v

The symmetry elements are:

and the symmetry group is denoted as C3v.z

x


x

b

c

x

b

c

Stereographic projections for C3v

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

e

C2v

C3

x

b

c C32= C3

-1

x

b

c

x

b

c

xz xzC3

x

b

c

xzC32


Combining the 6 operations for C3v leads to

C2v

x

b

c

e

C3

C32= C3

-1

xz

xzC3

xzC32

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

Also we see that the generators are xz and C3

Note that the operations do not all commute.

Thus xzC3 ≠ xzC3

Instead we get

xzC3 = xzC32

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


More on C3v

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.


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.

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

Lecture 4, January 14, 2011 aufbau principle atoms

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Transcript of Lecture 4, January 14, 2011 aufbau principle atoms