1. William A. Goddard, III, wag@wag.caltech.edu
Lecture 4, January 14, 2011
aufbau principle atoms
Nature of the Chemical Bond with applications to catalysis, materials science, nanotechnology, surface science, bioinorganic chemistry, and energy
William A. Goddard, III,
316 Beckman Institute, x3093
Charles and Mary Ferkel Professor of Chemistry, Materials Science, and Applied Physics, California Institute of Technology
Teaching Assistants: Wei-Guang Liu
Caitlin Scott

2. The excited states of H atomy z θ φ
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 2 – Z/r is independent of θ and φ which can be seen by noting that
2 = 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

3. The ground state of H atomx y z θ φ
h = - ½ 2 – 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=0
again ignoring normalization.
The radial wavefunction is nodeless, as expected
The angular function is a constant, which is clearly the best for KE.

4. The p excited angular states of H atom
φnlm = 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 plane
Also 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

5. 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?
For example we might rotate px by an angle a about the y axis to form px’. However multiplying, say by pz, leads to the integrand pzpx’ which clearly does not integrate to zero x z + -
px’ a z
pzpx’ .
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 a + - - x +

6. 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)

7. More d angular 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 x z + -
dz2-x2

8. More d angular 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
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 x z + -
dz2-x2

9. Summarizing we get 5 d angular functionsx z + -
dz2
Z20 = dz2 = [3 z2 – r2 ] m=0, ds
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 2p = 360º
When this number, m=0 we call it a s function
When m=1 we call it a p function
When m=2 we call it a d function
When m=3 we call it a f function
m = 1, dp
m = 2, dd
57º

10. 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

11. The real (Zlm) and complex (Ylm) momentum functions
Here the bar over m  negative

12. 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
Zr = 7.1
Zr = 2
Zr = 1.9
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)

13. 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
angular nodal planes
Size (a0)
radial nodal planes
total nodal planes
name 1s 2s 2p 3s 3p 3d 4s 4p 4d 4f ˉ R ˉ R ˉ R

14. Sizes hydrogen orbitals
Hydrogen orbitals 1s, 2s, 2p, 3s, 3p, 3d, 4s, 4p, 4d, 4f
Angstrom (0.1nm) 0.53, 2.12, , ˉ R
=(n2/Z) a0 = 0.53 n2 A
1.7 H
4.8 H C
H--H H H H H
0.74 H H H H

15. Hydrogen atom excited states
Energy
Enlm = - Z2/2n2 = -(1/2n2)h0 = 13.6 eV/n2
zero 4s
h0 = -0.9 eV 4p 4d 4f 3s
h0 = -1.5 eV 3p 3d 2s
h0 = -3.4 eV 2p 1s
-0.5 h0 = eV

16. Plotting of orbitals: line cross-section vs. contour

17. Contour plots of 1s, 2s, 2p hydrogenic orbitals
1s, no nodes
2s, one radial node
2p, one angular node

18. Contour plots of 3s, 3p, 3d hydrogenic orbitals
3p, one angular node one radial node
3s, two radial nodes
3d, two angular nodes

19. Contour plots of 4s, 4p, 4d hydrogenic orbtitals
4p, one angular node two radial nodes
4s, 3 radial nodes
4d, two angular nodes and one radial node

20. Contour plots of hydrogenic 4f orbitals
All seven 4f have three angular nodes and no radial nodes

21. He atom – describe using He+ orbitals
With 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[(Φ1sa)(Φ1sb)]= Φ1s(1)Φ1s(2) (ab-ba)
EHe= 2<1s|h|1s> + J1s,1s
two one-electron energies one Coulomb repulsion

22. He atom – describe using He+ orbitals
With 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[(Φ1sa)(Φ1sb)]= Φ1s(1)Φ1s(2) (ab-ba)
EHe= 2<1s|h|1s> + J1s,1s
First lets review the energy for He+. Writing Φ1s = exp(-zr) we determine the optimum z for He+ as follows
<1s|KE|1s> = + ½ z2 (goes as the square of 1/size)
<1s|PE|1s> = - Zz (linear in 1/size)

23. He atom – describe using He+ orbitals
With 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[(Φ1sa)(Φ1sb)]= Φ1s(1)Φ1s(2) (ab-ba)
EHe= 2<1s|h|1s> + J1s,1s
First lets review the energy for He+. Writing Φ1s = exp(-zr) we determine the optimum z for He+ as follows
<1s|KE|1s> = + ½ z2 (goes as the square of 1/size)
<1s|PE|1s> = - Zz (linear in 1/size)
Applying the variational principle, the optimum z must satisfy dE/dz = z - Z = 0 leading to z = Z,
This value of z leads to KE = ½ Z2, PE = -Z2, E=-Z2/2 = -2 h0.
write PE=-Z/R0, so that the average radius is R0=1/z

24. J1s,1s = e-e energy of He atom – He+ orbitals
Now consider He atom: EHe = 2(½ z2) – 2Zz + J1s,1s
How can we estimate J1s,1s
Assume that each electron moves on a sphere
With the average radius R0 = 1/z
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 z
A rigorous calculation (notes chapter 3, appendix 3-C page 6)
Gives J1s,1s = (5/8) z

25. The optimum atomic orbital for He atom
He atom: EHe = 2(½ z2) – 2Zz + (5/8)z
Applying the variational principle, the optimum z must satisfy dE/dz = 0 leading to
2z - 2Z + (5/8) = 0
Thus z = (Z – 5/16) =
KE = 2(½ z2) = z2
PE = - 2Zz + (5/8)z = -2 z2
E= - z2 = h0
Ignoring e-e interactions the energy would have been E = -4 h0
The exact energy is E = h0 (from memory, TA please check). Thus this simple approximation accounts for 98.1% of the exact result.

26. Interpretation: The optimum atomic orbital for He atom
ΨHe(1,2) = Φ1s(1)Φ1s(2) with Φ1s = exp(-zr)
We find that the optimum z = (Z – 5/16) =
With this value of z, the total energy is E= - z2 = h0
This wavefunction can be interpreted as two electrons moving independently in the orbital Φ1s = exp(-zr) 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 =1.69
The total energy is just the sum of the individual energies.
Ionizing the 2nd electron, the 1st electron readjusts to z = Z = 2
With E(He+) = -Z2/2 = - 2 h0. thus the ionization potential (IP) is h0 = 23.1 eV (exact value = 24.6 eV)

27. Now lets add a 3rd electron to form Li
ΨLi(1,2,3) = A[(Φ1sa)(Φ1sb)(Φ1sg)]
Problem with either g = a or g = b, 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[(Φ1sa)(Φ1sb) )(Φ2sa)] or
ΨLi(1,2,3) = A[(Φ1sa)(Φ1sb) )(Φ2pza)] (or 2px or 2py)
First consider Li+ with ΨLi(1,2,3) = A[(Φ1sa)(Φ1sb)]
Here Φ1s = exp(-zr) with z = Z = 2.69 and
E = -z2 = h0. Since the E (Li2+) = -9/2 = -4.5 h0 the IP = h0 = 74.1 eV
The size of the 1s orbital is R0 = 1/z = a0 = 0.2A

28. Consider adding the 3rd electron to the 2p orbital
ΨLi(1,2,3) = A[(Φ1sa)(Φ1sb) )(Φ2pza)] (or 2px or 2py)
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 = eV 1s
0.2A 2p
2.12A

29. Consider adding the 3rd electron to the 2s orbital
ΨLi(1,2,3) = A[(Φ1sa)(Φ1sb) )(Φ2pza)] (or 2px or 2py)
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 = h0 = eV
0.2A 1s
2.12A 2s
R~0.2A

30. Li atom excited states Energy MO picture State picture zero
DE = 2.2 eV
17700 cm-1
564 nm
1st excited state
h0 = -3.4 eV
(1s)2(2p) 2p
h0 = -5.6 eV
(1s)2(2s) 2s
Ground state
Exper
671 nm
DE = 1.9 eV
h0 = 74.1 eV 1s

31. Aufbau principle for atoms
Energy 14 10 4f
Kr, 36 4d 6 4p
Zn, 30 2 10 4s
Ar, 18 3d 6 3p 2 3s
Ne, 10 6
Including shielding, 2s<2p
3s<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 2p 2 2s
He, 2 2 1s

32. Kr, 36
Zn, 30
Ar, 18
Ne, 10
He, 2

33. 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

34. Many-electron configurations
General aufbau ordering
Particularly stable

35. 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

39. Transition metals; consider [Ar] plus one electron
[IP4s = (Zeff4s )2/2n2 = 4.34 eV  Zeff4s = 2.26; 4s<4p<3d
IP4p = (Zeff4p )2/2n2 = 2.73 eV  Zeff4p = 1.79;
IP3d = (Zeff3d )2/2n2 = 1.67 eV  Zeff3d = 1.05;
IP4s = (Zeff4s )2/2n2 = eV  Zeff4s = 3.74; 4s<3d<4p
IP3d = (Zeff3d )2/2n2 = eV  Zeff3d = 2.59;
IP4p = (Zeff4p )2/2n2 = 8.73 eV  Zeff4p = 3.20;
IP3d = (Zeff3d )2/2n2 = eV  Zeff3d = 4.05; 3d<4s<4p
IP4s = (Zeff4s )2/2n2 = eV  Zeff4s = 5.04;
IP4p = (Zeff4p )2/2n2 = eV  Zeff4p = 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 4
Thus charged system prefers 3d vs 4s

40. Transition metals; consider Sc0, Sc+, Sc2+
3d: IP3d = (Zeff3d )2/2n2 = eV  Zeff3d = 4.05;
4s: IP4s = (Zeff4s )2/2n2 = eV  Zeff4s = 5.04;
4p: IP4p = (Zeff4p )2/2n2 = eV  Zeff4p = 4.47;
Sc++
(3d)(4s): IP4s = (Zeff4s )2/2n2 = eV  Zeff4s = 3.89;
(3d)2: IP3d = (Zeff3d )2/2n2 = eV  Zeff3d = 2.85;
(3d)(4p): IP4p = (Zeff4p )2/2n2 = 9.66 eV  Zeff4p = 3.37;
Sc+
(3d)(4s)2: IP4s = (Zeff4s )2/2n2 = 6.56 eV  Zeff4s = 2.78;
(4s)(3d)2: IP3d = (Zeff3d )2/2n2 = 5.12 eV  Zeff3d = 1.84;
(3d)(4s)(4p): IP4p = (Zeff4p )2/2n2 = 4.59 eV  Zeff4p = 2.32; 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.

41. Implications on transition metals
The simple Aufbau principle puts 4s below 3d
But 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

42. Transtion metal orbitals

43. 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

44. Consider the ground state of B: [Be](2p)1x
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 z
2pz
2py
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

45. Consider the ground state of C: [Be](2p)2z x
Ignore the [Be] core then
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.
(2px)2 z x
(2pz)2 z x
(2px)(2pz)
Which state is better? The difference is in the electron-electron repulsion: 1/r12
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

46. 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)
(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)]

47. Consider the symmetric and antisymmetric combinations of (x)(y)
(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 p/2.
But for Φ(1,2)s the probability of φ2 = φ1 is comparable to that of being out of phase by p/2.
Thus the best combination is Φ(1,2)a
Combining with the spin parts we get
[φx(1) φy(2) + φy(1) φx(2)](ab-ba) or spin = S = 0
[φx(1) φy(2) - φy(1) φx(2)](ab+ba), also aa and bb or spin = S = 1
Thus for 2 electrons in orthogonal orbitals, high spin is best because the electrons can never be at same spot at the same time

48. 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

49. 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)

50. Summarizing the energies for C atom
E(1S) = 2hpp + Jxx + 2Kxy
3Kxy
E(1D) = 2hpp +Jxx - Kxy = 2hpp +Jxy + Kxy
2Kxy
E(3P) = 2hpp + Exy – Kxy = 2hpp +Jxy - Kxy

51. 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)

52. Summary ground state for C atom
Ψ(1,2,3,4,5,6)xz= A[(1sa)(1sb)(2sa)(2sb)(2pxa)(2pza)] =
= A[(1s)2(2s)2(2pxa)(2pza)] =A[(Be)(2pxa)(2pza)] =
= A[(xa)(za)] = which we visualize as z x
(2px)(2pz) z x
Ψ(1,2,3,4,5,6)xy = A[(xa)(ya)]
which we visualize as
Ψ(1,2,3,4,5,6)yz = A[(ya)(za)]
which we visualize as
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

53. Consider the ground state of N: [Be](2p)3
Ignore the [Be] core then
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[(xa)(ya)(za)] = [Axyz]aaa to be the ground state. (here Axyz is the antisymmetric combination of x(1)y(2)x(3)]
The four symmetric spin functions are
aaa, aab + aba + baa, abb + bba + bab, bbb
With Ms = 3/2, 1/2, -1/2 -3/2, which refer to as S=3/2 or quartet
Since there is only one xyz state = 2L+1 with L=0, we denote it as L=0, leading to the 4S state.

54. Energy of the ground state of N: A[(xa)(ya)(za)] = [Axyz]aaa
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) z x
Pictorial representation of the N ground state
E(2P)
E(2D)
E(4S)
4Kxy
2Kxy
=3hpp + 2Jxy + Jxx + 1Kxy
TA’s check this
=3hpp + 3Jxy + 1Kxy = 3hpp + 2Jxy + Jxx - 1Kxy
Since Jxy = Jxz = Jyz
and Kxy= Kxz = Kyz
=3hpp + 3Jxy - 3Kxy

55. Comparison with experiment
E(1S)
2Kxy
E(1D)
4Kxy
E(3P)
N P As Sb Bi
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)

56. 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
(2px)2(2py)(2pz) z x z x
(2pz)2(2px)2

57. Summary ground state for O atom
Ψ(1,2,3,4,5,6)xz= A[(1sa)(1sb)(2sa)(2sb)(2pya)(2pyb)(2pxa)(2pza)]
= A[(1s)2(2s)2(2pya)(2pyb)(2pxa)(2pza)] =
= A[(ya)(yb)(xa)(za)] = which we visualize as z x
(2py)2(2px)(2pz)
Ψ(1,2,3,4,5,6)yz = A[(xa)(xb)((ya)(za)]
which we visualize as z x
(2px)2(2py)(2pz) z x
Ψ(1,2,3,4,5,6)xy = A[(za)(zb)(((xa)(ya)]
which we visualize as
(2pz)2(2px)(2py)
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

58. Calculating energies for O atom
(2py)2(2px)(2pz)
The energy of Ψ(1,2,3,4,5,6)xz= A[(1sa)(1sb)(2sa)(2sb)(2pya)(2pyb)(2pxa)(2pza)]
= A[[Be](ya)(yb)(xa)(za)] 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

59. Comparison of O states with C states
Ne (2p)6 1S
C (2p)2 3P
O (2p)4 3P
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 z x
Thus holes in O map to electrons in C

60. Summarizing the energies for O atom
O S Se Te Po
3Kxy
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)
E(1D)
2Kxy
E(3P)

61. 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[(1sa)(1sb)(2sa)(2sb)(2pya)(2pyb)(2pxa)(2pza)]
= A[(1s)2(2s)2(2pya)(2pyb)(2pxa)(2pza)] =
= A{[Be](xa)(xb)(ya)(yb)(za)}
which we visualize as z x

62. Calculating energies for F atom
(2px)2(2py)2(2pz)
The energy of
Ψ(1-9)z= A{[Be](2pxa)(2pxb)(2pya)(2pyb)(2pza)] is
Exz = 5hpp + Jxx+ Jyy+ (4Jxy – 2Kxy) + (2Jxz – Kxz) + (2Jyx– Kyz)
Check: 5 electrons, therefore 5x4/2 = 10 coulomb interactions
3 up-spin electrons, therefore 3x2/2 = 3 exchange interactions
2 down-spin electrons, therefore 2x1/2 = 1 exchange interaction
Other ways to group energy terms
Same energy for other two components of 2P state

63. Comparison of F states with B states
Ne (2p)6 1S
B (2p)1 2P
F (2p)5 2P
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 z x z x
Thus holes in F map to electrons in B

64. Calculating energies for F atom
(2px)2(2py)2(2pz)
The energy of
Ψ(1-9)z= A{[Be](2pxa)(2pxb)(2pya)(2pyb)(2pza)] is
Exz = 5hpp + Jxx+ Jyy+ (4Jxy – 2Kxy) + (2Jxz – Kxz) + (2Jyx– Kyz)
Check: 5 electrons, therefore 5x4/2 = 10 coulomb interactions
3 up-spin electrons, therefore 3x2/2 = 3 exchange interactions
2 down-spin electrons, therefore 2x1/2 = 1 exchange interaction
Other ways to group energy terms
Same energy for other two components of 2P state

65. 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](xa)(xb)(ya)(yb)(za)(zb)]
which we visualize as z x
Ne (2p)6 1S

66. Calculating energy for Ne atom
(2px)2(2py)2(2pz)2
The energy of
Ψ(1-9)z= A{[Be](2pxa)(2pxb)(2pya)(2pyb)(2pza)(2pzb)} is
Exz = 6hpp +Jxx+Jyy+ Jzz + (4Jxy – 2Kxy) + (4Jxz – 2Kxz) + (4Jyx– 2Kyz)
Check: 6 p electrons, therefore 6x5/2 = 15 coulomb interactions
3 up-spin electrons, therefore 3x2/2 = 3 exchange interactions
3 down-spin electrons, therefore 3x2/2 = 3 exchange interaction
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

67. Summary of ground states of Li-Ne
Li 2S (2s)1
N 4S (2p)3
Be 1S (2s)2
O 3P (2p)4
Ignore (2s)2
B 2P (2p)1
F 2P (2p)5
C 3P (2p)2
Ne 1S (2p)6

68. Bonding H atom to He
Starting with the ground state of He, (1s)2 = A(He1sa)(He1sb) and bringing up an H atom (H1sa), leads to
HeH: A[(He1sa)(He1sb)(H1sa)]
But properties of A (Pauli Principle) tell us that the H1s must get orthogonal to the He 1s since both have an a spin.
A[(He1sa)(He1sb)(θa)]
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 R

69. Bonding H atom to Ne
Starting with the ground state of Ne, (1s)2(2s)2(2p)6
Ψ(Ne)= A{(2pxa)(2pxb)(2pya)(2pyb)(2pza)(2pzb)} (omitting the Be)
and bringing up an H atom (H1sa) along the z axis, leads to
A{(2px)2(2py)2(Ne2pza)(Ne2pzb)(H1sa)}
Where we focus on the Ne2pz orbital that overlaps the H atom
The properties of A (Pauli Principle) tell us that the H1s must get orthogonal to the Ne 2pz since both have an a spin.
θ = H1s – S Ne2pz R
θ has a nodal plane, increasing its KE. Smaller R  larger S  larger increase in KE. Get a repulsive interaction, no bond

70. Now consider Bonding H atom to all 3 states of Fz x
Bring H1s along z axis to F and consider all 3 spatial states.
F 2pz doubly occupied, thus H1s must get orthogonal  repulsive
A{(2pxa)1(2py)2(F2pza)(F2pzb)(H1sa)}
F 2pz doubly occupied, thus H1s must get orthogonal  repulsive
A{(2px)2(2pya)1(F2pza)(F2pzb)(H1sa)}
F 2pz singly occupied, Now H1s need not get orthogonal if it has opposite spin, can get bonding R

71. Now consider Bonding H atom to x2y2z1 state of F
Focus on 2pz and H1s singly occupied orbitals
(Pz H + H Pz)
(Pz H - H Pz)
energy R z x
3S+
Antibonding state (S=1, triplet)
[φpz(1) φH(2) - φH(1) φpz(2)](aa)
Just like H2.
Bonding state (S=0, singlet)
[φpz(1) φH(2) + φH(1) φpz(2)](ab-ba)
1S+
Full wavefunction for bond becomes
A{(F2px)2(F2py)2[(Fpz)(H)+(H)(Fpz)](ab-ba)}
Full wavefunction for antibond becomes
A{(F2px)2(F2py)2[(Fpz)(H)-(H)(Fpz)](aa)}

72. 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

73. 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?
antibond
bond
A{(F2pxa)(F2pxb)(Hxa) (F2py)2[(Fpz)(Hz)+(Hz)(Fpz)](ab-ba)}
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.

74. Now consider Bonding H atom to all 3 states of O
Bring H1s along z axis to O and consider all 3 spatial states.
O 2pz doubly occupied, thus H1s must get orthogonal  repulsive z x
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 2P R

75. Bonding H atom to x1y2z1 and x2y1z1 states of O
The full wavefunction for the bonding state
2Py
bond
A{(O2px)2(O2pya)1[(Opz)(H)+(H)(Opz)](ab-ba)}
2Px
bond
A{(O2pxa)1(O2py)2[(Opz)(H)+(H)(Opz)](ab-ba)} z x R 2P
(Pz H + H Pz)

76. 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 2Py state leads to repulsive interactions just as for NeH. No bond.
2Py
bond
antibond z x
A {(O2pxa)(O2pxb)(Hxa) (O2pya)1[(Opz)(H)+(H)(Opz)](ab-ba)}
Bonding a 2nd H along the x axis to the 2Px state leads to a covalent bond
2Px
bond
bond
A{(O2py)2[(Opx )(Hx)+(Hx)(Opx)](ab-ba)[(Opz)(Hz)+(Hz)(Opz)](ab-ba)}

77. Analize Bond in the ground state of H2O
A{(O2py)2[(Opx )(Hx)+(Hx)(Opx)](ab-ba)[(Opz)(Hz)+(Hz)(Opz)](ab-ba)}
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. z x θe Re
In fact the bond angle is far from 90º for H2O, but it does approach 90º for S  Se  Te

78. What is origin of large distorsion in bond angle of H2O
A{(O2py)2[(Opx )(Hx)+(Hx)(Opx)](ab-ba)[(Opz)(Hz)+(Hz)(Opz)](ab-ba)} z x
OHx bond
OHz bond
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.
Since the spin on Hx is a half the time and b the other half while the same is true for Hz, then ¼ the time both are a and ¼ the time both are b.
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

79. 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: z x θe Re
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).

80. 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.

81. 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
A{(O2pya)(O2pyb)(Hya) [(Opx )(Hx)+(Hx)(Opx)](ab-ba)[(Opz)(Hz)+(Hz)(Opz)](ab-ba)}
OHx bond
OHz bond
OHy antibond
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.

82. Now consider Bonding H atom to the ground state of N
Bring H1s along z axis to O and
N 2pz singly occupied, forms bond to Hz R z x
A{(N2pxa)(N2pya)[(Npz)(Hz)+(Hz)(Npz)](ab-ba)}
NHz bond
Two unpaired spins, thus get S=1, triplet state
Denote at 3S-

83. 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{(N2pya)[(Npx )(Hx)+(Hx)(Npx)](ab-ba)[(Npz)(Hz)+(Hz)(Npz)](ab-ba)}
bond
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. z x

84. 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)](ab-ba)[(Npx )(Hx)+(Hx)(Npx)](ab-ba)[(Npz)(Hz)+(Hz)(Npz)] (ab-ba)}
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

85. 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.

86. 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

87. 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, t12, 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 e G (that is both are symmetry elements) then R2 R1 is also a symmetry element, R2 R1 e 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) e G. Clearly eΨ is a symmetry element

88. The symmetry operations for the schrodinger equation form a group, continued
3. Associativity. If {R1,R2,R3 } e 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 e G then the inverse, (R1)-1 e 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.

89. 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 e G on any Ψi e 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.

90. 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

91. C2v The symmetry operators are
And the group {e, C2z, sxz, syz} is denoted as C2z
Czn denotes a rotation of 2p/n about the z axis (C for cyclisch, German for cyclic)
sxz denotes a reflection or mirror in the xz plane (s for spiegel, German for mirror)

92. 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. C2zsxz= syz
Combine these as below to show the relationships x y e
C2z
sxz
syz x y e
sxz
C2z
syz
C2v

93. The character table for C2v
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 and sxz 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)

94. 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 t (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 C2zsxz= syz , we see that the symmetry of syz is determined by those of C2z and sxz
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.

95. 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
N electron wavefunction, use A1,A2 etc
one electron orbitals: use a1,a2 etc
name
1-e N-e a1 a2 b1 b2

96. Symmetries for H2O, NH2, and CH2
Previously we discussed the wavefunctions for H2O, NH2, and CH2 from bonding H to two p orbitals
CH2
H2O
NH2
A{(O2py)2[(Opx )(Hx)+(Hx)(Opx)](ab-ba)[(Opz)(Hz)+(Hz)(Opz)](ab-ba)}
OHx bond
OHz bond
H2O
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)}

97. First do CH2 singlet CHL bond CHR bond
Ψ1=A{(C2py)0[(CpL )(HL)+(HL)(CpL)](ab-ba)[(CpR)(HR)+(HR)(CpR)](ab-ba)}
CHL bond
CHR bond
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 operations pR 2s z y HL HR pL
Applying sxz to the wavefunction leads to
Ψ2=A{(C2py)0[(CpR)(HR)+(HR)(CpR)](ab-ba)[(CpL)(HL)+(HL)(CpL)](ab-ba)}
Each term involves transposing two pairs of electrons, e.g., 13 and 24 as interchanging electrons. Since each interchange leads to a sign change we get that sxzΨ1 = Ψ2 = Ψ1
Thus interchanging a bond pair leaves Ψ invariant

98. Finish CH2 singlet CHL bond CHR bond pR 2s z y HL HR pL
Ψ1=A{(C2py)0[(CpL )(HL)+(HL)(CpL)](ab-ba)[(CpR)(HR)+(HR)(CpR)](ab-ba)}
CHL bond
CHR bond pR 2s z y HL HR pL
Applying C2z to the wavefunction also interchanges two bond pairs so that
C2zΨ1 = Ψ2 = Ψ1  A1 or A2 symmetry
Also sxzΨ1 = 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

99. Next consider NH2
The wavefunction for NH2 differs from that of CH2 only in having a singly occupied px orbital
A{(N2pxa)1[(NpL )(HL)+(HL)(NpL)](ab-ba)[(NpR)(HR)+(HR)(NpR)](ab-ba)}
NHR bond
NHL bond pR 2s z y 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, sxz does not. Thus Npx transforms as b1. (note it is not necessary to examine sxz since it is not a generator).
With one unpaired spin this state is S= ½ or doublet
Thus the symmetry of NH2 is 2B1

100. Now do OH2
The wavefunction for OH2 differs from that of NH2 only in having px orbital doubly occupied
A{(O2pxa)2[(OpL )(HL)+(HL)(OpL)](ab-ba)[(OpR)(HR)+(HR)(OpR)](ab-ba)}
OHR bond
OHL bond
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 HL y z pL HR pR 2s
Thus the symmetry of OH2 is 1A1

101. 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 s to indicate symmetric with respect to the plane of the molecule) is excited to the 2px orbital (denoted as p to indicate antisymmetric with respect to the plane)
A{(C2sa)1(2pxa)1[(CpL )(HL)+(HL)(CpL)](ab-ba)[(CpR)(HR)+(HR)(CpR)](ab-ba)}
CHR bond
CHL 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

102. Second example, C3v, with NH3 as the prototype
NHb bond
A{[(Npy )(Hy)+(Hy)(Npy)](ab-ba)[(Npx )(Hx)+(Hx)(Npx)](ab-ba)[(Npz)(Hz)+(Hz)(Npz)] (ab-ba)}
NHx bond
NHc bond z x
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.

103. C3v The symmetry elements are: zx
and the symmetry group is denoted as C3v.

104. Stereographic projections for C3v
Take the z axis out of the plane. The six symmetry operations are: x b c x b c x b c C3 e
C32= C3-1 x b c x b c x b c
sxzC32
sxz
sxzC3
C2v

105. Combining the 6 operations for C3v leads to
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 and C3 x b c e C3
C32= C3-1
sxz
sxzC3
sxzC32
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
C2v

106. More on C3v
The sxzC3 transformation corresponds to a reflection in the bz plane (which is rotated by C3 from the xz plane) and the sxzC32 transformation corresponds to a reflection in the cz plane (which is rotated by C32 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 sxz we say that they are in the same class. x b c e C3
C32= C3-1
sxz
sxzC3
sxzC32

107. The character table for C3v
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.

108. Application for C3v, NH3 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 we combined the 3 Valence bond wavefunctions in 3 pair functions and we denote what started as the 2s pair as slp
Consider first the effect of sxz. 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 sLP orbital is invariant.

109. Consider the effect of C3
Next consider the C3 symmetry operator. It does not change sLP. 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.
(135)
(13)
(35)
Thus the wavefunction for NH3 has 1A1 symmetry