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Transcript of William A. Goddard, III, [email protected]
EEWS-90.502-Goddard-L04 1© copyright 2009 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
William A. Goddard, III, [email protected] Professor at EEWS-KAIST and
Charles and Mary Ferkel Professor of Chemistry, Materials Science, and Applied Physics,
California Institute of Technology
Course number: KAIST EEWS 80.502 Room E11-101Hours: 0900-1030 Tuesday and Thursday
Senior Assistant: Dr. Hyungjun Kim: [email protected] of Center for Materials Simulation and Design (CMSD)
Teaching Assistant: Ms. Ga In Lee: [email protected] assistant: Tod Pascal:[email protected]
Lecture 7, September 24, 2009
EEWS-90.502-Goddard-L04 2© copyright 2009 William A. Goddard III, all rights reserved
Schedule changes
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
L8: Sept. 29, as scheduled
L9: Sept. 30, new replaces Oct 6
L10: Oct. 1, as scheduled
L11: Oct. 13, as scheduled
L12: Oct. 14, new replaces Oct 8
L13: Oct. 15, as scheduled
EEWS-90.502-Goddard-L04 3© copyright 2009 William A. Goddard III, all rights reserved
Last time
EEWS-90.502-Goddard-L04 4© copyright 2009 William A. Goddard III, all rights reserved
Contour plots of 3s, 3p, 3d hydrogenic orbitals
EEWS-90.502-Goddard-L04 5© copyright 2009 William A. Goddard III, all rights reserved
The ground state of He atomPut both electrons in 1s orbitals
ΨHe(1,2) = A[(Φ1s)(Φ1sΦ1s(1)Φ1s(2) (
Φ1s = exp(- r)
EHe = 2(½ 2) – 2Z(5/8)whereJ1s,1s = (5/8)
Applying the variational principle, dE/d = 0
get = (Z – 5/16) = 1.6875
E= 2(-½2) = - 2 = -2.8477 h0
Interpretation: two electrons move 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 2-0.31=1.69
The total energy is just the sum of the individual energies.
EEWS-90.502-Goddard-L04 6© copyright 2009 William A. Goddard III, all rights reserved
Now 3rd electron to form Li
ΨLi(1,2,3) = A[(Φ1s)(Φ1s(Φ1sby Pauli principle
Thus the 3rd electron must go into 2s or 2p atomic orbital
ΨLi(1,2,3) = A[(Φ1s)(Φ1s)(Φ2s
ΨLi(1,2,3) = A[(Φ1s)(Φ1s)(Φ2pz(or 2px or 2py)
Li+: Φ1sexp(-r) with = Z-0.3125 = 2.69
R1s = 1/ = 0.372 a0 = 0.2A
EEWS-90.502-Goddard-L04 7© copyright 2009 William A. Goddard III, all rights reserved
Add 3rd electron to the 2p orbital
ΨLi(1,2,3) = A[(Φ1s)(Φ1s)(Φ2pz(or 2px or 2py)
The 2p orbital sees effective charge of Zeff = 3 – 2 = 1,
since it goes to zero at z=0, so that there is no shielding of 1s
Get size: R2p = n2/Zeff = 4 a0 = 2.12A
Energy: e2p = -(Zeff)2/2n2 = -1/8 h0 = -3.40 eV
0.2A
1s
2.12A
2p
EEWS-90.502-Goddard-L04 8© copyright 2009 William A. Goddard III, all rights reserved
Add the 3rd electron to the 2s orbital
ΨLi(1,2,3) = A[(Φ1s)(Φ1s)(Φ2pz(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
Size: R2s = n2/Zeff = 3.1 a0 = 1.65A
Energy: e2s = -(Zeff)2/2n2 = -0.205 h0 = 5.57 eV
0.2A
1s
2.12A2s
R~0.2A
EEWS-90.502-Goddard-L04 9© copyright 2009 William A. Goddard III, all rights reserved
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 nm
E = 1.9 eV
EEWS-90.502-Goddard-L04 10© copyright 2009 William A. Goddard III, all rights reserved
He, 2
Ne, 10
Ar, 18
Zn, 30Kr, 36
Aufbau principle for atoms
Particularly stable atoms, closed shells
Xe, 54
Rn, 86
Uuo, 118
EEWS-90.502-Goddard-L04 11© copyright 2009 William A. Goddard III, all rights reserved
Many-electron configurations
General aufbau
ordering
Particularly stable
EEWS-90.502-Goddard-L04 12© copyright 2009 William A. Goddard III, all rights reserved
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
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 R(2s Li)/R(2s Ne) ~ 4.64/1.28 = 3.6
EEWS-90.502-Goddard-L04 13© copyright 2009 William A. Goddard III, all rights reserved
General trends along a column of the periodic table
As we go down a colum
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
EEWS-90.502-Goddard-L04 14© copyright 2009 William A. Goddard III, all rights reserved
EEWS-90.502-Goddard-L04 15© copyright 2009 William A. Goddard III, all rights reserved
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
IP3d = (Zeff3d )2/2n2 = 10.17 eV Zeff
3d = 2.59;
IP4p = (Zeff4p )2/2n2 = 8.73 eV Zeff
4p = 3.20;
IP3d = (Zeff3d )2/2n2 = 24.75 eV Zeff
3d = 4.05; 3d<4s<4p
IP4s = (Zeff4s )2/2n2 = 21.58 eV Zeff
4s = 5.04;
IP4p = (Zeff4p )2/2n2 = 17.01 eV Zeff
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
EEWS-90.502-Goddard-L04 16© copyright 2009 William A. Goddard III, all rights reserved
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++
For high charge (Sc++) 3d more stable than 4s (3d)1
For neutral system (Sc) fill 4s: (3d)(4s)2 This is because increased charge makes the differential shielding for 4s vs 3d less important than the 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
EEWS-90.502-Goddard-L04 17© copyright 2009 William A. Goddard III, all rights reserved
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
The character of transition metals Sc-Cu columns depends on circumstances
EEWS-90.502-Goddard-L04 18© copyright 2009 William A. Goddard III, all rights reserved
Transtion metal orbitals
EEWS-90.502-Goddard-L04 19© copyright 2009 William A. Goddard III, all rights reserved
Next 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
EEWS-90.502-Goddard-L04 20© copyright 2009 William A. Goddard III, all rights reserved
Consider the ground state of B: [Be](2p)1
Ψ(1,2,3,4,5) = A[Φ
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
EEWS-90.502-Goddard-L04 21© copyright 2009 William A. Goddard III, all rights reserved
New material
EEWS-90.502-Goddard-L04 22© copyright 2009 William A. Goddard III, all rights reserved
Consider the ground state of C: [Be](2p)2
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.
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
EEWS-90.502-Goddard-L04 23© copyright 2009 William A. Goddard III, all rights reserved
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 symmetric, thus must 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)]
EEWS-90.502-Goddard-L04 24© copyright 2009 William A. Goddard III, all rights reserved
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 = 0
[φx(1) φy(2) - φy(1) φx(2also andor spin = 1
EEWS-90.502-Goddard-L04 25© copyright 2009 William A. Goddard III, all rights reserved
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
EEWS-90.502-Goddard-L04 26© copyright 2009 William A. Goddard III, all rights reserved
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)
EEWS-90.502-Goddard-L04 27© copyright 2009 William A. Goddard III, all rights reserved
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
EEWS-90.502-Goddard-L04 28© copyright 2009 William A. Goddard III, all rights reserved
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)
EEWS-90.502-Goddard-L04 29© copyright 2009 William A. Goddard III, all rights reserved
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)]
which we visualize as
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
EEWS-90.502-Goddard-L04 30© copyright 2009 William A. Goddard III, all rights reserved
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[(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.
EEWS-90.502-Goddard-L04 31© copyright 2009 William A. Goddard III, all rights reserved
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
EEWS-90.502-Goddard-L04 32© copyright 2009 William A. Goddard III, all rights reserved
Comparison with experiment
E(1S)
E(1D)
E(3P)
4Kxy
2Kxy
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)
EEWS-90.502-Goddard-L04 33© copyright 2009 William A. Goddard III, all rights reserved
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
EEWS-90.502-Goddard-L04 34© copyright 2009 William A. Goddard III, all rights reserved
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)]
which we visualize as
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)
EEWS-90.502-Goddard-L04 35© copyright 2009 William A. Goddard III, all rights reserved
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)]
EEWS-90.502-Goddard-L04 36© copyright 2009 William A. Goddard III, all rights reserved
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
EEWS-90.502-Goddard-L04 37© copyright 2009 William A. Goddard III, all rights reserved
Summarizing the energies for O atom
E(1S)
E(1D)
E(3P)
2Kxy
3Kxy
O S Se Te Po
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)
EEWS-90.502-Goddard-L04 38© copyright 2009 William A. Goddard III, all rights reserved
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(1,2,3,4,5,6,7)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)]
which we visualize asz
x
EEWS-90.502-Goddard-L04 39© copyright 2009 William A. Goddard III, all rights reserved
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
EEWS-90.502-Goddard-L04 40© copyright 2009 William A. Goddard III, all rights reserved
Calculating energies for F atom
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
(2px)2(2py)2(2pz)The energy of
Ψ(1-9)z= A{[Be](2px)(2px)(2py)(2py)(2pz)]
EEWS-90.502-Goddard-L04 41© copyright 2009 William A. Goddard III, all rights reserved
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
EEWS-90.502-Goddard-L04 42© copyright 2009 William A. Goddard III, all rights reserved
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
EEWS-90.502-Goddard-L04 43© copyright 2009 William A. Goddard III, all rights reserved
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
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
(2px)2(2py)2(2pz)2
The energy of
Ψ(1-9)z= A{[Be](2px)(2px)(2py)(2py)(2pz)(2pz)}
EEWS-90.502-Goddard-L04 44© copyright 2009 William A. Goddard III, all rights reserved
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
EEWS-90.502-Goddard-L04 45© copyright 2009 William A. Goddard III, all rights reserved
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 or alternatively the Pauli Principle tells us that the H1s must get orthogonal to the He 1s since both have an spin.
Consequently H1s must get a nodal plane, increasing its KE.
The smaller R the larger increase in KE. Get a repulsive interaction, no bond
EEWS-90.502-Goddard-L04 46© copyright 2009 William A. Goddard III, all rights reserved
Bonding H atom to Ne
Starting with the ground state of Ne, (1s)2(2s)2(2p)6 Ψ(Ne)= A{[Be](2px)(2px)(2py)(2py)(2pz)(2pz)}
and bringing up an H atom (H1s) along the z axis, leads to
A{[Be](2px)2(2py)2(Ne2pz)(Ne2pz)(H1s}
Where we focus on the Ne2pz orbital that overlaps the H atom
R
Consequently H1s must get a nodal plane, increasing its KE.
The smaller R the larger increase in KE.
Get a repulsive interaction, no bond
The properties of A or alternatively the Pauli Principle tells us that the H1s must get orthogonal to the Ne 2pz since both have an spin.
EEWS-90.502-Goddard-L04 47© copyright 2009 William A. Goddard III, all rights reserved
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