Review: The H2 Molecule
R0
R
E
a
b
~ A  ψ a  ψb
ψ
2 protons
two 1s states each
 4 states total
2 anti-bonding states
-13.6 eV

y
a
b
2 bonding states
~ S  ψ a  ψb
ψ
• Result: H2 covalent bond
• Directional; typical of molecules
alternative
representations
H  H
1s
1s

The N-atom “Hydrogen Solid”
Chemical Bonding  Continuous Bands
1s1 N-atom solid  N electrons
R0
E
N states
unoccupied
R
overlap of states
discrete  continuous
N anti-bonding states
1s
N states
occupied
N bonding states
H2 molecule: N = 2
2N states
Silicon – A ‘real’ N-atom Solid
N-atom solid  4N relevant electrons
Si: #14 1s22s22p6 3s23p2
[3(sp3)4] hybrid orbital composed 3s: 2N states
[Ne]
of 3s and all 3p orbitals:
Hybridization:
consider just 2 atoms
R0
anti-bonding
3
3p
3s
1
6 states
2 states
3p
3s
4N states
unoccupied
Eg
1 3
bonding
4 states
(+ 4)
N-atom Solid
 Continuous Bands
8N states
R
overlap of states
discrete  continuous
E
4 states
(+ 4)
3p: 6N states
4N states
occupied
E
 0  R0
R
4N anti-bonding states
“sp3”
8N states
4N bonding states
Lithium: A Simple Metal
Li #3
1s22s1 N-atom solid
R0
R
E
N states
unoccupied
N states
occupied
anti-bonding
overlap of states
discrete  continuous
2s
2N states
1s
2N states
bonding
E
 0  R0
R
All states occupied, independent
Magnesium: A Metal?
Mg: #12 1s22s22p6 3s2
N atom solid, 2N electrons
[Ne]
R0
R
3s and 3p overlap to create a
band with 8N states; only 2N
states occupied  yes, a metal
E
3p
2N states
occupied
6N states
anti-bonding
3s
2N states
bonding
E
 0  R0
R
metal requires a partially
occupied band??
Ionic Solid Example: LiF


Li: #3 1s22s1
F: #9 1s22s22p5
1s2
2p6
 Li+1 + e F + e-  F-1
Energy of bonding for a hypothetical ion pair
E = Eionization + Ecoulombic
Eion ( Li) 
Eion ( F ) 
13.6

Li
Z eff
2
13.6


Eion
13.6Z 2

n2
2

5.4 eV
> 0 requires energy

-3.7 eV
< 0 releases energy

-7.2 eV
2
F
Z eff
22

2
+1 -1
ECoul
Z1Z 2 e 2

4 o R
Zeff < Zact due to shielding
Epair = 5.4 - 3.7 - 7.2 eV = 5.5 eV
N-atom Pair Solid of LiF
Li: #3 1s22s1
F: #9 1s22s22p5


e: Li(2s)  F(2p) Li+1 + e-
1s2
2p6
 E(F2p) < E(Li2s)
6N electrons
R0
R
LiF is a non-metal
E
Li 2s 2N states
Eg
6N states
occupied
E
 0  R0
R
F 2p
6N states
Thoughts on how to
transform it a metal?
Summary: MO/LCAO Approach
• N atom solid
– bonding and anti-bonding states
– isolated states  bands due to exclusion principle
• Metal
– no energy gap between occupied and unoccupied states
– many need to consider orbitals of slightly higher energy
• Semi-conductor
– hybrid orbitals  bands
– ‘small’ bandgap between occupied and unoccupied
• Ionic
qualitative distinction
– electron transfer from electropositive to electronegative ion
– orbitals  bands
– ‘large’ bandgap between occupied and unoccupied states
Metals: Electron in a Box
2
Ey ( x) 
y ( x)  V ( x)y ( x)
2
2m  x

2
V(x) = 0 for | x | ≤ L/2
inside
V(x) =  for | x | > L/2
outside


V(x)
solve for y(x) within the box
2
Ey ( x) 
y ( x)
2m  2 x

-L/2
2
or
y 0e
ikx
where
use boundary conditions to get A, B, k (or y0)
y
 / y 
L
2
L

/2  0
L/2
E  E (k )
solutions are of the form
y ( x)  A cos kx  B sin kx
0

2mE ½
k

Metals: Electron in a Box

2mE ½
k
y ( x)  A cos kx  B sin kx
y

 / y 
L
2
L

/2  0
 /   0  A cosk /  B sin k / 
A = B = 0 trivial solution
y  /   0  A cos k /   B sin  k / 
there are no values of k that
subtract: 2 B sin k /   0
make these both true for
add: 2 A cosk /   0
arbitrary, non-zero A and B
y
L
L
2
k is quantized, not continuous
L
2
L
2
L
2
L
2
2
L
2
L
2
 2 solution sets
{0, , 2, 3, etc….}


kL
nπ

n
'
π
k

sin k / 2  0 
set 1: A = 0
n = even
n
2
L
n = principle quantum number
y ( x)  B sin kx
kL
nπ
 ( n'  ½ ) π k n 
cos k L / 2  0 
set 2: B = 0
n = odd
2
L
y ( x)  A coskx
{/2, 3/2, 5/2, etc….}

L

Metals: Electron in a Box
nπx

set 1: y ( x)  B sin 
 L 
n = even

2mE ½
k

nπx
n = odd

 L 
to solve for A and B, normalize according to
kn 
nπ
L
set 2: y ( x)  A cos
L/2
y * ( x)y ( x)dx  1
L / 2
use
 cos
2
 A B
zdz  ½ z  ¼ sin 2 z
kn
y n ( x) 
2
y n( x) 
2
2
nπx
/ L sin 
 n = even
L


nπx
/ L cos
 n = odd
L


eigenfunctions
/L
general wave equation
 2π x
f ( x)  A cos




2L
n 
n
wave-length
kn 
2π
n
wave-vector