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5.04 Principles of Inorganic Chemistry II ��
Fall 2008
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5.04, Principles of Inorganic Chemistry II
Prof. Daniel G. Nocera
Lecture 8: N-Dimensional Cyclic Systems
This lecture will provide a derivation of the LCAO eigenfunctions and eigenvalues of
N total number of orbitals in a cyclic arrangement. The problem is illustrated below:
There are two derivations to this problem.
Polynomial Derivation
The Hückel determinant is given by,
x 1
1 x 1
1 x O
1 O O
O O O
DN(x) =
= 0
O O O
where
x=
α −E
β
O O 1
O x 1
1 x
From a Laplace expansion one finds,
DN(x) = xDn–1(x) – DN–2(x)
where
D1(x) = x
D2 (x) =
x 1
1 x
= x2 − 1
5.04, Principles of Inorganic Chemistry II
Prof. Daniel G. Nocera
Lecture 8
Page 1 of 6
With these parameters defined, the polynomial form of DN(x) for any value of N can
be obtained,
D3(x) = xD2(x) – D1(x) = x(x2–1) – x = x(x2–2)
D4(x) = xD3(x) – D2(x) = x2(x2–2) – (x2–1)
and so on
The expansion of DN(x) has as its solution,
x = −2 cos
2π
j
N
(j = 0, 1, 2, 3 ... N − 1)
and substituting for x,
E = α + 2β cos
2π
j
N
(j = 0, 1, 2, 3 ... N − 1)
Standing Wave Derivation
An alternative approach to solving this problem is to express the wavefunction
directly in an angular coordinate, θ
For a standing wave of λ about the perimeter of a circle of circumference c,
c
ψ j = sin θ
λ
The solution to the wave function must be single valued ∴ a single solution must be
obtained for ψ at every 2nπ or in analytical terms,
5.04, Principles of Inorganic Chemistry II
Prof. Daniel G. Nocera
Lecture 8
Page 2 of 6
c
c
ψ = sin (θ + 2π ) = sin θ
λ
λ
= sin
c
λ
θ • cos
c
λ
2π + sin
must go to 1
λ
2π
•
c
λ
2π m
j
N
c
c
θ = sin θ
λ
λ
2π = 2π j (j = 0, 1, 2 ... N − 1)
condition for an integral
number of λ’s about the
circumference of a circle
c
=j
λ
Thus the amplitude of ψj at atom m is, (where
ψ j(m) = sin
cos
must go to 0
iff
∴
c
c
λ
= j and θ =
2π
m)
N
(j = 0, 1, 2 ... N − 1)
Within the context of the LCAO method, ψj may be rewritten as a linear combination
in φm with coefficients cjm. Thus the amplitude of ψj at m is equivalent to the
coefficient of φm in the LCAO expansion,
N
ψ j = ∑ c jmφm
m=1
where c jm = sin
2π m
j
N
(j = 0, 1, 2 ... N − 1)
The energy of each MO, ψj, may be determined from a solution of Schrödinger’s
equation,
Hψ j = E jψ j
H − Ej ψ j = 0
N
H − E j ∑ c jmφm = 0
m
The energy of the φm orbital is obtained by left–multiplying by φm,
N
φm H − E j ∑ c jmφm = 0
m
5.04, Principles of Inorganic Chemistry II
Prof. Daniel G. Nocera
Lecture 8
Page 3 of 6
but the Hückel condition is imposed; the only terms that are retained are those
involving φm, φm+1, and φm-1. Expanding,
α
1
β
0
⎡c φ H φ
−
c E φ φ ⎤ +
⎡c
⎤
m
jm j m
m ⎥
j(m +1) φm H φ
m +1 −
c j(m +1)E
j φm φm +1 ⎥
⎢⎣ jm m
⎢
⎣
⎦
⎦
+
⎡c j(m–1) φm H φm–1 − c j(m–1)E j φm φm–1 ⎤ =
0
⎢⎣
⎥⎦
β
0
Evaluating the integrals,
[
]
α c jm − c jm E j + β c j(m+1) + c j(m−1) = 0
[
]
α c jm + β c j(m+1) + c j(m−1) = c jmE j
Substituting for cjm,
α
sin
⎛
2π
m
2π (m + 1)
2π (m − 1) ⎞⎟
2π
m
j +
β
⎜⎜
sin
j +
sin
j⎟
=
E j sin
j
N
N
N
N
⎝
⎠
Dividing by sin
2πm
j,
N
⎛
α
+
β
⎜⎜
sin
⎝
2π (m + 1)
2π (m − 1) ⎞⎟
j +
sin
j ⎟
N
N
⎠
=
E
j
2π m
sin
j
N
Making the simplifying substitution, κ =
Ej = α +
(
2π
j
N
)
β sin κ (m + 1) + sin κ (m − 1)
sin κm
⎛ sin κm • cos κ + sin κ • cos κm + sin κm • cos κ − sin κ • cos κm ⎞
⎟
E j = α + β ⎜⎜
⎟
sin κm
⎝
⎠
Ej = α + 2βcosκ
E j = α + 2β cos
2π
j
N
(j = 0, 1, 2 ... N − 1)
5.04, Principles of Inorganic Chemistry II
Prof. Daniel G. Nocera
Lecture 8
Page 4 of 6
Let’s look at the simplest cyclic system, N = 3
N = 3, so E j = α + 2β cos
2π
j where j = 0, 1, 2
N
E0 = α + 2
β
2π
= α −β
3
4π
= α + 2β cos
= α −β
3
E1 = α + 2β cos
E2
Continuing with our approach (LCAO) and using Ej to solve for the eigenfunction,
we find…
ψ
j =
∑
eijθ φm
m
⎧ N
⎪± for N even
⎪ 2
for j = 0, ± 1, ± 2 ... ⎨
⎪±
(N − 1)
for N odd
⎪⎩
2
Using the general expression for ψj, the eigenfunctions are:
i(0)
i(0)0
ψ0 = e
φ1 + e
i(1)
ψ +1 = ei(1)0φ1 + e
2π
3
2π
3
i( −1)
ψ −1 = ei(−1)0φ1 + e
i(0)
φ2 + e
i(1)
φ2 + e
2π
3
4π
3
φ3
4π
3
φ3
i( −1)
φ2 + e
4π
3
φ3
Obtaining real components of the wavefunctions and normalizing,
ψ 0 = φ1 + φ2 + φ3 →
ψ0 =
ψ +1 + ψ −1 = 2φ1 − φ2 − φ3 → ψ 1 =
ψ +1 − ψ −1 = φ2 − φ3 →
5.04, Principles of Inorganic Chemistry II
Prof. Daniel G. Nocera
ψ2 =
1
3
1
6
1
2
(φ
+ φ2 + φ3
1
(2φ
1
(φ
2
)
− φ2 − φ3
− φ3
)
)
Lecture 8
Page 5 of 6
Summarizing on a MO diagram where α is set equal to 0,
1
2
1
–2
2
2
–1
E/
6
0
1
1
2
6
1
3
5.04, Principles of Inorganic Chemistry II
Prof. Daniel G. Nocera
Lecture 8
Page 6 of 6