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5.04, Principles of Inorganic Chemistry II
Lecture 9: N-Dimensional Cyclic Systems
Polynomial Derivation
0 N-
where N is the total number of orbitals
The Hückel determinant is given by,
(^3 ) 1
N-
x 1 1 x 1 1 x
D (^) N (x) =
1
= 0 where x = ^ - E 1 x 1 1 x
From a Laplace expansion one finds,
D (^) N (x) = xD (^) n-1 (x) – D (^) N-2 (x)
where
D 1 (x) = x
x 1 D 2 (x) = = x^2 1 1 x
with these defined, the polynomial form of DN (x) for any value of N can be obtained,
D 3 (x) = xD 2 (x) - D 1 (x) = x(x^2 -1) – x = x(x^2 –2) D 4 (x) = xD 3 (x) – D 2 (x) = x^2 (x^2 -2) – (x^2 -1)
and so on
5.04, Principles of Inorganic Chemistry II Lecture 9
MIT Department of Chemistry
The expansion of DN (x) has as its solution,
2 x = 2cos j (j = 0,1,2,3...N 1) N
and substituting for x,
E = + 2 cos 2 j (j = 0,1,2,3...N 1) N
Standing Wave Derivation
An alternative approach to solving this problem is to express the wave function directly in an angular coordinate,
m+1 m m-
N- N-
For a standing wave of about the perimeter of a circle of circumference c,
= sin c j
The solution to the wave function must be single valued a single solution must be obtained for at every 2n…
= sin c^ ( + 2 ) = sin c = sin c^ cos c^2 + sin c^2 cos c^ = sin c
must go to 1 must go to 0
if c
2 = 2 j (j = 0, 1, 2... N 1)
condition for an integral number of ’s about the circumference of a circle
c
= j
5.04, Principles of Inorganic Chemistry II Lecture 9
Evaluating the appropriate integrals,
C jm C jmE j + (C j(m+1) + C j(m1)) = 0
C jm + (C j(m+1) + C j(m1) ) = C jmE j
Substituting for Cjm,
sin 2 m^ j + sin 2 (m^ +^ 1)^ j + sin 2 (m^ ^ 1)^ j = E^2 m j sin^ j N (^) N N (^) N
2 m Dividing by sin j, N
(^2) (m + 1) 2 (m 1) sin j + sin j + ^ N^ N^ = E j sin 2 mj N
Making the simplifying
(sin (m + 1) + sin (m 1))
E (^) j= + sin m
(^) sin m cos + sin cos m + sin m cos sin cos m Ej = + sin^ m
Ej = + 2cos
E^2
j =^ ^ +^2 ^ cos^ j^ (j^ =^ 0,1,2...N^ ^ 1) N
5.04, Principles of Inorganic Chemistry II Lecture 9
Let’s look at the simplest cyclic system, N = 3
3 N = 3... E = + 2 cos 2 ^ j, j = 0,1,
j N
E 0 = + 2
E = + 2 cos 2 = 1 3 4 E 2 = + 2 cos = 3
Continuing with our approach (LCAO) and using Ej to solve for the eigenfunction, we find…
(^) N ± for N even
j = eijm for j = 0, ± 1, ± 2... 2
m (^) ± (N^ ^ 1) for N odd (^2)
where ^ ~
( ^ ) and^ ^
j j j j (^ j +^ ^ j) =^ cos j
2i 2
Using the general expression for j , the eigenfunctions are:
i(0) 2 ^ i(0)^4 = e^ i(0)0 + e 3 + e 3 0 1 2 3 i(1)^2 ^ i(1)^4 = e^ i(1)0 + e 3 + e 3
- 1 1 2 3 i(1)^2 ^ i(1)^4 = e^ i( 1)0^ + e 3 + e 3 1 1 2 3
Obtaining real components, and normalizing,
+ 1 1 1 2 3 1 (^2 ^1 ^2 ^3 )
+ 1 1 2 3 2 (^2 ^3 )
5.04, Principles of Inorganic Chemistry II Lecture 9
