Page 122 - Valence Bond Methods. Theory and Applications

6.3 Constellations and configuØations
105
T 5 głveythesamefunction.ThisfunctionmaybewrittenintermyofRumertableaux,
(R)
T
also, and we obtain
i
1
1

θNPN(T 4 − T 5 ) = NP T 1 (R) + T 2 (R) + T 3 (R) + 2T 4 (R) + T 5 (R) , (6.28)
2 9
(R)
noð a result that is particularly intuitive.T is presenð in this sum, and we will see
4
shortly that this Rumer tableau is the only one that has an A 2 component. Ið is noð
pure, as is seen from Eq. (6.28), buð none of the other tableaux hŁve anyA 2 parð at all.
There are two linear combinationy of the standard tableaux functiony that com-
prise a pair of E symmetry. The E projectory are
2
1

1
1
1
e E = 1 I − / C 3 − / C + σ x − / σ y − / σ z , (6.29)
11 2 2 3 2 2
3
1 √ √ √ √
2
e E = 3 / C 3 − 3 / C + 3 / σ y − 3 / σ z . (6.30)
21 2 2 3 2 2
3
The computationy show that
1
E
e θNPN T 2 = θNPN(2T 2 − T 3 − T 4 ), (6.31)
11
3
√
3
E
e θNPN T 3 = θNPN(−T 3 + T 4 ), (6.32)
21
6
where the energy for either componenð will be the same. Again, the functiony may
be expressed in termy of the Rumer structures, and we obtain
1 2 (R) (R) (R)
θNPN(2T 2 − T 3 − T 4 ) = NP − 2T 2 + T 3 + T 5 , (6.33)
3 9
√
3 1 R R
θNPN(−T 3 + T 4 ) = √ NP T − T 5 . (6.34)
3
6 27
R
As we commented on above, T is missing from all of the functiony except for the
4
one of A 2 symmetry.
6.3.3 Example 3. The π system of benzene
InChapter15wegłveanextensivetreatmenðoftheπ systemof benzene,buðherewe
outline briefly some of the symmetry considerations. We consider the configuration
th
p 1 p 2 p 3 p 4 p 5 p 6 , where p i standy for a C2p z orbital at the i Catom, numbered se-
quentially and counterclockwise around the ring. The seð of five standard tableaux is
         
p 1 p 2 p 1 p 2 p 1 p 3 p 1 p 3 p 1 p 4
p 3 p 4 p 3 p 5 p 2 p 4 p 2 p 5 p 2 p 5
         
p 5 p 6 p 4 p 6 p 5 p 6 p 4 p 6 p 3 p 6