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348 10MOLECULAR SYMMETRY
by the order of the group (h = 8).
Row 3: ψ(B2g) = 1
8 (pA − pC − pB + pD − pC + pA − pB + pD)
= 1
4 (pA − pB − pC + pD)
Row 5: ψ(B3g) = 1
8 (pA − pC + pB − pD − pC + pA + pB − pD)
= 1
4 (pA + pB − pC − pD)
Row 7: ψ(Au) = 1
8 (pA + pC − pB − pD + pC + pA − pB − pD)
= 1
4 (pA − pB + pC − pD)
Row 9: ψ(B1u) = 1
8 (pA + pC + pB + pD + pC + pA + pB + pD)
= 1
4 (pA − pB + pC − pD)
(c) �e 2px orbitals, shown in the right-hand half of Fig. 10.19, are analysed
in the same way. �e characters are found by noting that under the E
operation the four px orbitals are notmoved so χ(E) = 4. Similarly under
the Cz2 operation the orbitals are all moved so χ(Cz2) = 0. Using this
approach the characters are written down as
E Cz2 C y2 Cx2 i σ x y σ yz σ zx
4 0 0 0 0 4 0 0
Using the decomposition formula, or simply by inspection of the charac-
ter table, this is decomposed to Ag + B1g + B2u + B3u . Use of the projec-
tion operator then gives
Row E Cz2 C y2 Cx2 i σ x y σ yz σ zx
1 e�ect on pA pA −pC −pB pD −pC pA −pB pD
2 characters for Ag 1 1 1 1 1 1 1 1
3 row 1 × row 2 pA −pC −pB pD −pC pA −pB pD
4 characters for B1g 1 1 −1 −1 1 1 −1 −1
5 row 1 × row 4 pA −pC pB −pD −pC pA pB −pD
6 characters for B2u 1 −1 1 −1 −1 1 1 −1
7 row 1 × row 6 pA pC −pB −pD pC pA −pB −pD
8 characters for B1u 1 −1 −1 1 −1 1 −1 1
9 row 1 × row 8 pA pC pB pD pC pA pB pD
Hence the SALCs are
ψ(Ag) = 1
4 (pA − pB − pC + pD) ψ(B1g) = 1
4 (pA + pB − pC − pD)
ψ(B2u) = 1
4 (pA − pB + pC − pD) ψ(B3u) = 1
4 (pA + pB + pC + pD)