Answer on Question#67438 – Chemistry | Inorganic Chemistry
1,4-C6H4Cl2, D2h, h = 2
E = 3, C2 = -1, i = -3, σ = 1
Ag=1/8(36−4+12+4)=6B1g=1/8(36+4+12−4)=6B2g=1/8(36+4−12+4)=4B3g=1/8(36−4−12−4)=2Au=1/8(36−4−12−4)=2B1u=1/8(36+4−12+4)=4B2u=1/8(36+4+12−4)=6B3u=1/8(36−4+12+4)=6Γ=6Ag+6B1g+4B2g+2B3g+2Au+4B1u+6B2u+6B3u
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