Q108 · CSIR NET Chemistry, June 2011
Paper: CSIR NET June 2011 · Subject: Physical Chemistry · Chapter: Group Theory · Topic: Reducible Representations · Marks: 2 · Difficulty: Medium
Given the character table of the point group $\mathrm{C}_{3 \mathrm{V}}$
Consider the reducible representation, $\Gamma$
Its irreducible components are
| E | $2 \mathrm{C}_{3}$ | $3 \sigma_{\mathrm{v}}$ | ||
|---|---|---|---|---|
| $\mathrm{A}_{1}$ | 1 | 1 | 1 | Z |
| $\mathrm{A}_{2}$ | 1 | 1 | -1 | |
| E | 2 | -1 | 0 | (x,y) |
| E | $2 \mathrm{C}_{3}$ | $3 \sigma_{\mathrm{v}}$ | |
|---|---|---|---|
| $\Gamma$ | 6 | 3 | 0 |
(a)$\mathrm{E}+2 \mathrm{A}_{1}+2 \mathrm{A}_{2}$
(b)$2 \mathrm{E}+\mathrm{A}_{1}+\mathrm{A}_{2}$
(c)$3 \mathrm{A}_{1}+3 \mathrm{A}_{2}$
(d)2E + 2A
Answer
Answer: A ✓ checked by 4AB · confidence high
Explanation
Γ = (6, 3, 0): n(A₁) = (6 + 2×3 + 0)/6 = 2; n(A₂) = (6 + 6 − 0)/6 = 2; n(E) = (12 − 6 + 0)/6 = 1. Γ = 2A₁ + 2A₂ + E.
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