Q110 · CSIR NET Chemistry, June 2011
Paper: CSIR NET June 2011 · Subject: Physical Chemistry · Chapter: Group Theory · Topic: Character Tables · Marks: 2 · Difficulty: Medium
The character table of the $\mathrm{C}_{2 \mathrm{v}}$ point group is given below:
1 atom of cis-butadiene and $\sigma_{\mathrm{v}}$ is the molecular plane) belong to
| $\mathrm{C}_{2 \mathrm{V}}$ | E | $\mathrm{C}_{2}$ | $\sigma_{\mathrm{v}}$ | $\sigma^{\prime}{ }_{\mathrm{v}}$ |
|---|---|---|---|---|
| $\mathrm{A}_{1}$ | 1 | 1 | 1 | 1 |
| $\mathrm{A}_{2}$ | 1 | 1 | -1 | -1 |
| $\mathrm{B}_{1}$ | 1 | -1 | 1 | -1 |
| $\mathrm{B}_{2}$ | 1 | -1 | -1 | 1 |
(a)$\mathrm{A}_{1}$ and $\mathrm{A}_{2}$ respectively
(b)Both $\mathrm{A}_{2}$
(c)Both $\mathrm{B}_{2}$
(d)$\mathrm{B}_{1}$ and $\mathrm{B}_{2}$ respectively.
Answer
Answer: C ✓ checked by 4AB · confidence high
Explanation
For p orbitals perpendicular to the molecular plane (σv), C₂ and σv change their sign while σv′ only swaps symmetric pairs: characters (1, −1, −1, 1) = B₂ for both combinations (c).
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