Q17 · CSIR NET Chemistry, June 2016
Paper: CSIR NET June 2016 · Subject: Physical Chemistry · Chapter: Group Theory · Topic: Character Tables · Marks: 2 · Difficulty: Hard
The character table for the $D_{4}$ point group is provided below:
For this point group, the correct statement among the following is:
| $\mathrm{D}_{3}$ | E | $2 \mathrm{C}_{3}$ | 3C | ||
|---|---|---|---|---|---|
| $\mathrm{A}_{1}$ | 1 | 1 | 1 | $\mathrm{x}^{2}+\mathrm{y}^{2}, \mathrm{z}^{2}$ | |
| $\mathrm{A}_{2}$ | 1 | 1 | -1 | $\mathrm{z}, \mathrm{R}_{\mathrm{z}}$ | |
| E | 2 | -1 | 0 | $(\mathrm{x}, \mathrm{y}),\left(\mathrm{R}_{\mathrm{x}}, \mathrm{R}_{\mathrm{y}}\right)$ | $\left(\mathrm{x}^{2}-\mathrm{y}^{2}, \mathrm{xy}\right),(\mathrm{xz}, \mathrm{yz})$ |
(a)it is possible to have a totally symmetric normal mode of vibration which is IR-active
(b)all IR-active normal modes are necessarily Raman inactive
(c)all Raman-active normal modes are necessarily IR-active
(d)it is possible to have a pair of IR-active normal modes that are degenerate.
Answer
Answer: D ✓ checked by 4AB · confidence high
Explanation
In D₄, z belongs to A₂ and (x, y) to E, so A₁ (the totally symmetric mode) is never IR-active (a false); A₁ is Raman-active but not IR-active (c false); E is both IR- and Raman-active (b false). The doubly degenerate E modes are IR-active, so (d) is correct.
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