Q102 · CSIR NET Chemistry, December 2011
Paper: CSIR NET December 2011 · Subject: Physical Chemistry · Chapter: Group Theory · Topic: Group Theory – General · Marks: 2 · Difficulty: Medium
For $\mathrm{H_2O}$ molecule, the electronic transition from the ground state to an excited state of $\mathrm{B_1}$ symmetry is
| $\mathrm{C_{2v}}$ | E | $\mathrm{C_2}$ | $\sigma_v$ | $\sigma_v'$ | |
|---|---|---|---|---|---|
| $\mathrm{A_1}$ | 1 | 1 | 1 | 1 | $z, z^2, x^2, y^2$ |
| $\mathrm{A_2}$ | 1 | 1 | -1 | -1 | xy |
| $\mathrm{B_1}$ | 1 | -1 | 1 | -1 | x,xz |
| $\mathrm{B_2}$ | 1 | -1 | -1 | 1 | y,yz |
(a)not allowed
(b)allowed with $\mathrm{x}$ polarization
(c)allowed with $\mathrm{y}$ polarization
(d)allowed with $\mathrm{z}$ polarization
Answer
Answer: B ✓ checked by 4AB · confidence high
Explanation
H₂O is C₂v with A₁ ground state; the transition to B₁ is allowed by the B₁ component of the dipole, which is x (with the molecule in the yz plane): x-polarised.
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