Discussion
Assign the spectral lines given in the table to different vibrational modes of A2B type molecule :
| \(\bar{v}\) (cm−1) | Description | |
|---|---|---|
| (1) | 590 | IR-active (PQR branches) |
| (2) | 1290 | Raman active, polarized IR-active (PR branches) |
| (3) | 2220 | Raman active, depolarized IR-active (PR branches) |
Answer: A
The assignment is 1 bending, 2 symmetric stretching, 3 asymmetric stretching.
The PR band contours of modes 2 and 3 and the PQR contour of mode 1 show the molecule is linear: stretching modes are parallel bands (P and R only) and the bending mode is a perpendicular band with a Q branch. Bending requires the least energy, so the lowest frequency, 590 cm−1, is the bend.
Of the two stretches, the totally symmetric one gives a polarized Raman line, so 1290 cm−1 (polarized) is the symmetric stretch, and the depolarized 2220 cm−1 band is the asymmetric stretch, which also lies higher in energy. These values match nitrous oxide, N2O, a linear A2B molecule without a centre of symmetry, which is why both stretches appear in both spectra. The other options misplace the bend or swap the stretches.
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