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Question

Assign the spectral lines given in the table to different vibrational modes of A2B type molecule :

\(\bar{\upsilon}\) (cm–1)Description
(1) 590IR-active (PQR branches)
(2) 1290Raman active, polarized
IR-active (PR branches)
(3) 2220Raman active, depolarized
IR-active (PR branches)

The correct answer is

1→ bending, 2→ symmetric stretching, 3→ asymmetric stretching

The data fit a linear, unsymmetrical A2B molecule such as N2O (N–N–O): it has no centre of symmetry, so all three fundamentals can be both IR and Raman active, and it has 3N – 5 = 4 vibrations (two stretches and a doubly degenerate bend).

Bending force constants are much smaller than stretching force constants, so the bend has the lowest wavenumber. For a linear molecule the bend changes the dipole perpendicular to the axis and gives a perpendicular band with P, Q and R branches. So line 1 (590 cm–1, PQR) is the bending mode.

Stretching vibrations change the dipole along the molecular axis and give parallel bands with only P and R branches, consistent with lines 2 and 3.

A polarized Raman line belongs to the totally symmetric vibration, i.e. the symmetric stretch, so line 2 (1290 cm–1) is the symmetric stretching mode.

The remaining stretch, line 3 (2220 cm–1), is the asymmetric stretch; the asymmetric stretch of a triatomic always lies at higher wavenumber than the symmetric stretch (for N2O about 2224 versus 1285 cm–1, with the bend at 589 cm–1).

Therefore 1 is bending, 2 is symmetric stretching and 3 is asymmetric stretching.

Hence, 1→ bending, 2→ symmetric stretching, 3→ asymmetric stretching.

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