The Selection Rules For Pure Rotational Transitions

Also when making rotational transition a molecule (at least for a moment) should have dipole which vibrates with frequency of electromagnetic field. The polar molecule has an alternative dipole moment when rotating and transition dipole moment can't be equal to 0.

Example HCl

| symmetric vibrations | = 0 !
| bending vibrations |
| antisymmetric vibrations |

High rotation changes the initial spherical symmetry and therefore there is a (weak) dipole moment.

Homonuclear diatomic molecules don't have rotational spectra at all; the same is true for spherical tops. It's clear one can find vibrating dipole when molecule is vibrating, that leads to the interaction by using electromagnetic field. High rotations can be met also in spherical tops which have dipole moments but these moments are quite weak, however. The conservation of kinetic moment gives us the following selection rules for a linear molecule:

ΔJ = ± 1 | ΔMJ = 0, ± 1

The dipole moment for a symmetric top lays always parallel to molecular axis; and so it's not possible to change rotation around molecular axis:

ΔK = 0

For absorption spectrum, J + 1 ← J one will have:

n [cm-1] = B(J + 1)(J + 2) - BJ(J + 1)

n= 2B(J + 1) where J = 0, 1, 2...

The separation between two lines is the same, Dn = n(J) -n(J − 1) = 2B. One can obtain the binding length r from it since I = µr2. The centrifugal expansion when having high rotations gives the following:

n = B(J + 1)(J + 2) - D(J + 1)2(J + 2)2 - BJ(J + 1) + DJ2(J + 1)2

n = 2B(J + 1) - 4D(J + 1)3

i.e. the line width
Dn = n (J) - n (J − 1)

Dn = 2B - 4D(3J2 + 9J + 7)

The centrifugal constant D is connected with vibration (wavenumber nS of vibration) for diatomic molecules by using the following expression:

D = 4B³/ns²

The transition intensity depends certainly on the dipole moment transition value and the amount of occupied electronic states of initial and resultant states.