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The rotational levels are degenerate. Just as there are three p orbitals for l = 1, for J = 1, there are 3 degenerate rotational states. The degeneracy for a given value of J is 2J + 1. The Boltzmann factor gets modified due to this degeneracy as follows
NJ/NJ′ = [(2J + 1)/(2J′ + 1)] e-ΔE/kBT (13.11)
The implication of this is that the rotational population of the J = 1 level is often more than the population of the J = 0 levels since their degeneracies are 3 and 1 respectively. When molecules rotate with great speeds, they cannot be treated as rigid any more. There are distortions due to centrifugal and other forces. The modification of rotational energies by considering the centrifugal distortion alone is
EJ ( in cm-1) = B J(J +1) – DJ2(J+1)2 (13.12)
Where the centrifugal distortion constant D is given by
D = h3/(32π4 I2 r2 kc) cm-1 (13.13)
The only new term in Eq (13.13) is the force constant k which will be discussed when we study molecular vibrations. |
13.3 |
Rotational Spectra of Polyatomics |
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Linear molecules such as OCS and HC≡CCl have spectra similar to diatomics. In diatomics as well as linear triatomics, IA = IB; IC = 0. IA, IB and IC are the three moments of inertia of molecules along three independent axes of rotation. Just as any translation can be decomposed into three independent components along three axes such as x, y and z, any rotation can be decomposed into rotations along three axes A, B, and C. The way to choose these axes is to have the simplest values of IA, IB and IC. Since triatomics are heavier than the constituent diatomics, their moments of inertia are larger and the values of rotational constants, B, are smaller, in the range of 1 cm-1. The value of IA or IB determined from the B value gives the total length of the triatomic. To determine the two bond lengths in the linear triatomic, we need to determine the moment of inertia IA' of an isotope of the triatomic. From two values of IA and IA' , we can determine the two bond lengths. |
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The rotational spectra of asymmetric molecules for whom IA ≠ IB ≠ IC can be quite complicated. For symmetric tops, two moments of inertia are equal ie.,
IA = IB ≠ IC ; IC ≠ 0 (13.14) |
