Examples: CH3I, C6H6, NH3
If we designate with I|| the molecular axis and with I^
the inertia moment perpendicular to the molecular axis then for
I|| > I^(for
instance, benzene) the top will rotate like a plate (oblate) and for I||<
I^ it will move like cigar (prolate CH3I).
The energy is given by (classical physics):
E = (JA² + JB²)/2I^ + JC²/2I||
The transition to quantum mechanics will be much easier if we take into account the fact that energy values corresponding to J2 and chosen axis are obtained simultaneously. Since
J² = JA² + JB² + JC²
we obtain energy values:
E = J²/2I^ + JC² · [(2I||)-1− (2I^)-1]
If now we substitute squared kinetic moment J² with eigenvalues of
quantum mechanical kinetic moment operator, J² →
J(J+1)h² and designate the eigenvalues of
corresponding components on molecular axis with K (JCy®
Kh y) we obtain the quantum
mechanical energy levels:
EJK = B · J(J+1) + (A - B)K2 | K = 0, ±1, ±2, ..., ±J | |
B = |
A = |
J = 0, 1, 2, ... |
Extreme cases of the symmetric top rotation | |
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a) around molecular axis | b) perpendicular to molecular axis |
The K sign obviously doesn't have influence on energy since it depends on K2. Since kinetic moment may take 2J + 1 values (MJ = 0, ± 1, ± 2, ± 3, ... ± J), the symmetric top level degeneracy is only 2(2J + 1)-fold. The exception is K = 0 because this level has (2J + 1)-fold degeneracy. K-sign shows direction of rotation, but nevertheless the rotational energy doesn't depend on direction of rotation.
For one extreme case when |K| = J, J is parallel to molecular axis as much as possible and molecule rotates mainly around its axis and energy levels essentially depend on the rotational constant A (resp. I||):
EJ,K = J = BJ + AJ2 ≈ AJ2
For another extreme case, K = 0, the molecule rotates perpendicular to its axis. Therefore it doesn't have kinetic moment in the molecular axis direction. The rotational energy is then given by: