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In pyramidal ammonia, the hydrogen symmetry orbitals span $A_1\oplus E$ of the C3v molecular point group. Nitrogen $2s$ and $2p_z$ both have $A_1$ symmetry and $(2p_x,2p_y)$ span $E$. In the planar D3h molecular point group, these become $A_1^{\prime }$, $A_2^{\prime \prime }$ and $E^{\prime }$, respectively, while the hydrogen combinations span $A_1^{\prime }\oplus E^{\prime }$. Thus the planar out-of-plane lone pair is a symmetry-isolated nonbonding orbital; pyramidalization permits mixing with the other $a_1$ functions.^[ref]

Chloroform has the C3v molecular point group. Its three identical ${}^{35}\mathrm{Cl}$ atoms are interchanged by three proper rotations, so $\sigma =3$. The three mirror operations do not increase the rotational symmetry number to six. Isotopic identity matters for this count.

Using the supplied wavenumber form of $k_{\mathrm{B}}$,All are far below $298\mathrm{K}$, justifying the high-temperature nonlinear rigid rotor expression. ThenandThe calculation treats the molecule as rigid and separates electronic, vibrational and nuclear-spin contributions from the requested rotational term.