 Solution

$I_{C_4}\simeq 2.43\times 10^{-45}\mathrm{kg}{\mathrm{m}}^2=146.6m_{\mathrm{u}}{\mathring{\mathrm{A}}}^{,2}$. The parallel-band progression gives $2\tilde{B}=0.42{\mathrm{cm}}^{-1}$, so $\tilde{B}=0.210{\mathrm{cm}}^{-1}$. The perpendicular Q-feature spacing givesFor an oblate symmetric top, $I_{C_4}>I_{\perp }$, hence $\tilde{C}<\tilde{B}$ and $\tilde{C}=0.115{\mathrm{cm}}^{-1}$. The rotational constant relation then givesThe estimated constants neglect centrifugal distortion and the change of rotational constants in the excited vibration.