 Solution

Use rovibrational spectroscopy combination differences so that the band origin cancels. With the assignments above, $R(2)$ and $P(4)$ end at the same upper level $J^{\prime }=3$, givingLikewise $R(3)$ and $P(3)$ start at the same lower level $J=3$, soOnly the printed wavenumbers are needed:Thus $B^{\prime }-B^{\prime \prime }\simeq 0.00429{\mathrm{cm}}^{-1}$. The rotational constants differ only slightly, so the average moment of inertia changes little on vibrational excitation. The upper value is slightly larger, corresponding to a slightly smaller effective moment of inertia. This sign is possible for a bending excitation; the intuition that excitation must always lengthen a diatomic bond is not a general rule for a polyatomic bend.

The rounded printed values are internally consistent: fitting all six positions to the same elementary model gives approximately ${\tilde{\nu }}_{\ast }=711.9945{\mathrm{cm}}^{-1}$, $B^{\prime \prime }=1.47797{\mathrm{cm}}^{-1}$ and $B^{\prime }=1.48203{\mathrm{cm}}^{-1}$, with residuals below $0.002{\mathrm{cm}}^{-1}$. The displayed combination-difference values have more digits than their physical accuracy; the meaningful scale is about $1.48{\mathrm{cm}}^{-1}$.