 Solution

Let $J$ label the lower rotational state and write the elementary band model asHere $B^{\prime \prime },B^{\prime }$ are lower and upper rotational constants, in inverse centimetres; a constant term associated with upper-state vibrational angular momentum can be included in ${\tilde{\nu }}_{\ast }$. The P branch has $J^{\prime }=J-1$, and the R branch has $J^{\prime }=J+1$.

In this model the six printed labels are assigned asThe low-frequency series steps away from the central origin towards increasing lower $J$. The first two high-frequency R positions are approximately $714.96$ and $717.93{\mathrm{cm}}^{-1}$, so the labels at $720.91$ and $723.90$ correspond to $R(2)$ and $R(3)$; they are not the first two rotational transitions.

There is a physical limitation to taking this schematic model literally. A strict $l=0\to |l|=1$ bending band has $J^{\prime }\ge 1$, so a genuine $P(1)$ transition to $J^{\prime }=0$ is forbidden. The printed $709.04$ feature has the formal $P(1)$ position of the elementary model. The numerical analysis below follows the printed line sequence; it does not infer a physically allowed $J^{\prime }=0$ bending level from that feature.