= Solution
This is a <rovibrational spectrum>: the vibrational excitation is accompanied by quantized <molecular rotation>. The high-wavenumber R branch has $\Delta J=+1$, and the low-wavenumber P branch has $\Delta J=-1$. In the simplest common-$B$ approximation,
$$
\widetilde\nu_R(J)=\widetilde\nu_0+2B(J+1),\qquad
\widetilde\nu_P(J)=\widetilde\nu_0-2BJ.
$$
Consecutive lines within either branch are therefore separated by roughly $2B$, while the gap between P(1) and R(0) is roughly $4B$. There is no Q branch: for this diatomic parallel $\Sigma\to\Sigma$ transition, the electric-dipole rotational selection rule allows $\Delta J=\pm1$, and the $\Delta J=0$ line strength vanishes.
The intensity envelope comes from rotational populations, approximately $(2J+1)e^{-hcBJ(J+1)/(k_BT)}$, together with the rotational line strengths. The paired peaks arise from the two chlorine <isotopologues>, $\mathrm H^{35}Cl$ and $\mathrm H^{37}Cl$: their different <reduced masses> shift both rotational and vibrational spacings. The more abundant $^{35}\mathrm{Cl}$ gives the stronger member. The changing spacing away from the band centre reflects $B_1\ne B_0$ and, at higher $J$, centrifugal distortion; it is not evidence for additional vibrational fundamentals.
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