The selection rules allow us to know the rotational and vibrational transitions allowed. The integral that determines the selection rules is:

$$\label{1-1} \int{\psi^{\ast}\hat{\mu} \psi'}dq$$

where $\mu$ is the electric dipole moment. $\psi$ and $\psi'$ are the two states involved by the transition. Applying the Born-Oppenheimer approximation the wave function can be separated into the product of the nuclear wave function times the electronic $\psi =\psi_N \psi_{ele}$. Since the vibrational-rotary transition does not change the electronic state $\psi'=\psi'_N \psi'_{ele}$. Substituting into the integral (\ref{1-1}):

$$\label{2} \int{\psi^{\ast}\hat{\mu} \psi'}dq=\int\int{\psi_{ele}^{\ast}\psi_{ N}^{\ast}\hat{\mu}\psi_{ele}\psi'_{N}dq_{ele}}dq_N$$

The electric dipole moment $\mu$ depends on the internuclear distance. Expanding $\mu$ in Taylor series around $R_e$

$$\label{3} \mu (R)=\mu (R_e)+\mu'(R_e)(R-R_e)+\frac{1}{2}\mu''(R_e)( R-R_e)^2+......$$

Substituting into (\ref{2}) the expansion (\ref{3}) and the expressions for $\psi_N$ and $\psi'_{N}$ and evaluating the integral, we obtain the following selection rules \begin{ equation}\label{4} \Delta J=\pm 1

$$\label{5} \Delta v =0,\pm 1, (\pm 2,\pm 3....)$$

The transition with $\Delta v=\pm 1$ is called the fundamental. Transitions with much less likely $\Delta v=\pm 2, \pm 3 ....$ are known as overtones.

True or false? (a) Vibration-rotation absorption bands in diatomic molecules always have $\Delta v=1$. (b) For pure rotational absorption spectra of diatomic molecules, only lines with $\Delta J=+1$ appear. (c) Because only lines with $\Delta J=+1$ are allowed in pure rotation spectra of diatomic molecules, such spectra contain only one line.}