 Solution

In the Condon approximation the intensity from $v^{\prime \prime }=0$ to $v^{\prime }$ contains the Franck-Condon factorThe lower ground-state vibrational wavefunction is localized near its equilibrium bond length. If the two equilibrium lengths are similar, it overlaps strongly with the upper $v^{\prime }=0$ wavefunction, but its overlap with highly excited upper vibrational states is usually small or suffers oscillatory cancellation. If the equilibrium lengths differ enough that a vertical excitation reaches the outer portion of the upper well, high-$v^{\prime }$ transitions can be strong while the $0\to 0$ overlap is small. Thus the two ends of a long progression are not normally both intense. This is an overlap argument, not a ban on either set of transitions; anharmonicity, a varying electronic transition moment and unusual wavefunctions can modify the intensity envelope.