The transitions between quantum states follow selection rules.
https://en.wikipedia.org/wiki/Selection_rule
During any transition, energy, momentum and quantum angular momentum (spin) must be preserved. Those are the hard rules. Also the incoming EM wave has to create an electric field oscillation that overlaps with the electronic waveform oscillation that would occur during the transition (they couple). This overlap can be significant for a given transition and the EM wave frequency matching its energy difference, which makes for a percievable probability of absorption, or very low, in which case the transiton is forbidden. How strictly forbidden depends on the circumstances (neighbouring atoms, collisions, ...).
The usual transitions absorb the complete energy of an incoming photon. Then the difference in spin quantum number between the two states must be 1, the spin of a single photon. Some variable amount of energy always goes into translational energy, pushing the molecule that is hit. That's why absorption lines always have a width. The momentum of the photon is retained in that push, so the molecule has to fly off in an angle that matches excess energy and the momentum.
Raman IR transitions grab a small amount of energy off of a higher energy EM wave / photon (or add to it). The spin difference between the states must therefore be zero, and the momentum and direction of the leaving photon must also be taken into account, which dictates the energy difference.
Crazier transitions are possible. Two photons could be absorbed at the same time. Very low probability unless you have an extremely bright light source, but this allows totally different transitions, because now the same wavelenght has double the energy, and you have the spin from two photons.
So, answering your actual question, there are several possibilities to absorb a photon with an energy that does not exactly match the given transition.
(And then of course you have the intrinsic bandwidth of the transition, due to brownian motion.)