Reaction of radiation with Nitrogen-14

The reaction releases around 4MeV of energy, which is vastly more than the chemical binding energy of nitrogen or the cyanide radial. This means the reaction splits the original $N_2$ molecule into two separate atoms of nitrogen and carbon 14.

The process that creates the carbon 14 nucleus is:

$$ {}^{14}N_7 + n to {}^{14}C_6 + p + mathrm{energy} $$

The reaction is with thermal neutrons so the energy of the neutron is negligible. That means the energy released is equal to the difference in the binding energies of the nuclei. The mass defect of ${}^14N_7$ is $0.108506u$ while the mass defect of $^{}14C_6$ is $0.113027u$ so the carbon 14 nucleus energy is lower by $0.004521$ amu or about $4.2$ MeV. The proton will carry away most of this energy because it's much lighter than the carbon 14 nucleus, but the KE of the carbon nucleus will still be around $0.28$ MeV.