= Solution
Both coordinated frequencies are substantially lower than that of free N$_2$. In a local harmonic model,
$$
\widetilde\nu={1\over2\pi c}\sqrt{k\over\mu},\qquad
\boxed{{k_A\over k_{\rm free}}\simeq\left({2021\over2359}\right)^2=0.734,
\quad {k_B\over k_{\rm free}}\simeq\left({2010\over2359}\right)^2=0.726.}
$$
Thus coordination lowers the approximate N–N stretching force constant by about \b[27%], with B marginally more activated than A by this measure. Donation into N$_2$ pi-antibonding orbitals provides the natural explanation: the M–N$_2$ bond has an important backbonding component and N–N loses some multiple-bond character.
The same qualitative red shift occurs for CO when metal-to-CO backbonding populates CO $\pi^*$ orbitals. N$_2$ is generally a poorer pi acceptor than CO, so low-valent electron-rich <metal> fragments are particularly helpful for stabilizing its complexes. A numerical claim that N$_2$ always shifts more or less than CO would require a matched <metal>/ligand comparison. Neither the two frequencies nor the local-force approximation establish integer N–N <bond orders>, complete N–N cleavage, or catalytic <nitrogen> conversion; they establish significant weakening of intact coordinated N$_2$.
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