= Solution
Choose $z$ along M–N–N. The occupied dinitrogen $3\sigma_g$ orbital, with amplitude on the terminal N atoms, overlaps an empty <metal> orbital of sigma symmetry to give \b[N$_2$-to-metal <sigma donation>]. Filled <metal> $d_{xz}$ and $d_{yz}$ orbitals separately overlap the two empty N$_2$ $\pi_g^*$ orbitals, giving \b[metal-to-N$_2$ <pi backbonding>]. The lower-energy combination of each interacting pair is bonding and the higher is antibonding; each donation channel starts with two <electrons> in the occupied fragment orbital. The lower panel of the original orbital diagram shows these three channels.
\Image[/past-exam-of-the-chemistry-b-course-of-the-university-of-cambridge/2022/paper-2-orbitals.png]
{title=Orbital-symmetry channels for sigma donation and two perpendicular pi-backbonding interactions}
As for a terminal <carbonyl ligand>, the donor and acceptor interactions reinforce one another: <sigma donation> helps make <electron density> available at the <metal>, while <pi backbonding> strengthens M–N and weakens N–N. For conventional <electron counting> an intact end-on neutral N$_2$ ligand is an L-type two-electron donor; counting the backbonded <electrons> a second time as extra donated ligand <electrons> would double count them.
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