= Solution
The <metal> interacts with the \b[extra in-plane bonding system of <benzyne>], not by replacing the ordinary out-of-plane aromatic pi system. Take the ring and <metal> to lie in the $xz$ plane, with the metal-to-C–C-midpoint direction along $z$ and C1–C2 along $x$. The <carbon> orbitals directed toward the <metal> form an occupied symmetric combination and an empty antisymmetric combination.
The occupied in-plane $\pi$ combination donates into an empty <metal> orbital of matching symmetry. An occupied <metal> $d_{xz}$-type <orbital overlaps> with the in-plane $\pi^*$ combination, providing back-donation. These are the same symmetry principles as <alkene> binding, rotated into the plane of the ring. The <carbon> $p_y$ orbitals perpendicular to the ring continue to participate in its aromatic system. \b[The strained formal triple bond therefore does not supply two independent, ordinary out-of-plane <alkene> interactions.]
Strong population of the in-plane acceptor can be described as reduction of the aryne to a dianionic, metallacyclopropene-like ligand. That description emphasizes two M–C <sigma bonds> and loss of the extra C–C <pi bond>. It is particularly useful for early-metal aryne complexes. The lower panels of the orbital schematic distinguish the interacting in-plane orbitals from the perpendicular aromatic set.
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