 Solution

In the chromium complex, Cr(0) is $d^6$ and receives substantial donation from four carbonyl ligands. Its occupied orbitals can give pi backbonding into the boron-rich acceptor orbitals of the B${}_2$N${}_2$ ring, while the nitrogen-rich occupied ring orbitals donate to the metal. This favors binding across the ring face, approximately ${\eta }^4$, with all four ring atoms participating. A nearly planar ring maintains the p-orbital alignment needed for that interaction and accounts for comparable Cr–B and Cr–N distances. A four-electron ring donor also gives the useful count $6_d+8_{\mathrm{CO}}+4_{\mathrm{ring}}=18$.

Ti(IV) in TiCl${}_4$ is $d^0$ and strongly electron deficient. It is a good acceptor for the two nitrogen lone pairs, but has no filled $d$ set for comparable backbonding into the boron acceptors. chromium therefore favors a face-bound pi complex, while titanium favors N,N sigma coordination. Donation makes the nitrogen environments more nearly tetrahedral and permits the 2,4-Diazadiboretidine ring to pucker. The borons are not directly bound in the same way, explaining $d(\mathrm{Ti}\cdots \mathrm{B})>d(\mathrm{Ti}-\mathrm{N})$. The distinction follows from electron availability, rather than simply calling every short metal–ring contact a separate ordinary sigma bond.