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Each cyanide contributes charge $-1$, so the formal nickel oxidation state is $+2$ and the metal has $d^8$. The requested diagram retains only the five principally metal-$3d$ molecular orbitals; the mainly ligand bonding molecular orbitals lie below them and are omitted.

In the square, $d_{x^2-y^2}$ points directly towards all four ligand donors and gives the highest, strongly antibonding molecular orbital, $B_{1g}$. The $d_{z^2}$ function also has an allowed $A_{1g}$ interaction, but its in-plane overlap is smaller. Including the allowed nickel $4s$ mixing gives a principally $d$ $A_{1g}$ level with relatively weak antibonding character. The $B_{2g}$ and $E_g$ metal functions have no ligand partner in the sigma bond donor basis and are nonbonding in this minimal model. The useful qualitative ordering is thereforeThe three nonbonding functions are shown together because the sigma-only model does not determine their finer splitting. Additional electrostatic and pi backbonding interactions can change the ordering of the lower $d$ levels; molecular symmetry alone does not fix these energy gaps.^[ref]

In the tetrahedron, the ligand $T_2$ combinations interact with the metal $T_2$ functions, producing higher $d$-derived antibonding molecular orbitals. The metal $E$ pair has no sigma donor partner and remains nonbonding:The diagram fills both arrangements with eight metal electrons; the square is shown in its fully paired state.

water lies below pyridine and cyanide ligand in the spectrochemical series. At fixed metal and dioxolene charges, replacement by water decreases ${\Delta }_o$ and favours the high-spin limits: Co(II) remains $S=3/2$; Co(III) can become $S=2$ if the splitting falls below the pairing threshold. Replacement by cyanide increases ${\Delta }_o$ and favours Co(III), $S=0$, and Co(II), $S=1/2$. The two unchanged semiquinonate ligands would then give the Co(II) form a combined independent-spin moment of $3{\mu }_B$. The ranking predicts the direction of the change; two substituted donors alone do not guarantee a Co(III) spin crossover, since all six donors determine its splitting. A shift of the valence-tautomer equilibrium is also possible and is separate from this fixed-charge comparison.

The products are hexanenitrile, $\mathrm{C}{\mathrm{H}}_3(\mathrm{C}{\mathrm{H}}_2{)}_4\mathrm{C}\equiv \mathrm{N}$, and ethyl triphenylmethyl ether, $\mathrm{P}{\mathrm{h}}_3\mathrm{C}-\mathrm{O}-\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_3$. The nitrile carbon comes from cyanide, so replacing chloride in the five-carbon chain gives a six-carbon product.

For the primary chloride, the carbon-end lone pair of $\mathrm{C}{\mathrm{N}}^{-}$ attacks the carbon carrying Cl from the back side while the C–Cl bond pair moves onto Cl. The single concerted nucleophilic substitution step forms the C–C bond and releases $\mathrm{C}{\mathrm{l}}^{-}$. Attack through the carbon end gives the nitrile, not an isocyanide.

For the triphenylmethyl chloride, the C–Cl bond first undergoes heterolysis to $\mathrm{P}{\mathrm{h}}_3{\mathrm{C}}^{+}$ and $\mathrm{C}{\mathrm{l}}^{-}$. An ethanol oxygen lone pair then attacks the carbocation, giving $\mathrm{P}{\mathrm{h}}_3\mathrm{C}-{\mathrm{O}}^{+}(\mathrm{H})\mathrm{Et}$. Finally pyridine accepts that O–H proton, with the O–H bond pair returning to oxygen, giving the neutral ether and pyridinium chloride.

A coordination complex with four carbon-bound cyanide ligands and a formal nickel(II), $d^8$, centre. Square planar molecular geometry permits a low-spin complex with no unpaired electrons.