 Solution

The ligand field leaves one low sigma bond orbital, two low pi bond orbitals and one low delta bond orbital. Their closed-shell occupation isIt requires eight metal d electrons, or four per Mo, hence Mo(II). With eight $\mathrm{C}{\mathrm{l}}^{-}$ ligands, charge balance givesOxidizing removes an electron from a Mo–Mo bonding orbital; reducing populates a higher orbital with Mo–Mo antibonding or strong metal–ligand antibonding character. Thus the $4-$ state has the particularly favourable filled bonding configuration in this model. This is an orbital explanation of a preferred charge, not a claim that charge alone determines solution redox stability independently of counterions or solvent.

The eclipsed geometry also follows naturally from the occupied delta bond: it aligns the four-lobed $d_{xy}$ functions. Twisting the chloride squares toward a staggered geometry reduces this overlap and weakens that contribution to the quadruple bond.