= 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 is
$$
\boxed{(a_{1g})^2(e_u)^4(b_{2g})^2=\sigma^2\pi^4\delta^2.}
$$
It requires eight metal d electrons, or four per Mo, hence Mo(II). With eight $\mathrm{Cl^-}$ ligands, charge balance gives
$$
\boxed{2(+2)+8(-1)=-4.}
$$
Oxidizing 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>.
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