= Solution
Eight chloride ions contribute charge $-8$. For an overall $4-$ charge, the two metals together contribute $+4$, so each is formally Mo(II). <Molybdenum> is in group six, giving four d electrons per centre and \b[eight metal-based electrons in total].
In the ligand-perturbed configuration developed below, they occupy $\sigma^2\pi^4\delta^2$, with no Mo–Mo antibonding electron. Therefore the formal <bond order> is
$$
\boxed{\frac{N_b-N_a}{2}=\frac{8-0}{2}=4.}
$$
This is one <sigma bond>, two <pi bonds> and one <delta bond>. Before including ligands, the two ideal delta channels are degenerate and the allocation of their last two electrons is not determined by the bare energy sketch alone; the ligands remove that ambiguity and select the surviving low delta channel. The eight-electron bonding count still gives formal order four.
The occupied bonding groups are experimentally meaningful: molybdenum X-ray emission resolves three metal–metal bonding energy groups in this ion, alongside a deeper metal–chloride bonding feature.<Direct evidence for the quadruple metal-metal bond in the octachlorodimolybdenum(II) ion using X-ray emission spectroscopy by Derek E. Haycock and others>[Primary X-ray emission report, abstract and pages 262–263]{ref}
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