Past exam of the chemistry A course of the University of Cambridge / 2024 / Paper 1 / A5 / a / iv / 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 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 isThis 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.^[ref]