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A delta bond involves sideways orbital overlap with two nodal planes containing the internuclear axis. In suitable metal-metal bonds, d-orbital overlap can supply this component in addition to sigma and pi bonding.

Ignoring ligand perturbations initially, end-on $d_{z^2}$ overlap gives the largest sigma bond splitting, side-on $d_{xz},d_{yz}$ overlap gives the two pi bonds, and the two four-lobed overlaps give weaker delta bonds. A useful qualitative ordering isThe original diagram shows the ten orbitals, including both members of each $e$ level.

For the strictly isolated collinear metal pair, cylindrical symmetry makes $d_{xy}$ and $d_{x^2-y^2}$ related by rotation, so the two delta bonding levels are degenerate, as are their antibonding partners. The displayed D4h molecular point group labels anticipate the ligand environment, which can split those different irreps. Symmetry supplies labels and degeneracies; the relative separation of unrelated levels is a qualitative overlap model, not a numerical prediction.

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]

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.

A formal fourfold bond. In the familiar d-metal example, one sigma bond, two pi bonds and one delta bond are occupied, giving the configuration ${\sigma }^2{\pi }^4{\delta }^2$ and formal bond order four.