Past exam of the chemistry A course of the University of Cambridge / 2024 / Paper 1 / A5 / a

The ideal eclipsed ion has D4h molecular point group symmetry. Put the Mo–Mo axis along $z$, with the metal atoms at $(0,0,\pm c)$, and orient the four chloride positions around each metal along $\pm x,\pm y$.

There is a $C_4(z)$ axis, perpendicular twofold axes, a horizontal plane through the midpoint, vertical mirror planes and an inversion centre at the bond midpoint. The eclipsed arrangement makes the two chloride squares coincide in projection along $z$. These labels refer to the ideal molecular geometry, rather than small crystal-packing distortions.

Let $d_{\mu ,A}$ and $d_{\mu ,B}$ be corresponding real 4d orbitals on the upper and lower metals, with identical global-axis phases. Their symmetry-adapted linear combinations arewith the normalization shown in the negligible-overlap approximation; including overlap changes the denominator but not the symmetry. Inversion exchanges the centres and each local d orbital has even intrinsic parity, so the plus combinations are gerade and the minus combinations ungerade.The reversal of bonding phase for $d_{xz},d_{yz}$ is important: the facing lobes have opposite signs when the two atomic orbitals share the same global phase convention, so their difference is bonding. Each $e$ irrep is two-dimensional. Thuswith total dimension ten. Rotating $x,y$ by $45^{\circ }$ interchanges the names of the two delta families; all labels here retain the chloride-aligned axes fixed above.

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]