20Organometallic Chemistry: Bonding and Ligands

Argued. The $\pi^*$ orbitals of CO are antibonding between C and O: populating them weakens the C–O bond, lowers its force constant, and hence the stretching wavenumber, which varies as the square root of the force constant (Chapter 6). $\sigma$ donation removes electrons from an orbital nearly non-bonding (slightly antibonding) for C–O, which by itself would raise the wavenumber a little: the observed decrease shows that back-donation dominates. ∎

Proof. The $e_g$ orbitals have $\sigma$ symmetry only. The $t_{2g}$ orbitals ($d_{xy}$, $d_{xz}$, $d_{yz}$) match a $t_{2g}$ combination of ligand $\pi$ orbitals: for $d_{xy}$, the four ligands in the $xy$ plane each offer a $\pi$ orbital perpendicular to their M–L axis in that plane, and the combination with alternating signs has the symmetry of $d_{xy}$ (the remaining combinations transform as $t_{1g}$, $t_{1u}$ and $t_{2u}$ and find no $d$ partner). Two orbitals of the same symmetry mix and repel: the lower is pushed down, the upper up. A $\pi$ acceptor offers empty orbitals above the $t_{2g}$ set, which is pushed down: $\Delta_{\mathrm o}$ increases. A $\pi$ donor offers filled orbitals below it, which push it up: $\Delta_{\mathrm o}$ decreases. The $e_g$ level is unaffected in both cases. ∎

Argued. Removing electrons from the $\pi$ (bonding) orbital and adding them to the $\pi^*$ (antibonding) orbital both lower the C–C bond order. Mixing $\pi^*$ character rehybridises the carbons towards $sp^3$, whose bonds to the substituents point away from the metal, towards the metallacyclopropane geometry. ∎

Argued. The $\pi$ orbitals of the two rings combine into symmetry-adapted pairs $a_{1g}$, $a_{2u}$, $e_{1g}$, $e_{1u}$, $e_{2g}$, $e_{2u}$ (from the five $\pi$ orbitals of each ring, with zero, one and two nodal planes through the axis). The filled ring $e_{1g}$ combination overlaps strongly with $d_{xz}$, $d_{yz}$ (one nodal plane through the axis each): they form bonding orbitals, mostly ring, and antibonding $e_{1g}^\ast$, mostly metal, pushed high. $d_{xy}$, $d_{x^2-y^2}$ (two nodal planes) meet only the empty ring $e_{2g}$ orbitals and are slightly stabilised by back-donation; $d_{z^2}$ points at the hole in the middle of each ring and hardly overlaps. Six electrons fill $e_{2g}$ and $a_{1g}$; with the twelve electrons of the bonding ring combinations, the count is 18. ∎

Proof. With $z$ along the Re–Re axis and the Cl atoms along $\pm x$ and $\pm y$ on each Re, $d_{x^2-y^2}$ points at the chlorides and is used in Re–Cl bonding. The other four $d$ orbitals of each metal overlap pairwise: $d_{z^2}$ with $d_{z^2}$ along the axis ($\sigma$), $d_{xz}$ with $d_{xz}$ and $d_{yz}$ with $d_{yz}$ side by side ($\pi$, twice), $d_{xy}$ with $d_{xy}$ face to face ($\delta$, the weakest). Eight electrons fill the four bonding combinations. Rotating one unit by $45^\circ$ about $z$ turns its $d_{xy}$ into $d_{x^2-y^2}$, which has zero overlap with the other $d_{xy}$ by symmetry: the $\delta$ bond, and with it the eclipsed preference, is lost. The other three overlaps are cylindrically symmetric or unchanged in magnitude. ∎

Argued. $\ce{CH3}$ is a tetrahedral fragment missing one bond: one $sp^3$-like orbital with one electron, one short of an octet. $\ce{Mn(CO)5}$ is an octahedron missing one ligand: the empty site holds one hybrid orbital pointing out; with $7 + 10 = 17$ electrons it is one short of 18, and its single electron sits in that hybrid. Removing two or three ligands (or bonds) gives two or three such orbitals and electrons, with the same counting on both sides: $\ce{Fe(CO)4}$ has 16 electrons ($8 + 8$), two short of 18; $\ce{Co(CO)3}$ has 15 ($9 + 6$), three short. ∎