Question

The sum of bond orders of metal-metal bonds in \([\mathrm{Os_2Cl_8}]^{2-}\), \([\mathrm{Re_2Cl_8}]^{2-}\), \([\mathrm{W_2(NMe_2)_6}]\) and \([\mathrm{Mo(C_5H_5)(CO)_2}]_2\) is (in integer).

Hint:

Work out each metal's oxidation state and d-electron count (or 18-electron deficit for the carbonyl/Cp dimer), then use the M-M MO filling order \(\sigma\pi\pi\delta\,|\,\delta^*\pi^*\pi^*\sigma^*\) to get each bond order before adding them.

Answer 1

Correct Answer: 13

Each of these four dimers is a textbook case of a metal-metal multiple bond, and the trick is counting the relevant electrons correctly for each bonding scheme (crystal-field d-count for the halide dimers, 18-electron count for the carbonyl/Cp dimer).

1. $[\mathrm{Re_2Cl_8}]^{2-}$: Overall charge $-2$ with 8 chloride ($-1$ each) fixes Re at $+3$, a $d^4$ ion. Stacking two $d^4$ centres face to face gives 8 electrons, which is exactly enough to fill one $\sigma$, two $\pi$, and one $\delta$ bonding orbital, nothing left over for antibonding orbitals. That is a full quadruple bond, bond order $4$, the historic Re-Re bond that first proved metal-metal quadruple bonding exists.
2. $[\mathrm{Os_2Cl_8}]^{2-}$: Same charge bookkeeping puts Os at $+3$, but Os sits one group to the right of Re in the same row, so $\mathrm{Os^{3+}}$ is $d^5$, one electron richer than $\mathrm{Re^{3+}}$. Two $d^5$ centres supply 10 electrons: 8 fill the bonding set (same as above) and the extra 2 must go into the antibonding $\delta^*$ level. Losing one bonding pair's worth of order, bond order $= 4 - 1 = 3$.
3. $\mathrm{W_2(NMe_2)_6}$: This is a staggered, ethane-shaped $\mathrm{M_2L_6}$ molecule with six anionic amide ligands and a neutral overall complex, so each W is $+3$. Neutral tungsten contributes 6 valence electrons, so $\mathrm{W^{3+}}$ is $d^3$. Two $d^3$ centres give 6 electrons, filling $\sigma^2\pi^4$ with no $\delta$ overlap available in the staggered geometry, a clean $\mathrm{W\equiv W}$ triple bond, order $3$.
4. $[\mathrm{Mo(C_5H_5)(CO)_2}]_2$: Use the 18-electron rule on one $\mathrm{CpMo(CO)_2}$ unit. Neutral-atom counting gives Mo = 6 electrons, $\eta^5$-Cp (radical) = 5 electrons, two CO = 4 electrons, total 15. Each metal is 3 electrons short of 18, and the only source is the metal-metal bond, so the bond must be a triple bond (order 3) to supply those 3 electrons to each Mo.

Adding the four bond orders: $3 + 4 + 3 + 3 = 13$.

Let's summarize:

• Halide-bridged $d^n$-$d^n$ dimers (Re, Os) get their bond order from how many of the four bonding MOs ($\sigma,\pi,\pi,\delta$) fill before electrons spill into $\delta^*$.
• Ligand-supported organometallic dimers (W-amide, Mo-Cp-carbonyl) get their bond order from whatever multiplicity is needed to complete the 18-electron count at each metal.

So the sum of all four metal-metal bond orders is $\boxed{13}$.