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

Step 1: Recall how metal-metal bond order is found from d-electron count.
For an eclipsed \(\mathrm{M_2X_8}\)-type dimer (the classic quadruple-bond framework used by Re, W, Mo, Os and similar dimetal species), the metal-metal molecular orbitals fill in the order \(\sigma < \pi = \pi < \delta < \delta^* < \pi^* = \pi^* < \sigma^*\). The bond order is half the difference between the number of electrons in bonding MOs (\(\sigma, \pi, \pi, \delta\)) and antibonding MOs. For simple carbonyl/cyclopentadienyl dimers that obey the 18-electron rule, the metal-metal bond order instead comes from how many electron pairs are needed to bring each metal centre up to 18 valence electrons.

Step 2: Bond order in \([\mathrm{Re_2Cl_8}]^{2-}\).
The charge balance is \(2\mathrm{Re} + 8(-1) = -2\), so each Re is \(+3\). Re (group 7, \([\mathrm{Xe}]4f^{14}5d^56s^2\)) as \(\mathrm{Re^{3+}}\) is \(d^4\). Two \(d^4\) centres give 8 metal d electrons, which exactly fill the four bonding MOs \(\sigma^2\pi^4\delta^2\) with none left for the antibonding set. Bond order \(= 4\) (the famous Re-Re quadruple bond, Cotton's classic result).

Step 3: Bond order in \([\mathrm{Os_2Cl_8}]^{2-}\).
By the same charge balance each Os is \(+3\). Os (\([\mathrm{Xe}]4f^{14}5d^66s^2\)) as \(\mathrm{Os^{3+}}\) is \(d^5\), so two centres give 10 d electrons. These fill \(\sigma^2\pi^4\delta^2\) (8 electrons, bonding) and then \(\delta^{*2}\) (2 electrons, antibonding). Bond order \(=\dfrac{(8-2)}{2}=3\).

Step 4: Bond order in \(\mathrm{W_2(NMe_2)_6}\).
This is an ethane-like, staggered \(\mathrm{M_2L_6}\) dimer. Each W is formally \(+3\) (six anionic amide ligands, neutral overall complex), so \(\mathrm{W^{3+}}\) is \(d^3\) (neutral W is \(d^6\)). Two \(d^3\) centres give 6 electrons, which fill \(\sigma^2\pi^4\) exactly, the classic \(\mathrm{W \equiv W}\) triple bond seen in \(\mathrm{W_2(OR)_6}\)/\(\mathrm{W_2(NR_2)_6}\) chemistry. Bond order \(= 3\).

Step 5: Bond order in \([\mathrm{Mo(C_5H_5)(CO)_2}]_2\).
Count valence electrons on one \(\mathrm{CpMo(CO)_2}\) fragment by the neutral (radical) method: Mo contributes 6, \(\mathrm{Cp}\) (\(\eta^5\), neutral radical donor) contributes 5, and each CO contributes 2, giving \(6+5+2(2)=15\) electrons. To reach the 18-electron count, the Mo-Mo bond must supply 3 electrons to each metal, so the bond order is 3 (an \(\mathrm{Mo \equiv Mo}\) triple bond, the same situation as the well known \([\mathrm{CpMo(CO)_2}]_2\) dimer).

Step 6: Add the four bond orders.
\[ 3\;(\mathrm{Os_2Cl_8^{2-}}) + 4\;(\mathrm{Re_2Cl_8^{2-}}) + 3\;(\mathrm{W_2(NMe_2)_6}) + 3\;([\mathrm{CpMo(CO)_2}]_2) = 13 \]
Final Answer:
The sum of the metal-metal bond orders is \[ \boxed{13} \]