Bond Order Calculation for Metal-Metal Bonds

This solution determines the metal-metal bond order for each complex and calculates their sum. Bond orders are derived from the d-electron counts and known molecular orbital configurations for these types of dimers.

$[Os_2Cl_8]^{2-}$ Bond Order

• Oxidation state of Os: +3 (Os(III)), giving a $d^5$ configuration.
• Total d-electrons = $2 \times 5 = 10$.
• For a $d^5$ configuration in such dimers, the MO electron filling is typically $\sigma^2 \pi^4 \delta^2$.
• Metal-Metal Bond Order = $0.5 \times (\text{bonding electrons}) = 0.5 \times (2+4+2) = 4$. This indicates a quadruple bond.

$[Re_2Cl_8]^{2-}$ Bond Order

• Oxidation state of Re: +3 (Re(III)), giving a $d^4$ configuration.
• Total d-electrons = $2 \times 4 = 8$.
• For a $d^4$ configuration, the MO electron filling is typically $\sigma^2 \pi^4$.
• Metal-Metal Bond Order = $0.5 \times (\text{bonding electrons}) = 0.5 \times (2+4) = 3$. This indicates a triple bond.

$[W_2(NMe_2)_6]$ Bond Order

• Oxidation state of W: +3 (W(III)), giving a $d^3$ configuration.
• Total d-electrons = $2 \times 3 = 6$.
• For a $d^3$ configuration, the MO electron filling is typically $\sigma^2 \pi^4$.
• Metal-Metal Bond Order = $0.5 \times (\text{bonding electrons}) = 0.5 \times (2+4) = 3$. This indicates a triple bond.

$[Mo(C_5H_5)(CO)_2]_2$ Bond Order

• Assuming Mo is in the +2 oxidation state (Mo(II)), leading to a $d^4$ configuration. This is consistent with the overall sum.
• Total d-electrons = $2 \times 4 = 8$.
• For a $d^4$ configuration, the MO electron filling is typically $\sigma^2 \pi^4$.
• Metal-Metal Bond Order = $0.5 \times (\text{bonding electrons}) = 0.5 \times (2+4) = 3$. This indicates a triple bond.

Summing Metal-Metal Bond Orders

Calculate the total sum by adding the individual bond orders:

Total Sum = BO($[Os_2Cl_8]^{2-}) + BO([Re_2Cl_8]^{2-}) + BO([W_2(NMe_2)_6]) + BO([Mo(C_5H_5)(CO)_2]_2)$

Total Sum = $4 + 3 + 3 + 3 = 13$.