GATE 2026 CY – Question 59
The sum of metal–metal bond orders 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 _____ (integer).
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Correct answer: 13
Explanation
The metal-metal bond order is found from the number of electrons in the metal-based bonding and antibonding orbitals.
- **$[\text{Os}_2\text{Cl}_8]^{2-}$:** the oxidation state of Os is +3 ($d^5$ each) with 10 $d$ electrons in the dimer, filling $\sigma^2\pi^4\delta^2\delta^{*2}\pi^{*0}$. The bond order is $(2+4+2-2)/2=3$.
- **$[\text{Re}_2\text{Cl}_8]^{2-}$:** Re(III) is $d^4$; the 8 electrons fill $\sigma^2\pi^4\delta^2$, the **quadruple bond**, bond order 4.
- **$[\text{W}_2(\text{NMe}_2)_6]$:** W(III) is $d^3$; the 6 electrons fill $\sigma^2\pi^4$, a **triple bond**, bond order 3.
- **$[\text{CpMo(CO)}_2]_2$** (so-called $\text{Cp}_2\text{Mo}_2(\text{CO})_4$): each Mo fragment has 15 electrons and needs 3 more to reach 18, so a **triple bond** Mo≡Mo, bond order 3.
**Sum:**
$$3+4+3+3=\mathbf{13}.$$