The compound $\left[\mathrm{Re}_{2}\left(\mathrm{Me}_{2} \mathrm{PPh}\right)_{2} \mathrm{Cl}_{4}\right](\mathrm{M})$ having a configuration of $\sigma^{2} \pi^{4} \delta^{2} \delta^{*2}$ can be oxidized to $\mathrm{M}^{+}$ and $\mathrm{M}^{2+}$. The formal metal-metal order in $\mathrm{M}, \mathrm{M}^{+}$ and $\mathrm{M}^{2+}$ respectively, are
(a)3.0, 3.5 and 4.0
(b)3.5, 4.0 and 3.0
(c)4.0, 3.5 and 3.0
(d)3.0, 4.0 and 3.5 $\left[\mathrm{Re}_{2}\left(\mathrm{Me}_{2} \mathrm{PPh}\right)_{2} \mathrm{Cl}_{4}\right]=(\mathrm{M})$
Answer
Answer: A ✓ checked by 4AB · confidence high
Explanation
σ²π⁴δ²δ*²: order (8 − 2)/2 = 3.0; removing one δ* electron gives 3.5; removing both gives 4.0.