Considering the quadrupolar nature of M–M bond in $[\mathrm{Re_2Cl_8}]^{2-}$, the M–M bond order in $[\mathrm{Re_2Cl_4(PMe_2Ph)_4}]^{+}$ and $[\mathrm{Re_2Cl_4(PMe_2Ph)_4}]$ respectively are
(a)3.0 and 3.0
(b)3.0 and 3.5
(c)3.5 and 3.5
(d)3.5 and 3.0
Answer
Answer: D ✓ checked by 4AB · confidence high
Explanation
[Re₂Cl₈]²⁻: Re(III) d⁴–d⁴, 8 metal electrons, σ²π⁴δ², bond order 4.
[Re₂Cl₄(PMe₂Ph)₄]⁺: Re₂⁵⁺, 9 metal electrons, σ²π⁴δ²δ*¹, bond order (8 − 1)/2 = 3.5.
[Re₂Cl₄(PMe₂Ph)₄]: Re(II) d⁵–d⁵, 10 metal electrons, σ²π⁴δ²δ*², bond order (8 − 2)/2 = 3.0.
Bond orders are 3.5 and 3.0 respectively — option (d).