The hepticities 'x' and 'y' of the arene moieties in the diamagnetic complex $\left[\left(\eta^{\mathrm{x}}-\mathrm{C}_{6} \mathrm{H}_{6}\right) \mathrm{Ru}\left(\eta^{\mathrm{y}}-\mathrm{C}_{6} \mathrm{H}_{6}\right)\right]$ respectively are
(a)6 and 6
(b)4 and 4
(c)4 and 6
(d)6 and 2
Answer
Answer: C ✓ checked by 4AB · confidence high
Explanation
Two η⁶ rings would give 8 + 12 = 20 electrons; one ring slips to η⁴ so that (η⁶-C₆H₆)Ru(η⁴-C₆H₆) is an 18-electron diamagnetic complex.