The number of bonding molecular orbitals and the number of available skeletal electrons in $\left[\mathrm{B}_{6} \mathrm{H}_{6}\right]^{2-}$, respectively, are:
(a)7 and 14
(b)6 and 12
(c)18 and 12
(d)11 and 14
Answer
Answer: A ✓ checked by 4AB · confidence high
Explanation
closo-[B₆H₆]²⁻: n + 1 = 7 bonding skeletal MOs filled by 6×2 + 2 = 14 skeletal electrons.