JEE Advanced 2017 · Paper 1, Chemistry question 11

Official key: marks to all (a printing inconsistency in this question).

The intended count, from molecular orbital configurations (with $s$-$p$ mixing for $\ce{B2}$, $\ce{C2}$, $\ce{N2}$):

Species	Unpaired electrons
$\ce{H2}$	0
$\ce{He2+}$	1
$\ce{Li2}$	0
$\ce{Be2}$	0 (bond order 0)
$\ce{B2}$	2
$\ce{C2}$	0
$\ce{N2}$	0
$\ce{O2-}$	1
$\ce{F2}$	0

That gives 6 diamagnetic species. The count is 5 if $\ce{Be2}$, which does not exist as a stable molecule, is excluded.

Answer: 6 (marks to all)