Answer
See the answer below.
Work Step by Step
$O_2^{2-}$ has 14 valence electrons so the configuration is:
$[core]\ (\sigma_{2s})^2\ (\sigma_{2s}^*)^2\ (\sigma_{2p})^2\ (\pi_{2p})^4\ (\pi_{2p}^*)^4$
$O_2$ has 12 valence electrons so the configuration is:
$[core]\ (\sigma_{2s})^2\ (\sigma_{2s}^*)^2\ (\sigma_{2p})^2\ (\pi_{2p})^4\ (\pi_{2p}^*)^2$
A) Since there are no unpaired electrons, the ion is diamagnetic while the molecule has two and is paramagnetic.
B) Net bonds 1 $\sigma$ for both and 1 $\pi$ for the molecule.
C) Since $B.O.=(N.\ Bonding -N.\ Antibonding)/2$
This ion has a bond order if 1 and the molecule, 2.
D) Since bond length decreases with bond order, the molecule has a shorter bond.
