According to Molecular Orbital Theory,

A] The molecular orbital energy configuration of ${C}_{2}^{2 -}$ is
${\sigma}_{1s}^{2}, {\sigma}_{1s}^{*2}, {\sigma}_{2s}^{2}, {\sigma}_{2s}^{*2}, {\pi}_{2{p}_{x}}^{2} = {\pi}_{2{p}_{y}}^{2}, {\sigma}_{2{p}_{z}}^{2}$
In the MO of ${C}_{2}^{2 -}$ there is no unpaired electron hence it is diamagnetic
B] Bond order of ${O}_{2}^{2 +}$ is 3 and ${O}_{2}$ is 2 therefore bond length of ${O}_{2}$ is greater than ${O}_{2}^{2 +}$
C] The molecular orbital energy configuration of ${N}_{2}^{+}$ is
${\sigma}_{1s}^{2}, {\sigma}_{1s}^{*2}, {\sigma}_{2s}^{2}, {\sigma}_{2s}^{*2}, {\pi}_{2{p}_{x}}^{2} = {\pi}_{2{p}_{y}}^{2}, {\sigma}_{2{p}_{z}}^{1}$
Bond order of ${N}_{2}^{+} = \dfrac{1}{2}\left(9 - 4\right) = 2.5$
The molecular orbital energy configuration of ${N}_{2}^{-}$ is
${\sigma}_{1s}^{2}, {\sigma}_{1s}^{*2}, {\sigma}_{2s}^{2}, {\sigma}_{2s}^{*2}, {\pi}_{2{p}_{x}}^{2} = {\pi}_{2{p}_{y}}^{2}, {\sigma}_{2{p}_{z}}^{2}, {\pi}_{2{p}_{x}}^{*1} = {\pi}_{2{p}_{y}}^{*}$
Bond order of ${N}_{2}^{-} = \dfrac{1}{2}\left(10 - 5\right) = 2.5$
D] $H{e}_{2}^{+}$ has less energy as compare to two isolated He atoms