According to Molecular Orbital Theory,

A] The molecular orbital energy configuration of ${\mathrm{C}}_2^{2-}$ is
${\mathrm{\sigma }}_{1\mathrm{s}}^2,\mathrm{ }{\mathrm{\sigma }}_{1\mathrm{s}}^{\mathrm{*}2},\mathrm{ }{\mathrm{\sigma }}_{2\mathrm{s}}^2,\mathrm{ }{\mathrm{\sigma }}_{2\mathrm{s}}^{\mathrm{*}2},\mathrm{ }{\mathrm{\pi }}_{2{\mathrm{p}}_{\mathrm{x}}}^2={\mathrm{\pi }}_{2{\mathrm{p}}_{\mathrm{y}}}^2\mathrm{ },\mathrm{ }{\mathrm{\sigma }}_{2{\mathrm{p}}_{\mathrm{z}}}^2$
In the MO of ${\mathrm{C}}_2^{2-}$ there is no unpaired electron hence it is diamagnetic
B] Bond order of ${\mathrm{O}}_2^{2+}$ is 3 and ${\mathrm{O}}_2$ is 2 therefore bond length of ${\mathrm{O}}_2$ is greater than ${\mathrm{O}}_2^{2+}$
C] The molecular orbital energy configuration of ${\mathrm{N}}_2^{+}$ is
${\mathrm{\sigma }}_{1\mathrm{s}}^2,\mathrm{ }{\mathrm{\sigma }}_{1\mathrm{s}}^{\mathrm{*}2},\mathrm{ }{\mathrm{\sigma }}_{2\mathrm{s}}^2,\mathrm{ }{\mathrm{\sigma }}_{2\mathrm{s}}^{\mathrm{*}2},\mathrm{ }{\mathrm{\pi }}_{2{\mathrm{p}}_{\mathrm{x}}}^2={\mathrm{\pi }}_{2{\mathrm{p}}_{\mathrm{y}}}^2\mathrm{ },\mathrm{ }{\mathrm{\sigma }}_{2{\mathrm{p}}_{\mathrm{z}}}^1$
Bond order of ${\mathrm{N}}_2^{+}=\frac{1}{2}\left(9-4\right)=2.5$
The molecular orbital energy configuration of ${\mathrm{N}}_2^{-}$ is
${\mathrm{\sigma }}_{1\mathrm{s}}^2,\mathrm{ }{\mathrm{\sigma }}_{1\mathrm{s}}^{\mathrm{*}2},\mathrm{ }{\mathrm{\sigma }}_{2\mathrm{s}}^2,\mathrm{ }{\mathrm{\sigma }}_{2\mathrm{s}}^{\mathrm{*}2},\mathrm{ }{\mathrm{\pi }}_{2{\mathrm{p}}_{\mathrm{x}}}^2={\mathrm{\pi }}_{2{\mathrm{p}}_{\mathrm{y}}}^2\mathrm{ },{\mathrm{\sigma }}_{2{\mathrm{p}}_{\mathrm{z}}}^2\mathrm{ },{\mathrm{\pi }}_{2{\mathrm{p}}_{\mathrm{x}}}^{\mathrm{*}1}={\mathrm{\pi }}_{2{\mathrm{p}}_{\mathrm{y}}}^{\mathrm{*}}$
Bond order of ${\mathrm{N}}_2^{-}=\frac{1}{2}\left(10-5\right)=2.5$
D] $\mathrm{H}{\mathrm{e}}_2^{+}$ has less energy as compare to two isolated He atoms