The diamagnetic species among the following is 

To be diamagnetic, no unpaired electron must be present. 

Hint Write molecular orbital electronic configuration of each species.

For all options given,

${\mathrm{\sigma }}_{1\mathrm{s}}^2{\mathrm{\sigma }}_{1\mathrm{s}}^{\star 2}{\mathrm{\sigma }}_{2\mathrm{s}}^2{\mathrm{\sigma }}_{2\mathrm{s}}^{*2}{\mathrm{\sigma }}_{2{\mathrm{p}}_{\mathrm{z}}}^2{\mathrm{\pi }}_{2{\mathrm{p}}_{\mathrm{x}}}^2={\mathrm{\pi }}_{2{\mathrm{p}}_{\mathrm{y}}}^2$ is common. 

After that, ${\mathrm{O}}_2^{+}:{\mathrm{\pi }}_{2{\mathrm{p}}_{\mathrm{x}}}^{*1}$,

${\mathrm{O}}_2^{-}:{\mathrm{\pi }}_{2{\mathrm{p}}_{\mathrm{x}}}^{*2}={\mathrm{\pi }}_{2{\mathrm{p}}_{\mathrm{y}}}^{*1}$,

and ${\mathrm{O}}_2:{\mathrm{\pi }}_{2{\mathrm{p}}_{\mathrm{x}}}^{*1}={\mathrm{\pi }}_{2{\mathrm{p}}_{\mathrm{y}}}^{*1}$,

all contain unpaired electron. They are paramagnetic, but ${\mathrm{O}}_2^{2-}:{\mathrm{\pi }}_{2{\mathrm{p}}_{\mathrm{x}}}^{*2}={\mathrm{\pi }}_{2{\mathrm{p}}_{\mathrm{y}}}^{\star 2}$ has no unpaired electrons. 

$\therefore$ It is diamagnetic.