Among the following, the paramagnetic compound is:

In ${\mathrm{Na}}_2{\mathrm{O}}_2$, we have ${\mathrm{O}}_2^{2-}$ ion. Number of valence elctrons of the two oxygen in ${\mathrm{O}}_2^{2-}$ ion $=8\times 2+2=18$ which are present as follows: $\sigma 1s^2,{\sigma }^{\ast }1s^2,\sigma 2s^2,{\sigma }^{\ast }2s^2,\sigma 2p_x^2,\pi 2p_y^2=\pi 2p_z^2,{\pi }^{\ast }2p_y^2=$ ${\pi }^{\ast }2p_z^2$
$\therefore$ Number of unpaired electrons $=0$, hence, ${\mathrm{O}}_2^{2-}$ is diamagnetic.
No. of valence electrons of all atoms in ${\mathrm{O}}_3=6\times 3=18$. Thus, it also does not have any unpaired electron, hence, it is diamagnetic.
No. of valence electrons of all atom in ${\mathrm{N}}_2\mathrm{O}$ $=2\times 5+6=16$. Hence, here also all electrons are paired. So, it is diamagnetic.
In ${\mathrm{KO}}_2$, we have ${\mathrm{O}}_2^{-}$. So, No. of valence electrons of all atoms in ${\mathrm{O}}_2^{-}=2$ $\times 6+1=13$.
Thus, it has one unpaired electron, hence it is paramagnetic.