Assuming 2s-2p mixing is not operative, the paramagnetic species among the following are :

${\mathrm{Be}}_2,{\mathrm{B}}_2,{\mathrm{C}}_2,{\mathrm{N}}_2,{\mathrm{O}}_2,{\mathrm{O}}_2^{+},\mathrm{NO}$

If  2s-2p mixing is not operative, the increasing order of Molecular orbitals will be :

$\sigma 1s^2,{\sigma }^{\ast }1s^2,\sigma 2s^2,{\sigma }^{\ast }2s^2,\sigma 2{\mathrm{P}}_{\mathrm{x}}^2\left[\begin{array}{l}\pi 2{\mathrm{p}}_{\mathrm{y}}^1\\ \pi 2{\mathrm{p}}_z^1\end{array}\right]$

Considering this ${\mathrm{C}}_2,{\mathrm{O}}_2,{\mathrm{O}}_2^{+}$ NO become diamagnetic,. Only ${\mathrm{C}}_2,{\mathrm{O}}_2,{\mathrm{O}}_2^{+}$  NO would be paramagnetic with electronic configuration as above.

Thus, the correct answer is 4.