In the molecular orbital diagram for the molecular ion, ${\mathrm{N}}_2^{+}$, the number of electrons in the ${\sigma }_{2\mathrm{p}}$ molecular orbital is

In molecular orbitals, electrons are arranged as follows:

$\mathrm{\sigma }1{\mathrm{s}}^2, {\mathrm{\sigma }}^{\ast }1{\mathrm{s}}^2 , \mathrm{\sigma }2{\mathrm{s}}^2, {\mathrm{\sigma }}^{\ast }2{\mathrm{s}}^2, \mathrm{\pi }2{{\mathrm{p}}_{\mathrm{y}}}^2=\mathrm{\pi }2{{\mathrm{p}}^2}_{\mathrm{z}}, \mathrm{\sigma }2{{\mathrm{p}}^1}_{\mathrm{x}}$.

Hence, the correct option is option (B) 1.