• # question_answer Compare the relative stabilities of $O_{2}^{-}$ and $N_{2}^{+}$ and comment on their magnetic (paramagnetic or diamagnetic) behaviour.

M.O. Electronic configuration of$O_{2}^{-}\,=KK\,\sigma _{2s}^{2}\,{{\sigma }^{*2}}_{2}\sigma _{2{{p}_{z}}}^{2}\,\pi _{2{{p}_{x}}}^{2}\,\pi _{2{{p}_{y}}}^{2}\,\pi _{2{{p}_{x}}}^{*2}\,\pi _{2{{p}_{y}}}^{*1}$ Bond order$=\frac{1}{2}(8-5)=\frac{3}{2}=1\cdot 5$ M.O. Electronic configuration of$N_{2}^{+}=KK\,\sigma _{2s}^{2}\sigma _{2s}^{*2}\,\pi _{2{{p}_{x}}}^{2}\,\pi _{2{{p}_{y}}}^{2}\,\sigma _{2\,{{p}_{Z}}}^{1}$ Bond order $=\frac{1}{2}\,(7-2)=\frac{5}{2}\,=2\cdot 5$ As bond order of $N_{2}^{+}>$ bond order of $O_{2}^{-},$ therefore, $N_{2}^{+}$ is more stable than $O_{2}^{-}$. Each of them contains unpaired electron, hence both are paramagnetic.