question_answer

Compare the relative stabilities of \[O_{2}^{-}\] and \[N_{2}^{+}\] and comment on their magnetic (paramagnetic or diamagnetic) behaviour.

Answer:

                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.