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.
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