A) \[ N\operatorname{O}\to {{\operatorname{NO}}^{+}}\]
B) \[{{N }_{2}}\to {{\operatorname{N}}_{2}}^{+}\]
C) \[{{C}_{2}}\to {{C}_{2}}^{+}\]
D) \[{{O}_{2}}\to {{O}_{2}}^{+}\]
Correct Answer: A
Solution :
\[1.NO\xrightarrow{}N{{O}^{+}}\] \[15{{e}^{-}}\] \[14{{e}^{-}}\] \[2.\underset{14}{\mathop{{{N}_{2}}}}\,\xrightarrow{\,}\underset{13}{\mathop{N_{2}^{+}}}\,\] \[3.\underset{12}{\mathop{{{C}_{2}}}}\,\xrightarrow{\,}\underset{11}{\mathop{C_{2}^{+}}}\,\] \[4.\underset{16}{\mathop{{{O}_{2}}}}\,\xrightarrow{\,}\underset{15}{\mathop{O_{2}^{+}}}\,\] In \[N{{O}^{+}}\] all electrons are paired, hence, it is diamagnetic.
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