question_answer

What is the oxidation state of iron in \[{{[Fe{{({{H}_{2}}O)}_{5}}NO]}^{2+}}\]?

A)  0                     

B)  +1

C)  +2                   

D)  +3

Correct Answer: B

Solution :

\[{{[Fe{{({{H}_{2}}O)}_{5}}NO]}^{2+}}\] This is a typical case of complex formed by electron exchange. NO changes to \[N{{O}^{+}}\]by loss of electron and this electron is gained by \[F{{e}^{2+}}\]which changes to \[F{{e}^{+}}\]                 \[NO\xrightarrow{{}}N{{O}^{+}}+{{e}^{-}}\]                 \[F{{e}^{2+}}-{{e}^{-}}\xrightarrow{{}}F{{e}^{+}}\] Thus, oxidation state of iron = + 1 \[F{{e}^{2+}}=Ar\,\,4s\,\,3{{d}^{6}}\] \[F{{e}^{2+}}=Ar\,\,4s\,\,3{{d}^{7}}\] Three unpaired electrons + 1 oxidation state is confirmed by magnetic moment of iron\[=\sqrt{3\,(3+2)}\] \[=\sqrt{15}\,BM\] and also by diamagnetic character of NO (which is only possible when it has no unpaired electron).