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 \[\text{N}{{\text{O}}^{\text{+}}}\]by loss of electron and this electron is gained by \[\text{F}{{\text{e}}^{2+}}\]which changes to\[\text{F}{{\text{e}}^{\text{+}}}\] \[NO\xrightarrow{{}}N{{O}^{+}}+{{e}^{-}}\] \[F{{e}^{2+}}+{{e}^{-}}\xrightarrow{{}}F{{e}^{+}}\] Thus, oxidation state of iron = + 1 \[F{{e}^{2+}}=Ar\,\,4s\,3{{d}^{6}}\]                                 \[F{{e}^{+}}=Ar\,4s\,\,3{{d}^{7}}\]                                                           Three unpaired electrons + 1 oxidation state is conirmed 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 noun paired electron).