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).You need to login to perform this action.
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