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).You need to login to perform this action.
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