KCET · Chemistry · Coordination Compounds
The oxidation state of \(\mathrm{Fe}\) in the brown ring complex: \(\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{NO}\right] \mathrm{SO}_{4}\) is
- A \(+3\)
- B 0
- C \(+2\)
- D \(+1\)
Answer & Solution
Correct Answer
(D) \(+1\)
Step-by-step Solution
Detailed explanation
Let the oxidation state of \(\mathrm{Fe}\) in
\[
\begin{array}{cc}
& {\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{NO}\right]^{2+}} \\
\Rightarrow & \mathrm{x}+0+1=2 \\
\therefore & \mathrm{x}=+1
\end{array}
\]
Here \(\mathrm{NO}\) exists as nitrosyl ion \(\left(\mathrm{NO}^{+}\right)\).
\[
\begin{array}{cc}
& {\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{NO}\right]^{2+}} \\
\Rightarrow & \mathrm{x}+0+1=2 \\
\therefore & \mathrm{x}=+1
\end{array}
\]
Here \(\mathrm{NO}\) exists as nitrosyl ion \(\left(\mathrm{NO}^{+}\right)\).
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