Search any question & find its solution
Question:
Answered & Verified by Expert
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
Options:
Solution:
1352 Upvotes
Verified Answer
The correct answer is:
$+1$
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 $\left(\mathrm{NO}^{+}\right)$.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.