Search any question & find its solution
Question:
Answered & Verified by Expert
A brown ring is formed in the ring test for $\mathrm{NO}_3^{-}$ion. It is due to the formation of
Options:
Solution:
2186 Upvotes
Verified Answer
The correct answer is:
$\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_5(\mathrm{NO})\right]^{2+}$
$\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_5(\mathrm{NO})\right]^{2+}$
When freshly prepared solution of $\mathrm{FeSO}_4$ is added to an aqueous solution containing $\mathrm{NO}_3^{-}$ion, it leads to the formation of a brown coloured complex. This is known as brown ring test for nitrate ion.
$$
\begin{aligned}
&\mathrm{NO}_3^{-}+3 \mathrm{Fe}^{2+}+4 \mathrm{H}^{+} \longrightarrow \mathrm{NO}+3 \mathrm{Fe}^{3+}+2 \mathrm{H}_2 \mathrm{O} \\
&{\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}+\mathrm{NO} \longrightarrow\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_5(\mathrm{NO})\right]^{2+}+\mathrm{H}_2 \mathrm{O}}
\end{aligned}
$$
Brown ring
$$
\begin{aligned}
&\mathrm{NO}_3^{-}+3 \mathrm{Fe}^{2+}+4 \mathrm{H}^{+} \longrightarrow \mathrm{NO}+3 \mathrm{Fe}^{3+}+2 \mathrm{H}_2 \mathrm{O} \\
&{\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}+\mathrm{NO} \longrightarrow\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_5(\mathrm{NO})\right]^{2+}+\mathrm{H}_2 \mathrm{O}}
\end{aligned}
$$
Brown ring
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.