Question

In the ring test of $N{O}_3^{-}$ ion, $F{e}^{2+}$ ion reduces nitrate ion to nitric oxide, which combines with $F{e}^{2+}\left(aq\right)$ ion to form a brown complex. Write the reaction involved in the formation of brown ring

Answer 1

Brown-ring test

The test is carried out by adding dilute ferrous sulphate solution to an aqueous solution containing nitrate ion $N{O}_3^{-}$, and then adding concentrated sulphuric acid along the sides of the test tube.

Formation of a brown ring at the interface between the solution and sulphuric acid layers takes place and it indicates the presence of nitrate ion in solution.

Reactions involved:

Reactions involved in the formation of brown ring test:

$N{O}_3^{-}+3F{e}^{2+}+4H^{+}$
$\downarrow$
$NO+3F{e}^{3+}+2H_{2}O$

$\left[Fe\right(H_{2}O{)}_6{]}^{2+}+NO$
$\downarrow$
$\left[Fe\right(H_{2}O{)}_5\left(NO\right){]}^{2+}+H_{2}O$

Thus, brown ring appears due to the formation of $\left[Fe\right(H_{2}O{)}_5\left(NO\right){]}^{2+}$ .

Additional Information:

The oxidation state of Fe in brown ring complex$\left[Fe\right(H_{2}O{)}_5\left(NO\right){]}^{2+}$ is +1