Question

In the 'ring test' of ${N{O}_3}^{-}$ ion, there is the formation of a brown color ring which has a formula:

• A. $\left[Fe\right(H_{2}O{)}_{4}NOS{O}_4{]}^{2+}$
• B. $\left[Fe\right(H_{2}O{)}_{5}NO{]}^{2+}$
• C. $\left[Fe\right(H_{2}O{)}_{5}NO]S{O}_4$
• D. $\left[Fe\right(H_{2}O{)}_{4}NO{]}^{3+}$

Answer 1

The correct option is C $\left[Fe\right(H_{2}O{)}_{5}NO]S{O}_4$

$N{O}_3^{-}$ gives brown ring test
A common nitrate test, known as the brown ring test can be performed by adding iron(II) sulfate to a solution of a nitrate, then slowly adding concentrated sulfuric acid such that the sulfuric acid forms a layer above the aqueous solution. A brown ring will form at the junction of the two layers, indicating the presence of the nitrate ion. Note that the presence of nitrite ions will interfere with this test.
The overall reaction is the reduction of the nitrate ion by iron(II) which is oxidised to iron(III) and formation of a nitrosyl complex.
$N{O}_3^{-}+3F{e}^{2+}+4H^{+}\to 3F{e}^{3+}+NO+2H_{2}O$
$\left[Fe\right(H_{2}O{)}_6{]}^{2+}+NO\to \left[Fe\right(H_{2}O{)}_5\left(NO\right){]}^{2+}$ brown ring