Question

The chemical formula of brown ring formed in the brown ring test is .

• A. $\left(Fe\right(H_{2}O{)}_5\mathrm{NO}){\mathrm{SO}}_4$
• B. $\mathrm{Fe}(\mathrm{NO}{)}_2{\mathrm{SO}}_4$
• C. ${\mathrm{FeSO}}_4.{\mathrm{NO}}_3$

Answer 1

The correct option is A $\left(Fe\right(H_{2}O{)}_5\mathrm{NO}){\mathrm{SO}}_4$

To determine the presence of nitrate ion in any solution, 'brown ring test' is conducted.

Step 1: Solution to be tested is taken in a test-tube
Step 2: Iron(II) sulphate is added to it
Step 3: Slowly concentrated sulphuric acid (H₂SO₄) is added such that the acid added forms a layer below the aqueous solution.
If a ring brown in colour is formed at the junction of the 2 layers, it indicates the presence of nitrate ion.
Its chemical formula is $\left[\mathrm{Fe}\right(H_{2}O{)}_5\left(\mathrm{NO}\right)]{\mathrm{SO}}_4.$

This can also be written as $[FeS{O}_4.NO]$. The equations for the same are given below.

$2{\mathrm{HNO}}_3+3H_2{\mathrm{SO}}_4+6{\mathrm{FeSO}}_4\to 3{\mathrm{Fe}}_2({\mathrm{SO}}_4{)}_3+2\mathrm{NO}+4H_{2}O[\mathrm{Fe}\left(H_{2}O{)}_6\right]{\mathrm{SO}}_4+\mathrm{NO}\to \left[\mathrm{Fe}\right(H_{2}O{)}_5\left(\mathrm{NO}\right)]{\mathrm{SO}}_4+H_{2}O$