Question

What is the formula of the compound formed during Brown ring test of nitrate radical ?

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

Answer 1

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

The formula of the compound formed during brown ring test of nitrate radical is pentaaquanitrosyliron (II) sulphate $\left(\right[Fe\left(H_{2}O{)}_{5}NO\right]S{O}_4)$ .
Brown ring test is be performed by adding iron (II) sulfate to the solution. When concentrated sulfuric acid is slowly added, a brown ring is formed on the test tube, which will indicate the presence of the nitrate ion. The test follows several phases. These have been written as balanced chemical equations:
$Zn(N{O}_3{)}_2+2H_{2}S{O}_4\to 2ZnS{O}_4+4HN{O}_3$
$4FeS{O}_4+2H_{2}S{O}_4\to 2F{e}_2(S{O}_4{)}_3+2H_2$ These two reactions happen simultaneously.
$HN{O}_3\to 2H_{2}O+3O_2+4NO$
The Nitric Acid decomposes in the intense heat produced by the high concentrate Sulphuric Acid used.
$2F{e}_2(S{O}_4{)}_3+2H_2+4NO\to 4[Fe\left(H_{2}O{)}_{5}NO\right]S{O}_4+2H_{2}S{O}_4$ The $\left[Fe\right(H_{2}O{)}_{5}NO]S{O}_4$ forms a brown ring in the middle of the solution produced by the reaction, making it easy to identify the presence of nitrates in the water.