Question

What is the action of concentrated nitric acid on acidified ferrous sulphate solution?

Answer 1

Hint: In the given question we are given Nitrate ions when treated with Acid (Sulfuric acid) and ferrous sulphate solution. The absorption of NO leads to a formation of a brown ring, which very much resembles the brown ring test.

Complete answer:
The brown ring test is used to detect the presence of nitrate ions. This is done by addition of Ferrous sulphate to nitric acid solution. Nitrate salts are soluble in water and act as an oxidizer.
Fe (II) compounds are found to be unstable and in the presence of an oxidizer can disproportionate into two ions. The reaction is given as:
$3F{e^{ + 2}} \to Fe + 2F{e^{ + 3}}$
The formation of $F{e^{ + 3}}$ cannot be stopped in the presence of Nitrate ions which is the oxidiser. But to stabilize this we usually add conc. sulfuric acid or a piece of iron to back the equilibrium backward towards the formation of $F{e^{ + 2}}$. The oxidation reaction can be given as:
$F{e^{ + 2}} \to F{e^{ + 3}} + {e^ - }$
This $F{e^{ + 3}}$ reacts with the sulphate ions to form $F{e_2}{(S{O_4})_3}$which is a brown colour solid formed.
Further the electrons are utilized as: $3{e^ - } + 4{H^ + } + 3NO_3^ - \to NO + 2{H_2}O$
The NO formed is absorbed by the ferrous sulphate solution to give a nitrosyl complex $[Fe{({H_2}O)_5}NO]S{O_4}$. This complex gives a brown coloured ring at the junction. The name of the complex is Penta aqua nitrosyl iron(I) sulphate and the nature is paramagnetic.
So by adding ferrous sulphate to nitrate solution, we’ll oxidize $F{e^{ + 2}} \to F{e^{ + 3}}$ and lead to the formation of brown colour compound Iron (III) Sulphate.

Note:
In the nitrosyl complex the iron is in $ + 1$ oxidation state and the complex formed is an outer sphere complex. One electron is transferred/paired due to the presence of the strong ligand NO.