Question

Cold ferrous sulphate solution on absorption of NO develops brown colour due to the formation of:
(A) Paramagnetic $\left[ {Fe{{\left( {{H_2}O} \right)}_5}\left( {NO} \right)} \right]S{O_4}$
(B) Diamagnetic $\left[ {Fe{{\left( {{H_2}O} \right)}_5}\left( {{N_3}} \right)} \right]S{O_4}$
(C) Paramagnetic $\left[ {Fe{{\left( {{H_2}O} \right)}_5}\left( {N{O_3}} \right)} \right]{\left( {S{O_4}} \right)_2}$
(D) Diamagnetic $\left[ {Fe{{\left( {{H_2}O} \right)}_4}\left( {S{O_4}} \right)} \right]N{O_3}$

Answer 1

Hint: The brown ring test is done to check for the presence of nitrate anion. The nitrate anion is treated with $FeS{O_4}$ and ${H_2}S{O_4}$. Later there is absorption of NO leading to the formation of this nitrosyl complex which gives the brown coloured ring at the end.

Complete step by step answer:
-According to the question the reactant is cold ferrous sulphate (cold $FeS{O_4}$) and on absorption of NO there is formation of brown colour complex. This resembles the brown ring test.
-We will now discuss about the brown ring test.
A brown ring test is done to determine the presence of nitrate ion in any solution and is also known as the nitrate test. It is done by the addition of ferrous sulphate ($FeS{O_4}$) to the nitrate solution. Since all the nitrate salts are soluble in water, their presence cannot be detected through wet chemistry. The nitrate anion is an oxidizer and hence it is detected by the tests due to this property.
-We will now discuss the procedure of the brown ring test and the reactions involved.
We will take the nitrate solution and add ferrous sulphate ($FeS{O_4}$) to it. After this we will slowly add concentrated sulphuric acid (conc. ${H_2}S{O_4}$) in a manner that it forms a layer below the aqueous solution. As a result a brown coloured ring is formed at the junction of the two layers, hence indicating the presence of nitrate anion.
The overall reaction involved in the Brown Ring test is:
 $2HN{O_3} + 3{H_2}S{O_4} + 6FeS{O_4} \to 3Fe{(S{O_4})_3} + 2NO + 4{H_2}O$
 $\left[ {Fe{{({H_2}O)}_6}} \right]S{O_4} + NO \to \left[ {Fe{{({H_2}O)}_5}(NO)} \right]S{O_4} + {H_2}O$

In these reactions the nitrate anion ($NO_3^ - $) is reduced to nitric oxide (NO) by the ferrous sulphate ($FeS{O_4}$) which itself is oxidised to $F{e^{ + 3}}$. In the second reaction NO is absorbed by the ferrous sulphate solution leading to the formation of a nitrosyl complex $\left[ {Fe{{\left( {{H_2}O} \right)}_5}\left( {NO} \right)} \right]S{O_4}$ which gives the brown coloured ring and also the nitric oxide (NO) is reduced to $N{O^ - }$ ion. This nitrosyl complex is known as penta aqua nitrosyl iron(I) sulphate and it is paramagnetic by nature.
-Hence we can finally say that cold ferrous sulphate solution on absorption of NO develops brown colour due to the formation of $\left[ {Fe{{\left( {{H_2}O} \right)}_5}\left( {NO} \right)} \right]S{O_4}$ complex.
So, the correct answer is “Option A”.

Note: We should note that when ligands are attached to a transition metal to form a coordination complex and the electrons in the d orbital of the transition metal ion split into high energy and low energy orbitals. We should note that, when certain wavelengths are absorbed in this process, subtractive colour mixing occurs and the coordination complex solution becomes coloured.