Question

The brown ring complex $\left[ {{\text{Fe}}{{\left( {{{\text{H}}_{\text{2}}}{\text{O}}} \right)}_{\text{5}}}{\text{N}}{{\text{O}}^ + }} \right]{\text{S}}{{\text{O}}_{\text{4}}}$ has oxidation number of ${\text{Fe}}$ as:

Answer 1

The number which is assigned to an element in the chemical formula is known as the oxidation number. The oxidation number represents the number of electrons gained or lost by the atom of that element in the chemical compound.

Complete step by step answer:
We are given a brown ring complex $\left[ {{\text{Fe}}{{\left( {{{\text{H}}_{\text{2}}}{\text{O}}} \right)}_{\text{5}}}{\text{N}}{{\text{O}}^ + }} \right]{\text{S}}{{\text{O}}_{\text{4}}}$. We have to calculate the oxidation number of the ${\text{Fe}}$ atom in the brown ring complex.
There are three ligands as follows:
1.${{\text{H}}_{\text{2}}}{\text{O}}$ is a neutral ligand. Thus, the charge on ${{\text{H}}_{\text{2}}}{\text{O}}$ ligand is zero.
2.${\text{N}}{{\text{O}}^ + }$ is a positive ligand. Thus, the charge on ${\text{N}}{{\text{O}}^ + }$ ligand is $+ {\text{1}}$.
3.${\text{S}}{{\text{O}}_{\text{4}}}$ is a negative ligand. Thus, the charge on ${\text{S}}{{\text{O}}_{\text{4}}}$ ligand is $- {\text{2}}$.
There is no charge on the complex. Thus, the complex is neutral.
Consider that the charge on the ${\text{Fe}}$ is x.
Thus,
$\left( {{\text{O}}{\text{.N}}{\text{. of Fe}}} \right) + \left( {{\text{5}} \times {\text{O}}{\text{.N}}{\text{. of }}{{\text{H}}_{\text{2}}}{\text{O}}} \right) + \left( {{\text{O}}{\text{.N}}{\text{. of N}}{{\text{O}}^ + }} \right) + \left( {{\text{O}}{\text{.N}}{\text{. of S}}{{\text{O}}_{\text{4}}}} \right) = {\text{Charge on the complex}}$
$\left( x \right) + \left( {{\text{5}} \times {\text{0}}} \right) + \left( { + 1} \right) + \left( { - {\text{2}}} \right) = {\text{0}}$
$x = + 2 - 1$
$x = + 1$
Thus, the brown ring complex $\left[ {{\text{Fe}}{{\left( {{{\text{H}}_{\text{2}}}{\text{O}}} \right)}_{\text{5}}}{\text{N}}{{\text{O}}^ + }} \right]{\text{S}}{{\text{O}}_{\text{4}}}$ has oxidation number of ${\text{Fe}}$ as $+ 1$.

Additional Information:
The absorption of nitric acid is done by the aqueous solution of ferrous sulphate $\left( {{\text{FeS}}{{\text{O}}_{\text{4}}}} \right)$. In the reaction, a brown coloured solution is obtained. The reaction is as follows:
$\left[ {{\text{Fe}}{{\left( {{{\text{H}}_{\text{2}}}{\text{O}}} \right)}_{\text{6}}}} \right]{\text{S}}{{\text{O}}_{\text{4}}}{\text{ + NO }} \to {\text{ }}\left[ {{\text{Fe}}{{\left( {{{\text{H}}_{\text{2}}}{\text{O}}} \right)}_{\text{5}}}{\text{N}}{{\text{O}}^ + }} \right]{\text{S}}{{\text{O}}_{\text{4}}}$
The brown coloured solution is due to the formation of the complex $\left[ {{\text{Fe}}{{\left( {{{\text{H}}_{\text{2}}}{\text{O}}} \right)}_{\text{5}}}{\text{N}}{{\text{O}}^ + }} \right]{\text{S}}{{\text{O}}_{\text{4}}}$. The complex formed is unstable and can readily dissociate on heating.

Note:
The brown ring test is done to determine the presence of nitrate ion in any chemical compound. Take aqueous solution of nitrate. Add iron sulphate to it. Add concentrated sulphuric acid slowly. At the junction of two layers, a brown coloured ring is observed. The brown ring indicates the presence of nitrate ions.