Question

# The acidic aqueous solution of ferrous ion forms a brown complex in the presence of $\text{NO}_{3}^{-}$, by the following two steps:${{\left( \text{Fe}{{({{\text{H}}_{2}}\text{O)}}_{6}} \right)}^{2+}}\,\text{+ NO}_{3}^{-}\ +\text{ }{{\text{H}}^{+}}\text{ }\to \text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ + }{{\left( \text{Fe}{{({{\text{H}}_{2}}\text{O)}}_{6}} \right)}^{3+}}\ \text{+ }{{\text{H}}_{2}}\text{O}$${{\left( \text{Fe}{{({{\text{H}}_{2}}\text{O)}}_{6}} \right)}^{2+}}\,\text{+ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\to \text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\ \text{+ }{{\text{H}}_{2}}\text{O}$Complete and balance the equations.

Hint: To check the presence of nitrate ion in the solution, a test is done commonly called ring test. In this test, iron (II) sulphate is added to the solution in which the presence of nitrate ion is to be checked. The formation of the brown coloured ring, confirms the presence of nitrate ion.

-The given reaction in the question is performed to check if the nitrate ion is present in the solution or not.
-In the ring test, the complex of ferrous (II) sulphate is added to the solution consisting nitrate ion (as given in the question).
-Along with the ferrous (II) sulphate, the concentrated sulfuric acid is also added slowly to the solution due to which they yield Hexa aqua iron (II) ion and nitric oxide.
-The balanced reaction will be:

$6{{\left( \text{Fe}{{({{\text{H}}_{2}}\text{O)}}_{6}} \right)}^{2+}}\,\text{+ 2NO}_{3}^{-}\ +\text{ 6}{{\text{H}}^{+}}\text{ }\to \text{ 2NO + 6}{{\left( \text{Fe}{{({{\text{H}}_{2}}\text{O)}}_{6}} \right)}^{3+}}\ \text{+ 4}{{\text{H}}_{2}}\text{O}$

-Here, nitric oxide is formed which is brown.
-Now, in the second reaction, nitric oxide will react with the iron complex. Also, we can see that one molecule of water is released so in the ligand one molecule will decrease.
-So, the product formed will have one ligand that is nitric oxide so the reaction will be:

${{\left( \text{Fe}{{({{\text{H}}_{2}}\text{O)}}_{6}} \right)}^{2+}}\,\text{+ NO }\to \text{ }{{\left( \text{Fe}{{({{\text{H}}_{2}}\text{O)}}_{5}}\text{NO} \right)}^{2+}}\text{ + }{{\text{H}}_{2}}\text{O}$

-To balance the equation, one should remember that the no. of atoms at the product as well as at the reactant side should be the same.
-Also, the net charge of the molecule should be the same at both the reactant and product side.

Note: In the reaction, the nitrate ion is reduced to the nitric oxide due to decrease in the oxidation state or loss of the electrons whereas iron (II) is oxidised to the iron (III) due to increase in the oxidation state or gain of the electron.