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The brown ring complex \[\left[ {{\rm{Fe}}{{\left( {{{\rm{H}}_{\rm{2}}}{\rm{O}}} \right)}_{\rm{5}}}{\rm{N}}{{\rm{O}}^{\rm{ + }}}} \right]{\rm{S}}{{\rm{O}}_{\rm{4}}}\] has oxidation number of \[{\rm{Fe}}\] as:
A ) +1
B ) +2
C ) +3
D ) +4

Answer Verified Verified
Hint: In a neutral molecule, the sum of the oxidation states of all the elements/ions is zero. The oxidation states of water, nitrosonium ion and sulphate ion are 0, +1, -2 respectively.

Complete step by step answer:
The brown ring test for nitrate ion is a qualitative test that confirms the presence of nitrate ion. In the ring test for \[{\rm{NO}}_3^ - \] ion, a freshly prepared ferrous sulphate solution is added to the solution containing nitrate ion. A ring containing a brown coloured complex is formed. The formation of this brown coloured complex confirms the ring test for the presence of nitrate ion. The brown coloured complex is \[\left[ {{\rm{Fe}}{{\left( {{{\rm{H}}_2}{\rm{O}}} \right)}_5}{\rm{NO}}} \right]{\rm{S}}{{\rm{O}}_4}\].
Write the balanced chemical equation for the brown ring test for nitrate ions.

${\rm{NO}}_3^ - {\rm{ + 3F}}{{\rm{e}}^{2 + }}{\rm{ + 4 }}{{\rm{H}}^ + }{\rm{ }} \to {\rm{ NO + 3F}}{{\rm{e}}^{3 + }}{\rm{ + 2 }}{{\rm{H}}_2}{\rm{O}}$
${\left[ {{\rm{Fe}}{{\left( {{{\rm{H}}_2}{\rm{O}}} \right)}_6}} \right]^{2 + }}{\rm{ + NO + SO}}_4^{2 - }{\rm{ }} \to {\rm{ }}\left[ {{\rm{Fe}}{{\left( {{{\rm{H}}_2}{\rm{O}}} \right)}_5}{\rm{NO}}} \right]{\rm{S}}{{\rm{O}}_4}{\rm{ + }}{{\rm{H}}_2}{\rm{O}}$

In the first step, ferrous ions react with nitrate ions in presence of protons to form ferric ions and nitric oxide. In the second step, nitric oxide reacts with hydrated ferrous ions to form a brown ring complex.

Let X be the oxidation number of \[{\rm{Fe}}\] in the brown ring complex \[\left[ {{\rm{Fe}}{{\left( {{{\rm{H}}_{\rm{2}}}{\rm{O}}} \right)}_{\rm{5}}}{\rm{N}}{{\rm{O}}^{\rm{ + }}}} \right]{\rm{S}}{{\rm{O}}_{\rm{4}}}\].

In a neutral molecule, the sum of the oxidation states of all the elements/ions is zero. The oxidation states of water, nitrosonium ion and sulphate ion are 0, +1 and -2 respectively. Calculate the oxidation number of \[{\rm{Fe}}\] in the brown ring complex \[\left[ {{\rm{Fe}}{{\left( {{{\rm{H}}_{\rm{2}}}{\rm{O}}} \right)}_{\rm{5}}}{\rm{N}}{{\rm{O}}^{\rm{ + }}}} \right]{\rm{S}}{{\rm{O}}_{\rm{4}}}\].

$\left[ {{\rm{Fe}}{{\left( {{{\rm{H}}_{\rm{2}}}{\rm{O}}} \right)}_{\rm{5}}}{\rm{N}}{{\rm{O}}^{\rm{ + }}}} \right]{\rm{S}}{{\rm{O}}_{\rm{4}}}$
$X + 5\left( 0 \right) + 1 - 2 = 0$
X - 1 = 0
X = + 1
Therefore, the oxidation number of Fe is +1 .

Hence, the option A ) is the correct answer.

Note: The brown ring test for nitrate ion is a qualitative test, that confirms the presence of nitrate ion by forming a brown coloured complex \[\left[ {{\rm{Fe}}{{\left( {{{\rm{H}}_2}{\rm{O}}} \right)}_5}{\rm{NO}}} \right]{\rm{S}}{{\rm{O}}_4}\].