
The number of ligands in the complex compound formed when nitrate ion is poured in the solution of ferrous sulfate and sulphuric acid for the conformation of nitrate ion is:
Answer
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Hint: The number of ligands in the complex compound formed when nitrate ion is poured in the solution of ferrous sulfate and sulphuric acid for the conformation of nitrate ion is:
Complete Step by step answer: As the question clearly states that the compound is formed during the confirmation of the nitrate ion (${\text{NO}}_3^ - $) in a solution, with ferrous sulfate (${\text{FeS}}{{\text{O}}_4}$) and concentrated sulphuric acid (${{\text{H}}_2}{\text{S}}{{\text{O}}_4}$).
Hence the compound is brown ring, which is also the name of the conformation test(brown ring test)
The reaction involved in the brown ring test are as follows;
${\text{2HN}}{{\text{O}}_3} + 3{{\text{H}}_2}{\text{S}}{{\text{O}}_4} + 6{\text{FeS}}{{\text{O}}_4} \to 3{\text{F}}{{\text{e}}_2}{\left( {{\text{S}}{{\text{O}}_4}} \right)_3}{\text{ + 2NO + 4}}{{\text{H}}_2}{\text{O}}$
And ${\text{[Fe}}{\left( {{{\text{H}}_2}{\text{O}}} \right)_6}]{\text{S}}{{\text{O}}_4} + {\text{NO = [Fe}}{\left( {{{\text{H}}_2}{\text{O}}} \right)_5}{\text{NO}}]{\text{S}}{{\text{O}}_4} + {{\text{H}}_2}{\text{O}}$
The formula of the brown ring complex is ${\text{[Fe}}{\left( {{{\text{H}}_2}{\text{O}}} \right)_5}{\text{NO}}]{\text{S}}{{\text{O}}_4}$ and its IUPAC name is PentaaquuonitrosoIron(I)sulfate
The question asks the number of ligands this brown complex has, it is evident from the chemical formula of the brown ring that, it has 7 ligands[ 5(${{\text{H}}_2}{\text{O}}$), 1 (NO), and 1( ${\text{S}}{{\text{O}}_4}$)]
Hence, the answer is 7 ligands.
Note: The number of ligands present in the compound need not be equal to the coordination number, a ligand present in the inner shell{ligands present inside parenthesis[] in a chemical formula} of the metal ion can only be considered as a contributor to the coordination number. For Eg: in the brown ring number of ligands in 7, but the coordination of Fe is 6.
Complete Step by step answer: As the question clearly states that the compound is formed during the confirmation of the nitrate ion (${\text{NO}}_3^ - $) in a solution, with ferrous sulfate (${\text{FeS}}{{\text{O}}_4}$) and concentrated sulphuric acid (${{\text{H}}_2}{\text{S}}{{\text{O}}_4}$).
Hence the compound is brown ring, which is also the name of the conformation test(brown ring test)
The reaction involved in the brown ring test are as follows;
${\text{2HN}}{{\text{O}}_3} + 3{{\text{H}}_2}{\text{S}}{{\text{O}}_4} + 6{\text{FeS}}{{\text{O}}_4} \to 3{\text{F}}{{\text{e}}_2}{\left( {{\text{S}}{{\text{O}}_4}} \right)_3}{\text{ + 2NO + 4}}{{\text{H}}_2}{\text{O}}$
And ${\text{[Fe}}{\left( {{{\text{H}}_2}{\text{O}}} \right)_6}]{\text{S}}{{\text{O}}_4} + {\text{NO = [Fe}}{\left( {{{\text{H}}_2}{\text{O}}} \right)_5}{\text{NO}}]{\text{S}}{{\text{O}}_4} + {{\text{H}}_2}{\text{O}}$
The formula of the brown ring complex is ${\text{[Fe}}{\left( {{{\text{H}}_2}{\text{O}}} \right)_5}{\text{NO}}]{\text{S}}{{\text{O}}_4}$ and its IUPAC name is PentaaquuonitrosoIron(I)sulfate
The question asks the number of ligands this brown complex has, it is evident from the chemical formula of the brown ring that, it has 7 ligands[ 5(${{\text{H}}_2}{\text{O}}$), 1 (NO), and 1( ${\text{S}}{{\text{O}}_4}$)]
Hence, the answer is 7 ligands.
Note: The number of ligands present in the compound need not be equal to the coordination number, a ligand present in the inner shell{ligands present inside parenthesis[] in a chemical formula} of the metal ion can only be considered as a contributor to the coordination number. For Eg: in the brown ring number of ligands in 7, but the coordination of Fe is 6.
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