Question

Aqueous copper sulphate solution (blue in colour) gives:
(a) a green precipitate with aqueous potassium fluoride, and
(b) a bright green solution with aqueous potassium chloride,
Then the molecular formula of the given copper sulphate compound is:
A.\[[{\text{Cu}}{({{\text{H}}_2}{\text{O}})_3}({\text{CN}})]{\text{S}}{{\text{O}}_4}.{{\text{H}}_2}{\text{O}}\]
B.\[[{\text{Cu}}{({{\text{H}}_2}{\text{O}})_3}{\text{S}}{{\text{O}}_4}].{{\text{H}}_2}{\text{O}}\]
C.\[[{\text{Cu}}{({{\text{H}}_2}{\text{O}})_4}]{\text{S}}{{\text{O}}_4}.{{\text{H}}_2}{\text{O}}\]
D.None of these

Answer 1

Hint: For this question we basically need to calculate the number of atoms that water will have in the formula for copper sulphate. We need to check the number of fluoride and chloride ions formed in the respective reaction. The equal number of water atoms will be in the formula. The charge on the complex formed will be balance by sulphate ions

Complete step by step solution:
Copper sulphate in aqueous medium firms a complex with water. In this question we have to calculate the number of water molecules that will be present in the complex. Let us write the complete reactions for the given statement to get an idea of the number of water molecules. Let us assume that x molecules of water are present in aqueous copper sulphate :
Reaction of aqueous copper sulphate with aqueous potassium fluoride:
\[{[{\text{Cu}}{({{\text{H}}_2}{\text{O}})_x}]^y} + {\text{KF}} \to {[{\text{Cu}}{{\text{F}}_4}]^{2 - }}\]
Reaction of aqueous copper sulphate with aqueous potassium chloride:
\[{[{\text{Cu}}{({{\text{H}}_2}{\text{O}})_x}]^y} + {\text{KCl}} \to {[{\text{CuC}}{{\text{l}}_4}]^{2 - }}\]
Since four atoms of fluorine and chlorine are there in the respective reactions, this indicates that there must be 4 molecules of water that are replaced by 4 ions of fluoride and chloride.
The oxidation state of copper is 2 in both the above formed complexes.
In \[{[{\text{Cu}}{{\text{F}}_4}]^{2 - }}\] oxidation state of copper is \[ + 4 \times - 1 = - 2\]
Hence, the oxidation state of copper is \[ = - 2 + 4 = 2\]. We can do similarly in the other complex as charge on chloride is the same as fluoride.
Now since the oxidation state of copper is \[ + 2\] and water is a neutral ligand having no charge, the overall charge on the complex becomes\[ + 2\]. This charge will be balanced by the negative charge of the sulphate ion which has \[ - 2\] charge.
 Hence, the formula will be \[[{\text{Cu}}{({{\text{H}}_2}{\text{O}})_4}]{\text{S}}{{\text{O}}_4}.{{\text{H}}_2}{\text{O}}\]

Note: The extra water molecule that is attached with the complex in each of the given options is merely for the balancing of the chemical equation from which these complexes are formed. Otherwise water is neutral and will detach from the water on its own.