Aqueous CuSO_4 solution gives green precipitation of (X) with aqueous…

The correct answer is option 2. \([CuF_4]^{2-}\).

The formation of a green precipitate \([CuF_4]^{2-}\) when copper sulfate (\(CuSO_4\)) reacts with potassium fluoride (\(KF\)) can be explained by the following chemical reaction:

\(CuSO_4 \, (aq) + 4KF \, (aq) \rightarrow [CuF_4]^{2-} \, (s) + K_2SO_4 \, (aq)\)

Let us break down the reaction step by step:

1. Copper sulfate (\(CuSO_4\)):

In the aqueous solution, copper sulfate dissociates into copper ions (\(Cu^{2+}\)) and sulfate ions (\(SO_4^{2-}\)):

\(CuSO_4 \, (aq) \rightarrow Cu^{2+} \, (aq) + SO_4^{2-} \, (aq)\)

2. Potassium fluoride (\(KF\)):

Potassium fluoride dissociates into potassium ions (\(K^+\)) and fluoride ions (\(F^-\)):

\(4KF \, (aq) \rightarrow 4K^+ \, (aq) + 4F^- \, (aq)\)

3. Formation of \([CuF_4]^{2-}\) Precipitate:

When copper ions (\(Cu^{2+}\)) from copper sulfate react with fluoride ions (\(F^-\)) from potassium fluoride, a complex ion \([CuF_4]^{2-}\) is formed, leading to the green precipitate:

\(Cu^{2+} \, (aq) + 4F^- \, (aq) \rightarrow [CuF_4]^{2-} \, (s) \)

The green precipitate formed is \([CuF_4]^{2-}\), and it is insoluble in water, resulting in a solid appearing in the reaction mixture.

4. Formation of \(K_2SO_4\):

The remaining ions, potassium ions (\(K^+\)) and sulfate ions (\(SO_4^{2-}\)), combine to form potassium sulfate (\(K_2SO_4\)):

\(4K^+ \, (aq) + SO_4^{2-} \, (aq) \rightarrow K_2SO_4 \, (aq) \)

In summary, the reaction between copper sulfate and potassium fluoride results in the formation of a green precipitate \([CuF_4]^{2-}\). The correct formula for \(X\) is 2. \([CuF_4]^{2-}\). The complex ion \([CuF_4]^{2-}\) is responsible for the observed green color in the precipitate.