Practicing Success
Aqueous \(CuSO_4\) solution gives green precipitation of \((X)\) with aqueous \(KF\). Correct formula of \(X\) is: |
\([CuF_6]^{3-}\) \([CuF_4]^{2-}\) \([CuF_2]\) \([CuF_5]^+\) |
\([CuF_4]^{2-}\) |
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 (\(CuSO_4\)): 2. Potassium fluoride (\(KF\)): 3. Formation of \([CuF_4]^{2-}\) Precipitate: 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 \(K_2SO_4\): \[ 4K^+ \, (aq) + SO_4^{2-} \, (aq) \rightarrow K_2SO_4 \, (aq) \] 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. |