Aqueous \(CuSO_4\) solution gives green precipitation of \((X)\) with aqueous \(KF\). Correct formula of \(X\) is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → [CuF4]2−
Blue aqueous \( \ce{CuSO4} \) contains \( \ce{[Cu(H2O)4]^{2+}} \). With aqueous KF the water ligands are replaced by \( \ce{F^{-}} \), and a green precipitate of \( \ce{K2[CuF4]} \) forms:
\( \ce{[Cu(H2O)4]^{2+}} \) + \( \ce{4F^{-}} \) → \( \ce{[CuF4]^{2-}} \) + \( \ce{4H2O} \)
Copper stays +2: x + 4(−1) = −2 gives x = +2, with coordination number 4.
Why the other options are incorrect
Option (1) [CuF6]3− — Incorrect. x + 6(−1) = −3 gives Cu as +3; nothing in the reaction oxidises \( \ce{Cu^{2+}} \).
Option (3) [CuF2] — Incorrect. A neutral \( \ce{CuF2} \) unit is not the fluoro complex formed with excess \( \ce{F^{-}} \); \( \ce{Cu^{2+}} \) here takes four ligands.
Option (4) [CuF5]+ — Incorrect. x + 5(−1) = +1 gives Cu as +6, which is impossible for copper.