Question

Aqueous copper sulphate solution (blue in colour) gives (X), a green precipitate with aqueous potassium fluoride and (Y), a bright green solution with aqueous potassium chloride. The compounds X and Y is

	X	Y
a	$[CuCl]^{2-}$	$[Cu(H_2O)_4]^{2+}$
b	$ [CuCl_4]^{2-}$	$[CuF_4]^{2-}$
c	$[CuF_4]^{2-}$	$ [CuCl_4]^{2-}$
d	$[Cu(H_2O)_4]^{2-}$	$[CuF_4]^{2-}$

• A. a
• B. b
• C. c
• D. d

Answer 1

Answer: (C)

Aqueous $CuSO_4$ solution contains $[Cu(H_2O)_4]^{2+}$ ions which are blue in colour. When $KF$ is added, $H_2O$ being weak field ligand can be replaced by $F^-$ forming $[CuF_4]^{2-}$ ions which are green in colour.

$[Cu(H_2O)_4]^{2+} + \underset{[\text{From} \, KF_{(aq)}]}{4F^-} \to \underset{\overset{X}{\text{(Green ppt.)}}}{[CuF_4]^{2-}} + 4H_2O$

When $KCl$ is added, $Cl^-$ ligands replace $H_2O$ forming

$[CuCl_4]^{2-}$ ions which are bright green in colour.

$[Cu(H_2O)_4]^{2+} + \underset{\text{[From} KCl(_{(aq)}]}{4Cl^-} \to \underset{\text{(Bright green solution)}}{\underset{Y}{[CuCl_4]^{2-}} + 4H_2O}$