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	$[CuCl{]}^{2-}$	$[Cu(H_{2}O{)}_4{]}^{2+}$
b	$[CuC{l}_4{]}^{2-}$	$[Cu{F}_4{]}^{2-}$
c	$[Cu{F}_4{]}^{2-}$	$[CuC{l}_4{]}^{2-}$
d	$[Cu(H_{2}O{)}_4{]}^{2-}$	$[Cu{F}_4{]}^{2-}$

• A. a
• B. b
• C. c
• D. d

Answer 1

Answer: (C)

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

$[Cu(H_{2}O{)}_4{]}^{2+}+\underset{[\text{From}K{F}_{(aq)}]}{4F^{-}}\to \underset{\stackrel{X}{\text{(Green ppt.)}}}{[Cu{F}_4{]}^{2-}}+4H_{2}O$

When $KCl$ is added, $C{l}^{-}$ ligands replace $H_{2}O$ forming

$[CuC{l}_4{]}^{2-}$ ions which are bright green in colour.

$[Cu(H_{2}O{)}_4{]}^{2+}+\underset{\text{[From}KCl{(}_{(aq)}]}{4C{l}^{-}}\to \underset{\text{(Bright green solution)}}{\underset{Y}{[CuC{l}_4{]}^{2-}}+4H_{2}O}$