What is the coordination entity formed when excess of aqueous $KCN$ is added to an aqueous solution of copper sulphate? Why is it that no precipitate of copper sulphide is obtained when $H_{2}S_{(g)}$ is passed through this solution?

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) The reaction between copper sulphate and excess potassium cyanide is: $CuSO_{4(aq)} + 4KCN_{(aq)} \to K_{2}[Cu(CN)_{4}]_{(aq)} + K_{2}SO_{4(aq)}$.
In ionic form: $[Cu(H_{2}O)_{4}]^{2+} + 4CN^{-} \to [Cu(CN)_{4}]^{2-} + 4H_{2}O$.
The coordination entity formed is the tetracyanocuprate$(II)$ ion,$[Cu(CN)_{4}]^{2-}$,which exists as the salt $K_{2}[Cu(CN)_{4}]$.
$K_{2}[Cu(CN)_{4}]$ is a highly stable complex that does not dissociate to provide free $Cu^{2+}$ ions in the solution.
Since the concentration of free $Cu^{2+}$ ions is extremely low,the ionic product does not exceed the solubility product of $CuS$,and therefore no precipitate of copper sulphide is obtained when $H_{2}S_{(g)}$ is passed through the solution.

Explore More

Similar Questions

Which of the following complex ions is the most stable?

The molar ionic conductances of the octahedral complexes:
$(I)$ $PtCl_4 \cdot 5NH_3$
$(II)$ $PtCl_4 \cdot 4NH_3$
$(III)$ $PtCl_4 \cdot 3NH_3$
$(IV)$ $PtCl_4 \cdot 2NH_3$
Follow the order $-$

The aqueous solution of which of the following complexes has the least conductivity under identical conditions?

$K_4[Fe(CN)_6]$ reacts with ozone to give:

Which of the following will not give a precipitate with $AgNO_3$?

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo