CuSO_4 reacts with KCN solution and forms | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) When $CuSO_4$ reacts with an excess of $KCN$ solution,it first forms a precipitate of $Cu(CN)_2$,which is unstable and decomposes to form $CuCN$ a

Vedclass · Accepted Answer

$K_3[Cu(CN)_4]$

Vedclass · Answer

(D) When $CuSO_4$ reacts with an excess of $KCN$ solution,it first forms a precipitate of $Cu(CN)_2$,which is unstable and decomposes to form $CuCN$ and $(CN)_2$. $2CuSO_4 + 4KCN 	o 2CuCN + (CN)_2 + 2K_2SO_4$ $CuCN$ then reacts with excess $KCN$ to form the stable complex $K_3[Cu(CN)_4]$. $CuCN + 3KCN 	o K_3[Cu(CN)_4]$ The overall reaction is: $2CuSO_4 + 10KCN 	o 2K_3[Cu(CN)_4] + (CN)_2 + 2K_2SO_4$ Thus,the final product is $K_3[Cu(CN)_4]$.

$CuSO_4$ reacts with $KCN$ solution and forms

Similar Questions

The molar ionic conductance of the octahedral complexes:
$(i)$ $PtCl_4 \cdot 5 NH_3$ $(ii)$ $PtCl_4 \cdot 4 NH_3$ $(iii)$ $PtCl_4 \cdot 3 NH_3$ $(iv)$ $PtCl_4 \cdot 2 NH_3$
Follow the order:

Assertion : Low spin complexes have lesser number of unpaired electrons.
Reason : $[FeF_6]^{3-}$ is a low spin complex.

Which of the following ions is the most stable?

Which of the following will $NOT$ give the characteristic test for $Fe^{2+}$ or $Fe^{3+}$ ions?

The strongest ligand among the following is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module