Question

Copper sulphate when dissolved in excess of KCN gives
A. $Cu{{(CN)}_{2}}$
B. ${{[Cu{{(CN)}_{4}}]}^{3-}}$
C. ${{[Cu{{(CN)}_{4}}]}^{2-}}$
D. CuCN

Answer 1

Hint: The molecular formula of copper sulphate is $CuS{{O}_{4}}$ . When it is dissolved in potassium cyanide solution it will react with KCN and forms an unstable compound called cupric cyanide with a molecular formula $Cu{{(CN)}_{2}}$ .

Complete answer:
- In the question it is given that what is the product is going to form when copper sulphate is dissolved in excess amount of KCN.
- At the initial stage copper sulphate reacts with few moles of potassium cyanide (KCN) and forms cupric cyanide as the product.
- The chemical reaction of copper sulphate with KCN is as follows.
\[CuS{{O}_{4}}+2KCN\to {{K}_{2}}S{{O}_{4}}+Cu{{(CN)}_{2}}\]
- In the above chemical reaction one mole of copper sulphate reacts with two moles of potassium cyanide and forms one mole of cupric cyanide and one mole of potassium sulphate as the products.
- The unstable cupric cyanide undergoes decomposition reaction and forms cuprous cyanide as the product and the chemical reaction is as follows.
\[Cu{{(CN)}_{2}}\to CuCN+{{(CN)}_{2}}\]
- If we are going to add an excess amount of KCN to the cuprous cyanide, it will undergo a complex reaction and the chemical reaction is as follows.
\[CuCN+3KCN\to {{K}_{3}}[Cu{{(CN)}_{4}}]\] .
- Therefore the formed product when Copper sulphate when dissolved in excess of KCN is ${{[Cu{{(CN)}_{4}}]}^{3-}}$

- So, the correct option is B.

Note: Generally Cupric cyanide is an unstable compound and decomposes to cuprous cyanide by liberating cyanide. The formed cuprous cyanide reacts with an excess amount of potassium cyanide and forms potassium tetracyanocuprate (III) as the product.