Question

${\mathrm{CuSO}}_4$ dissolves in excess of $\mathrm{KCN}$ to give ----------?

Answer 1

${\mathbf{CuSO}}_{\mathbf{4}}$ when dissolves in excess of $\mathbf{KCN}$:

• Copper sulfate when dissolves in Potassium cyanide solution it reacts with potassium cyanide to form Potassium sulfate and Cupric cyanide, given as:
  $\underset{\left(\mathrm{Copper}\mathrm{sulfate}\right)}{\mathrm{CuS}{\mathrm{O}}_4}+\underset{\left(\mathrm{Potassium}\mathrm{cyanide}\right)}{2\mathrm{KCN}}\to \mathrm{}\underset{\left(\mathrm{Potassium}\mathrm{sulfate}\right)}{{\mathrm{K}}_2\mathrm{S}{\mathrm{O}}_4}\underset{\left(\mathrm{Cupric}\mathrm{cyanide}\right)}{+\mathrm{Cu}{\left(\mathrm{CN}\right)}_2}$
• But, the formed product, Cupric cyanide is unstable, hence decomposes to form Cuprous cyanide and Cyanogen gas.
  $\underset{\left(\mathrm{Cupric}\mathrm{cyanide}\right)}{2\mathrm{Cu}{\mathrm{(}\mathrm{CN}\mathrm{)}}_2}\mathrm{}\to \underset{\left(\mathrm{Cuprous}\mathrm{cyanide}\right)}{2\mathrm{CuCN}}+\underset{(\mathrm{Cyanogen}\mathrm{gas})}{{\left(\mathrm{CN}\right)}_2}$
• Here, the oxidation state of Copper decreases from $+2\mathrm{to}+1,$which is called reduction, hence Cupric cyanide is reduced to Cuprous cyanide.
• Hence 1 mole of Cuprous cyanide undergoes a reaction with 3 moles of potassium cyanide to form the Copper cyanide complex compound, given as:
  $\underset{(\mathrm{Cuprous}\mathrm{cyanide})}{\mathrm{CuCN}}+\underset{(\mathrm{potassium}\mathrm{cyanide})}{3\mathrm{KCN}}\mathrm{\to }\underset{(\mathrm{Copper}\mathrm{cyanide}\mathrm{complex})}{{\left[\mathrm{Cu}{\left(\mathrm{CN}\right)}_4\right]}^{-3}}$

Thus, ${\mathrm{CuSO}}_4$ dissolves in excess of $\mathrm{KCN}$ to give Copper cyanide complex $[\mathrm{Cu}{\left(\mathrm{CN}\right)}_4{]}^{3-}$