If (Cu^{2+}) ions is treated with excess of (KCN), the compl... | Chemistry | SolveeIt

If $Cu^{2+}$ ions is treated with excess of $KCN$ , the complex formed is

$K_2[Cu (CN)_4]$

$K_3[Cu(CN)_4]$

$K_4[Cu (CN)_6]$

All of these

Answer

$K_3[Cu(CN)_4]$

Explanation

$Cu^{2+}$ is reduced to $Cu^{+}$ which forms complex $K_{3}[Cu(CN)_{4}]$ $2CuSO_{4}+10KCN \to 2K_{3}[Cu(CN)_{4}] + 2K_{2}SO_{4}+(CN)_{2}$

Chemistry Question on d -and f -Block Elements