Question

In quantitative analysis of second group in lab. $H_2S$ gas is passed in acidic medium for ppt. When $Cu^{+2}$ and $Cd^{+2}$ react with $KCN$, than in which of the following condition, ppt will not be formed due to relative stability:

• A. $K_2[Cu(CN)_4] – More\ stable$
  $K_2[Cd(CN)_4] – Less \ stable$
• B. $K_2[Cu(CN)_4] – Less \ stable$
  $K_2[Cd(CN)_4] – More\ stable$
• C. $K_3[Cu(CN)_4] – More\ stable$
  $K_2[Cd(CN)_4] – Less \ stable$
• D. $K_3[Cu(CN)_4] – Less\ stable$
  $K_3[Cd(CN)_4] – More \ stable$

Answer 1

Answer: (C)

$K_3[Cu(CN)_4] – More\ stable$
$K_2[Cd(CN)_4] – Less \ stable$

Answer 2

$K_3[Cu(CN)_4]$ is more stable than $K_2[Cd(CN)_4]$, which means that the complex with $Cu^{2+}$ is more stable and less likely to precipitate compared to the complex with $Cd^{2+}$.

So, the correct option is (C): $K_3[Cu(CN)_4] – More\ stable$, $K_2[Cd(CN)_4] – Less \ stable$