Question

$Pb(CN{)}_2$ $+\text{excess}KCN$ $\to \text{product}$

Identify the product for the given reaction.

• A. $\left[Pb\right(CN{)}_4{]}^{2-}$
• B. $\left[Pb\right(CN{)}_6{]}^{2-}$
• C. $Pb$
• D. No reaction takes place

Answer 1

Answer: (D)

No reaction takes place

$Pb(CN{)}_2+\text{excess}KCN\to \text{no reaction}$

Due to the inert pair effect, $Pb\left(II\right)$ is more stable than $Pb\left(IV\right)$. Hence, $Pb\left(IV\right)$$readilyreducesto$Pb(II)$.Hence,duringthereaction$\displaystyle [Pb(CN)_6]^{2-}$ is not formed.

The formation of $\left[Pb\right(CN{)}_4{]}^{2-}$ and $\left[Pb\right(CN{)}_6{]}^{2-}$ is not possible as Pb is not a transition element and it does not form complex compounds as $5d$ orbital in it is completely filled.

Hence, $Pb(CN{)}_2$ and $KCN$ undergoes no reaction.