Question

Write a balanced equation for the reaction of${\mathbf{\text{Hg(H}}}_{\mathbf{\text{2}}}{\mathbf{\text{O)}}}_{\mathbf{\text{4}}}^{\mathbf{\text{2 +}}}$in aqueous $\mathbf{\text{KCN}}$.

Short answer

The balanced equation for the reaction of $\text{Hg}{\left({\text{H}}_{\text{2}}\text{O}\right)}_{\text{4}}^{\text{2 +}}$in aqueous KCN is –

$\mathrm{Hg}{\left({\mathrm{H}}_2\mathrm{O}\right)}_4^{2+}\left(\mathrm{aq}\right)+4{\mathrm{CN}}^{-}\left(\mathrm{aq}\right)\rightleftharpoons \mathrm{Hg}\left(\mathrm{CN}{)}_4^{2-}\right(\mathrm{aq})+4{\mathrm{H}}_2\mathrm{O}(\mathrm{l})$

Step-by-step solution

Concept Introduction

A chemical equation is a representation of a chemical process in the form of components. A balanced chemical equation is one in which the number of atoms in all of the molecules is equal on both sides of the equation.

Balanced Equation for  Hg(H2O)42 + 

First mention that the coordination number for $\text{Hg(II)}$is $4$. Thus, when $\text{Hg}{\left({\text{H}}_{\text{2}}\text{O}\right)}_{\text{4}}^{\text{2 +}}$reacts with $\mathrm{KCN}$, there can't be more than ${\text{4 CN}}^{\text{-}}$ ions coordinated to $\text{Hg(II)}$ .

The reaction is as follows –

$$\begin{gathered} \mathrm{Hg}{\left({\mathrm{H}}_2\mathrm{O}\right)}_4^{2+}\left(\mathrm{aq}\right)+4{\mathrm{CN}}^{-}\left(\mathrm{aq}\right)\rightleftharpoons \mathrm{Hg}\left(\mathrm{CN}{)}_4^{2-}\right(\mathrm{aq})+4{\mathrm{H}}_2\mathrm{O}(\mathrm{l}) \\ {\mathrm{K}}_{\mathrm{f}}=\frac{\left[\mathrm{Hg}(\mathrm{CN}{)}_4^{2-}\right]}{\left[\mathrm{Hg}{\left({\mathrm{H}}_2\mathrm{O}\right)}_4^{2+}\right]\times {\left[{\mathrm{CN}}^{-}\right]}^4} \end{gathered}$$

Liquid and solid compounds are not included in in K_f expression.

Therefore, the balanced equation is:

$\text{Hg}{\left({\text{H}}_{\text{2}}\text{O}\right)}_{\text{4}}^{\text{2 +}}{\text{(aq) + 4CN}}^{\text{-}}\text{(aq)}\iff {\text{Hg(CN)}}_{\text{4}}^{\text{2 -}}{\text{(aq) + 4H}}_{\text{2}}\text{O(l)}$.