Q19.95P - Chemistry: Molecular Nature Of Matter And Change

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  $\mathbf{\text{KCN}}$.

Step-by-Step Solution

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 –

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

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 ( H 2 O ) 4 2 + 

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:

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