Question

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

Short answer

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

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

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  ZnH2O42 + 

First mention that the coordination number for $\text{Zn(II)}$is $4$. Thus, when $\text{Zn}{\left({\text{H}}_{\text{2}}\text{O}\right)}_{\text{4}}^{\text{2 +}}$reacts with $\text{NaCN}$, there can't be more than role="math" localid="1663626610069" ${\text{4 CN}}^{\text{-}}$ ions around $\text{Zn(II)}$.

The reaction is as follows –

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

Liquid and solid compounds are not included in the equilibrium expression.

Therefore, the balanced equation is :

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