Question

At one time, it was common to write the formula for copper chloride as ${\mathbf{Cu}}_{\mathbf{2}}{\mathbf{Cl}}_{\mathbf{2}}$, instead of $\mathbf{CuCl}$, analogously to ${\mathbf{Hg}}_{\mathbf{2}}{\mathbf{Cl}}_{\mathbf{2}}$for mercury(I) chloride. Use electron configurations to explain why ${\mathbf{Hg}}_{\mathbf{2}}{\mathbf{Cl}}_{\mathbf{2}}\mathbf{}\mathbf{and}\mathbf{}\mathbf{CuCl}$are both correct.

Short answer

The electron configuration of ${\mathrm{Hg}}^{+}\mathrm{is}\left[\mathrm{Xe}\right]4{\mathrm{f}}^{14}5{\mathrm{d}}^{10}6{\mathrm{s}}^1$and therefore two Hg ions can react through unpaired 6s electrons and form ${\mathrm{Hg}}_2{\mathrm{Cl}}_2$

The electron configuration of ${\mathrm{Cu}}^{+}\mathrm{is}\left[\mathrm{Ar}\right]3{\mathrm{d}}^{10}$there are no unpaired electrons, so the usual formula is$\mathrm{CuCl}$ .

Step-by-step solution

Given Data

Compounds are

$$\begin{gathered} {\mathrm{Hg}}_2{\mathrm{Cl}}_2 \\ \mathrm{CuCl} \end{gathered}$$

Electronic Configuration

Electron configuration is the distribution of electrons in atomic or molecular orbitals of an atom, molecule, or other physical structure.

Explanation

The electron configuration of ${\mathrm{Hg}}^{+}\mathrm{is}\left[\mathrm{Xe}\right]4{\mathrm{f}}^{14}5{\mathrm{d}}^{10}6{\mathrm{s}}^1$. We can see that there is one unpaired electron in the 6s subshell, so the two ${\mathrm{Hg}}^{+}$ions can react with their two unpaired 6s electrons and form ${\mathrm{Hg}}_2{\mathrm{Cl}}_2$.

The electron configuration of ${\mathrm{Cu}}^{+}\mathrm{is}\left[\mathrm{Ar}\right]3{\mathrm{d}}^{10}$. We can see that there are no unpaired electrons, so this compound usually exists as CuCl, because two ${\mathrm{Cu}}^{+}$ions cannot bond to each other because they don't have unpaired electrons.