# What are the charge numbers for the cation and the anion in lead(II) sulfate?

Jul 28, 2016

$2$

#### Explanation:

The charge number is simply the ratio between the net charge of an ion and the elementary charge, $e$.

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \text{charge number" = "net charge"/"elementary charge} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

As you know, the elementary charge is the charge carried by a single proton. Alternatively, the elementary charge can be thought of as the absolute value of the charge carried by a single electron, which is equal to $- e$.

Now, a formula unit of lead(II) sulfate, ${\text{PbSO}}_{4}$, an ionic compound, contains

• one lead(II) cation, $1 \times {\text{Pb}}^{2 +}$
• one sulfate anion, $1 \times {\text{SO}}_{4}^{2 -}$

You're interested in finding the charge number of the lead(II) cation. A cation that carries a $2 +$ charge has a deficit of $2$ electrons. This is equivalent to saying that it has an excess of $2$ protons.

The net charge of the lead(II) cation will thus be twice the charge carried by a single proton, i.e. the elementary charge

${\text{net charge Pb}}^{2 +} = 2 \times e = 2 e$

This means that the charge number is equal to

${\text{charge number Pb}}^{2 +} = \frac{2 \textcolor{red}{\cancel{\textcolor{b l a c k}{e}}}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{e}}}} = 2$

As you can see, the charge number is equivalent to the valence of the ion.

Similarly, the sulfate anion carries an overall $2 -$ charge, which means that it has an excess of $2$ electrons. The net charge of the sulfate anion will thus be twice the charge carried by a single electron

${\text{net charge SO}}_{4}^{2 -} = 2 \times \left(- e\right) = - 2 e$

The charge number of the sulfate anion will thus be

${\text{charge number SO}}_{4}^{2 -} = \frac{- 2 \textcolor{red}{\cancel{\textcolor{b l a c k}{e}}}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{e}}}} = - 2$