Question

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

Answer 1

`2`

Explanation:

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

`color(blue)(|bar(ul(color(white)(a/a)"charge number" = "net charge"/"elementary charge"color(white)(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, `"PbSO"_4`, an ionic compound, contains

• one lead(II) cation, `1 xx "Pb"^(2+)`
• one sulfate anion, `1 xx "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

`"net charge Pb"^(2+) = 2 xx e = 2e`

This means that the charge number is equal to

`"charge number Pb"^(2+) = (2color(red)(cancel(color(black)(e))))/color(red)(cancel(color(black)(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

`"net charge SO"_4^(2-) = 2 xx (-e) = -2e`

The charge number of the sulfate anion will thus be

`"charge number SO"_4^(2-) = (-2color(red)(cancel(color(black)(e))))/color(red)(cancel(color(black)(e))) = -2`