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

1 Answer
Jul 28, 2016

Answer:

#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#