Question

Oxidation number of Pb in $PbS{O_4}$ is?

Answer 1

Hint: Oxidation number is also known as oxidation state of an element. Oxidation number is defined as the number of electrons an atom has donated or gained to form bonds with other compounds. Oxidation state of any ion is defined as the amount of charge present on that ion and for the element present in a molecule, oxidation number is calculated using substitution method which we will apply to find the oxidation state of lead (Pb) in $PbS{O_4}$ .

Complete answer:
We can define the oxidation number of any atom as the number of electrons it loses or gains to make bonds with different atoms in a molecule.
An atom can show multiple oxidation states according to the atoms it is bonded with. The oxidation number can be negative, positive or zero (when the atom is neutral and there is no charge on it).
For the atom present in a molecule, oxidation number is calculated using a substitution method which we will apply to find the oxidation state of lead (Pb) in $PbS{O_4}$ .
Now, we know that the oxidation number of oxygen (O) is $ - 2$ and that of Sulphur (S) is $ + 6.$ The given molecule is neutral and hence, we will equate the charges to find the oxidation number of Lead (Pb).
Let the oxidation number of Pb be x.
$x + 6 + 4( - 2) = 0$
$x = + 2$

Note:
Oxidation number changes when an atom undergoes oxidation or reduction reaction. In an oxidation reaction an atom loses electrons and its oxidation number becomes more positive due to a positive charge on it. In a reduction reaction an atom gains electrons and its oxidation number becomes more negative or less positive due to a negative charge on it.