Question

Find the oxidation number of S in ${\rm{PbS}}{{\rm{O}}_{\rm{4}}}$.

Answer 1

Hint: We know that oxidation number is the number of electrons that an atom loses or gains to result in a chemical bond. Oxidation number is also termed as oxidation state.

Complete step by step answer:
To calculate the oxidation number of S in ${\rm{PbS}}{{\rm{O}}_{\rm{4}}}$, we first write the dissociation of ${\rm{PbS}}{{\rm{O}}_{\rm{4}}}$.
Lead sulfate dissociates to form lead ion and sulphate ion. The reaction is as follows:

${\rm{PbS}}{{\rm{O}}_{\rm{4}}} \to {\rm{P}}{{\rm{b}}^{2 + }} + {\rm{S}}{{\rm{O}}_{\rm{4}}}^{ - 2}$

Now, we take the oxidation state of S in sulphate ion as x. The sulphate ion carries -2 charge and there are four oxygen atoms in the ion. Now we add the oxidation state of all atoms in the sulphate ion to calculate the oxidation state of sulphur. The oxidation number of oxygen is -2.
So, for sulphate ion, the summation of the oxidation state of all atoms is -2.

$\begin{array}{c}x + 4\left( { - 2} \right) = - 2\\x - 8 = - 2\\x = 6\end{array}$

Hence, the oxidation state of sulphur in ${\rm{PbS}}{{\rm{O}}_{\rm{4}}}$ is 6.

Additional information:
1) If a compound exists in elemental form (only one type of atoms present), the oxidation number of the element is always zero.
2) For ions, the charge indicates the oxidation number. For example, the oxidation number of chloride ions is -1.

Note:
Oxygen has three oxidation states, -1, -2 and +2. Students might confuse which oxidation state to be taken to calculate the oxidation state of S in ${\rm{PbS}}{{\rm{O}}_{\rm{4}}}$. In peroxides, oxidation state of oxygen in -1, in ${{\rm{F}}_{\rm{2}}}{\rm{O}}$, the oxidation state of oxygen is +2 and in all other compounds, the oxidation state of oxygen is -2. So, we should take the -2 oxidation state for oxygen in ${\rm{PbS}}{{\rm{O}}_{\rm{4}}}$.