Question

Calculate the oxidation state of $Pb$ in $P{b_3}{O_4}$

Answer 1

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:
We take the oxidation state of $Pb$ in $P{b_3}{O_4}$ ion as x. The oxidation number of oxygen is -2.
So, for $Pb$ ion, we can calculate the oxidation state as follows:

$$x\left( 3 \right) + 4\left( { - 2} \right) = 0 \\\ \Rightarrow 3x = 8 \\\ x = \dfrac{8}{3} \\\$$

Hence, the oxidation state of lead in $P{b_3}{O_4}$ is $\dfrac{8}{3}$.

Additional Information:
Oxidation number is the number of electrons that an atom loses or gains to form a chemical bond. If we know the type of compound, we will easily know the oxidation number of compounds.

1. If a compound exists in elemental form (only one type of atoms present), the oxidation number of the element is always zero.
   For ions, the charge indicates the oxidation number. For example, the oxidation number of chloride ions is –1. Also if the compound is neutral, the addition of oxidation numbers of all atoms or ions 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 ${\text{Pb}}$ in ${\text{P}}{{\text{b}}_{\text{3}}}{{\text{O}}_{\text{4}}}$. In peroxides, oxidation state of oxygen in -1, in ${{\text{F}}_{\text{2}}}{\text{O}}$, the oxidation state of oxygen is +2 and in all other compounds, the oxidation state of oxygen is -2. So, we should take -2 oxidation state for oxygen in ${\text{P}}{{\text{b}}_{\text{3}}}{{\text{O}}_{\text{4}}}$.