Question

The order of the oxidation state of the phosphorus atom in ${{\rm{H}}_{\rm{3}}}{\rm{P}}{{\rm{O}}_{\rm{2}}}$,${{\rm{H}}_{\rm{3}}}{\rm{P}}{{\rm{O}}_{\rm{4}}}$,${{\rm{H}}_{\rm{3}}}{\rm{P}}{{\rm{O}}_3}$and ${{\rm{H}}_4}{{\rm{P}}_2}{{\rm{O}}_6}$ is:
(A) ${{\rm{H}}_{\rm{3}}}{\rm{P}}{{\rm{O}}_{\rm{4}}} > {{\rm{H}}_{\rm{3}}}{\rm{P}}{{\rm{O}}_{\rm{2}}} > {{\rm{H}}_{\rm{3}}}{\rm{P}}{{\rm{O}}_3} > {{\rm{H}}_3}{{\rm{P}}_2}{{\rm{O}}_6}$
(B) ${{\rm{H}}_{\rm{3}}}{\rm{P}}{{\rm{O}}_{\rm{4}}}{\rm{ < }}{{\rm{H}}_{\rm{3}}}{\rm{P}}{{\rm{O}}_{\rm{2}}}{\rm{ < }}{{\rm{H}}_{\rm{3}}}{\rm{P}}{{\rm{O}}_3}{\rm{ < }}{{\rm{H}}_3}{{\rm{P}}_2}{{\rm{O}}_6}$
(C) ${{\rm{H}}_{\rm{3}}}{\rm{P}}{{\rm{O}}_{\rm{2}}} > {{\rm{H}}_{\rm{3}}}{\rm{P}}{{\rm{O}}_3} > {{\rm{H}}_4}{{\rm{P}}_2}{{\rm{O}}_6}{\rm{ > }}{{\rm{H}}_{\rm{3}}}{\rm{P}}{{\rm{O}}_{\rm{4}}}$
(D) ${{\rm{H}}_{\rm{3}}}{\rm{P}}{{\rm{O}}_{\rm{4}}}{\rm{ > }}{{\rm{H}}_4}{{\rm{P}}_2}{{\rm{O}}_6}{\rm{ > }}{{\rm{H}}_{\rm{3}}}{\rm{P}}{{\rm{O}}_3} > {{\rm{H}}_{\rm{3}}}{\rm{P}}{{\rm{O}}_2}$

Answer 1

As we know that, the redox reaction is a disproportionation reaction in which oxidation and reduction occur simultaneously. The oxidation represents the loss of electrons from molecules and reduction represents the gain of electrons.

Complete answer
The oxidation state of any atom is the state when that atom loses its electrons. In the above question, there are four compounds in which phosphorus loses its electrons. As we know that phosphorus is ${\rm{1}}{{\rm{5}}^{{\rm{th}}}}$ group element and it has five electrons in its outermost shell so it can lose a maximum of five electrons.
Let’s calculate the oxidation of phosphorus
1. In ${{\rm{H}}_{\rm{3}}}{\rm{P}}{{\rm{O}}_{\rm{2}}}$
Suppose x is the oxidation state of phosphorus-

{{\rm{H}}_{\rm{3}}}{\rm{P}}{{\rm{O}}_{\rm{2}}}$$ $$\Rightarrow ( + 1) \times 3 + x + 2 \times \left( { - 2} \right) = 0\\\ \Rightarrow x = + 1

Here the formal charge of hydrogen is $+1$ and formal charge of oxygen is $- 2$.
2. In ${{\rm{H}}_{\rm{3}}}{\rm{P}}{{\rm{O}}_{\rm{4}}}$
Suppose x is the oxidation state of phosphorus

{{\rm{H}}_{\rm{3}}}{\rm{P}}{{\rm{O}}_{\rm{4}}}$$ $$\Rightarrow ( + 1) \times 3 + x + 4 \times \left( { - 2} \right) = 0\\\ \Rightarrow x = + 5

3. In ${{\rm{H}}_{\rm{3}}}{\rm{P}}{{\rm{O}}_3}$
Suppose x is the oxidation state of phosphorus in

{{\rm{H}}_{\rm{3}}}{\rm{P}}{{\rm{O}}_3}$$ $$\Rightarrow ( + 1) \times 3 + x + 3 \times \left( { - 2} \right) = 0\\\ \Rightarrow x = + 3

4. In ${{\rm{H}}_4}{{\rm{P}}_2}{{\rm{O}}_6}$
Suppose x is the oxidation state of phosphorus in

{{\rm{H}}_4}{{\rm{P}}_2}{{\rm{O}}_6}$$ $$\Rightarrow ( + 1) \times 4 + 2x + 6 \times \left( { - 2} \right) = 0\\\ \Rightarrow x = + 4

So, the oxidation state of phosphorous is maximum in four compounds is $+ 5$
Therefore, order the oxidation state of phosphorus in all four compounds is,
${{\rm{H}}_{\rm{3}}}{\rm{P}}{{\rm{O}}_{\rm{4}}}{\rm{ > }}{{\rm{H}}_4}{{\rm{P}}_2}{{\rm{O}}_6}{\rm{ > }}{{\rm{H}}_{\rm{3}}}{\rm{P}}{{\rm{O}}_3} > {{\rm{H}}_{\rm{3}}}{\rm{P}}{{\rm{O}}_2}$

**Hence, the correct option is option (D).

Note: **
Any compound in which the central element exists with its highest oxidation state, will have the maximum oxidizing property as in ${{\rm{H}}_{\rm{3}}}{\rm{P}}{{\rm{O}}_{\rm{2}}}$, ${{\rm{H}}_{\rm{3}}}{\rm{P}}{{\rm{O}}_{\rm{4}}}$, ${{\rm{H}}_{\rm{3}}}{\rm{P}}{{\rm{O}}_3}$ and ${{\rm{H}}_4}{{\rm{P}}_2}{{\rm{O}}_6}$ the most oxidizing agent is ${{\rm{H}}_{\rm{3}}}{\rm{P}}{{\rm{O}}_{\rm{4}}}$. Oxidizing agent has the property to oxidize the other compound by reducing itself. On the other hand, reducing agents have the property to reduce other compounds by oxidizing itself.