Question

The correct order for reducing nature of $H_{3}P{O}_4,H_{3}P{O}_3$ and $H_{3}P{O}_2$ is :

• A. $H_{3}P{O}_2<H_{3}P{O}_3<H_{3}P{O}_4$
• B. $H_{3}P{O}_4<H_{3}P{O}_3<H_{3}P{O}_2$
• C. $H_{3}P{O}_3<H_{3}P{O}_4<H_{3}P{O}_2$
• D. $H_{3}P{O}_4<H_{3}P{O}_2<H_{3}P{O}_3$

Answer 1

Answer: (B)

$H_{3}P{O}_4<H_{3}P{O}_3<H_{3}P{O}_2$

$H_{3}P{O}_4<H_{3}P{O}_3<H_{3}P{O}_2$
+5 +3 +1

As oxidation state decreases, species have more tendency to oxidise then to reduce. So, they have higher reducing nature.