JEE Chemistry 2023 Question with Solution

Step-by-step solution

Standard Method

Given: hypophosphoric acid is $\text{H}_4\text{P}_2\text{O}_6$.

Find: the oxidation state of phosphorus.

Use the usual oxidation states: hydrogen is $+1$ and oxygen is $-2$.

Let the oxidation state of each phosphorus atom be $x$. Then for $\text{H}_4\text{P}_2\text{O}_6$,

$$4(+1) + 2x + 6(-2) = 0$$

So,

$$4 + 2x - 12 = 0$$ $$2x = 8$$ $$x = +4$$

Therefore, the oxidation state of phosphorus is $+4$.

Using the Structure

The solution shows the structure of hypophosphoric acid and states that the oxidation state of phosphorus in $\text{H}_4\text{P}_2\text{O}_6$ is $+4$.

This is consistent with the oxidation number calculation for the compound.

Common mistakes

• Assuming the oxidation state from the name alone is incorrect because the name does not directly give the oxidation number. Always use the molecular formula and oxidation state rules.

• Taking total oxidation state of two phosphorus atoms as the oxidation state of one phosphorus is wrong. After finding the combined value for both phosphorus atoms, divide by $2$ to get the oxidation state of each phosphorus atom.

• Using oxygen as $-1$ here is incorrect because hypophosphoric acid does not contain a peroxide linkage. Use oxygen as $-2$ and hydrogen as $+1$.