1. Idea used

A species can act as both oxidising and reducing agent if the central atom is in an intermediate oxidation state. Then:

• it can be oxidised further  act as a reducing agent,
• it can be reduced further  act as an oxidising agent.

If the element is already in its highest oxidation state, it cannot be oxidised further, so it cannot act as a reducing agent. Such a species may only act as an oxidising agent or may be redox-inactive in that sense.

2. Find oxidation states in each compound

(A) $H_3PO_4$

Let oxidation state of P be $x$.

$3(+1) + x + 4(-2) = 0$ $3 + x - 8 = 0$ $x = +5$

Phosphorus is in the maximum common oxidation state $+5$. So it cannot be oxidised further. Hence $H_3PO_4$ cannot act as a reducing agent, and therefore cannot act as both oxidising and reducing agent.

(B) $H_2SO_3$

Let oxidation state of S be $x$.

$2(+1) + x + 3(-2) = 0$ $2 + x - 6 = 0$ $x = +4$

Sulfur in $+4$ is an intermediate oxidation state.

• It can be oxidised to $+6$ (as in $H_2SO_4$)  so $H_2SO_3$ can act as a reducing agent.
• It can be reduced to lower oxidation states such as $0$ or $-2$  so it can act as an oxidising agent.

Thus $H_2SO_3$ can act as both.

(C) $H_2O_2$

Let oxidation state of O be $x$.

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

Oxygen is in the intermediate oxidation state $-1$.

• It can be reduced to $-2$ (in $H_2O$)  oxidising agent.
• It can be oxidised to $0$ (in $O_2$)  reducing agent.

Therefore $H_2O_2$ acts as both oxidising and reducing agent.

(D) $HNO_2$

Let oxidation state of N be $x$.

$(+1) + x + 2(-2) = 0$ $1 + x - 4 = 0$ $x = +3$

Nitrogen in $+3$ is an intermediate oxidation state.

• It can be oxidised to $+5$  reducing agent.
• It can be reduced to lower oxidation states such as $+2$, $0$, or $-3$  oxidising agent.

So $HNO_2$ can act as both.

3. Option-wise conclusion

• A: $H_3PO_4$  cannot act as both
• B: $H_2SO_3$  can act as both
• C: $H_2O_2$  can act as both
• D: $HNO_2$  can act as both

4. Final answer

The compound that cannot act both as oxidising and reducing agent is:

$\boxed{H_3PO_4}$

So, Option A is correct.