To determine the role of hydrogen peroxide ($H_2O_2$) in its reactions with potassium periodate ($KIO_4$) and hydroxylamine ($NH_2OH$), we need to analyze the change in oxidation states of the elements involved in each reaction.

Key concepts about Hydrogen Peroxide ($H_2O_2$):

• The oxidation state of oxygen in $H_2O_2$ is -1.
• When $H_2O_2$ acts as an oxidizing agent, its oxygen atoms are reduced from -1 to -2 (usually forming water, $H_2O$).
• When $H_2O_2$ acts as a reducing agent, its oxygen atoms are oxidized from -1 to 0 (forming oxygen gas, $O_2$).

Step 1: Reaction of $H_2O_2$ with $KIO_4$

1. First, let's determine the oxidation state of Iodine (I) in $KIO_4$. Let the oxidation state of I be $x$. The oxidation state of K is +1 and O is -2. $(+1) + x + 4(-2) = 0$ $1 + x - 8 = 0$ $x = +7$ Iodine is in its highest possible oxidation state (+7). Therefore, it cannot be further oxidized; it can only be reduced. This means $KIO_4$ must act as an oxidizing agent.

2. Since $KIO_4$ acts as an oxidizing agent, $H_2O_2$ must act as a reducing agent.

3. The reaction proceeds as follows: $KIO_4 + H_2O_2 \rightarrow KIO_3 + H_2O + O_2$

4. Let's verify the changes in oxidation states:

   • Oxygen in $H_2O_2$: The oxidation state changes from -1 to 0 in $O_2$. This is an increase in oxidation state, which is oxidation. A substance that gets oxidized is a reducing agent. So, $H_2O_2$ is a reducing agent.
   • Iodine in $KIO_4$: The oxidation state changes from +7 to +5 in $KIO_3$. This is a decrease in oxidation state, which is reduction. A substance that gets reduced is an oxidizing agent.

Conclusion for the first reaction: $H_2O_2$ acts as a reducing agent.

Step 2: Reaction of $H_2O_2$ with $NH_2OH$

1. First, let's determine the oxidation state of Nitrogen (N) in hydroxylamine ($NH_2OH$). Let the oxidation state of N be $y$. The oxidation state of H is +1 and O is -2. $y + 2(+1) + (-2) + (+1) = 0$ $y + 3 - 2 = 0$ $y = -1$ Nitrogen is in a low oxidation state (-1) and can be easily oxidized. Hydroxylamine is generally known as a reducing agent.

2. Since $NH_2OH$ is a reducing agent, it will be oxidized. Therefore, $H_2O_2$ must act as an oxidizing agent.

3. A possible reaction is: $2NH_2OH + H_2O_2 \rightarrow N_2 + 4H_2O$

4. Let's verify the changes in oxidation states:

   • Oxygen in $H_2O_2$: The oxidation state changes from -1 to -2 in $H_2O$. This is a decrease in oxidation state, which is reduction. A substance that gets reduced is an oxidizing agent. So, $H_2O_2$ is an oxidizing agent.
   • Nitrogen in $NH_2OH$: The oxidation state changes from -1 to 0 in $N_2$. This is an increase in oxidation state, which is oxidation. A substance that gets oxidized is a reducing agent.

Conclusion for the second reaction: $H_2O_2$ acts as an oxidizing agent.

Step 3: Final Conclusion

• In its reaction with $KIO_4$, $H_2O_2$ acts as a reducing agent.
• In its reaction with $NH_2OH$, $H_2O_2$ acts as an oxidizing agent.

Therefore, the correct description of the roles of hydrogen peroxide is: reducing agent, oxidising agent. This corresponds to option A.