Which of the following act as an oxidising agent?

Options

A. `HNO_3`
B. `K_2Cr_2O_7`
C. `KMnO_4`
D. All of these

Explanation

To determine which of the given compounds act as oxidizing agents, we need to analyze their oxidation states. Here’s the step-by-step solution:

### Step 1: Identify the oxidation states of the elements in each compound.

1. **Nitric Acid (HNO3)**:
- The oxidation state of Nitrogen (N) in HNO3 is +5.
- This is calculated as follows:
- H is +1, O is -2 (there are three O atoms, contributing -6), so:
- Let the oxidation state of N be x.
- The equation is: \( x + 1 + (-6) = 0 \)
- Solving gives \( x = +5 \).

2. **Potassium Dichromate (K2Cr2O7)**:
- The oxidation state of Chromium (Cr) in K2Cr2O7 is +6.
- The calculation is:
- K is +1 (two K atoms contribute +2), O is -2 (seven O atoms contribute -14), so:
- Let the oxidation state of Cr be y.
- The equation is: \( 2 + 2y + (-14) = 0 \)
- Solving gives \( 2y = 12 \) → \( y = +6 \).

3. **Potassium Permanganate (KMnO4)**:
- The oxidation state of Manganese (Mn) in KMnO4 is +7.
- The calculation is:
- K is +1, O is -2 (four O atoms contribute -8), so:
- Let the oxidation state of Mn be z.
- The equation is: \( 1 + z + (-8) = 0 \)
- Solving gives \( z = +7 \).

### Step 2: Determine if they can act as oxidizing agents.

- An oxidizing agent is a substance that can accept electrons and is reduced in the process. Typically, compounds with high oxidation states can act as oxidizing agents because they have a tendency to gain electrons and be reduced to a lower oxidation state.

- In this case:
- HNO3 (N at +5)
- K2Cr2O7 (Cr at +6)
- KMnO4 (Mn at +7)

All three compounds have elements in high oxidation states (+5, +6, and +7 respectively), indicating that they can act as oxidizing agents.

### Conclusion:
Since all three compounds exhibit maximum oxidation states, they can all act as oxidizing agents.

**Final Answer**: Option D: All of these.

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