Which of the following represents oxidation?
a. `NO_(2)^(ө)rarrN_(2), b. VO_(2)^(ө)rarrVO_(3)^(ө)`
c. `ClO^(ө)rarrCl^(ө), d. CrO_(4)^(2-)rarrcr_(2)O_(7)^(2-)`

Explanation

To determine which of the given reactions represents oxidation, we need to analyze the changes in oxidation states for each reaction. Oxidation can be defined as an increase in oxidation number, loss of electrons, or gain of oxygen.

Let's analyze each option step by step:

### Step 1: Analyze Reaction a: \( NO_2^{−} \rightarrow N_2 \)

1. **Determine oxidation states:**
- In \( NO_2^{−} \):
- Oxygen (O) has an oxidation state of -2.
- Let the oxidation state of nitrogen (N) be \( x \).
- The equation for the oxidation state is: \( x + 2(-2) = -1 \) (since it is an anion).
- Solving gives \( x - 4 = -1 \) → \( x = +3 \).
- In \( N_2 \):
- The oxidation state of nitrogen is 0.

2. **Change in oxidation state:**
- From +3 (in \( NO_2^{−} \)) to 0 (in \( N_2 \)): **Decrease in oxidation state**.

### Conclusion for Reaction a:
- This reaction represents **reduction**, not oxidation.

---

### Step 2: Analyze Reaction b: \( VO_2^{−} \rightarrow VO_3^{−} \)

1. **Determine oxidation states:**
- In \( VO_2^{−} \):
- Let the oxidation state of vanadium (V) be \( x \).
- The equation is: \( x + 2(-2) = -1 \).
- Solving gives \( x - 4 = -1 \) → \( x = +3 \).
- In \( VO_3^{−} \):
- The equation is: \( x + 3(-2) = -1 \).
- Solving gives \( x - 6 = -1 \) → \( x = +5 \).

2. **Change in oxidation state:**
- From +3 (in \( VO_2^{−} \)) to +5 (in \( VO_3^{−} \)): **Increase in oxidation state**.

### Conclusion for Reaction b:
- This reaction represents **oxidation**.

---

### Step 3: Analyze Reaction c: \( ClO^{−} \rightarrow Cl^{−} \)

1. **Determine oxidation states:**
- In \( ClO^{−} \):
- Let the oxidation state of chlorine (Cl) be \( x \).
- The equation is: \( x + (-2) = -1 \).
- Solving gives \( x - 2 = -1 \) → \( x = +1 \).
- In \( Cl^{−} \):
- The oxidation state of chlorine is -1.

2. **Change in oxidation state:**
- From +1 (in \( ClO^{−} \)) to -1 (in \( Cl^{−} \)): **Decrease in oxidation state**.

### Conclusion for Reaction c:
- This reaction represents **reduction**, not oxidation.

---

### Step 4: Analyze Reaction d: \( CrO_4^{2−} \rightarrow Cr_2O_7^{2−} \)

1. **Determine oxidation states:**
- In \( CrO_4^{2−} \):
- Let the oxidation state of chromium (Cr) be \( x \).
- The equation is: \( x + 4(-2) = -2 \).
- Solving gives \( x - 8 = -2 \) → \( x = +6 \).
- In \( Cr_2O_7^{2−} \):
- The equation is: \( 2x + 7(-2) = -2 \).
- Solving gives \( 2x - 14 = -2 \) → \( 2x = 12 \) → \( x = +6 \).

2. **Change in oxidation state:**
- From +6 (in both \( CrO_4^{2−} \) and \( Cr_2O_7^{2−} \)): **No change in oxidation state**.

### Conclusion for Reaction d:
- This reaction represents **neither oxidation nor reduction**.

---

### Final Answer:
The only reaction that represents oxidation is **b. \( VO_2^{−} \rightarrow VO_3^{−} \)**.

---