A : `N_(2)` is more stable than `O_(2)` .
R : Bond order of `N_(2)` is more than that of `O_(2)` .

Options

A. If both Assertion & Reason are true and the reason is the correct explanation of the assertion , then mark (1)
B. If both Assertion & Reason are true but the reason is not the correct explanation of the assertion , then mark (2)
C. If Assertion is true statement but Reason is false , then mark (3)
D. If both Assertion and Reason are false statements , then mark (4)

Answer

To analyze the given statements about the stability of \(N_2\) and \(O_2\), we will evaluate both the assertion (A) and the reason (R) step by step.

### Step 1: Understanding the Assertion
The assertion states that \(N_2\) is more stable than \(O_2\). Stability in molecules can often be assessed by examining the bond strength, which is related to the bond order.

**Hint:** Consider how bond order relates to bond strength and stability.

### Step 2: Understanding Bond Order
Bond order is defined as the number of bonding pairs of electrons between two atoms. Generally, the higher the bond order, the stronger the bond, and thus the more stable the molecule.

**Hint:** Recall the formula for bond order:
\[ \text{Bond Order} = \frac{1}{2} \times (\text{Number of bonding electrons} - \text{Number of antibonding electrons}) \]

### Step 3: Calculate Bond Order for \(N_2\)
1. \(N_2\) has a total of 14 electrons.
2. The electron configuration for \(N_2\) is:
- \(\sigma_{1s}^2\)
- \(\sigma^*_{1s}^2\)
- \(\sigma_{2s}^2\)
- \(\sigma^*_{2s}^2\)
- \(\pi_{2p_x}^2\)
- \(\pi_{2p_y}^2\)
- \(\sigma_{2p_z}^2\)

3. Count the bonding and antibonding electrons:
- Bonding electrons: 10 (from \(\sigma\) and \(\pi\) orbitals)
- Antibonding electrons: 4 (from \(\sigma^*\) orbitals)

4. Calculate bond order:
\[ \text{Bond Order of } N_2 = \frac{1}{2} \times (10 - 4) = \frac{1}{2} \times 6 = 3 \]

**Hint:** Remember that a triple bond corresponds to a bond order of 3.

### Step 4: Calculate Bond Order for \(O_2\)
1. \(O_2\) has a total of 16 electrons.
2. The electron configuration for \(O_2\) is:
- \(\sigma_{1s}^2\)
- \(\sigma^*_{1s}^2\)
- \(\sigma_{2s}^2\)
- \(\sigma^*_{2s}^2\)
- \(\sigma_{2p_z}^2\)
- \(\pi_{2p_x}^2\)
- \(\pi_{2p_y}^2\)
- \(\pi^*_{2p_x}^1\)
- \(\pi^*_{2p_y}^1\)

3. Count the bonding and antibonding electrons:
- Bonding electrons: 10
- Antibonding electrons: 6

4. Calculate bond order:
\[ \text{Bond Order of } O_2 = \frac{1}{2} \times (10 - 6) = \frac{1}{2} \times 4 = 2 \]

**Hint:** A double bond corresponds to a bond order of 2.

### Step 5: Conclusion
From the calculations:
- The bond order of \(N_2\) is 3, indicating a triple bond, which is stronger and thus more stable.
- The bond order of \(O_2\) is 2, indicating a double bond, which is weaker than a triple bond.

Thus, the assertion that \(N_2\) is more stable than \(O_2\) is correct, and the reason that the bond order of \(N_2\) is greater than that of \(O_2\) is also correct. Therefore, both the assertion and reason are true, and the reason correctly explains the assertion.

**Final Answer:** Both assertion and reason are true, and the reason is the correct explanation for the assertion.