Assertion :- Bond order of `O_(2)` and BN is same .
Reason `O_(2) and BN` are isoelectronic

Options

A. If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion.
B. If both Assertion & Reason are True but Reason is not a correct explanation of the Assertion.
C. If Assertion is True, but the Reason is False.
D. If both Assertion & Reason are false

Answer

To solve the question, we need to analyze both the assertion and the reason provided.

### Step 1: Understanding the Assertion
The assertion states that the bond order of \( O_2 \) (oxygen) and \( BN \) (boron nitride) is the same.

**Hint:** Recall the formula for bond order, which is given by:
\[ \text{Bond Order} = \frac{1}{2} \left( N_b - N_a \right) \]
where \( N_b \) is the number of electrons in bonding molecular orbitals and \( N_a \) is the number of electrons in antibonding molecular orbitals.

### Step 2: Calculate the Bond Order for \( O_2 \)
Using molecular orbital theory, the electronic configuration of \( O_2 \) is:
\[ \sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \pi_{2p_x}^*^1 \pi_{2p_y}^*^1 \]

- Total electrons in bonding orbitals: \( 10 \) (from \( \sigma_{1s}, \sigma_{2s}, \sigma_{2p_z}, \pi_{2p_x}, \pi_{2p_y} \))
- Total electrons in antibonding orbitals: \( 6 \) (from \( \sigma_{1s}^*, \sigma_{2s}^*, \pi_{2p_x}^*, \pi_{2p_y}^* \))

Now, we can calculate the bond order:
\[ \text{Bond Order of } O_2 = \frac{1}{2} (10 - 6) = \frac{1}{2} \times 4 = 2 \]

**Hint:** Remember that the bond order indicates the strength and stability of the bond.

### Step 3: Calculate the Bond Order for \( BN \)
Next, we calculate the bond order for boron nitride (\( BN \)). The electronic configuration of \( BN \) is:
\[ \sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^1 \]

- Total electrons in bonding orbitals: \( 8 \) (from \( \sigma_{1s}, \sigma_{2s}, \sigma_{2p_z}, \pi_{2p_x} \))
- Total electrons in antibonding orbitals: \( 2 \) (from \( \sigma_{1s}^*, \sigma_{2s}^* \))

Now, we can calculate the bond order:
\[ \text{Bond Order of } BN = \frac{1}{2} (8 - 2) = \frac{1}{2} \times 6 = 3 \]

**Hint:** Compare the bond orders calculated for both molecules.

### Step 4: Conclusion on the Assertion
The bond order of \( O_2 \) is \( 2 \) and the bond order of \( BN \) is \( 3 \). Therefore, the assertion that the bond order of \( O_2 \) and \( BN \) is the same is **false**.

### Step 5: Understanding the Reason
The reason states that \( O_2 \) and \( BN \) are isoelectronic.

**Hint:** Isoelectronic species have the same number of total electrons.

### Step 6: Calculate the Total Electrons
- For \( O_2 \): Each oxygen atom has \( 8 \) electrons, so \( O_2 \) has \( 8 + 8 = 16 \) electrons.
- For \( BN \): Boron has \( 5 \) electrons and nitrogen has \( 7 \) electrons, so \( BN \) has \( 5 + 7 = 12 \) electrons.

Since \( O_2 \) has \( 16 \) electrons and \( BN \) has \( 12 \) electrons, they are **not** isoelectronic.

### Final Conclusion
- The assertion is **false**.
- The reason is also **false**.

Thus, the correct answer is that the assertion is false and the reason is false.