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Assertion Bond order for N(2)^(o+) and N...

Assertion Bond order for `N_(2)^(o+)` and `N_(2)^(Theta)` are same (i.e2.5)
Reasoning `N_(2)^(o+)` is more stable than `N_(2)^(Theta)` .

A

If both `(A)` and `(R )` are correct and `(R )` is the correct
explanation of `(A)`

B

If both `(A)` and `(R )` are correct and `(R )` is not the correct
explanation of `(A)`

C

If `(A)` is correct but `(R )` is incorrect

D

If `(A)` is incorrect but `(R )` is correct

Text Solution

AI Generated Solution

The correct Answer is:
To solve the question regarding the assertion and reasoning about the bond order of \( N_2^+ \) and \( N_2^- \), we will follow these steps: ### Step 1: Determine the Electron Configuration - For \( N_2^+ \) (which has 13 electrons): - The electron configuration is: \[ \sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \sigma_{2p_z}^1 \] - For \( N_2^- \) (which has 15 electrons): - The electron configuration is: \[ \sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \sigma_{2p_z}^2 \pi_{2p_x}^*^1 \] ### Step 2: Calculate the Bond Order - Bond order is calculated using the formula: \[ \text{Bond Order} = \frac{\text{Number of bonding electrons} - \text{Number of antibonding electrons}}{2} \] - For \( N_2^+ \): - Bonding electrons = 9 (from \( \sigma_{1s}, \sigma_{2s}, \pi_{2p_x}, \pi_{2p_y}, \sigma_{2p_z} \)) - Antibonding electrons = 2 (from \( \sigma_{1s}^* \) and \( \sigma_{2s}^* \)) - Bond Order = \( \frac{9 - 2}{2} = \frac{7}{2} = 3.5 \) - For \( N_2^- \): - Bonding electrons = 10 (from \( \sigma_{1s}, \sigma_{2s}, \pi_{2p_x}, \pi_{2p_y}, \sigma_{2p_z} \)) - Antibonding electrons = 2 (from \( \sigma_{1s}^* \) and \( \sigma_{2s}^* \), and \( \pi_{2p_x}^* \)) - Bond Order = \( \frac{10 - 2}{2} = \frac{8}{2} = 4 \) ### Step 3: Analyze the Assertion and Reasoning - The assertion states that the bond order for \( N_2^+ \) and \( N_2^- \) is the same (2.5). This is incorrect based on our calculations. - The reasoning states that \( N_2^+ \) is more stable than \( N_2^- \). This is true since \( N_2^+ \) has a higher bond order, indicating greater stability. ### Conclusion - Therefore, the assertion is false, while the reasoning is true. The correct answer is that the assertion is incorrect, but the reasoning is correct.

To solve the question regarding the assertion and reasoning about the bond order of \( N_2^+ \) and \( N_2^- \), we will follow these steps: ### Step 1: Determine the Electron Configuration - For \( N_2^+ \) (which has 13 electrons): - The electron configuration is: \[ \sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \sigma_{2p_z}^1 \] ...
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  17. Assertion : N(2) and NO^(+) both are diamagnetic substances. Reason ...

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