Q7 / 160 · AP EAMCET 2021 (23 Aug Shift 2) · Chemistry
Which pair among the following does not have the same bond order?
- A$\mathrm{CN}^{-}$and $\mathrm{NO}^{-}$
- B$\mathrm{CN}^{-}$and $\mathrm{CO}$
- C$\mathrm{O}_2^{2-}$ and $\mathrm{B}_2$
- Correct answerD$\mathrm{O}_2^{+}$and $\mathrm{NO}^{+}$
Solution
We know that,
For $\mathrm{O}_2^{+}$
Molecular orbital configuration of $\mathrm{O}_2^{+}$is $\sigma \mathrm{l} s^2 \sigma^* 1 s^2 \sigma 2 s^2 \sigma^* 2 s^2 \sigma 2 p_z^2 \pi 2 p_x^2 \approx \pi 2 p_y^2 \pi^* 2 p_x^1$
Number of electrons in bonding orbital $=10$
Number of electrons in antibonding orbital $=5$
Bond order of $\mathrm{NO}_2^{+}=\frac{10-5}{2}=2.5$
For $\mathrm{NO}^{+}$
Molecular orbital configuration is $\sigma 1 s^2 \sigma^* 1 s^2 \sigma 2 s^2 \sigma^* 2 s^2 \pi 2 p_x^2 \pi 2 p_y^2 \sigma 2 p_z^2$
Number of electrons in bonding orbital $=10$
Number of electrons in antibonding orbital $=4$
Bond order $=\frac{10-4}{2}=3$
So, $\mathrm{O}_2^{+}$and $\mathrm{NO}^{+}$does not have same bond order.
For $\mathrm{O}_2^{+}$
Molecular orbital configuration of $\mathrm{O}_2^{+}$is $\sigma \mathrm{l} s^2 \sigma^* 1 s^2 \sigma 2 s^2 \sigma^* 2 s^2 \sigma 2 p_z^2 \pi 2 p_x^2 \approx \pi 2 p_y^2 \pi^* 2 p_x^1$
Number of electrons in bonding orbital $=10$
Number of electrons in antibonding orbital $=5$
Bond order of $\mathrm{NO}_2^{+}=\frac{10-5}{2}=2.5$
For $\mathrm{NO}^{+}$
Molecular orbital configuration is $\sigma 1 s^2 \sigma^* 1 s^2 \sigma 2 s^2 \sigma^* 2 s^2 \pi 2 p_x^2 \pi 2 p_y^2 \sigma 2 p_z^2$
Number of electrons in bonding orbital $=10$
Number of electrons in antibonding orbital $=4$
Bond order $=\frac{10-4}{2}=3$
So, $\mathrm{O}_2^{+}$and $\mathrm{NO}^{+}$does not have same bond order.Hence, the correct answer is option: 'D'.