Chemical Bonding Result Question 3
3. Among the following species, the diamagnetic molecule is
(2019 Main, 9 April II)
(a) $\mathrm{CO}$
(b) $\mathrm{B} _2$
(c) $\mathrm{NO}$
(d) $\mathrm{O} _2$
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Answer:
Correct Answer: 3. ( a )
Solution:
- Key Idea: Magnetic nature can be detected by molecular orbital theory. Presence of unpaired electrons means paramagnetic and absence of unpaired electrons means diamagnetic in nature.
Among the given options, $\mathrm{CO}$ is a diamagnetic molecule. It can be proved by molecular orbital $(MO)$ theory. The electronic configuration of given diatomic molecules are given below.
- $\mathrm{CO}$ (Number of electrons $=14$ )
Electronic configuration $=\sigma 1 s^2, \sigma^* 1 s^2, \sigma 2 s^2, \sigma^* 2 s^2$, $\sigma 2 p_z^2, \pi 2 p_x^2 \approx \pi 2 p_y^2$
Since, there is no unpaired electron in the $\mathrm{CO}$ molecule, so it is diamagnetic.
- $\mathrm{NO}$ (Number of electrons $=15$ )
Electronic configuration $=\sigma 1 s^2, \sigma^{\circ} 1 s^2, \sigma 2 s^2$, $\sigma^* 2 s^2, \sigma 2 p_z^2, \pi 2 p_x^2 \approx \pi 2 p_y^2, \pi^* 2 p_x^1 \approx \pi^* 2 p_y^0$
Since, $NO$ has one unpaired electron in $\pi^* 2 p_x^1$ orbital, so it is paramagnetic.
- $\mathrm{B}_2$ (Number of electrons $=10$ )
Electronic configuration $=\sigma l s^2, \sigma^* 1 s^2, \sigma 2 s^2, \sigma^* 2 s^2, \pi 2 p_x^1 \approx \pi p_y^1$
Since, two unpaired electrons are present in $\pi 2 p_x^1$ and $\pi 2 p_y^1$ orbital. So, it is paramagnetic.
- $\mathrm{O}_2$ (Number of electrons $=16$ )
Electronic configuration $=\sigma 1 s^2, \sigma^* 1 s^2, \sigma 2 s^2, \sigma^* 2 s^2$, $\sigma 2 p z^2, \pi 2 p_x^2 \approx \pi 2 p_y^2, \pi^* 2 p_x^1 \approx \pi^* 2 p_y^1$
Since, two unpaired electrons are present in $\pi^* 2 p_x^1$ and $\pi^* 2 p_y^1$ orbital. So, it is also paramagnetic.
BETA
