Chemical Bonding Result Question 9
9. Which of the following species is not paramagnetic?
(2017 Main)
(a) $\mathrm{NO}$
(b) $\mathrm{CO}$
(c) $\mathrm{O} _2$
(d) $\mathrm{B} _2$
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Answer:
Correct Answer: 9. ( b )
Solution:
- To identify the magnetic nature we need to check the molecular orbital configuration. If all orbitals are fully occupied, species is diamagnetic while when one or more molecular orbitals is/are singly occupied, species is paramagnetic.
(a) $\mathrm{NO}(7+8=15)-\sigma 1 s^2, \sigma^* 1 s^2, \sigma 2 s^2, \sigma^* 2 s^2$,
$\pi 2 p_x^2=\pi 2 p_y^2, \pi 2 p_z^2, \pi^* 2 p_x^1=\pi^* 2 p_y^0$
One unpaired electron is present. Hence, it is paramagnetic.
(b) $\mathrm{CO}(6+8=14)-\sigma l s^2, \sigma^* 1 s^2, \sigma 2 s^2, \sigma^* 2 s^2$,
$\pi 2 p_x^2=\pi 2 p_y^2, \sigma 2 p_z^2$
No unpaired electron is present. Hence, it is diamagnetic.
(c) $\mathrm{O} _2(8+8=16)-\sigma 1 s^2, \sigma^* 1 s^2, \sigma 2 s^2, \sigma^* 2 s^2, \sigma 2 p_z^2$,
$\pi 2 p_x^2=\pi 2 p_x^2, \pi^* 2 p_x^1=\pi^* 2 p_y^1$
Two unpaired electrons are present.
Hence, it is paramagnetic.
(d) $\mathrm{B} _2(5+5)-\sigma l s^2, \sigma^* 1 s^2, \sigma 2 s^2, \sigma^* 2 s^2, \pi 2 p_x^1=\pi 2 p_y^1$
Two unpaired electrons are present.
Hence, it is paramagnetic.
BETA
