Chemical Bonding Result Question 10
10. Assuming $2 s-2 p$ mixing is not operative, the paramagnetic species among the following is
(2014 Adv.)
(a) $\mathrm{Be} _2$
(b) $\mathrm{B} _2$
(c) $\mathrm{C} _2$
(d) $\mathrm{N} _2$
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Answer:
Correct Answer: 10. ( c )
Solution:
- PLAN: This problem can be solved by using the concept involved in molecular orbital theory. Write the molecular orbital electronic configuration keeping in mind that there is no $2 s-2 p$ mixing, then if highest occupied molecular orbital contain unpaired electron then molecule is paramagnetic otherwise diamagnetic.
Assuming that no $2 s - 2 p$ mixing takes place the molecular orbital electronic configuration can be written in the following sequence of energy levels of molecular orbitals
$\sigma 1 s, \stackrel{*}{\sigma} 1 s, \sigma 2 s, \stackrel{*}{\sigma} 2 s, \sigma 2 p_z, \pi 2 p_x \equiv \pi 2 p_y, \stackrel{*}{\pi} 2 p_x \equiv \stackrel{*}{\pi} 2 p_y, \stackrel{*}{\sigma} 2 p_z$
(a) $\mathrm{Be}_2 \rightarrow \sigma 1 s^2, \stackrel{*}{\sigma} 1 s^2, \sigma 2 s^2, \stackrel{*}{\sigma} 2 s^2$ (diamagnetic)
(b) $\mathrm{B}_2 \rightarrow \sigma 1 s^2, \stackrel{*}{\sigma} 1 s^2, \sigma 2 s^2, \stackrel{*}{\sigma} 2 s^2, \sigma 2 p_z^2, \stackrel{\pi 2 p_x^0}{\pi 2 p_y^0}$ (diamagnetic)
(c) $\mathrm{C}_2 \rightarrow \sigma 1 s^2, \stackrel{*}{\sigma} 1 s^2, \sigma 2 s^2, \stackrel{*}{\sigma} 2 s^2, \sigma 2 p_z^2, \stackrel{\pi 2 p_x^1}{\pi 2 p_y^1}$,
$ \stackrel{*}{\pi} 2 p_x^0, \stackrel{*}{\sigma} 2 p_z^0 \text { (paramagnetic) }$
$ \stackrel{*}{\pi} 2 p_y^0 $
(d) $ \mathrm{N}_2 \rightarrow \sigma 1 s^2, \stackrel{*}{\sigma} 1 s^2, \sigma 2 s^2, \stackrel{*}{\sigma} 2 s^2, \sigma 2 p_z^2, \stackrel{\pi 2 p_x^2}{\pi 2 p_y^2}, $
$ \stackrel{*}{\pi} 2 p_x^0, \sigma * 2 p_z^0 \text { (diamagnetic) } $
$ \stackrel{*}{\pi} 2 p_y^0 $
Hence, (c) is the correct choice.
BETA
