Chemical Bonding And Molecular Structure Question 365
Question: Which of the following is paramagnetic
Options:
A) $ B_2 $
B) $ C_2 $
C) $ N_2 $
D) $ F_2 $
Show Answer
Answer:
Correct Answer: A
Solution:
Paramagnetic property arise through unpaired electron. $ B_2 $ molecule have the unpaired electron so it show paramagnetism.
$ B_2\to $ $ \underset{\text{(2 unpaired electron)}}{\mathop{\sigma 1s^{2}{{\sigma }^{*}}1s^{2},\sigma 2s^{2}{{\sigma }^{*}}2s^{2},\pi 2p_x^{1}=\pi 2p_y^{1}}}, $
$ C_2\to $ $ \underset{\text{(No unpaired electron)}}{\mathop{\sigma 1s^{2}{{\sigma }^{*}}1s^{2},\sigma 2s^{2}{{\sigma }^{*}}2s^{2},\pi 2p_x^{2}.\pi 2p_y^{2}}}, $
$ N_2\to $ $ \underset{\text{(No unpaired electron)}}{\mathop{\sigma 1s^{2}{{\sigma }^{*}}1s^{2},\sigma 2s^{2}{{\sigma }^{*}}2s^{2},\sigma 2p_x^{2},\pi 2p_y^{2}\pi 2p_z^{2}}}, $
$ F_2\to $ $ \underset{\text{(No unpaired electron)}}{\mathop{\sigma s^{2},{{\sigma }^{*}}1s^{2},\sigma 2s^{2},{{\sigma }^{*}}2s^{2},\sigma 2p_x^{2},\pi 2p_y^{2},\pi 2p_z^{2},}}, $
$ {{\pi }^{*}}2p_y^{2},{{\pi }^{*}}2p_z^{2} $
So only $ B_2 $ exist unpaired electron and show the paramagnetism.
