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.



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