Which of the following is paramagnetic?

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Question

(A) Paramagnetic property arises due to the presence of unpaired electrons.$B_2$: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^1

Vedclass · Accepted Answer

$B_2$

Vedclass · Answer

(A) Paramagnetic property arises due to the presence of unpaired electrons.$B_2$: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^1 = \pi 2p_y^1$ (contains $2$ unpaired electrons).$C_2$: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2, \pi 2p_y^2$ (no unpaired electrons).$N_2$: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_y^2, \pi 2p_z^2, \sigma 2p_x^2$ (no unpaired electrons).$F_2$: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_x^2, \pi 2p_y^2, \pi 2p_z^2, \pi^* 2p_y^2, \pi^* 2p_z^2$ (no unpaired electrons).Therefore,only $B_2$ has unpaired electrons and exhibits paramagnetism.

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