The magnetic behaviour of Li_2O,Na_2O_2 and K | vedclass

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Question

(D) $Li_2O$ contains the oxide ion $O^{2-}$,which has the electronic configuration $[He] 2s^2 2p^6$. Since all electrons are paired,it is diamagnetic.

Vedclass · Accepted Answer

diamagnetic,diamagnetic and paramagnetic

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(D) $Li_2O$ contains the oxide ion $O^{2-}$,which has the electronic configuration $[He] 2s^2 2p^6$. Since all electrons are paired,it is diamagnetic.$Na_2O_2$ contains the peroxide ion $O_2^{2-}$. The molecular orbital configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^2 = \pi^* 2p_y^2$. All electrons are paired,so it is diamagnetic.$KO_2$ contains the superoxide ion $O_2^-$. The molecular orbital configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^2 = \pi^* 2p_y^1$. It has one unpaired electron in the $\pi^*$ orbital,making it paramagnetic.Therefore,the correct sequence is diamagnetic,diamagnetic,and paramagnetic.

The magnetic behaviour of $Li_2O$,$Na_2O_2$ and $KO_2$,respectively,are

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