Among the following,the paramagnetic compound | vedclass

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(D) Paramagnetism is due to the presence of unpaired electrons.$1$. $Na_2O_2$ contains the peroxide ion $O_2^{2-}$. Its molecular orbital configuratio

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$KO_2$

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(D) Paramagnetism is due to the presence of unpaired electrons.$1$. $Na_2O_2$ contains the peroxide ion $O_2^{2-}$. Its molecular orbital configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^2 = \pi^* 2p_y^2$. It has $0$ unpaired electrons,so it is diamagnetic.$2$. $O_3$ (Ozone) is diamagnetic as all electrons are paired.$3$. $N_2O$ is diamagnetic as all electrons are paired.$4$. $KO_2$ contains the superoxide ion $O_2^-$. Its 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 $1$ unpaired electron,making it paramagnetic.Therefore,$KO_2$ is the paramagnetic compound. The correct option is $(D)$.

Among the following,the paramagnetic compound is

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