When burnt in excess of oxygen,sodium forms a | vedclass

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(B) When sodium is burnt in excess of oxygen,it forms sodium peroxide $(Na_2O_2)$,so $X = Na_2O_2$. In $Na_2O_2$,the peroxide ion is $O_2^{2-}$,which

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$X$ is diamagnetic and $Y$ is paramagnetic in nature

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(B) When sodium is burnt in excess of oxygen,it forms sodium peroxide $(Na_2O_2)$,so $X = Na_2O_2$. In $Na_2O_2$,the peroxide ion is $O_2^{2-}$,which has the electronic configuration $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^2 = \pi^* 2p_y^2$. Since all electrons are paired,$Na_2O_2$ is diamagnetic. When potassium is burnt in excess of oxygen,it forms potassium superoxide $(KO_2)$,so $Y = KO_2$. In $KO_2$,the superoxide ion is $O_2^-$,which has one unpaired electron in the $\pi^* 2p$ orbitals. Therefore,$KO_2$ is paramagnetic. Thus,$X$ is diamagnetic and $Y$ is paramagnetic.

When burnt in excess of oxygen,sodium forms a compound $X$ and potassium forms a compound $Y$. The magnetic natures of $X$ and $Y$ respectively are

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