When burnt in excess of oxygen, sodium forms | 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 $O_2^{2-}$, wh

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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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