Number of paramagnetic oxides among the follo | vedclass

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(A) To determine the paramagnetism,we analyze the electronic configuration of the oxide ions present in each compound:$Li_{2}O$: Contains $O^{2-}$ ion

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$1$

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(A) To determine the paramagnetism,we analyze the electronic configuration of the oxide ions present in each compound:$Li_{2}O$: Contains $O^{2-}$ ion. Configuration: $1s^{2} 2s^{2} 2p^{6}$ (diamagnetic).$CaO$: Contains $O^{2-}$ ion. Configuration: $1s^{2} 2s^{2} 2p^{6}$ (diamagnetic).$Na_{2}O_{2}$: Contains peroxide ion $O_{2}^{2-}$. Configuration: $(\sigma 1s)^{2} (\sigma^{*} 1s)^{2} (\sigma 2s)^{2} (\sigma^{*} 2s)^{2} (\sigma 2p_{z})^{2} (\pi 2p_{x})^{2} (\pi 2p_{y})^{2} (\pi^{*} 2p_{x})^{2} (\pi^{*} 2p_{y})^{2}$ (diamagnetic).$KO_{2}$: Contains superoxide ion $O_{2}^{-}$. Configuration: $(\sigma 1s)^{2} (\sigma^{*} 1s)^{2} (\sigma 2s)^{2} (\sigma^{*} 2s)^{2} (\sigma 2p_{z})^{2} (\pi 2p_{x})^{2} (\pi 2p_{y})^{2} (\pi^{*} 2p_{x})^{2} (\pi^{*} 2p_{y})^{1}$. Due to the presence of one unpaired electron in the $\pi^{*}$ orbital,it is paramagnetic.$MgO$: Contains $O^{2-}$ ion (diamagnetic).$K_{2}O$: Contains $O^{2-}$ ion (diamagnetic).Only $KO_{2}$ is paramagnetic. Therefore,the total number of paramagnetic oxides is $1$.

Number of paramagnetic oxides among the following given oxides is ..... .
$Li_{2}O, CaO, Na_{2}O_{2}, KO_{2}, MgO$ and $K_{2}O$

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Number of paramagnetic oxides among the following given oxides is ..... .$Li_{2}O, CaO, Na_{2}O_{2}, KO_{2}, MgO$ and $K_{2}O$