Chemistry - CHEMICAL BONDING Question with Solution | TestHub

The electron distribution in${\text{O}}_2^{-}$is :

${\mathrm{\sigma }}_{1\mathrm{s}}^2{\mathrm{ }\mathrm{\sigma }}_{2\mathrm{s}}^{\mathrm{*}2}\mathrm{ }{\mathrm{\sigma }}_{2\mathrm{s}\mathrm{ }}^2{\mathrm{\sigma }}_{2\mathrm{s}\mathrm{ }}^{\mathrm{*}2}{\mathrm{\sigma }}_{2\mathrm{p}\mathrm{x}\mathrm{ }}^2{\mathrm{\sigma }}_{2\mathrm{p}\mathrm{y}\mathrm{ }}^2{\mathrm{\sigma }}_{2\mathrm{p}\mathrm{z}\mathrm{ }}^2{\mathrm{\sigma }}_{2\mathrm{p}\mathrm{y}\mathrm{ }}^{\mathrm{*}2}{\mathrm{\sigma }}_{2\mathrm{p}\mathrm{z}\mathrm{ }}^{\mathrm{*}1}$

$\therefore \mathrm{ }{\mathrm{O}}_2^{2-}$is diamagnetic and${\mathrm{O}}_2^{-}$is paramagnetic. In$\mathrm{A}\mathrm{l}{\mathrm{O}}_2^{-}$, oxide ions$\left({\mathrm{O}}^{2-}\right)$are present

and oxide ion is diamagnetic.

${\mathrm{N}\mathrm{O}}_2$has one unpair electron hence paramagnetic but$\mathrm{N}{\mathrm{O}}_2^{+}$has no unpair electron therefore diamagnetic