Chemistry - CHEMICAL BONDING Question with Solution | TestHub

The species that contain unpaired electrons are paramagnetic.

Electronic configuration of ${\mathrm{Li}}_2$ molecule is ${\left(\sigma 1 \mathrm{s}\right)}^2{\left({\sigma }^{*}1 \mathrm{s}\right)}^2{\left(\sigma 2 \mathrm{s}\right)}^2$.

The electronic configuration of ${\mathrm{B}}_2$ molecule is

${\left(\mathrm{\sigma }1\mathrm{s}\right)}^2{\left({\mathrm{\sigma }}^{\star }1\mathrm{s}\right)}^2{\left(\mathrm{\sigma }2\mathrm{s}\right)}^2{\left({\mathrm{\sigma }}^{\star }2\mathrm{s}\right)}^2\left(\mathrm{\pi }2{{\mathrm{p}}_{\mathrm{x}}}^1=\mathrm{\pi }2{{\mathrm{p}}_{\mathrm{y}}}^1\right)$

The presence of unpaired electrons in the two $\pi$ bonding molecular orbitals explains its paramagnetic nature.

The electronic configuration of ${\mathrm{C}}_2$  $\mathrm{KK}\sigma 2s^2{\sigma }^{*}2s^2\pi 2p_x^2\pi 2p_y^2$

${\mathrm{O}}_2: \mathrm{KK}{\left({\mathrm{\sigma }}_{2 \mathrm{s}}\right)}^2{\left({\mathrm{\sigma }}_{2 \mathrm{s}}\right)}^2{\left({\mathrm{\sigma }}_{2{\mathrm{p}}_{\mathrm{z}}}\right)}^2{\left({\mathrm{\pi }}_{2{\mathrm{p}}_{\mathrm{x}}}\right)}^2={\left({\mathrm{\pi }}_{2{\mathrm{p}}_{\mathrm{y}}}\right)}^2{\left(\mathrm{\pi }{*}_{2{\mathrm{p}}_{\mathrm{x}}}\right)}^1={\left(\mathrm{\pi }{*}_{2{\mathrm{p}}_{\mathrm{y}}}\right)}^1$

Hence, ${\mathrm{O}}_2$ molecule is paramagnetic.

The electronic configuration of ${\mathrm{C}}_2^{-}$ is ${\left(\mathrm{\sigma }1\mathrm{s}\right)}^2{\left({\mathrm{\sigma }}^{\star }1\mathrm{s}\right)}^2{\left(\mathrm{\sigma }2\mathrm{s}\right)}^2{\left({\mathrm{\sigma }}^{\star }2\mathrm{s}\right)}^2\left(\mathrm{\pi }2{{\mathrm{p}}_{\mathrm{x}}}^2=\mathrm{\pi }2{{\mathrm{p}}_{\mathrm{y}}}^2\right)\mathrm{\sigma }2{{\mathrm{p}}_{\mathrm{z}}}^1$

Hence, it is paramagnetic substance.

${\mathrm{O}}_2^{2-}: \mathrm{KK}{\left({\mathrm{\sigma }}_{2 \mathrm{s}}\right)}^2{\left({\mathrm{\sigma }}_{2 \mathrm{s}}\right)}^2{\left({\mathrm{\sigma }}_{2{\mathrm{p}}_{\mathrm{z}}}\right)}^2{\left({\mathrm{\pi }}_{2{\mathrm{p}}_{\mathrm{x}}}\right)}^2={\left({\mathrm{\pi }}_{2{\mathrm{p}}_{\mathrm{y}}}\right)}^2{\left(\mathrm{\pi }{*}_{2{\mathrm{p}}_{\mathrm{x}}}\right)}^2={\left(\mathrm{\pi }{*}_{2{\mathrm{p}}_{\mathrm{y}}}\right)}^2$

${\mathrm{O}}_2^{+}: \mathrm{KK}{\left({\mathrm{\sigma }}_{2 \mathrm{s}}\right)}^2{\left({\mathrm{\sigma }}_{2 \mathrm{s}}\right)}^2{\left({\mathrm{\sigma }}_{2{\mathrm{p}}_{\mathrm{z}}}\right)}^2{\left({\mathrm{\pi }}_{2{\mathrm{p}}_{\mathrm{x}}}\right)}^2={\left({\mathrm{\pi }}_{2{\mathrm{p}}_{\mathrm{y}}}\right)}^2{\left(\mathrm{\pi }{*}_{2{\mathrm{p}}_{\mathrm{x}}}\right)}^1=\left(\mathrm{\pi }{*}_{2{\mathrm{p}}_{\mathrm{y}}}\right)$

Hence, the above species is paramagnetic.