Chemistry - CHEMICAL BONDING Question with Solution | TestHub

In${\mathrm{O}}_2$, molecule, last two electrons are present in anti bonding
orbitals.
${\mathrm{O}}_2\left(8+8=16\right)=\mathrm{\sigma }1 {\mathrm{s}}^2,{\mathrm{\sigma }}^{*}1 {\mathrm{s}}^2,\mathrm{\sigma }2 {\mathrm{s}}^2,{\mathrm{\sigma }}^{*}2 {\mathrm{s}}^2, \mathrm{\sigma }2{{\mathrm{p}}_{\mathrm{z}}}^2,\mathrm{\pi }2{{\mathrm{p}}_{\mathrm{x}}}^2=\mathrm{\pi }2{{\mathrm{p}}_{\mathrm{y}}}^2,{\mathrm{\pi }}^{*}2{{\mathrm{p}}_{\mathrm{x}}}^1={\mathrm{\pi }}^{*}2{{\mathrm{p}}_{\mathrm{y}}}^1$
Removal of one electron from the${\mathrm{O}}_2$, molecule gives${\mathrm{O}}_2^{+}$in which the number of anti bonding electrons is one less and hence,$\mathrm{BO}$increases. Thus, removal of one electron from${\mathrm{O}}_2$stabilises the molecule.