Chemistry - CHEMICAL BONDING Question with Solution | TestHub

The electronic configuration of fluorine atom is $1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^5$. The fluorine molecule has $18$ electrons. The $\mathrm{s}-\mathrm{p}$ mixing occurs when the $\mathrm{s}$ and $\mathrm{p}$ orbitals have similar energies. The energy difference between $2\mathrm{s} \text{and} 2\mathrm{p}$ orbitals in $\mathrm{O}, \mathrm{F}, \text{and} \mathrm{Ne}$ is greater than that in $\mathrm{Li}, \mathrm{Be}, \mathrm{B}, \mathrm{C}, \text{and} \mathrm{N}$. Hence, the molecular orbital electronic configuration of ${\mathrm{F}}_2$ is $\mathrm{\sigma }\left(1{\mathrm{s}}^2\right)\mathrm{\sigma }*\left(1{\mathrm{s}}^2\right)\mathrm{\sigma }\left(2{\mathrm{s}}^2\right)\mathrm{\sigma }*\left(2{\mathrm{s}}^2\right)\mathrm{\sigma }\left(2{{\mathrm{p}}_{\mathrm{z}}}^2\right)\mathrm{\pi }\left(2{{\mathrm{p}}_{\mathrm{x}}}^2\right)=\mathrm{\pi }\left(2{{\mathrm{p}}_{\mathrm{y}}}^2\right)\mathrm{\pi }*\left(2{{\mathrm{p}}_{\mathrm{x}}}^2\right)=\mathrm{\pi }*\left(2{{\mathrm{p}}_{\mathrm{y}}}^2\right)\mathrm{\sigma }*\left(2{\mathrm{p}}_{\mathrm{z}}\right)$