What is bond order of $\mathrm{F}_2$ molecule?

Question

Aicharya · Accepted Answer

The electronic configuration of $\mathrm{F}_2$ $\begin{aligned} & (\sigma 1 \mathrm{~s})^2\left(\sigma^* 1 \mathrm{~s}ight)^2(\sigma 2 \mathrm{~s})^2\left(\sigma^* 2 \mathrm{~s}ight)^2\left(\sigma 2 \mathrm{p}_{\mathrm{z}}ight)^2\left(\pi 2 \mathrm{p}_{\mathrm{x}}ight)^2 \ & \left(\pi 2 \mathrm{p}_{\mathrm{y}}ight)^2\left(\pi^* 2 \mathrm{p}_{\mathrm{x}}ight)^2\left(\pi^* 2 \mathrm{p}_{\mathrm{y}}ight)^2 \ & 	ext { Bond order of } \mathrm{F}_2 	ext { molecule }=\frac{\mathrm{N}_{\mathrm{b}}-\mathrm{N}_{\mathrm{a}}}{2}=\frac{10-8}{2}=1 \end{aligned}$

What is bond order of $\mathrm{F}_2$ molecule?

Solution