JEE Main · 2021 · 26 Feb 2021 · Shift 2 · Chemical Bonding and Molecular Structure

Bond order is half the excess of bonding over antibonding electrons. $\mathrm{N}{\mathrm{e}}_2$ has $20$ electrons that fill the bonding and antibonding orbitals equally, $10$ against $10$, so its bond order is $0$ and the molecule does not exist. ${\mathrm{N}}_2$ has $10$ bonding and $4$ antibonding electrons, giving $\frac{10-4}{2}=3$. ${\mathrm{F}}_2$ has $10$ bonding and $8$ antibonding electrons, giving $\frac{10-8}{2}=1$. ${\mathrm{O}}_2$ has $10$ bonding and $6$ antibonding electrons, giving $\frac{10-6}{2}=2$. So $\mathrm{N}{\mathrm{e}}_2$ goes with $0$, ${\mathrm{N}}_2$ with $3$, ${\mathrm{F}}_2$ with $1$ and ${\mathrm{O}}_2$ with $2$.