Chemistry - CHEMICAL BONDING Question with Solution | TestHub

Bond order = $\frac{1}{2}$(number of electrons in bonding MO $-$number of electrons in anti-bonding MO)

(a) ${\mathrm{Ne}}_2=$ Total ${\mathrm{e}}^{\mathrm{\Theta }}=20$

B.O. $=\frac{10-10}{2}=0$

(b) ${\mathrm{N}}_2=$ Total ${\mathrm{e}}^{⊝}=14$

B.O. $=\frac{10-4}{2}=3$

(c) ${\mathrm{F}}_2=$ Total ${\mathrm{e}}^{⊝}=18$

B.O. $=\frac{10-8}{2}=1$

(d) ${\mathrm{O}}_2=$ Total ${\mathrm{e}}^{⊝}=16$

B.O. $=\frac{10-6}{2}=2$