Question

Which of the following has the highest bond order ?

• A. $N_2$
• B. $O_2$
• C. $H{e}_2$
• D. $H_2$

Answer 1

Answer: (A)

Molecular orbital configuration of

$N_2=14e^{-}=\sigma 1s^2,{}^{\ast }1s^2,\sigma 2s^2,{\sigma }^{\ast }2s^2,$

$\sigma 2p_x^2,\pi 2p_y{}^2,\pi 2p_z{}^2$

Bond order $=\frac{N_b-N_a}{2}=\frac{10-4}{2}=3$

Similarly for $O_2$ molecule, bond order $=2$

For $H{e}_2$ molecule, bond order $=0$

For $H_2$ molecule bond order $=1$ So, $N_2$ has highest bond order.