Question

Bond order of $N_2$ molecule is

• A. $3$
• B. $2$
• C. $1$
• D. $0$

Answer 1

Answer: (A)

The molecular configuration of $N_2$ is $\sigma 1s^2,\stackrel{\ast }{\sigma }1s^2\sigma 2s^2\stackrel{\ast }{\sigma }2s^2\{\begin{array}{c}\pi 2p_y^2\\ \pi 2p_x^2\end{array}\stackrel{\ast }{\sigma }2{p_z}^2$ $\{\begin{array}{c}\stackrel{\ast }{\pi }2p_y\\ \stackrel{\ast }{\pi }2p_x\end{array}\stackrel{\ast }{\sigma }2p_z$ number of bonding electrons $Bondorder=\frac{-numberofantibondingelectrons}{2}$ $=\frac{10-4}{2}$ $\therefore$ $=3$