Calculate the bond order of $N_2, O_2, O_2^+$ and $O_2^-$.
For $N_2$: Total electrons $= 14$. Configuration: $(\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\sigma 2p_z)^2$. Bond Order $(BO) = \frac{1}{2}(10 - 4) = 3$.
For $O_2$: Total electrons $= 16$. Configuration: $(\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\sigma 2p_z)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\pi^* 2p_x)^1 (\pi^* 2p_y)^1$. $BO = \frac{1}{2}(10 - 6) = 2$.
For $O_2^+$: Total electrons $= 15$. Configuration: $(\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\sigma 2p_z)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\pi^* 2p_x)^1$. $BO = \frac{1}{2}(10 - 5) = 2.5$.
For $O_2^-$: Total electrons $= 17$. Configuration: $(\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\sigma 2p_z)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\pi^* 2p_x)^2 (\pi^* 2p_y)^1$. $BO = \frac{1}{2}(10 - 7) = 1.5$.
For $O_2$: Total electrons $= 16$. Configuration: $(\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\sigma 2p_z)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\pi^* 2p_x)^1 (\pi^* 2p_y)^1$. $BO = \frac{1}{2}(10 - 6) = 2$.
For $O_2^+$: Total electrons $= 15$. Configuration: $(\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\sigma 2p_z)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\pi^* 2p_x)^1$. $BO = \frac{1}{2}(10 - 5) = 2.5$.
For $O_2^-$: Total electrons $= 17$. Configuration: $(\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\sigma 2p_z)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\pi^* 2p_x)^2 (\pi^* 2p_y)^1$. $BO = \frac{1}{2}(10 - 7) = 1.5$.
Explore More
Similar Questions
Match List-$I$ with List-$II$ based on the bond order of the molecules.
List-$I$ List-$II$ $(a)$ $Ne_2$ $(i)$ $1$ $(b)$ $N_2$ $(ii)$ $2$ $(c)$ $F_2$ $(iii)$ $0$ $(d)$ $O_2$ $(iv)$ $3$
Choose the correct answer from the options given below:
| List-$I$ | List-$II$ |
| $(a)$ $Ne_2$ | $(i)$ $1$ |
| $(b)$ $N_2$ | $(ii)$ $2$ |
| $(c)$ $F_2$ | $(iii)$ $0$ |
| $(d)$ $O_2$ | $(iv)$ $3$ |
Choose the correct answer from the options given below:
Which of the following molecules has two unpaired $e^-$ with a bond order equal to $2$?
Medium
View SolutionAccording to $MO$ theory,the molecule which contains only $\pi$-bonds between the atoms is
Which of the following statements is not correct according to molecular orbital theory $:-$
Medium
View SolutionDistinguish between bonding molecular orbitals and antibonding molecular orbitals.
Difficult
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo