Arrange the following in the increasing order | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) According to Molecular Orbital Theory $(MOT)$,the bond order is calculated as: $	ext{Bond Order} = \frac{1}{2} (N_b - N_a)$.For $O_{2}^{+}$ ($15$

Vedclass · Accepted Answer

$O_{2}^{2-} < O_{2}^{-} < O_{2} < O_{2}^{+}$

Vedclass · Answer

(A) According to Molecular Orbital Theory $(MOT)$,the bond order is calculated as: $	ext{Bond Order} = \frac{1}{2} (N_b - N_a)$.For $O_{2}^{+}$ ($15$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1$. Bond order = $\frac{1}{2}(10 - 5) = 2.5$.For $O_{2}$ ($16$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1 = \pi^* 2p_y^1$. Bond order = $\frac{1}{2}(10 - 6) = 2.0$.For $O_{2}^{-}$ ($17$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^2 = \pi^* 2p_y^1$. Bond order = $\frac{1}{2}(10 - 7) = 1.5$.For $O_{2}^{2-}$ ($18$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^2 = \pi^* 2p_y^2$. Bond order = $\frac{1}{2}(10 - 8) = 1.0$.Thus,the increasing order of bond order is: $O_{2}^{2-} < O_{2}^{-} < O_{2} < O_{2}^{+}$.

Arrange the following in the increasing order of their bond order: $O_{2}, O_{2}^{+}, O_{2}^{-}, O_{2}^{2-}$

Similar Questions

The lowest $O-O$ bond length in the following molecules is:

Which of the following species is paramagnetic?

In which of the following ionisation processes does the bond energy increase and the magnetic behaviour change from paramagnetic to diamagnetic?

In oxygen $(O_2)$ and carbon $(C_2)$ molecules,the bonding is:

Which of the following species contain unpaired electrons: $KO_2, AlO_2^-, BaO_2$,and $NO_2^+$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module