The correct bond order in the following speci | vedclass

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

Question

(B) According to Molecular Orbital Theory $(MOT)$,the bond order is calculated as: $	ext{Bond Order} = \frac{N_b - N_a}{2}$,where $N_b$ is the number

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) According to Molecular Orbital Theory $(MOT)$,the bond order is calculated as: $	ext{Bond Order} = \frac{N_b - N_a}{2}$,where $N_b$ is the number of bonding electrons and $N_a$ is the number of antibonding electrons.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 = $\frac{10 - 7}{2} = 1.5$.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 = $\frac{10 - 5}{2} = 2.5$.For $O_{2}^{2+}$ ($14$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2$. Bond order = $\frac{10 - 4}{2} = 3.0$.Thus,the increasing order of bond order is $O_{2}^{-} < O_{2}^{+} < O_{2}^{2+}$.

The correct bond order in the following species is

Similar Questions

What is the bond order in $N_{2}$ molecule?

The correct order of energies of molecular orbitals of $N_2$ molecule is

The correct order of bond dissociation energy among $N_2, O_2, O_2^-$ is shown in which of the following arrangements?

Which of the following pairs of species are paramagnetic?

Paramagnetism is exhibited by molecules:

Vedclass Test Series

Exam Paper Generator

Online Exam Module