Which of the following species has a bond ord | vedclass

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

Question

(B) The bond order is calculated using the formula: $	ext{Bond Order} = \frac{N_b - N_a}{2}$,where $N_b$ is the number of bonding electrons and $N_a$

Vedclass · Accepted Answer

$O_{2}^{-}$

Vedclass · Answer

(B) The bond order is calculated using the formula: $	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 $O_{2}^{+}$ (total $15$ electrons): Configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1$. Bond order $= \frac{10 - 5}{2} = 2.5$.For $O_{2}^{-}$ (total $17$ electrons): Configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^2 = \pi^* 2p_y^1$. Bond order $= \frac{10 - 7}{2} = 1.5$.For $O_{2}^{2-}$ (total $18$ electrons): Bond order $= \frac{10 - 8}{2} = 1.0$.For $O_{2}$ (total $16$ electrons): Bond order $= \frac{10 - 6}{2} = 2.0$.Thus,$O_{2}^{-}$ has a bond order of $1.5$.

Which of the following species has a bond order of $1.5$?

Similar Questions

Give the order of relative stability for $N_2, N_2^+, N_2^-,$ and $N_2^{2+}$.

In which of the following transformations,the bond order has increased and the magnetic behaviour has changed?

In which of the following processes,the bond order increases and paramagnetic character changes to diamagnetic one?

Assuming $2s-2p$ mixing is $NOT$ operative,the paramagnetic species among the following is:

Among the following,the paramagnetic compound is

Vedclass Test Series

Exam Paper Generator

Online Exam Module