During the change of O_{2} to O_{2}^{-},the i | vedclass

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

Question

(C) The electronic configuration of $O_{2}$ ($16$ electrons) is: $\sigma 1s^{2} \sigma^{*} 1s^{2} \sigma 2s^{2} \sigma^{*} 2s^{2} \sigma 2p_{z}^{2} \p

Vedclass · Accepted Answer

$\pi^{*} 2 p_{x}$

Vedclass · Answer

(C) The electronic configuration of $O_{2}$ ($16$ electrons) is: $\sigma 1s^{2} \sigma^{*} 1s^{2} \sigma 2s^{2} \sigma^{*} 2s^{2} \sigma 2p_{z}^{2} \pi 2p_{x}^{2} = \pi 2p_{y}^{2} \pi^{*} 2p_{x}^{1} = \pi^{*} 2p_{y}^{1}$.When an electron is added to $O_{2}$ to form $O_{2}^{-}$,the incoming electron enters the lowest energy vacant or half-filled molecular orbital,which is the $\pi^{*}$ antibonding orbital.Thus,the electron enters the $\pi^{*} 2p_{x}$ or $\pi^{*} 2p_{y}$ orbital.Therefore,option $(C)$ is correct.

During the change of $O_{2}$ to $O_{2}^{-}$,the incoming electron goes to the orbital:

Similar Questions

Two homo-diatomic molecules $X_2$ and $Y_2$ have the same bond order. Then,select the $CORRECT$ statement.

In $O_2^-$,$O_2$,and $O_2^{2-}$,the total number of antibonding electrons respectively are:

What is the bond order of $N_2$?

Paramagnetism is exhibited by molecules:

Which of the following is paramagnetic as well as it has a fractional bond order?

Vedclass Test Series

Exam Paper Generator

Online Exam Module