What is the bond order of O_2^{2-}? | vedclass

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

Question

(C) The total number of electrons in $O_2^{2-}$ is $8 + 8 + 2 = 18$.According to the Molecular Orbital Theory,the electronic configuration of $O_2^{2-

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(C) The total number of electrons in $O_2^{2-}$ is $8 + 8 + 2 = 18$.According to the Molecular Orbital Theory,the electronic configuration of $O_2^{2-}$ is: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^2 = \pi^* 2p_y^2$.Number of bonding electrons $(N_b)$ = $10$.Number of antibonding electrons $(N_a)$ = $8$.Bond Order = $\frac{1}{2} (N_b - N_a) = \frac{1}{2} (10 - 8) = \frac{2}{2} = 1$.

List-$I$ (Molecule / ion)	List-$II$ (Bond order)
$1$. $NO$	$i$. $1.5$
$2$. $CO$	$ii$. $2.0$
$3$. $O_2^{-}$	$iii$. $2.5$
$4$. $O_2$	$iv$. $3.0$

What is the bond order of $O_2^{2-}$?

Similar Questions

Two $\pi$ and half $\sigma$ bonds are present in

Match List-$I$ with List-$II$ and select the correct option:
List-$I$ (Molecule / ion) List-$II$ (Bond order)
$1$. $NO$ $i$. $1.5$
$2$. $CO$ $ii$. $2.0$
$3$. $O_2^{-}$ $iii$. $2.5$
$4$. $O_2$ $iv$. $3.0$

Given below are two statements:
Statement $(I)$: The correct sequence of bond lengths in the following species is:
$O_2^+ < O_2 < O_2^- < O_2^{2-}$
Statement $(II)$: The correct sequence of number of unpaired electrons in the following species is:
$O_2 > O_2^+ > O_2^- > O_2^{2-}$

The $\pi$ bond between $Cl$ and $O$ in $ClO_4^-$ is formed by the overlapping of . . . . . . orbitals.

$N_2$ and $O_2$ are converted into $N_2^+$ and $O_2^+$ respectively. Which of the following is not correct?

Vedclass Test Series

Exam Paper Generator

Online Exam Module