Question

Bond order of ${O_2}$, $O_2^ + $, $O_2^ - $ and $O_2^{2 - }$ is in the order ___________
A.$O_2^ - < O_2^{2 - } < {O_2} < O_2^ + $
B.$O_2^{2 - } < O_2^ - < {O_2} < O_2^ + $
C.$O_2^ + < {O_2} < O_2^ - < O_2^{2 - }$
D.${O_2} < O_2^ + < O_2^ - < O_2^{2 - }$

Answer 1

Hint:We have studied that bond order is defined as the difference between the number of bonds and antibonds. The bond number expresses the number of electron pairs or bonds between a pair of atoms. As we know oxygen exists in diatomic form. Here, the bond order comes into the picture.

Complete step by step answer:
Bond number is given by the expression,
${\text{Bond order}} = \dfrac{{{\text{Number of bonding electrons}} - {\text{Number of antibonding electrons}}}}{2}$
We know that oxygen exists in diatomic form.
The electronic configuration of oxygen atom, according to molecular orbital theory can be written as,
$\sigma 1{s^2}{\sigma ^*}1{s^2}\sigma 2{s^2}{\sigma ^*}2{s^2}\sigma 2{p^2}\pi 2p_x^2\pi 2p_y^2{\pi ^*}2p_x^1{\pi ^*}2p_y^1$
Therefore, bond order can be calculated as,
$B.O. = \dfrac{{8 - 4}}{2}$
$B.O. = 2$
We can write the electronic configuration of oxygen atom in $O_2^ + $ state as,
$\sigma 1{s^2}{\sigma ^*}1{s^2}\sigma 2{s^2}{\sigma ^*}2{s^2}\sigma 2{p^2}\pi 2p_x^2\pi 2p_y^2{\pi ^*}2p_x^1$
The bond order is calculated as,
$B.O. = \dfrac{{8 - 3}}{2}$
$B.O. = 2.5$
Likewise, the electronic configuration of oxygen atom in $O_2^ - $ state can be given as,
$\sigma 1{s^2}{\sigma ^*}1{s^2}\sigma 2{s^2}{\sigma ^*}2{s^2}\sigma 2{p^2}\pi 2p_x^2\pi 2p_y^2{\pi ^*}2p_x^2{\pi ^*}2p_y^1$
The bond order is calculated as,
$B.O. = \dfrac{{8 - 5}}{2}$
$B.O. = 1.5$
Lastly, the electronic configuration of $O_2^{2 - }$ is given as,
$\sigma 1{s^2}{\sigma ^*}1{s^2}\sigma 2{s^2}{\sigma ^*}2{s^2}\sigma 2{p^2}\pi 2p_x^2\pi 2p_y^2{\pi ^*}2p_x^2{\pi ^*}2p_y^2$
The bond order is calculated as,
$B.O. = \dfrac{{8 - 6}}{2}$
$B.O. = 1$
According to the theory of bond energy. The greater the bond order, the more stable the molecule is. Hence, accordingly we can observe that the highest is $O_2^ + $ followed by ${O_2}$ followed by $O_2^ - $ and the least stable is $O_2^{2 - }$ .
Hence, the correct answer is option C.

Note:
We must remember that bond order can also be explained as an index of bond strength and it is extensively used in valence bond theory. This term bond order is also used to determine whether the molecule is paramagnetic or diamagnetic. The bond order indicates the stability of a molecule. The higher the bond order, the more stable the molecule will be. Isoelectronic species of different elements have the same bond number.