Question

Select the correct statement about ${O_2}^ + $and ${O_2}^{}$
A. ${O_2}^ + $and ${O_2}^{}$ both are paramagnetic bond order of ${O_2}^ + $is greater than ${O_2}^{}$.
B. ${O_2}^ + $is paramagnetic and ${O_2}^{}$ is diamagnetic and bond order of ${O_2}^ + $is greater than that of ${O_2}^{}$.
C. ${O_2}^ + $is paramagnetic and ${O_2}^{}$ is diamagnetic and bond order of ${O_2}^ + $is less than that of ${O_2}^{}$
D. None of the above is correct.

Answer 1

Hint: We know that bond order is defined as the number of bonds between two atoms. The bond order is given by $\dfrac{{{N_b} - {N_a}}}{2}$. Here, ${N_a} = $ number of anti bonding electrons and ${N_b} = $number of bonding electrons. Diamagnetic compounds have all their electrons paired. Paramagnetic compounds have unpaired electrons .

Complete step by step answer:
${O_2}^{}$ has a total of sixteen electrons. The electronic configuration of ${O_2}^{}$ can be written as $\sigma 1{s^2},{\sigma ^*}1{s^2},\sigma 2{s^2},{\sigma ^*}2{s^2},\sigma 2p_x^2,\pi 2p_x^2,\pi 2p_y^2,{\pi ^*}2p_x^1,{\pi ^*}2p_y^1$. We can see that there are unpaired electrons in it. So it is paramagnetic in nature. The number of bonding electrons is equal to $10$ and the number of antibonding electrons is equal to $6$. So we can easily find out the bond order by using the formula, $\dfrac{{{N_b} - {N_a}}}{2}$.
$ \Rightarrow \dfrac{{{N_b} - {N_a}}}{2} = \dfrac{{10 - 4}}{2} = 2$, So the bond order is two.
Now the total number of electrons of ${O_2}^ + $is equal to fifteen. The electronic configuration of ${O_2}^ + $can be written as $\sigma 1{s^2},{\sigma ^*}1{s^2},\sigma 2{s^2},{\sigma ^*}2{s^2},\sigma 2p_x^2,\pi 2p_x^2,\pi 2p_y^2,{\pi ^*}2p_x^1,{\pi ^*}2p_y^0$. We can see that it has one unpaired electron in it. So it is also paramagnetic in nature. The number of bonding electrons is equal to ten and the number of antibonding electrons is equal to five. So we can easily find out the bond order by using the formula , $\dfrac{{{N_b} - {N_a}}}{2}$.
$ \Rightarrow \dfrac{{{N_b} - {N_a}}}{2} = \dfrac{{10 - 5}}{2} = 2.5$.
By the above explanation and calculation it is clear to us that both ${O_2}^ + $ and ${O_2}^{}$ are paramagnetic and the bond orders of ${O_2}^ + $ is $2.5$ and the bond order of ${O_2}^{}$ is $2$.
${O_2}^ + $ and ${O_2}^{}$ both are paramagnetic bond order of ${O_2}^ + $ is greater than ${O_2}^{}$.

So the correct answer of the given question is option: A.

Note:
Always remember that the bond order of a compound can be given by the formula $\dfrac{{{N_b} - {N_a}}}{2}$. Here, ${N_a} = $ number of anti bonding electrons and ${N_b} = $number of bonding electrons.The electronic configuration of ${O_2}$ is $\sigma 1{s^2},{\sigma ^*}1{s^2},\sigma 2{s^2},{\sigma ^*}2{s^2},\sigma 2p_x^2,\pi 2p_x^2,\pi 2p_y^2,{\pi ^*}2p_x^1,{\pi ^*}2p_y^1$and the electronic configuration of $O_2^ + $is $\sigma 1{s^2},{\sigma ^*}1{s^2},\sigma 2{s^2},{\sigma ^*}2{s^2},\sigma 2p_x^2,\pi 2p_x^2,\pi 2p_y^2,{\pi ^*}2p_x^1,{\pi ^*}2p_y^0$. Both of them are paramagnetic in nature.