Question

The pairs of species of oxygen and their magnetic behaviours are noted below. Which of the following presents the correct description ?
a.) $O_2^ - $,$O_2^{2 - }$ - Both diamagnetic
b.) ${O^ + }$,$O_2^{2 - }$ - Both paramagnetic
c.) $O_2^ + $,${O_2}$ - Both paramagnetic
d.) O ,$O_2^{2 - }$ - Both paramagnetic

Answer 1

Hint:Paramagnetic species have unpaired electrons while the diamagnetic have all electrons paired up. So, from above the Oxygen atom has eight as the total number of electrons. The unpaired or paired electrons are decided on the basis of filling up of orbitals.

Complete answer:
In the question we have been given about paramagnetic and diamagnetic species.
First, let us understand about these only.
The paramagnetic species are those which have unpaired electrons in any subshell of the orbital while the diamagnetic species have all the electrons paired up.
To classify these into whether paramagnetic or diamagnetic, we should know about their electronic configuration.

So, let us write the electronic configuration or specifically here molecular orbital configuration of all the above atoms as-
The first is $O_2^ - $:- $\sigma 1{s^2}{\sigma ^*}1{s^2}\sigma 2{s^2}{\sigma ^*}2{s^2}\sigma 2p_z^2\Pi 2p_x^2\Pi 2p_y^2{\Pi ^*}2p_x^2{\Pi ^*}2p_y^1$
It has one unpaired electron. Thus, paramagnetic in nature.
The next is $O_2^{2 - }$:- $\sigma 1{s^2}{\sigma ^*}1{s^2}\sigma 2{s^2}{\sigma ^*}2{s^2}\sigma 2p_z^2\Pi 2p_x^2\Pi 2p_y^2{\Pi ^*}2p_x^2{\Pi ^*}2p_y^2$
It has zero unpaired electrons Thus, diamagnetic in nature.
After this, the next one is ${O^ + }$:- $1{s^2}2{s^2}2p_x^12p_y^12p_z^1$
This was an atom. It has three unpaired electrons. Thus, paramagnetic in nature.
And after this is$O_2^ + $:- $\sigma 1{s^2}{\sigma ^*}1{s^2}\sigma 2{s^2}{\sigma ^*}2{s^2}\sigma 2p_z^2\Pi 2p_x^2\Pi 2p_y^2{\Pi ^*}2p_x^1$
It has one unpaired electron. Thus, paramagnetic in nature.
The next one is ${O_2}$molecule :- $\sigma 1{s^2}{\sigma ^*}1{s^2}\sigma 2{s^2}{\sigma ^*}2{s^2}\sigma 2p_z^2\Pi 2p_x^2\Pi 2p_y^2{\Pi ^*}2p_x^1{\Pi ^*}2p_y^1$
It has two unpaired electrons. Thus, paramagnetic in nature.
After this is one oxygen atom.
So, for O :- $1{s^2}2{s^2}2p_x^22p_y^12p_z^1$
It also has two unpaired electrons. Thus, paramagnetic in nature.
So, now as we are done with the configurations of all the molecules and even we have seen the nature of molecules also.
Thus we can say that $O_2^ - $,${O^ + }$,$O_2^ + $,${O_2}$ and O are all paramagnetic in nature.
So, the correct answer is “Option C”.

Note: It must be noted that if we have one atom say oxygen then we will talk about electronic configuration. And in this the subshells, shells and orbitals are filled accordingly while in case of molecules which consists of at least two atoms; we talk about its molecular orbitals.