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Question: Among \({O_2}\), \(O_2^ - \),\(O_2^{2 - }\), \(O_2^ + \), \(O_2^{2 + }\), the diamagnetic species wi...

Among O2{O_2}, O2−O_2^ - ,O22−O_2^{2 - }, O2+O_2^ + , O22+O_2^{2 + }, the diamagnetic species will be __________
A.O2,O2+{O_2},O_2^ +
B.O22−,O2−O_2^{2 - },O_2^ -
C.O2+,O22+O_2^ + ,O_2^{2 + }
D.O22−,O22+O_2^{2 - },O_2^{2 + }

Explanation

Solution

We know that elements are distinguished by their atomic properties. An atom contains a core of protons and neutrons and the electrons revolve around the core in different orbitals. We also know that different orbitals have different capacities. The electrons revolving in the orbitals have quantum numbers assigned. Diamagnetic and paramagnetic properties of an element in its state is determined by the quantum numbers assigned.

Complete step by step answer:
We should understand that diamagnetism is a quantum mechanical effect that occurs in all materials. When this effect contributes to magnetism, the material is called diamagnetic. Diamagnetic materials are repelled by magnetic force in the two directions.
In chemistry, the thumb rule is used to determine whether the particle atom, ion or molecule is paramagnetic or diamagnetic. If all the electrons in the atom or molecule are paired then the substance is diamagnetic. If the atom or molecule has unpaired electrons then the substance is paramagnetic.
If we look at the oxygen molecule, the two atoms of oxygen form a molecule.
The electronic configuration can be given as - 1s22s22p41{s^2}2{s^2}2{p^4} .
We know that one oxygen atom has 8 electrons hence, a molecule will have 16 electrons.
Using molecular orbital theory, we can write the electronic configuration of oxygen as,
σ1s2σ∗1s2σ2s2σ∗2s2σ2p2π2px2π2py2π∗2px1π∗2py1\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
In O2{O_2} state, the 2px12p_x^1 and 2py12p_y^1 orbitals have one unpaired electron each. Hence, this molecule will show paramagnetic properties.
To form O2−O_2^ - , one extra electron enters the 2px2{p_x} orbital. But the 2py2{p_y} orbital still has one unpaired electron. Hence, this molecule will show paramagnetic properties.
To form O22−O_2^{2 - } , one extra electron enters the 2px2{p_x} orbital and one more electron enters the 2py2{p_y} orbital. Since, all the orbitals have paired electrons. This molecule will show diamagnetic properties.
Likewise, in case of O2+O_2^ + , one electron is removed from the outermost orbital 2py2{p_y} . Hence this orbital becomes empty. But the 2px2{p_x} orbital has an unpaired electron. Hence, this molecule is paramagnetic.
Lastly, in case of O22+O_2^{2 + }, two electrons are removed from the outermost orbitals. One from 2px2{p_x} orbital and one from 2py2{p_y} orbital. Hence, the two orbitals are empty and all the other orbitals have paired electrons. Thus, this molecule is diamagnetic.
Hence, the correct answer to the question is option D.

Note:
We need to know that in case of paramagnetic and ferromagnetic materials, they too have diamagnetic properties but the diamagnetic forces are overruled by the paramagnetic forces. In orbital molecular theory, the magnetic forces are distinguished by the electrons in the orbitals. If there are unpaired electrons in the orbitals, then the substance is paramagnetic. If all the electrons are paired then the substance is diamagnetic.