
Among ${O_2}$, $O_2^ - $,$O_2^{2 - }$, $O_2^ + $, $O_2^{2 + }$, the diamagnetic species will be __________
A.${O_2},O_2^ + $
B.$O_2^{2 - },O_2^ - $
C.$O_2^ + ,O_2^{2 + }$
D.$O_2^{2 - },O_2^{2 + }$
Answer
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Hint: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 - $1{s^2}2{s^2}2{p^4}$ .
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,
$\sigma 1{s^2}{\sigma ^*}1{s^2}\sigma 2{s^2}{\sigma ^*}2{s^2}\sigma 2{p^2}\pi 2p_x^2\pi 2p_y^2{\pi ^*}2p_x^1{\pi ^*}2p_y^1$
In ${O_2}$ state, the $2p_x^1$ and $2p_y^1$ orbitals have one unpaired electron each. Hence, this molecule will show paramagnetic properties.
To form $O_2^ - $ , one extra electron enters the $2{p_x}$ orbital. But the $2{p_y}$ orbital still has one unpaired electron. Hence, this molecule will show paramagnetic properties.
To form $O_2^{2 - }$ , one extra electron enters the $2{p_x}$ orbital and one more electron enters the $2{p_y}$ orbital. Since, all the orbitals have paired electrons. This molecule will show diamagnetic properties.
Likewise, in case of $O_2^ + $ , one electron is removed from the outermost orbital $2{p_y}$ . Hence this orbital becomes empty. But the $2{p_x}$ orbital has an unpaired electron. Hence, this molecule is paramagnetic.
Lastly, in case of $O_2^{2 + }$, two electrons are removed from the outermost orbitals. One from $2{p_x}$ orbital and one from $2{p_y}$ 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.
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 - $1{s^2}2{s^2}2{p^4}$ .
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,
$\sigma 1{s^2}{\sigma ^*}1{s^2}\sigma 2{s^2}{\sigma ^*}2{s^2}\sigma 2{p^2}\pi 2p_x^2\pi 2p_y^2{\pi ^*}2p_x^1{\pi ^*}2p_y^1$
In ${O_2}$ state, the $2p_x^1$ and $2p_y^1$ orbitals have one unpaired electron each. Hence, this molecule will show paramagnetic properties.
To form $O_2^ - $ , one extra electron enters the $2{p_x}$ orbital. But the $2{p_y}$ orbital still has one unpaired electron. Hence, this molecule will show paramagnetic properties.
To form $O_2^{2 - }$ , one extra electron enters the $2{p_x}$ orbital and one more electron enters the $2{p_y}$ orbital. Since, all the orbitals have paired electrons. This molecule will show diamagnetic properties.
Likewise, in case of $O_2^ + $ , one electron is removed from the outermost orbital $2{p_y}$ . Hence this orbital becomes empty. But the $2{p_x}$ orbital has an unpaired electron. Hence, this molecule is paramagnetic.
Lastly, in case of $O_2^{2 + }$, two electrons are removed from the outermost orbitals. One from $2{p_x}$ orbital and one from $2{p_y}$ 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.
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