Which of the following is a diamagnetic ion?
(A) \[Z{n^{2 + }}\]
(B) $N{i^{2 + }}$
(C) $C{o^{2 + }}$
(D) $C{u^{2 + }}$
Answer
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Hint:There are two types of atoms. Paramagnetic and diamagnetic. When two electrons are paired in an orbital such that their total spin is equal to zero, then it is known as a diamagnetic atom. In other cases, if electrons are unpaired then the atom is called as paramagnetic.
Complete step by step answer:If we consider all the given options, we can find out whether the ions are paramagnetic or diamagnetic by knowing their electronic configuration.
(i) $Z{n^{2 + }}$
Atomic number of Zinc = $30$
$\therefore $ Numbers of electrons in $Z{n^ + } = 30 - 2 \Rightarrow 28$ electrons
Electronic configuration of $Z{n^ + }$
Or $1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^{10}}$
All the electrons in $Z{n^{2 + }}$ are paired. Therefore, the total spin value is zero and it is a diamagnetic ion.
(ii) $N{i^{2 + }}$
Atomic number of $Ni = 28$
$\therefore $ Number of electrons in $N{i^{2 + }} = 28 - 2 \Rightarrow 26$electrons
Electronic configuration of $N{i^{2 + }}$
Or $1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^8}$
In $3d$ orbital, two electrons are unpaired. Therefore, $N{i^{2 + }}$ is paramagnetic.
(iii) \[C{o^{2 + }}\]
Atomic number of $Co = 27$
Number of electrons in $C{o^{2 + }} = 27 - 2 \Rightarrow 25$ electrons
Electronic configuration of $C{o^{2 + }}$
Or $1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^7}$
In $3d$ orbital, three electrons are unpaired. Therefore $C{o^{2 + }}$ is paramagnetic.
Atomic number of $Cu = 29$
Number of electrons in $C{u^{2 + }} = 29 - 2 \Rightarrow 27$ electrons
Electronic configuration of $C{u^{2 + }}$
Or $1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^9}$
In $3d$ orbital, one electron is unpaired. Therefore, $C{u^{2 + }}$ is paramagnetic.
Hence, among all the options, only $Z{n^{2 + }}$ is diamagnetic ion.
Note:Even if only one unpaired electron is present in an atom and rest all electrons are paired. Still, the atom would be considered as paramagnetic as its total spin would not be equal to zero.
Complete step by step answer:If we consider all the given options, we can find out whether the ions are paramagnetic or diamagnetic by knowing their electronic configuration.
(i) $Z{n^{2 + }}$
Atomic number of Zinc = $30$
$\therefore $ Numbers of electrons in $Z{n^ + } = 30 - 2 \Rightarrow 28$ electrons
Electronic configuration of $Z{n^ + }$
Or $1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^{10}}$
All the electrons in $Z{n^{2 + }}$ are paired. Therefore, the total spin value is zero and it is a diamagnetic ion.
(ii) $N{i^{2 + }}$
Atomic number of $Ni = 28$
$\therefore $ Number of electrons in $N{i^{2 + }} = 28 - 2 \Rightarrow 26$electrons
Electronic configuration of $N{i^{2 + }}$
Or $1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^8}$
In $3d$ orbital, two electrons are unpaired. Therefore, $N{i^{2 + }}$ is paramagnetic.
(iii) \[C{o^{2 + }}\]
Atomic number of $Co = 27$
Number of electrons in $C{o^{2 + }} = 27 - 2 \Rightarrow 25$ electrons
Electronic configuration of $C{o^{2 + }}$
Or $1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^7}$
In $3d$ orbital, three electrons are unpaired. Therefore $C{o^{2 + }}$ is paramagnetic.
(iv) $C{u^{2 + }}$
Atomic number of $Cu = 29$
Number of electrons in $C{u^{2 + }} = 29 - 2 \Rightarrow 27$ electrons
Electronic configuration of $C{u^{2 + }}$
Or $1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^9}$
In $3d$ orbital, one electron is unpaired. Therefore, $C{u^{2 + }}$ is paramagnetic.
Hence, among all the options, only $Z{n^{2 + }}$ is diamagnetic ion.
Note:Even if only one unpaired electron is present in an atom and rest all electrons are paired. Still, the atom would be considered as paramagnetic as its total spin would not be equal to zero.
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