Question

Which of the following transition metal ions are diamagnetic?
A. $C{o^{2 + }}$
B. $N{i^{2 + }}$
C. $C{u^{2 + }}$
D. $Z{n^{2 + }}$

Answer 1

Hint:Transition metals are d block elements placed in group 3-11 of the Periodic Table. The d orbitals in the atoms of transition metals are partially filled. They occurred in different oxidation states. The transition metals form different cations with different ionic charges. These ions are formed by loss of s electrons rather than that of d electrons. The most common charge on transition metal ions is +2 because their atoms generally contain two valence electrons.

Complete answer:
As per Pauli’s Exclusion Principle, for an atom an orbital will be filled with only two electrons and two electrons present in the same orbital will have opposite spin.
When two electrons occupy an orbital and are considered as paired, their spin should be opposite. It means their net spin will be equal to zero. Atoms with such electrons will be considered as diamagnetic that means atoms with only paired electrons are diamagnetic in nature. Diamagnetic atoms will repel magnetic fields.
If even one orbital of an atom is filled with only a single electron, then the electron is unpaired making the atom paramagnetic in nature. It means that the atom with unpaired electrons is paramagnetic. When an atom is paramagnetic in nature, then it will be attracted by a magnetic field. The paramagnetic atoms will not act as a magnet in the absence of external magnetic fields.
Here, we have to identify the diamagnetic ion. We will do it by looking at the electronic configuration of each element in its +2 ion form.
Let us first write the electronic configuration of each element.
The atomic number of cobalt (Co) is 27 and its electronic configuration is given below:
$[1{s^2}2{s^2}2{p^2}3{s^2}3{p^6}3{d^7}4{s^2}]$
The atomic number of nickel (Ni) is 28 and its electronic configuration is given below:
$ [1{s^2}2{s^2}2{p^2}3{s^2}3{p^6}3{d^8}4{s^2}]$
The atomic number of copper (Cu) is 29 and its electronic configuration is given below:
$ [1{s^2}2{s^2}2{p^2}3{s^2}3{p^6}3{d^{10}}4{s^1}]$
The atomic number of zinc (Zn) is 30 and its electronic configuration is given below:
$ [1{s^2}2{s^2}2{p^2}3{s^2}3{p^6}3{d^{10}}4{s^2}]$
When transition metals form ions, they lose electrons and form cations. Transition metal atoms will lose the electrons in 4s orbital first rather than 3d orbital. The formation +2 ions for each transition metal is shown below along with electronic configuration of transition metal ions formed is given below. The electronic configuration each +2 ion is shown in brackets.
$ [Co \to C{o^{2 + }} + 2{e^ - }] ([1{s^2}2{s^2}2{p^2}3{s^2}3{p^6}3{d^7}])$
$ [Ni \to N{i^{2 + }} + 2{e^ - }] ([1{s^2}2{s^2}2{p^2}3{s^2}3{p^6}3{d^8}])$
$ [Cu \to C{u^{2 + }} + 2{e^ - }] ([1{s^2}2{s^2}2{p^2}3{s^2}3{p^6}3{d^9}])$
$ [Zn \to Z{n^{2 + }} + 2{e^ - }] ([1{s^2}2{s^2}2{p^2}3{s^2}3{p^6}3{d^{10}}])$
Now, observe electronic configuration of each transition metal ion given above. In case of cobalt, nickel and copper, the 3d orbital is partially filled as their 3d orbitals contain 7, 8 and 9 electrons respectively. Therefore, ions of cobalt, nickel and zinc contain 3, 2 and 1 unpaired electrons respectively. Therefore, $C{o^{2 + }}$, $N{i^{2 + }}$ and $C{u^{2 + }}$ are paramagnetic in nature.
In case of $Z{n^{2 + }}$ ion, the 3d orbital is completely filled; it means there are no unpaired electrons in $Z{n^{2 + }}$ ion. Therefore, $[Z{n^{2 + }}]$ is diamagnetic in nature.

Hence, the correct option is option D.

Note:
While writing electronic configuration of transition metals, students should keep in mind that the 4s orbital will be filled after filling the 3d orbital. During formation of ions, the atom will lose 4s electrons first, not 3d electrons. In transition metals, the 4s orbitals have higher energy than that of 3d orbitals. Hence, during ion formation the 4s electrons will be lost first.