Question

Which one of the following shows the highest magnetic moment?
A. $F{e^{2 + }}$
B. $C{o^{2 + }}$
C. $C{r^{3 + }}$
D. $N{i^{2 + }}$

Answer 1

Hint: Most of the substances show magnetic nature. They are either paramagnetic or diamagnetic.
Most of the compounds of transition elements are paramagnetic in nature as unpaired electrons in d-subshell are present.
Larger the number of unpaired electrons in a substance, greater will be its magnetic moment. So to solve this question we will be comparing the number of unpaired electrons in the outer shell.

Complete step by step answer:
$F{e^{2 + }}$ With outer electronic configuration of 3d6 contains 4 unpaired electrons.

$C{o^{2 + }}$With outer electronic configuration of 3d7 contains 3 unpaired electrons.

$C{r^{3 + }}$With outer electronic configuration of 3d5 contains 3 unpaired electrons.

$N{i^{2 + }}$With outer electronic configuration of 3d8 contains 2 unpaired electrons.

$F{e^{2 + }}$ Contains maximum number of unpaired electrons. Hence, it shows the highest magnetic moment.

So, the correct answer is Option A .

Additional Information:
Paramagnetic substance shows weak attraction towards the magnetic field and diamagnetic substance gets repelled by the magnetic field.
Paramagnetic behaviour arises due to presence of one or more singly occupied orbitals and diamagnetic behaviour arises due to presence of paired electrons in the orbitals.
The effective total magnetic moment is given by the expression
${\mu _{eff}} = \sqrt {n(n + 2)} B.M$
Where n is the number of unpaired electrons. From the above equation it is clear that as the value of n increases from 1-5 paramagnetic behaviour increases. After d5 configuration, there is a decrease in magnetic moment as the number of unpaired electrons decreases.

Note:
In case of iron, cobalt and nickel the unpaired electron spins are much more pronounced. As a result these elements are more magnetic when compared to the rest of the elements. And these elements are called ferromagnetic.