Question

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

Answer 1

Hint: Magnetic moment is defined as the magnetic strength and orientation of a magnet or other object which produces a magnetic field. Some examples of objects that have magnetic moments are electromagnets, permanent magnets, elementary particles, various molecules and many astronomical objects.

Complete step-by-step solution: -
The magnetic moment is given by $\sqrt{n(n+2)}$, where n is the number of unpaired electrons
For $C{{r}^{3+}}$ the number of unpaired electrons is $3(4s3{{d}^{3}})$ . Here n is 3, therefore the magnetic moment will be$\sqrt{3(3+2)}=3.873$. This formula is used for spin-only cases.
Magnetic moment normally refers to a system’s magnetic dipole moment, the component of the magnetic that can be represented by an equivalent magnetic dipole.
SI unit of a magnetic moment is $J/Tesla$ .
Electronic configuration of each ion is given as below
For $C{{r}^{3+}}$ it is $[Ar]3{{d}^{3}}$ 3 unpaired electrons are available.
 For ${{V}^{3+}}$ it is ${{[Ar]}^{18}}3{{d}^{2}}$ 2 unpaired electrons are available.
For $F{{e}^{3+}}$ it is ${{[Ar]}^{18}}3{{d}^{5}}$ 5 unpaired electrons are available.
For $C{{o}^{3+}}$ it is ${{[Ar]}^{18}}3{{d}^{6}}$ 4 unpaired electrons are available.
From the formula of magnetic moment, higher the number of unpaired electrons (n), more will be the amount of magnetic moment.
From the above options $F{{e}^{3+}}$ has the highest number of unpaired electrons.
So the correct option is C.

Note: Students magnetic moments are used in conjunction with electronic spectra to gain information about the oxidation number and stereochemistry of the central metal ion in coordination complexes. For first row transition metal ions in the free ion state, i.e. isolated ions in a vacuum, all 5 of the 3rd orbitals are degenerate.