Question

Calculate the spin-only magnetic moment of Fe. [ Atomic number of Fe = 26]

Answer 1

Hint: The magnetic moment contributes in orbital angular momentum and spin. Spin-only magnetic moment related to total number of the unpaired number of the electrons.
Spin only magnetic moment,$\mu = \sqrt {n\left( {n + 2} \right)} $ Bohr’s Magneton
Where n is the number of unpaired electrons present in the atom.

Complete step by step answer:
The electronic configuration of Fe atom is
$1{s^2},2{s^2},2{p^6},3{s^2},3{p^6},4{s^2},3{d^6}$
The condensed electron configuration will be
$\left[ {Ar} \right]4{s^2}3{d^6}$
For unpaired electrons, fill the orbitals in accordance to Hund’s rule. According to Hund’s rule, every orbital in a subshell is singly filled with one electron before any one orbital is doubly filled.
This makes presence of 4 unpaired electrons in the 3d-orbital of the Fe-atom.
Spin only magnetic moment,$\mu = \sqrt {n\left( {n + 2} \right)} $
$ = \sqrt {4\left( {4 + 2} \right)} $
= 4.89 Bohr’s Magneton.

Hence, the spin only magnetic moment of the Fe atom is 4.89 Bohr’s Magneton.

Additional Information: Pauli’s Exclusion Principle – No two electrons in an atom can have the same value for all four quantum numbers. In other words,
No more than two electrons can occupy the same orbital.
Two electrons in the same orbital must have opposite spins. Their spin multiplicity (2s+1) must equal to one.
Aufbau’s Principle – In the ground state of an atom or ion, electrons are filled in lower energy atomic orbitals before they occupy higher energy atomic orbitals. For determining energy of the level, (n+l) rule can be used.

Note: The number of unpaired electrons needs to be found out carefully which can be done by keeping in mind the Aufbau’s Principle, Hund’s rule and Pauli’s Exclusion Principle during filling of orbitals with electrons.