Question

Calculate the magnetic moment of $F{e^{3 + }}$ in ${\left[ {Fe{{\left( {CN} \right)}_6}} \right]^{3 + }}$ and in ${\left[ {Fe{{({H_2}O)}_6}} \right]^{3 - }}$ .

Answer 1

Magnetic moment is the magnetic strength and orientation of a magnetic or other object that produces a magnetic field. The direction of the magnetic moment points from the south to North Pole of the magnet (inside the magnet). The magnetic field of a magnetic dipole is proportional to its magnetic dipole. The SI unit of the magnetic dipole is $wb \times m$ . An SI unit of magnetic moment is $A \times {m^2}\,$ .

Complete answer:
Given, ${\left[ {Fe{{\left( {CN} \right)}_6}} \right]^{3 + }}$ and in ${\left[ {Fe{{({H_2}O)}_6}} \right]^{3 - }}$ we have to calculate the magnetic moment of $F{e^{3 + }}$ .
The electronic configuration of Fe is $[Ar]3{d^6}4{s^2}$
Here the Fe is given by $F{e^{3 + }}$ .
$F{e^{3 + }}$ it means iron donates three electrons and possesses the oxidation state of $+ 3.$
So the electronic configuration of $F{e^{3 + }}$ is $[Ar]3{d^5}$
In ${\left[ {Fe{{\left( {CN} \right)}_6}} \right]^{3 + }}$ $C{N^ - }$ is a strong field ligand. Thus the $F{e^{3 + }}$ show low spinning splitting. $F{e^{3 + }}$ is a ${d^5}$ and contains $n = 1$ unpaired electron. So, magnetic moment of ${\left[ {Fe{{\left( {CN} \right)}_6}} \right]^{3 + }}$ is given by
As we know that magnetic moment is
$\therefore u = \sqrt {n(n + 2)}$
Put the value of n
$\Rightarrow u = \sqrt {1(1 + 2)}$
Simplify
$\Rightarrow u = \sqrt 3$
$\Rightarrow u = 1.732BM$
So the magnetic moment of ${\left[ {Fe{{\left( {CN} \right)}_6}} \right]^{3 + }}$ is $1.732BM$
In ${\left[ {Fe{{({H_2}O)}_6}} \right]^{3 - }}$ ${H_2}O$ is a weak field ligand. Thus the $F{e^{3 + }}$ shows high spinning splitting. $F{e^{3 + }}$ is a ${d^5}$ and contains $n = 5$ unpaired electron. So, magnetic moment of ${\left[ {Fe{{\left( {CN} \right)}_6}} \right]^{3 + }}$ is given by
As we know that magnetic moment is
$\therefore u = \sqrt {n(n + 2)}$
Put the value of n
$\Rightarrow u = \sqrt {5(5 + 2)}$
Simplify
$\Rightarrow u = \sqrt {5 \times 7}$
$\Rightarrow u = \sqrt {35}$
$\Rightarrow u = 5.91BM$
So the magnetic moment of ${\left[ {Fe{{({H_2}O)}_6}} \right]^{3 - }}$ is $5.91BM$

Note:
The magnetic dipole moment $(\mu )$ is a vector field defined as $\mu = iA$ whose direction is perpendicular to A and determined by the right hand thumb rule. In these rules grip shows the direction of flow of current and thumbs show the direction of magnetic field. The magnetic moment is produced by two methods- motion electric charge and spin angular momentum.