Question

What can be inferred from the magnetic moment values of the following complex species?

$A)$ Example$:K_4\left[Mn\right(CN{)}_6]$
Magnetic moment $2.2BM$

$B)$ Example$:\left[Fe\right(H_{2}O{)}_6{]}^{2+}$
Magnetic moment$:5.3BM$

$C)$ Example$:K_2\left[MnC{l}_4\right]$
Magnetic moment$:5.9BM$

Answer 1

$A)$ Magnetic moment

Magnetic moment of a complex is the combination of spin and orbital magnetic moment, For transition metals, the magnetic moment is calculated from the spin-only formula.

Spin only magnetic moment

Rotation of the electrons about their own axes.

Spin only magnetic moment can be calculated by the following formula:

Magnetic moment $(M.M)$

$=\sqrt{n(n+2)}\dots ..\left(i\right)$

Where, n is number of unpaired electrons.

Calculation of unpaired electrons

For value $n=1,$
$(M.M)=\sqrt{1(1+2)}=$$\sqrt{3}=1.732$

For value $n=2,$
$(M.M)=\sqrt{2(2+2)}=$$\sqrt{8}=2.83$

For value $n=3,$
$(M.M)=\sqrt{3(3+2)}=$$\sqrt{15}=3.87$

For value $n=4,$
$(M.M)=\sqrt{4(4+2)}=$$\sqrt{24}=4.899$

For value $n=5,$
$(M.M)=\sqrt{5(5+2)}=$$\sqrt{35}=5.92$

Given$:(M.M)=2.2B.M$

Now from the equation $\left(i\right),$

$\sqrt{n(n+2)}=2.2$

For transition metals, the magnetic moment is calculated from the spin-only formula. Therefore, we can see from the above calculation that the given value is closest to

$n=1.$

Hence, here number of unpaired electron is one.

Inferred from the magnetic moment values

For this complex,$n=1,Mn$ is in the $+2$ oxidation state and Mn has $5$ electrons in the $d-$orbital in which four are paired and one is unpaired.

Hence, we can say that $C{N}^{-}$ is a strong field ligand that causes the pairing of electrons and this $K_4\left[Mn\right(CN{)}_6]$ complex is a low-spin complex.

$B)$ Magnetic moment

Magnetic moment of a complex is the combination of spin and orbital magnetic moment, For transition metals, the magnetic moment is calculated from the spin-only formula.

Spin only magnetic moment

Rotation of the electrons about their own axes.

Spin only magnetic moment can be calculated by the following formula:

Magnetic moment $(M.M)$

$=\sqrt{n(n+2)}\dots ..\left(i\right)$

Where, n is number of unpaired electrons.

Calculation of unpaired electrons

For value $n=1,$
$(M.M)=\sqrt{1(1+2)}=$$\sqrt{3}=1.732$

For value $n=2,$
$(M.M)=\sqrt{2(2+2)}=$$\sqrt{8}=2.83$

For value $n=3,$
$(M.M)=\sqrt{3(3+2)}=$$\sqrt{15}=3.87$

For value $n=4,$
$(M.M)=\sqrt{4(4+2)}=$$\sqrt{24}=4.899$

For value $n=5,$
$(M.M)=\sqrt{5(5+2)}=$$\sqrt{35}=5.92$

Given: $(M.M)=5.3B.M$

Now from the equation $\left(i\right),$

$\sqrt{n(n+2)}=5.3$

For transition metals, the magnetic moment is calculated from the spin-only formula. Therefore, we can see from the above calculation that the given value is closest to

$n=4.$

Hence, here number of unpaired electrons are $4.$

Inferred from the magnetic moment value

For this complex, $n=4,Fe$ is in the $+2$ oxidation state and Fe has $6$ electrons in the $d-$orbital in which four are paired and one is unpaired.

Hence, we can say that $H_{2}O$ is a weak field ligand which does not causes any pairing of electrons and this $\left[Fe\right(H_{2}O{)}_6{]}^{2+}$ complex is a high-spin complex.

$C)$ Magnetic moment

Magnetic moment of a complex is the combination of spin and orbital magnetic moment, For transition metals, the magnetic moment is calculated from the spin-only formula.

Spin only magnetic moment

Rotation of the electrons about their own axes.

Spin only magnetic moment can be calculated by the following formula:

Magnetic moment $(M.M)$

$=\sqrt{n(n+2)}\dots ..\left(i\right)$

Where, $n$ is number of unpaired electrons.

Calculation of unpaired electrons
For value $n=1,$
$(M.M)=\sqrt{1(1+2)}=$$\sqrt{3}=1.732$

For value $n=2,$
$(M.M)=\sqrt{2(2+2)}=$$\sqrt{8}=2.83$

For value $n=3,$
$(M.M)=\sqrt{3(3+2)}=$$\sqrt{15}=3.87$

For value $n=4,$
$(M.M)=\sqrt{4(4+2)}=$$\sqrt{24}=4.899$

For value $n=5,$
$(M.M)=\sqrt{5(5+2)}=$$\sqrt{35}=5.92$

Given: $(M.M)=5.9B.M$

Now from the equation $\left(i\right),$

$\sqrt{n(n+2)}=5.9$

For transition metals, the magnetic moment is calculated from the spin-only formula. Therefore, we can see from the above calculation that the given value is closest to
$n=5.$

Hence, here number of unpaired electrons are $5.$

Inferred from the magnetic moment value

For this complex $\left(K_2\right[MnC{l}_4\left]\right),n=5,Mn$ is in

The $+2$ oxidation state and $Mn$ has $5$ electrons in the d-orbital which are all unpaired. Since, tetrahedral complexes are always high spin and there is no pairing of electrons occur.