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

Solution:

Magnetic moment $(\mu)$ is given as $\mu=\sqrt{n(n+2)}$.

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

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

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

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

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

(i) K4[Mn(CN)6]

For in transition metals, the magnetic moment is calculated from the spin-only formula. Therefore,

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

We can see from the above calculation that the given value is closest to. Also, in this complex, Mn is in the +2 oxidation state. This means that Mn has 5 electrons in the d-orbital.

Hence, we can say that CN− is a strong field ligand that causes the pairing of electrons.

(ii) [Fe(H2O)6]2+

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

We can see from the above calculation that the given value is closest to. Also, in this complex, Fe is in the +2 oxidation state. This means that Fe has 6 electrons in the d-orbital.

Hence, we can say that H2O is a weak field ligand and does not cause the pairing of electrons.

(iii) K2[MnCl4]

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

We can see from the above calculation that the given value is closest to $n=5$. Also, in this complex, $M n$ is in the $+2$ oxidation state. This means that Mn has 5 electrons in the $d$-orbital.

Hence, we can say that Cl− is a weak field ligand and does not cause the pairing of electrons.