Magnetic moment (μ) is given as μ = \(\sqrt{n(n+2)}\)

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

For value n = 2, μ = \(\sqrt{2(2+2)}\) = \(\sqrt{8}\) = 2.83

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

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

For value n = 5, μ = \(\sqrt{5(5+2)}\) = \(\sqrt{35}\) = 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 n = 1. 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 n = 4. 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, Mn 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.