Arrange the following coordination compounds in the increasing order of magnetic moments. (Atomic numbers: Mn = 25; Fe = 26)
A. $$[FeF_6]^{3-}$$
B. $$[Fe(CN)_6]^{3-}$$
C. $$[MnCl_6]^{3-}$$ (high spin)
D. $$[Mn(CN)_6]^{3-}$$
Choose the correct answer from the options given below

We need to arrange the coordination compounds in increasing order of magnetic moments.

Magnetic moment $$\mu = \sqrt{n(n+2)}$$ BM, where $$n$$ is the number of unpaired electrons.

Determine the number of unpaired electrons in each complex.

A. $$[FeF_6]^{3-}$$: Fe is in +3 oxidation state, so $$Fe^{3+}$$ has configuration $$3d^5$$. $$F^-$$ is a weak field ligand, so no pairing occurs (high spin). Unpaired electrons = 5.

$$\mu = \sqrt{5(5+2)} = \sqrt{35} = 5.92$$ BM

B. $$[Fe(CN)_6]^{3-}$$: Fe is in +3 oxidation state, so $$Fe^{3+}$$ has $$3d^5$$. $$CN^-$$ is a strong field ligand, so maximum pairing occurs (low spin): $$t_{2g}^5 e_g^0$$. Unpaired electrons = 1.

$$\mu = \sqrt{1(1+2)} = \sqrt{3} = 1.73$$ BM

C. $$[MnCl_6]^{3-}$$ (high spin): Mn is in +3 oxidation state, so $$Mn^{3+}$$ has $$3d^4$$. $$Cl^-$$ is a weak field ligand and it is given as high spin: $$t_{2g}^3 e_g^1$$. Unpaired electrons = 4.

$$\mu = \sqrt{4(4+2)} = \sqrt{24} = 4.90$$ BM

D. $$[Mn(CN)_6]^{3-}$$: Mn is in +3 oxidation state, so $$Mn^{3+}$$ has $$3d^4$$. $$CN^-$$ is a strong field ligand (low spin): $$t_{2g}^4 e_g^0$$. Unpaired electrons = 2.

$$\mu = \sqrt{2(2+2)} = \sqrt{8} = 2.83$$ BM

Arrange in increasing order of magnetic moment.

$$B (1.73) < D (2.83) < C (4.90) < A (5.92)$$

The increasing order is: $$B < D < C < A$$

The correct answer is Option B.