Arrange the following coordination compounds in the increasing order of magnetic moments. (Atomic numbers: $\mathrm{Mn}=25;\mathrm{Fe}=26$ )
(A) ${\left[{\mathrm{FeF}}_6\right]}^{3-}$
(B) ${\left[\mathrm{Fe}(\mathrm{CN}{)}_6\right]}^{3-}$
(C) ${\left[{\mathrm{MnCl}}_6\right]}^{3-}$ (high spin)
(D) ${\left[\mathrm{Mn}(\mathrm{CN}{)}_6\right]}^{3-}$
 

(A) ${\left[{\mathrm{FeF}}_6\right]}^{3-}$
$$\begin{gathered} {\mathrm{Fe}}^{+3}\to 3{ \mathrm{d}}^{5}4{ \mathrm{s}}^0 \\ n= 5 \end{gathered}$$

(B) ${\left[\mathrm{Fe}(\mathrm{CN}{)}_6\right]}^{3-}$

$$\begin{gathered} {\mathrm{Fe}}^{+3}\to 3{ \mathrm{d}}^{5}4{ \mathrm{s}}^0 \\ n= 1 \end{gathered}$$

(C) ${\left[{\mathrm{MnCl}}_6\right]}^{3-}$ (high spin)

$$\begin{gathered} {\mathrm{Mn}}^{+3}\to 3{ \mathrm{d}}^{4}4{ \mathrm{s}}^0 \\ n = 4 \end{gathered}$$

(D) ${\left[\mathrm{Mn}(\mathrm{CN}{)}_6\right]}^{3-}$

$$\begin{gathered} {\mathrm{Mn}}^{+3}\to 3{ \mathrm{d}}^{4}4{ \mathrm{s}}^0 \\ n =2 \end{gathered}$$

As no. of unpaired electrons(n) increases, magnetic moments increases.

$\mu \Rightarrow A>C>D>B$