1. Find oxidation state and $d$-electron count of the metal in each complex

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

  Let oxidation state of Fe be $x$. $x+6(-1)=-3 \Rightarrow x=+3$ So, $\mathrm{Fe}^{3+}: 3d^5$

• For $\left[\mathrm{CoF}_6\right]^{3-}$: $x+6(-1)=-3 \Rightarrow x=+3$ So, $\mathrm{Co}^{3+}: 3d^6$

• For $\left[\mathrm{MnBr}_4\right]^{2-}$: $x+4(-1)=-2 \Rightarrow x=+2$ So, $\mathrm{Mn}^{2+}: 3d^5$

• For $\left[\mathrm{Mn}(\mathrm{CN})_6\right]^{3-}$: $x+6(-1)=-3 \Rightarrow x=+3$ So, $\mathrm{Mn}^{3+}: 3d^4$

2. Determine strong/weak field and high-spin/low-spin nature

• $\mathrm{CN}^-$ is a strong field ligand
• $\mathrm{F}^-$ and $\mathrm{Br}^-$ are weak field ligands
• Tetrahedral complexes are generally high spin

So:

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

Octahedral, strong field, $d^5$ low spin: $t_{2g}^5 e_g^0$ Number of unpaired electrons $n=1$

(ii) $\left[\mathrm{CoF}_6\right]^{3-}$

Octahedral, weak field, $d^6$ high spin: $t_{2g}^4 e_g^2$ Number of unpaired electrons $n=4$

(iii) $\left[\mathrm{MnBr}_4\right]^{2-}$

Tetrahedral, weak field, $d^5$ high spin: Number of unpaired electrons $n=5$

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

Octahedral, strong field, $d^4$ low spin: $t_{2g}^4 e_g^0$ Number of unpaired electrons $n=2$

3. Use spin-only magnetic moment formula

$\mu = \sqrt{n(n+2)}\, \text{BM}$

Thus:

• For $n=1$: $\mu=\sqrt{3}$
• For $n=2$: $\mu=\sqrt{8}$
• For $n=4$: $\mu=\sqrt{24}$
• For $n=5$: $\mu=\sqrt{35}$

Hence increasing order is: