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

For all complexes, let oxidation state of metal be $x$.

Since each complex has charge $-3$:

• For $[Fe(CN)_6]^{3-}$:
  $x+6(-1)=-3 \Rightarrow x=+3$ So $Fe^{3+}$ is $3d^5$.

• For $[FeF_6]^{3-}$:
  $x+6(-1)=-3 \Rightarrow x=+3$ So $Fe^{3+}$ is again $3d^5$.

• For $[CoF_6]^{3-}$:
  $x+6(-1)=-3 \Rightarrow x=+3$ So $Co^{3+}$ is $3d^6$.

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

2. Use ligand strength to decide high spin / low spin

• $CN^-$ is a strong-field ligand  low spin.
• $F^-$ is a weak-field ligand  high spin.

All are octahedral complexes.

3. Determine unpaired electrons in each complex

(i) $[Fe(CN)_6]^{3-}$

• Metal ion: $Fe^{3+} = d^5$
• Ligand: $CN^-$ strong field  low spin octahedral

Configuration: $t_{2g}^5 e_g^0$

In $t_{2g}^5$, there is 1 unpaired electron.

(ii) $[FeF_6]^{3-}$

• Metal ion: $Fe^{3+} = d^5$
• Ligand: $F^-$ weak field  high spin octahedral

Configuration: $t_{2g}^3 e_g^2$

This has 5 unpaired electrons.

(iii) $[CoF_6]^{3-}$

• Metal ion: $Co^{3+} = d^6$
• Ligand: $F^-$ weak field  high spin octahedral

Configuration: $t_{2g}^4 e_g^2$

Number of unpaired electrons = 4.

(iv) $[Mn(CN)_6]^{3-}$

• Metal ion: $Mn^{3+} = d^4$
• Ligand: $CN^-$ strong field  low spin octahedral

Configuration: $t_{2g}^4 e_g^0$

In $t_{2g}^4$, number of unpaired electrons = 2.

4. Final sequence

Thus, the numbers of unpaired electrons are: $1, 5, 4, 2$

This matches Option A.