1. Find oxidation state and d-electron count for each complex

We determine the metal oxidation state and then its $d^n$ configuration.

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

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

• B. $\left[\mathrm{FeF}_6\right]^{3-}$ $x+6(-1)=-3 \Rightarrow x=+3$ So, $\mathrm{Fe}^{3+}: 3d^5$.

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

• D. $\left[\mathrm{Cr}(\text{oxalate})_3\right]^{3-}$ Oxalate $= \mathrm{C_2O_4^{2-}}$. $x+3(-2)=-3 \Rightarrow x=+3$ So, $\mathrm{Cr}^{3+}: 3d^3$.

• E. $\left[\mathrm{Ni}(\mathrm{CO})_4\right]$ CO is neutral, so Ni oxidation state is $0$. Thus, $\mathrm{Ni}^0: 3d^8 4s^2$. In this complex, Ni attains effective $3d^{10}$ configuration.

2. Determine strong-field / weak-field nature and unpaired electrons

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

• $\mathrm{CN}^-$ is a strong-field ligand.
• Octahedral low-spin $d^5$: $t_{2g}^5 e_g^0$
• Number of unpaired electrons $=1$.

B. $\left[\mathrm{FeF}_6\right]^{3-}$

• $\mathrm{F}^-$ is a weak-field ligand.
• Octahedral high-spin $d^5$: $t_{2g}^3 e_g^2$
• Number of unpaired electrons $=5$.

C. $\left[\mathrm{CoF}_6\right]^{3-}$

• $\mathrm{F}^-$ is weak-field.
• Octahedral high-spin $d^6$: $t_{2g}^4 e_g^2$
• Number of unpaired electrons $=4$.

D. $\left[\mathrm{Cr}(\text{oxalate})_3\right]^{3-}$

• $\mathrm{Cr}^{3+}$ is $d^3$.
• In octahedral field: $t_{2g}^3 e_g^0$
• Number of unpaired electrons $=3$.

E. $\left[\mathrm{Ni}(\mathrm{CO})_4\right]$

• CO is a strong-field ligand.
• This is a tetrahedral-type carbonyl complex with all electrons paired.
• Number of unpaired electrons $=0$.

3. Arrange in increasing order

Thus,

• $E = 0$
• $A = 1$
• $D = 3$
• $C = 4$
• $B = 5$

So the correct order is: $E < A < D < C < B$

4. Match with options

This corresponds to Option D.

5. Comparison with stored answer

Stored correct answer: D

Our derived answer: D

Hence, they agree.