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

In a tetrahedral field, the lower-energy set is $e$ and the higher-energy set is $t_2$. So we must write configurations as $e^x t_2^y$.

2. Complex A: $\mathrm{TiCl_4}$

• Oxidation state of Ti: $x+4(-1)=0 \Rightarrow x=+4$
• Ti has atomic number $22$: $\mathrm{Ti}: [Ar]3d^24s^2$
• $\mathrm{Ti^{4+}}$ is: $3d^0$

Hence tetrahedral configuration: $e^0 t_2^0$ So, $A \to \text{III}$

3. Complex B: $[\mathrm{FeO_4}]^{2-}$

• Let oxidation state of Fe be $x$: $x+4(-2)=-2 \Rightarrow x=+6$
• Fe atomic number $26$: $\mathrm{Fe}: [Ar]3d^64s^2$
• $\mathrm{Fe^{6+}}$ is: $3d^2$

In tetrahedral splitting, two electrons occupy the lower $e$ set: $e^2 t_2^0$ So, $B \to \text{I}$

4. Complex C: $[\mathrm{FeCl_4}]^{-}$

• Let oxidation state of Fe be $x$: $x+4(-1)=-1 \Rightarrow x=+3$
• $\mathrm{Fe^{3+}}$ is: $3d^5$

For tetrahedral complexes with weak ligand $\mathrm{Cl^-}$, it is high spin. Filling gives: $e^2 t_2^3$ So, $C \to \text{IV}$

5. Complex D: $[\mathrm{CoCl_4}]^{2-}$

• Let oxidation state of Co be $x$: $x+4(-1)=-2 \Rightarrow x=+2$
• Co atomic number $27$: $\mathrm{Co}: [Ar]3d^74s^2$
• $\mathrm{Co^{2+}}$ is: $3d^7$

Tetrahedral with weak ligand $\mathrm{Cl^-}$ gives high spin configuration: $e^4 t_2^3$ So, $D \to \text{II}$

6. Final matching

$A \to \text{III}, \quad B \to \text{I}, \quad C \to \text{IV}, \quad D \to \text{II}$

This corresponds to Option B.

7. Comparison with stored answer

Stored correct answer = B.

Our derived answer = B. So they agree.