JEE Chemistry 2025 Question with Solution

Given: The complexes are $[Fe(NH_3)_6]^{3+}$, $[Fe(Cl)_6]^{3-}$, $[Fe(C_2O_4)_3]^{3-}$ and $[Fe(H_2O)_6]^{3+}$. We must identify the most stable complex ion, count the electrons in its $t_{2g}$ orbitals, and then determine the nature of vanadium oxide of the type $V_2O_x$.

Find: The correct option for the nature of the oxide.

From the given solution, the most stable complex is taken as $[Fe(C_2O_4)_3]^{3-}$ because oxalate is a bidentate ligand and stabilizes the iron(III) complex effectively.

For iron in the $+3$ oxidation state:

$$Fe^{3+} = [Ar] \, 3d^5$$

In the octahedral complex discussed in the solution, these five $d$-electrons are placed in the lower-energy $t_{2g}$ orbitals, giving:

$$t_{2g}^5$$

So, $X = 5$.

Now consider the oxide of vanadium of the type $V_2O_x$. The solution identifies this with vanadium oxide such as $V_2O_5$, which shows amphoteric behaviour.

Therefore, the correct option is D.

Given: We compare the complexes $[Fe(NH_3)_6]^{3+}$, $[Fe(Cl)_6]^{3-}$, $[Fe(C_2O_4)_3]^{3-}$ and $[Fe(H_2O)_6]^{3+}$.

Find: Which option correctly states the nature of vanadium oxide.

The solution states the following points:

1. $NH_3$ is treated as a strong field ligand.
2. $Cl^-$ and $H_2O$ are treated as weak field ligands.
3. Oxalate $C_2O_4^{2-}$ is treated as a strong field ligand and also as a bidentate ligand, which increases stability.
4. For all complexes, iron is in the $+3$ oxidation state with configuration:

$$[Ar] \, 3d^5$$

5. The most stable complex is identified as:

$$[Fe(C_2O_4)_3]^{3-}$$

6. The number of electrons in $t_{2g}$ is given as:

$$t_{2g}^5$$

Hence,

$$X = 5$$

For the second part, the solution says vanadium oxide of the form $V_2O_x$ is represented by oxides such as $V_2O_5$ and states that such an oxide is amphoteric.

So, the correct option is D (Amphoteric).