Question

‘X’ is the number of electrons in $ t_2g $ orbitals of the most stable complex ion among $ [Fe(NH_3)_6]^{3+} $, $ [Fe(Cl)_6]^{3-} $, $ [Fe(C_2O_4)_3]^{3-} $ and $ [Fe(H_2O)_6]^{3+} $. The nature of oxide of vanadium of the type $ V_2O_x $ is: 
 

• A. Acidic
• B. Neutral
• C. Basic
• D. Amphoteric

Hint:

The stability of a complex ion depends on the ligand field and the metal ion's charge. The presence of strong ligands, like oxalate, increases the stability of the complex.

Answer 1

The Correct Option is D

To solve this question, we need to analyze the coordination complexes provided and determine the number of electrons in the \( t_{2g} \) orbitals of the most stable complex ion.

Let's examine each of the given complexes:

1. \([Fe(NH_3)_6]^{3+}\):
   • Ligand: Ammonia (\( NH_3 \)) is a strong field ligand.
   • Iron's oxidation state: \( +3 \) (as the complex ion has a \( 3+ \) charge).
   • Electronic configuration of \( Fe^{3+} \): \( [Ar] \, 3d^5 \).
   • In a strong field ligand environment, pairing occurs due to splitting of the d-orbitals (\( t_{2g} \) and \( e_g \)).
   • The \( t_{2g} \) orbitals will be fully filled before \( e_g \), adding 4 electrons to the \( t_{2g} \):
2. \([Fe(Cl)_6]^{3-}\):
   • Ligand: Chloride (\( Cl^- \)) is a weak field ligand.
   • Iron's oxidation state: \( +3 \).
   • Electronic configuration of \( Fe^{3+} \): \( [Ar] \, 3d^5 \).
   • No pairing occurs in weak field ligands, resulting in:
3. \([Fe(C_2O_4)_3]^{3-}\):
   • Ligand: Oxalate (\( C_2O_4^{2-} \)) is a strong field ligand.
   • Iron's oxidation state: \( +3 \).
   • Electronic configuration of \( Fe^{3+} \): \( [Ar] \, 3d^5 \).
   • In a strong field ligand, pairing occurs:
4. \([Fe(H_2O)_6]^{3+}\):
   • Ligand: Water (\( H_2O \)) is a weak field ligand.
   • Iron's oxidation state: \( +3 \).
   • Electronic configuration of \( Fe^{3+} \): \( [Ar] \, 3d^5 \).
   • No pairing occurs:

From the analysis, the most stable complex in terms of maximum electrons in the \( t_{2g} \) is \([Fe(C_2O_4)_3]^{3-}\) with a configuration of \( t_{2g}^5 \).

Now, let's analyze the second part about the nature of the vanadium oxide \( V_2O_x \).

Vanadium can have multiple oxidation states, resulting in different oxides:

• If \( x=5 \) in \( V_2O_5 \), it is known to be amphoteric, meaning it can act both as an acid and a base.

The correct nature of oxide in this case is Amphoteric.

Answer 2

1. Identifying the most stable complex ion: The stability of the complex depends on factors like ligand field strength and the charge on the metal ion. Among the given complexes, the most stable complex is \( [Fe(C_2O_4)_3]^{3-} \) because oxalate (C\(_2\)O\(_4\)) is a bidentate ligand and provides a strong ligand field, stabilizing the iron(III) ion effectively.
2. Electron Configuration of Iron in \( [Fe(C_2O_4)_3]^{3-} \):
- Iron in the \( [Fe(C_2O_4)_3]^{3-} \) complex is in the \( +3 \) oxidation state, so its electron configuration is \( [Ar] 3d^5 \). This means it has 5 electrons in its \( 3d \)-orbitals. - For the octahedral \( [Fe(C_2O_4)_3]^{3-} \) complex, these 5 \( d \)-electrons will occupy the \( t_{2g} \) orbitals, as these orbitals are lower in energy in an octahedral field.
Thus, the number of electrons in the \( t_{2g} \) orbitals is \( \mathbf{5} \), and X = 5.
3. The nature of \( V_2O_x \):
- Vanadium oxides, such as \( V_2O_5 \), exhibit amphoteric properties. This means they can act as both acids and bases depending on the reaction conditions.
Therefore, the correct answer for the nature of vanadium oxide is amphoteric.
Thus, the correct answer is (4).