Question

In aqueous solution, Cr\(^{2+}\) is stronger reducing agent than Fe\(^{2+}\). This is because

• A. Cr$^{2+}$ ion is more stable than Fe$^{2+}$
• B. Cr$^{3+}$ ion with $d^3$ configuration has favourable crystal field stabilization energy
• C. Cr$^{3+}$ has half-filled configuration and hence more stable
• D. Fe$^{3+}$ in aqueous solution is more stable than Cr$^{3+}$
• E. Fe$^{2+}$ ion with $d^6$ configuration has favourable crystal field stabilization energy

Hint:

Greater stability of oxidized state ⇒ stronger reducing agent.

Answer 1

The Correct Option is B

Concept: Reducing strength and stability of oxidation states
A strong reducing agent:
• Easily loses electrons
• Gets oxidized to a more stable state ---

Step 1: Compare reactions
\[ \text{Cr}^{2+} \rightarrow \text{Cr}^{3+} \] \[ \text{Fe}^{2+} \rightarrow \text{Fe}^{3+} \] ---

Step 2: Electronic configurations
Cr: \[ \text{Cr}^{2+} = 3d^4 \] \[ \text{Cr}^{3+} = 3d^3 \] Fe: \[ \text{Fe}^{2+} = 3d^6 \] \[ \text{Fe}^{3+} = 3d^5 \] ---

Step 3: Stability analysis

• Cr$^{3+}$ (d$^3$) → very stable due to CFSE
• Fe$^{3+}$ (d$^5$) → half-filled stable but CFSE less in comparison ---

Step 4: Key reasoning

• Cr$^{2+}$ easily oxidizes → forms very stable Cr$^{3+}$
• Hence strong reducing agent --- Final Answer: \[ \boxed{\text{(B)}} \]