Question

Which of the following orders is correct for the property given?

• A. \( Cr < Mn < Fe \) - standard electrode potential value for \( M^{3+}/M^{2+} \)
• B. \( Cr^{2+} < Mn^{2+} < Fe^{2+} \) - magnetic moments
• C. \( VO_2^+ < Cr_2O_7^{2-} < MnO_4^- \) - oxidizing power
• D. \( Ti < V < Cr \) - first ionization enthalpy

Hint:

Oxidizing power increases with the oxidation state of the element. Permanganate (\( MnO_4^- \)) is the strongest oxidizing agent among transition metal oxides due to Mn in the \( +7 \) oxidation state.

Answer 1

The Correct Option is C

Step 1: Understanding Oxidizing Power in Transition Metal Oxides 
- The oxidizing power of a species depends on its ability to accept electrons and undergo reduction.
- Higher oxidation states and higher standard reduction potential (\( E^\circ \)) indicate a stronger oxidizing agent.
Step 2: Analyzing the Given Oxidizing Power Order 
1. \( VO_2^+ \) (Vanadyl Ion)
- Vanadium in \( VO_2^+ \) is in the \( +5 \) oxidation state.
- It has a relatively lower oxidizing power compared to chromate and permanganate.
2. \( Cr_2O_7^{2-} \) (Dichromate Ion)
- Chromium in \( Cr_2O_7^{2-} \) is in the \( +6 \) oxidation state.
- It is a stronger oxidizing agent than \( VO_2^+ \).
3. \( MnO_4^- \) (Permanganate Ion)
- Manganese in \( MnO_4^- \) is in the \( +7 \) oxidation state.
- It has the highest oxidizing power among the three.
Thus, the correct order of oxidizing power is: \[ VO_2^+<Cr_2O_7^{2-}<MnO_4^- \] 

Step 3: Evaluating the Given Options 
- Option (1): Incorrect, as the standard electrode potential trend for \( M^{3+}/M^{2+} \) does not follow this order. - Option (2): Incorrect, as magnetic moments depend on unpaired electrons, and Mn\(^{2+}\) has more than Fe\(^{2+}\). - Option (3): Correct, as the given order for oxidizing power matches the correct trend. - Option (4): Incorrect, as the first ionization enthalpy does not follow the given order. Thus, the correct answer is 

Option (3).