Question

In which of the following, ions are correctly arranged in the increasing order of oxidizing power?

• A. \(\text{Cr}_2\text{O}_7^{2-}<\text{MnO}_4^{-}<\text{VO}_2^{+}\)
• B. \(\text{VO}_2^{+}<\text{Cr}_2\text{O}_7^{2-}<\text{MnO}_4^{-}\)
• C. \(\text{VO}_2^{+}<\text{MnO}_4^{-}<\text{Cr}_2\text{O}_7^{2-}\)
• D. \(\text{MnO}_4^{2-}<\text{Cr}_2\text{O}_7^{2-}<\text{VO}_2^{+}\)

Hint:

\( \text{MnO}_4^{-} \) is a strong oxidizing agent due to Mn in the +7 oxidation state, whereas \( \text{VO}_2^{+} \) has the lowest oxidizing power among the three.

Answer 1

The Correct Option is B

Step 1: Understanding Oxidizing Power - The oxidizing power of an ion is related to its ability to accept electrons. A stronger oxidizer has a higher oxidation state and a greater tendency to gain electrons. 

Step 2: Order of Oxidizing Power - \(\text{MnO}_4^{-}\) has the highest oxidizing power, followed by \(\text{Cr}_2\text{O}_7^{2-}\), and \(\text{VO}_2^{+}\) has the weakest oxidizing power among these three. Hence, the correct order is \(\text{VO}_2^{+}<\text{Cr}_2\text{O}_7^{2-}<\text{MnO}_4^{-}\).