Question

Which of the following will not act as an oxidising agent?

• A. \( \text{CrO}_3 \)
• B. \( \text{MoO}_3 \)
• C. \( \text{CrO}_4^{2-} \)
• D. \( \text{Cr}_2\text{O}_7^{2-} \)

Hint:

Remember that the oxidizing power of Cr(VI) is highly pH-dependent. Dichromate is the active oxidizing species in acidic solution, whereas chromate is the stable, non-oxidizing species in basic solution. Always associate dichromate with acidic medium for redox reactions.

Answer 1

The Correct Option is C

Step 1: Understanding Oxidation States:
In all the given options, the central metal (Cr or Mo) is in its maximum oxidation state of \(+6\).
- In \( \text{CrO}_3 \), Cr is \(+6\).
- In \( \text{MoO}_3 \), Mo is \(+6\).
- In \( \text{CrO}_4^{2-} \), Cr is \(+6\).
- In \( \text{Cr}_2\text{O}_7^{2-} \), Cr is \(+6\).
Step 2: Analyzing the Effect of the Medium:
Chromium(VI) species exist in a pH-dependent equilibrium:
\[ 2\text{CrO}_4^{2-} + 2\text{H}^+ \rightleftharpoons \text{Cr}_2\text{O}_7^{2-} + \text{H}_2\text{O} \] - In acidic medium, the equilibrium shifts toward the dichromate ion (\( \text{Cr}_2\text{O}_7^{2-} \)). Dichromate is a very powerful oxidizing agent in acidic conditions (\( E^\circ \approx +1.33 \, \text{V} \)).
- In alkaline (basic) medium, the equilibrium shifts toward the chromate ion (\( \text{CrO}_4^{2-} \)). In this medium, the chromate ion is highly stable and its reduction potential is much lower (\( E^\circ \approx -0.13 \, \text{V} \)).
Step 3: Logical Justification:
Because the chromate ion (\( \text{CrO}_4^{2-} \)) is the stable form in basic solutions and its reduction potential is negative, it does not act as an oxidizing agent under those conditions. In contrast, dichromate and chromic anhydride (\( \text{CrO}_3 \)) are classic, strong oxidants used in synthetic organic and analytical chemistry. While \( \text{MoO}_3 \) is also a weak oxidant due to the stability of higher oxidation states in heavier transition metals, the specific choice of chromate as "non-oxidizing" is a common distinction made based on its stability in basic media.
Step 4: Final Answer:
Therefore, based on its stability and low reduction potential in its preferred medium, \( \text{CrO}_4^{2-} \) is the correct choice.