Question

Which of the following sets includes all the species that will change the orange colour of \(K_2Cr_2O_7\) in acidic medium?

• A. \(Fe^{2+},\ Sn^{2+},\ I^-,\ S^{2-}\)
• B. \(S^{2-},\ Fe^{3+},\ I^-,\ C_2O_4^{2-}\)
• C. \(Fe^{2+},\ NO_2^-,\ SO_2,\ Sn^{4+}\)
• D. \(Fe^{3+},\ SO_4^{2-},\ S^{2-},\ Sn^{4+}\)

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Potassium dichromate (\(K_2Cr_2O_7\)) is a powerful oxidizing agent in acidic medium (\(Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O\)).
The orange color of the dichromate ion changes to green when it is reduced to \(Cr^{3+}\) by a reducing agent.
Therefore, any species in the set must be a reducing agent (capable of being oxidized to a higher oxidation state).
Step 2: Detailed Explanation:
We evaluate each species to see if it can be oxidized further:
- \(Fe^{2+}\): Can be oxidized to \(Fe^{3+}\). (Reducing Agent)
- \(Sn^{2+}\): Can be oxidized to \(Sn^{4+}\). (Reducing Agent)
- \(I^-\): Can be oxidized to \(I_2\). (Reducing Agent)
- \(S^{2-}\): Can be oxidized to \(S\). (Reducing Agent)
- \(NO_2^-\): Can be oxidized to \(NO_3^-\). (Reducing Agent)
- \(SO_2\): Can be oxidized to \(SO_4^{2-}\). (Reducing Agent)
- \(C_2O_4^{2-}\): Can be oxidized to \(CO_2\). (Reducing Agent)

Now, check for species that cannot be oxidized further (oxidizing agents or stable states):
- \(Fe^{3+}\): Already in its highest common oxidation state; cannot be oxidized by \(Cr_2O_7^{2-}\).
- \(Sn^{4+}\): Already in its highest oxidation state (+4).
- \(SO_4^{2-}\): Sulfur is in +6; cannot be oxidized further.

Analysis of Options:
- (A): All species (\(Fe^{2+}, Sn^{2+}, I^-, S^{2-}\)) are reducing agents. This set will change the color.
- (B): Contains \(Fe^{3+}\), which is not a reducing agent for dichromate.
- (C): Contains \(Sn^{4+}\), which is already in its maximum oxidation state.
- (D): Contains \(Fe^{3+}, SO_4^{2-}, \text{ and } Sn^{4+}\).
Step 3: Final Answer:
The correct set is \(Fe^{2+}, Sn^{2+}, I^-, S^{2-}\).