Question

In the following reaction, identify X: $Cr_{2}O_{7}^{2-} + X \xrightarrow{H^{+}} Cr^{3+} + H_{2}O + \text{Oxidized product of X}$}

• A. $C_{2}O_{4}^{2-}$
• B. $SO_{4}^{2-}$
• C. $S^{2-}$
• D. $Fe^{2+}$

Hint:

Dichromate ion oxidizes $Fe^{2+}$ (green) to $Fe^{3+}$ (yellow) in acidic medium.

Answer 1

The Correct Option is D

Step 1: Concept
$Cr_{2}O_{7}^{2-}$ (Dichromate ion) is a strong oxidizing agent in acidic medium.
Step 2: Analysis
The reaction shows $Cr_{2}O_{7}^{2-}$ being reduced to $Cr^{3+}$. This must be accompanied by the oxidation of species X.
Step 3: Evaluation
$SO_4^{2-}$ is already in its highest oxidation state. $Fe^{2+}$ can be easily oxidized to $Fe^{3+}$. While $C_2O_4^{2-}$ and $S^{2-}$ can also be oxidized, in standard textbook reactions following this specific skeletal format in NTA papers, $X$ usually represents the Ferrous ($Fe^{2+}$) ion.
Step 4: Conclusion
Hence, X is $Fe^{2+}$.
Final Answer:(D)