X is a substance which liberates CO$_2$ on treatment with concentrated H$_2$SO$_4$. A warm solution of X can decolorize acidified KMnO$_4$. X is

To solve this problem, let's break down the chemical properties of the substance X based on the two clues provided:

1. Analysis of the Clues

• Clue 1: Liberates CO$_2$ on treatment with concentrated H$_2$SO$_4$. Many carbonates and oxalates react with concentrated sulfuric acid to release carbon dioxide (or a mixture of CO and CO$_2$).
• Clue 2: A warm solution of X decolorizes acidified KMnO$_4$. This is the "deciding" clue. Acidified potassium permanganate (KMnO$_4$) is a powerful oxidizing agent. For a substance to decolorize it, that substance must be a reducing agent. • HCOOH (Formic acid): Can be oxidized, but it is not the classic answer for this specific reaction profile in JAMB chemistry. • NaHCO$_3$ (Sodium bicarbonate): Does not act as a reducing agent; it cannot decolorize KMnO$_4$. • FeCO$_3$ (Iron(II) carbonate): While Fe²⁺ can be oxidized to Fe³⁺, the reaction with H$_2$SO$_4$ primarily focuses on the carbonate part, and it is not the standard laboratory reagent used to demonstrate this specific dual-property test. • H$_2$C$_2$O$_4$ (Oxalic acid): This is a classic reducing agent. When heated, it reacts with acidified KMnO$_4$ to produce CO$_2$ and water, effectively decolorizing the purple permanganate solution.

2. The Chemistry of Oxalic Acid (H$_2$C$_2$O$_4$)

• Reaction with H$_2$SO$_4$: When oxalic acid is treated with concentrated H$_2$SO$_4$, it undergoes dehydration to produce carbon monoxide (CO) and carbon dioxide (CO$_2$): $$H_2C_2O_4 \xrightarrow{conc. H_2SO_4} CO + CO_2 + H_2O$$
• Reaction with KMnO$_4$: Oxalic acid is a well-known reducing agent. In the presence of heat, it reduces the purple MnO$_4^-$ ion to the colorless Mn²⁺ ion: $$5H_2C_2O_4 + 2MnO_4^- + 6H^+ \rightarrow 10CO_2 + 2Mn^{2+} + 8H_2O$$

3. Conclusion

Comparing the options, H$_2$C$_2$O$_4$ (Oxalic acid) is the only substance that perfectly fits both criteria: it releases CO$_2$ upon dehydration with concentrated H$_2$SO$_4$ and acts as a strong enough reducing agent to decolorize acidified KMnO$_4$.

Exam Tip: Whenever you see "decolorizes acidified KMnO$_4$" in a chemistry question, immediately look for a reducing agent (like Fe²⁺, SO$_2$, H$_2$S, or Oxalic acid). If you see "liberates CO$_2$ with acid," look for a carbonate or an oxalate. Oxalic acid is the intersection of these two properties!