JEE Advance - Chemistry (1980 - No. 18)
Compound A is a light green crystalline solid. It gives the following tests :
(i) It dissolves in dilute sulphuric acid. No gas is produced.
(ii) A drop of KMnO4 is added to the above solution. The pink colour disappears.
(iii) Compound A is heated strongly. Gases B and C, with pungent smell, come out. A brown residue D is left behind.
(iv) The gas mixture (B and C) is passed into a dichromate solution. The solution turns green.
(v) The green solution from step (iv) gives a white precipitate E with a solution of barium nitrate.
(vi) Residue D from step (iii) is heated on charcoal in a reducing flame. It gives a magnetic substance E.
Name the compounds A, B, C, D and E.
Explanation
Let's analyze the given information step-by-step to identify the compounds.
(i) Dissolution in dilute sulfuric acid without gas evolution: This suggests that compound A is likely a metal compound that forms a soluble sulfate. The lack of gas rules out carbonates and many other reactive compounds.
(ii) Decolorization of KMnO₄: Potassium permanganate ($KMnO_4$) is a strong oxidizing agent. Its decolorization indicates that compound A, or a species derived from it in the acidic solution, is readily oxidized. This points towards a compound containing a metal in a lower oxidation state, capable of being oxidized to a higher oxidation state.
(iii) Strong heating producing pungent gases (B and C) and brown residue (D): The pungent gases suggest the presence of sulfur and/or nitrogen. The brown residue is indicative of a metal oxide. The fact that gases are produced implies that compound A might be a metal salt containing an anion which decomposes upon heating.
(iv) Gases B and C turn a dichromate solution green: Dichromate ($Cr_2O_7^{2-}$) solutions are orange. Their reduction to a green solution indicates that gases B and C are reducing agents. The green color is characteristic of chromium(III) ions ($Cr^{3+}$). This strongly suggests that one of the gases (B or C) is $SO_2$, which is capable of reducing dichromate.
(v) The green solution gives a white precipitate (E) with barium nitrate: The formation of a white precipitate with barium nitrate is a classic test for sulfate ions ($SO_4^{2-}$). The reaction is:
$$Ba^{2+}(aq) \ + \ SO_4^{2-}(aq) \ \rightarrow \ BaSO_4(s)$$
Therefore, gas B or C must be sulfur dioxide ($SO_2$), which is oxidized to sulfate ions ($SO_4^{2-}$) upon reaction with the dichromate solution.
(vi) Residue D gives a magnetic substance (E) upon reduction: This strongly suggests that residue D is a metal oxide of a transition metal, and the magnetic substance E is the corresponding metal. The reduction reaction can be generally represented as:
$$Metal Oxide \ + \ Reducing \ Agent \ \rightarrow \ Metal$$
Considering all the observations, a plausible scenario is that compound A is ferrous sulfate heptahydrate ($FeSO_4 \cdot 7H_2O$). This is a light green crystalline solid. Let's check if it fits the observations:
(i) & (ii): Ferrous sulfate dissolves in dilute sulfuric acid and gets oxidized by $KMnO_4$.
(iii): Upon strong heating, ferrous sulfate decomposes. The gases produced include sulfur dioxide ($SO_2$ - gas B) and sulfur trioxide ($SO_3$ - gas C). Both have pungent smells. The brown residue (D) is ferric oxide ($Fe_2O_3$).
(iv): $SO_2$ reduces dichromate to $Cr^{3+}$.
(v): $SO_2$ is oxidized to $SO_4^{2-}$, which forms a white precipitate of barium sulfate ($BaSO_4$) with barium nitrate.
(vi): Reduction of $Fe_2O_3$ on charcoal produces elemental iron (Fe - E), which is magnetic.
Therefore, the compounds are:
A: Ferrous sulfate heptahydrate ($FeSO_4 \cdot 7H_2O$)
B: Sulfur dioxide ($SO_2$)
C: Sulfur trioxide ($SO_3$)
D: Ferric oxide ($Fe_2O_3$)
E: Iron (Fe) (or Barium Sulfate ($BaSO_4$), depending on which step (v) or (vi) is considered)
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