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P Block Elements question

2024 · 31 Jan · Shift 2 · Q6
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P Block Elements question

2024 · 31 Jan · Shift 2 · Q6

JEE MainChemistryP Block ElementsMCQ+4 / −1
Given below are two statements: Statement I : S8\mathrm{S}_8S8​ solid undergoes disproportionation reaction under alkaline conditions to form S2−\mathrm{S}^{2-}S2− and S2O32−\mathrm{S}_2 \mathrm{O}_3{ }^{2-}S2​O3​2−. Statement II : ClO4−\mathrm{ClO}_4^{-}ClO4−​ can undergo disproportionation reaction under acidic condition. In the light of the above statements, choose the most appropriate answer from the options given below :
  1. A
    Statement I is correct but statement II is incorrect
  2. B
    Both statement I and statement II are incorrect
  3. C
    Statement I is incorrect but statement II is correct
  4. D
    Both statement I and statement II are correct
View written solutionFree

Correct answer: A

  1. Check Statement I

Elemental sulfur reacts with alkali on heating to give sulfide and thiosulfate. A standard reaction is: 4S+6OH−→2S2−+S2O32−+3H2O4S + 6OH^- \rightarrow 2S^{2-} + S_2O_3^{2-} + 3H_2O4S+6OH−→2S2−+S2​O32−​+3H2​O

Here sulfur in oxidation state 000 forms:

  • S2−S^{2-}S2− where oxidation state of sulfur is −2-2−2 (reduction)
  • S2O32−S_2O_3^{2-}S2​O32−​ which contains sulfur in mixed oxidation states (overall oxidation of sulfur derived from 000 to higher/lower states)

Thus the same element sulfur is simultaneously oxidized and reduced, so this is a disproportionation reaction.

Therefore, Statement I is correct.

  1. Check Statement II

In ClO4−ClO_4^-ClO4−​, chlorine is already in its highest oxidation state, i.e. x+4(−2)=−1⇒x=+7x + 4(-2) = -1 \Rightarrow x=+7x+4(−2)=−1⇒x=+7

Since chlorine is at its maximum oxidation state +7+7+7, it cannot be oxidized further. Disproportionation requires the same species to undergo both oxidation and reduction.

Hence ClO4−ClO_4^-ClO4−​ cannot undergo disproportionation under acidic condition.

Therefore, Statement II is incorrect.

  1. Conclusion
  • Statement I: Correct
  • Statement II: Incorrect

So the correct option is: A\boxed{A}A​

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