JEE Main · 2023 · Shift-IhardPB12-115

When Cu2+ ion is treated with KI, a white precipitate, X appears in solution. The solution is titrated with sodium…

p-Block Elements (Class 12) · Class 12 · JEE Main Previous Year Question

Question

When Cu2+\mathrm{Cu^{2+}} ion is treated with KI, a white precipitate, X appears in solution. The solution is titrated with sodium thiosulphate, the compound Y is formed. X and Y respectively are

Options
  1. a

    X=Cu2I2, Y=Na2S4O5\mathrm{X = Cu_2I_2,\ Y = Na_2S_4O_5}

  2. b

    X=Cu2I2, Y=Na2S4O6\mathrm{X = Cu_2I_2,\ Y = Na_2S_4O_6}

    ✓
  3. c

    X=CuI2, Y=Na2S4O3\mathrm{X = CuI_2,\ Y = Na_2S_4O_3}

  4. d

    X=CuI2, Y=Na2S4O6\mathrm{X = CuI_2,\ Y = Na_2S_4O_6}

Correct Answerb

X=Cu2I2, Y=Na2S4O6\mathrm{X = Cu_2I_2,\ Y = Na_2S_4O_6}

Detailed Solution

Strategy: Identify the products of the redox reaction between Copper(II) and Iodide, followed by iodometric titration.

Step 1: Formation of X When \ceCu2+\ce{Cu^{2+}} ions react with excess Iodide (\ceI−)(\ce{I^-}), Copper(II) is reduced to Copper(I) and Iodine is liberated: \ce2Cu2++4I−→Cu2I2(s)+I2\ce{2Cu^{2+} + 4I^- \to Cu2I2(s) + I2} X is \ceCu2I2\ce{Cu2I2} (a white precipitate).

Step 2: Formation of Y The liberated Iodine is titrated with sodium thiosulphate (\ceNa2S2O3)(\ce{Na2S2O3}): \ceI2+2Na2S2O3→Na2S4O6+2NaI\ce{I2 + 2Na2S2O3 \to Na2S4O6 + 2NaI} Y is \ceNa2S4O6\ce{Na2S4O6} (Sodium tetrathionate).

Answer: (Option) [b] \ceX=Cu2I2,Y=Na2S4O6\boxed{\text{Answer: (Option) [b] $\ce{X = Cu2I2, Y = Na2S4O6}$}}

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