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JEE Main202331 Jan 2023Morning ShiftChemistryp Block Elements (Group 15, 16, 17 & 18)Actual

When 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. AX = Cu 2 I 2 ,   Y = Na 2 S 4 O 5
  2. BX = Cu 2 I 2 ,   Y = Na 2 S 4 O 6
  3. CX = CuI 2 ,   Y = Na 2 S 4 O 3
  4. DX = CuI 2 ,   Y = Na 2 S 4 O 6

Correct answer

B. X = Cu 2 I 2 ,   Y = Na 2 S 4 O 6

Step-by-step solution

The reaction between Cu 2 + and KI will takes place as: Cu 2 + + 2 KI ⟶ CuI 2 ↓ Unstable + 2   K + I - is a strong reducing agent, it reduces Cu 2 + to Cu + and a white precipitate of Cu 2 I 2 is formed. 2 CuI 2 ⟶ Cu 2 I 2 ↓ ( White ) ' X ' + I 2 Sodium thiosulphate is used as reducing agent to titrate iodine, with iodide being the product of the reaction. 2 Na 2   S 2 O 3   +   I 2   →   2 NaI +   Na 2   S 4 O 6 ( Y ) So, the compound

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