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AP EAMCET202125 Aug 2021Evening ShiftChemistryp Block Elements (Group 13 & 14)Actual

The Product(s) of the reaction NaBH ₄+ I ₂ 'Products' is/are

Options

  1. AB ₂ H ₄ I ₂+2 Nal
  2. BB ₂ H ₆+ NaH + HI
  3. CB ₂ H ₆+2 NaI + H ₂
  4. D2 NaBH ₄ I

Correct answer

C. B ₂ H ₆+2 NaI + H ₂

Step-by-step solution

NaBH ₄ is the mild reducing agent. When sodium borohydride react with iodine, it will produced diboron, sodium iodide and hydrogen gas. It involves oxidation of sodium borohydride with iodine in diglyme gives diborane. This approach is common in industrial production of B ₂ H ₆ . Sodium borohydride 2 NaBH ₄ + Iodine I ₂ Diglyme Diborane B ₂ H ₆ + Sodium iodide 2 NaI + H ₂

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