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AP EAMCET · Chemistry · p Block Elements (Group 13 & 14)

The Product(s) of the reaction
\(\mathrm{NaBH}_4+\mathrm{I}_2 \longrightarrow\) 'Products' is/are

  1. A \(\mathrm{B}_2 \mathrm{H}_4 \mathrm{I}_2+2 \mathrm{Nal}\)
  2. B \(\mathrm{B}_2 \mathrm{H}_6+\mathrm{NaH}+\mathrm{HI}\)
  3. C \(\mathrm{B}_2 \mathrm{H}_6+2 \mathrm{NaI}+\mathrm{H}_2\)
  4. D \(2 \mathrm{NaBH}_4 \mathrm{I}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(\mathrm{B}_2 \mathrm{H}_6+2 \mathrm{NaI}+\mathrm{H}_2\)

Step-by-step Solution

Detailed explanation

\(\mathrm{NaBH}_4\) 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…
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