BCl₃ and NH₄ Cl were heated at 140° C to give compound X, which when treated with NaBH₄ gave another compound… · ABC26IN0465
Subject: Inorganic Chemistry · Chapter: Main Group Elements · Topic: Main Group Element – General · Exam: GATE 2005-2021 · Marks: 2 · Difficulty: Medium
$\mathrm{BCl}_{3}$ and $\mathrm{NH}_{4} \mathrm{Cl}$ were heated at $140^{\circ} \mathrm{C}$ to give compound X , which when treated with $\mathrm{NaBH}_{4}$ gave another compound Y . Compounds X and Y are
(a)$\mathrm{X}=\mathrm{B}_{3} \mathrm{N}_{3} \mathrm{H}_{3} \mathrm{Cl}_{3}$ and $\mathrm{Y}=\mathrm{B}_{3} \mathrm{N}_{3} \mathrm{H}_{6}$
(b)$\mathrm{X}=\mathrm{B}_{3} \mathrm{N}_{3} \mathrm{H}_{9} \mathrm{Cl}_{3}$ and $\mathrm{Y}=\mathrm{B}_{3} \mathrm{N}_{3} \mathrm{H}_{6}$
(c)$\mathrm{X}=\mathrm{B}_{3} \mathrm{N}_{3} \mathrm{H}_{3} \mathrm{Cl}_{3}$ and $\mathrm{Y}=\mathrm{B}_{3} \mathrm{N}_{3} \mathrm{H}_{12}$
(d)$\mathrm{X}=\mathrm{B}_{3} \mathrm{N}_{3} \mathrm{Cl}_{6}$ and $\mathrm{Y}=\mathrm{B}_{3} \mathrm{N}_{3} \mathrm{H}_{6}$
Answer
Answer: A ✓ checked by 4AB · confidence high
Explanation
3BCl₃ + 3NH₄Cl → B₃N₃H₃Cl₃ (X); NaBH₄ reduces it to borazine B₃N₃H₆ (Y) (a).