JEE Main · 2022 · Shift-ImediumPB11-023

The geometry around boron in the product 'B' formed from the following reaction isBF3 + NaH 450\,K A + NaF A + NMe3 B

p-Block Elements (Class 11) · Class 11 · JEE Main Previous Year Question

Question

The geometry around boron in the product 'B' formed from the following reaction isBF3+NaH→450 KA+NaFA+NMe3→B\mathrm{BF_3 + NaH \xrightarrow{450\,K} A + NaF \qquad A + NMe_3 \rightarrow B}

Options
  1. a

    trigonal planar

  2. b

    tetrahedral

    ✓
  3. c

    pyramidal

  4. d

    square planar

Correct Answerb

tetrahedral

Detailed Solution

Strategy: Follow the reaction sequence to identify compound B and then determine the geometry around Boron in B.

Step 1: Identifying Compound A\ceBF3+NaH−>[ext450K]A+NaF\ceNaH\ce{BF3 + NaH ->[ ext{450K}] A + NaF}\ce{NaH}provides\ceH−\ce{H^-}, which reduces\ceBF3\ce{BF3}to give boron hydrides. The product formed is diborane (A=\ceB2H6A = \ce{B2H6}).

Step 2: Identifying Compound B and its Geometry\ceA(B2H6)+NMe3−>B\ceNMe3\ce{A (B2H6) + NMe3 -> B}\ce{NMe3}(trimethylamine) is a Lewis base that donates its lone pair to the Lewis acid\ceBH3\ce{BH3}(a fragment of diborane):\ceB2H6+2NMe3−>2[BH3⋅NMe3]\ce{B2H6 + 2NMe3 -> 2[BH3 \cdot NMe3]}Product B is\ceH3B⋅N(CH3)3\ce{H3B \cdot N(CH3)3}. In this adduct, B forms 4 bonds (3 B–H + 1 B←\leftarrowN dative) — it issp3sp^3hybridised with tetrahedral geometry.Answer: (B) Tetrahedral\boxed{\text{Answer: (B) Tetrahedral}}

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