JEE Main 2022 · 25 Jul · Shift 1 · Chemistry Q70 · Dronacharya Exams

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ChemistrySingle correct+4 / -1Question 70 of 90

The geometry around boron in the product 'B' formed from the following reaction is

$\mathrm{BF_3+NaH\xrightarrow{450\ K}A+NaF}$
$\mathrm{A+NMe_3\rightarrow B}$

A

trigonal planar

B

tetrahedral

C

pyramidal

D

square planar

Answer and worked solution

Official answer: B

A is diborane; with $\mathrm{NMe_3}$ it gives $\mathrm{H_3B\cdot NMe_3}$ , tetrahedral boron.

(B)

Topics:Boranes VSEPR theory

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