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  • The geometry around boron in the product 'B' formed from the following reaction is <img alt="\begin{aligned} \mathrm{BF}_{3}+\mathrm{NaH} \stackrel{450 \mathrm{~K}}{\longrightarro

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

\begin{aligned} \mathrm{BF}_{3}+\mathrm{NaH} \stackrel{450 \mathrm{~K}}{\longrightarrow} \mathrm{A}+\mathrm{NaF} \\ \mathrm{A}+\mathrm{NMe}_{3} \rightarrow \mathrm{B} \end{aligned}

 

Option: 1

trigonal planar
 


Option: 2

tetrahedral
 


Option: 3

pyramidal
 


Option: 4

square planar


Answers (1)

best_answer

The Reaction will be

2 \mathrm{BF}_3+6 \mathrm{NaH} \longrightarrow \mathrm{B}_2 \mathrm{H}_6+6 \mathrm{NaF}

                                           \text{(A)}

\mathrm{B}_2 \mathrm{H}_6+2 \mathrm{NMe}_3 \rightarrow 2 \mathrm{BH}_3 \cdot \mathrm{NMe}_3

The geometry around Boron. In B is tetrahedral.

Hence, the correct answer is option (2).

Posted by

Kuldeep Maurya

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