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JEE Main 2022 (Online) 25th July Morning Shift

MCQ+4 / -12022

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


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

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