Chemistry · 2022
JEE Main · 25 July 2022, Shift 1 · Q70
The geometry around boron in the product ' B ' formed from the following reaction is BF_3+ NaH → 450 K A + NaF; A + NMe_3 → B
The geometry around boron in the product ' B ' formed from the following reaction is
$$\begin{gathered}
\mathrm{BF}_3+\mathrm{NaH} \xrightarrow{450 \mathrm{~K}} \mathrm{~A}+\mathrm{NaF} \\
\mathrm{~A}+\mathrm{NMe}_3 \rightarrow \mathrm{~B}
\end{gathered}
$$
Official answer
(2)
tetrahedral
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JEE Main 2022 Chemistry question, with the answer from NTA’s final answer key. Where our answers come from.