The geometry around boron in the product ' B ' formed from the following reaction is
$$
\begin{aligned}
& \mathrm{BF}_3+\mathrm{NaH} \xrightarrow{450 \mathrm{~K}} \mathrm{~A}+\mathrm{NaF} \\
& \mathrm{~A}+\mathrm{NMe}_3 \rightarrow \mathrm{~B}
\end{aligned}
$$
