The geometry around boron in the product ‘B’ formed from the following reaction is
\(\mathrm{BF}_3+\mathrm{NaH} \xrightarrow{450 \mathrm{~K}} A+\mathrm{NaF}\)
A + NMe3 → B
1
Trigonal planar
2
Tetrahedral
3
Pyramidal
4
Square planar
5
Question Not Attempted