The geometry around boron in the product ‘B’ formed from the following reaction is\( \displaystyle \text{BF}_3 + \text{NaH} \;\dfrac{450\text{K}}{\text{A} + \text{NMe}_3 \rightarrow \text{B}}\; \text{A} + \text{NaF}\)
Trigonal planar
Tetrahadral
Pyramidal
Square planar