In this question the methyl group CH3 is described as Me.Trimethylamine Me3N reacts with boron triflouride BF3 to form compound of Me3NBF3?

In Me₃N, the central atom N has 5 valence electrons, forming 3 N-H bonds and 1 lone pair of electrons (sp³ hybridization). According to VSEPR theory, the 4 electron pairs are arranged tetrahedrally in order to minimize the repulsion between the electron pairs, and thus the molecular shape is trigonal pyramidal. For the shape of Me₃N, all of the 4 options are collect.

In BF₃, the central atom B has 3 valence electrons, forming 3 B-F bonds (sp² hybridization). According to VSEPR theory, the 3 bond pairs are arranged as planar trigonally in order to minimize the repulsion between the electron pairs. This rules out options B and D, in which the 3 B-F bonds in BF₃ are NOT trigonal planar.

In the product Me₃NBF₃, the lone pair electrons on N form a N→F dative bond. Now, the B atom are attached with 3 B-F bonds and one N→F bonds. According to VSEPR theory, the 3 electron pairs are arranged tetrahedrally in order to minimize the repulsion between the electron pairs, and thus the molecular shape with respect to central atom B is tetrahedral. This rules out option A.

Conclusion: Option B and D are ruled out according to the molecular shape of BF₃ while option A is ruled out according to the molecular shape of Me₃NBF₃.

Hence, the answer is C.

The N has a non-bonding electron pair. The B has an empty orbital.

The N contributes both electrons for a covalent bond between N and B.

It is most commonly called a coordinate covalent bond. Another term is a dative bond.

This type of reaction can be understood in Lewis acid base terms, with the N being the electron-rich base and the electron-deficient B being the acid.