For ${\mathrm{BF}}_3$ molecule which of the following is true?

The center B in ${\mathrm{BF}}_3$ contains three bond pairs. It has a trigonal planner shape and is hence ${\mathrm{sp}}^2$ hybridized.

Because F can provide a lone pair of electrons to the unoccupied 2p orbitals of the central atom in ${\mathrm{BF}}_3$, there is back bonding between F atoms and the center (B).

Due to back bonding, the $\mathit{B}\mathbf{-}\mathit{F}$ bond length in ${\mathrm{BF}}_3$ is lower than what would be predicted. Back bonding results in a partial double bond character and uniform bond lengths.

Here, all the statements are correct. Hence, option D is correct.