In option $B$, Boron atom has empty $2 p$ orbital after formation of $BF _3$, so it can accept lone pair of fluorine atom ($F$ atom has $3$ lone pairs) and form $BF^{-} _4$.

In option $C$, there is no coordinate bonding but very strong Hydrogen bonding due to high electronegativity of Fluorine. $(C)$ is the correct answer.

In option $D$, Nitrogen has one lone pair left after forming $3$ covalent bonds with hydrogen, it forms coordinate bond by sharing that lone pair with $H$ ion.