c
In option \(A\), coordinate\(/\)dative bond is formed between lone pair of Oxygen atom and empty s orbital of \(H\) ion.

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.