Which of the following does not contain any co-ordinate bond?

H_{3} O^{+}

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B F_{4}^{-}

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H F_{2}^{-}

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N H_{4}^{+}

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Solution

The correct option is D $H F_{2}^{-}$
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 2p orbital after formation of BF3, so it can accept lone pair of fluorine atom (F atom has 3 lone pairs) and form BF4-
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.

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Similar questions

Which of the following conversions involve change in both shape and hybridisation?

a) CH4 -> C2H6

b) NH3 -> NH4+

c) BF3 -> BF4-

d) H2O ->H3O+

B F_{3} + F^{-} \to B F_{4}^{-}

H F + S b F_{5} \to H S b F_{6}

P C l_{3} + C l^{-} \to P C l_{4}^{-}

N H_{3} + H^{+} \to N H_{4}^{+}

H_{3} O + H^{+} \to H_{4} O^{+}

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