Question

Which of the following does not contain any co-ordinate bond?
A. \[{{\mathbf{H}}_3}{{\mathbf{O}}^ + }\]
B. \[{\mathbf{B}}{{\mathbf{F}}_4}^ - \]
C. \[{\mathbf{H}}{{\mathbf{F}}_2}^ - \]
D. \[{\mathbf{N}}{{\mathbf{H}}_4}^ + \]

Answer 1

Hint: To find out an accurate answer to this question we should first consider understanding the co-ordinate bond. Now, coordinate bond is also known as dative bond, it is a kind of covalent bond, the only difference is that sharing of electrons is done by only one atom.

Complete answer:
Now, to identify coordinate bonds there are a few important points we need to consider. The first point is there should be a lone pair available on one atom and an empty orbital on the other one. Now in coordinate bonding the electron shares are only given by one atom, a lone pair from one atom is shared between two, to form a bond which can be shown by an arrow $ ( \to ) $ , the arrow is put in the direction of donor to receiver.
Now by drawing the lewis structure of all four options we can find out which does not consist of a coordinate bond.
  \[{{\mathbf{H}}_3}{{\mathbf{O}}^ + }\] : shares a lone pair with empty 1s orbital of hydrogen to form H3O+
  \[{\mathbf{B}}{{\mathbf{F}}_4}^ - \] : Fluorine shares a lone pair electron with Boron's empty 2p orbital to form a coordinate bond.
 \[{\mathbf{H}}{{\mathbf{F}}_2}^ - \] : Now there is a hydrogen bonding between H and F. So there is no formation of coordinate bond.
 \[{\mathbf{N}}{{\mathbf{H}}_4}^ + \] : coordinate bond is formed between lone pairs of nitrogen to empty 1s orbital of hydrogen.

The correct answer is option C .

Note:
We need to draw the chemical structure of a compound if we wish to find out the
coordinate bond. We try to stabilise the structure by completing all atoms octet or duplet. Now HF cannot form a coordinate bond because it makes hydrogen bonding.