3.8 Co-ordinate Covalent Bond

A coordinate covalent bond (also called a dative covalent bond) is a special type of covalent bond in which one atom provides both electrons for the shared pair.

For a dative covalent bond to form, two conditions must be met:

• One atom must have a lone pair of electrons to donate (Donor).
• The second atom must have an unfilled (empty) orbital to accept the lone pair, making it an electron-deficient atom or molecule (Acceptor).

Examples of Coordinate Covalent Bonds

1. Hydronium Ion ($H_3{O}^{+}$)

A coordinate covalent bond is formed when a water molecule ($H_{2}O$) donates one of its lone pairs to an electron-deficient hydrogen ion ($H^{+}$, a proton).

$H_{2}O+H^{+}\to [H_{3}O{]}^{+}$

The oxygen atom in water provides both electrons to form a new O-H bond, resulting in the hydronium ion.

2. Ozone ($O_3$)

In the ozone molecule, the central oxygen atom forms a double bond with one oxygen atom and a coordinate covalent bond with the other. The central oxygen donates one of its lone pairs to the third oxygen atom.

3. Carbon Monoxide ($CO$)

In a carbon monoxide molecule, after forming a double bond, the oxygen atom donates a lone pair to the carbon atom to help it achieve a stable octet. This forms a coordinate covalent bond.

Concept Assessment Exercise 3.3

Question 1

Draw dot-and-cross diagrams to show the formation of a coordinate bond between:

i. Boron trichloride ($BC{l}_3$) and ammonia ($N{H}_3$) to form the compound $C{l}_{3}BN{H}_3$.

ii. Phosphine ($P{H}_3$) and a hydrogen ion ($H^{+}$) to form the ion $P{H}_4^{+}$.

Answer 1:

i. Formation of $C{l}_{3}BN{H}_3$:

In ammonia ($N{H}_3$), the nitrogen atom has one lone pair of electrons. In boron trichloride ($BC{l}_3$), the boron atom has an empty p-orbital and is electron-deficient (only 6 valence electrons). Nitrogen donates its lone pair to boron, forming a coordinate covalent bond.

Dot-and-cross representation:

${\text{Cl}}_3\text{B}+:{\text{NH}}_3\to {\text{Cl}}_3\text{B}\leftarrow :{\text{NH}}_3$

ii. Formation of $P{H}_4^{+}$:

In phosphine ($P{H}_3$), the phosphorus atom has one lone pair of electrons. The hydrogen ion ($H^{+}$) has an empty 1s orbital. Phosphorus donates its lone pair to the hydrogen ion to form the phosphonium ion.

Dot-and-cross representation:

${\text{H}}_3\text{P}+{\text{H}}^{+}\to [{\text{H}}_3\text{P}\to \text{H}{]}^{+}$

Question 2

Draw the displayed formulae of the products formed in Question 1. Show the coordinate bond by an arrow.

Answer 2:

i. $C{l}_{3}BN{H}_3$: The arrow points from the electron donor (nitrogen) to the electron acceptor (boron).

$\begin{array}{ccccccc}& \text{Cl}& & & \text{H}& & \\ & |& & & |& & \\ \text{Cl}& -& \text{B}& \leftarrow & \text{N}& -& \text{H}\\ & |& & & |& & \\ & \text{Cl}& & & \text{H}& & \end{array}$

ii. $P{H}_4^{+}$: The arrow points from the electron donor (phosphorus) to the electron acceptor (hydrogen).

${\left[\begin{array}{ccccc}& \text{H}& & & \\ & |& & & \\ \text{H}& -& \text{P}& \to & \text{H}\\ & |& & & \\ & \text{H}& & & \end{array}\right]}^{+}$