## 3.8 Coordinate 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 ($H_3O^+$) A coordinate covalent bond is formed when a water molecule ($H_2O$) donates one of its lone pairs to an electron-deficient hydrogen ion ($H^+$, a proton). $$ H_2O + H^+ \rightarrow [H_3O]^+ $$ 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 ($BCl_3$) and ammonia ($NH_3$) to form the compound $Cl_3BNH_3$. **ii.** Phosphine ($PH_3$) and a hydrogen ion ($H^+$) to form the ion $PH_4^+$. **Answer 1:** **i. Formation of $Cl_3BNH_3$:** In ammonia ($NH_3$), the nitrogen atom has one lone pair of electrons. In boron trichloride ($BCl_3$), 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:* $$ \text{Cl}_3\text{B} \; + \; :\text{NH}_3 \rightarrow \text{Cl}_3\text{B} \leftarrow :\text{NH}_3 $$ **ii. Formation of $PH_4^+$:** In phosphine ($PH_3$), 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:* $$ \text{H}_3\text{P} \; + \; \text{H}^+ \rightarrow [\text{H}_3\text{P} \rightarrow \text{H}]^+ $$ ### Question 2 Draw the displayed formulae of the products formed in Question 1. Show the coordinate bond by an arrow. **Answer 2:** **i. $Cl_3BNH_3$:** The arrow points from the electron donor (nitrogen) to the electron acceptor (boron). $$ \begin{matrix} & \text{Cl} & & & \text{H} \\ & | & & & | \\ \text{Cl} & - & \text{B} & \leftarrow & \text{N} & - & \text{H} \\ & | & & & | \\ & \text{Cl} & & & \text{H} \end{matrix} $$ **ii. $PH_4^+$:** The arrow points from the electron donor (phosphorus) to the electron acceptor (hydrogen). $$ \left[ \begin{matrix} & \text{H} & \\ & | & \\ \text{H} & - & \text{P} & \rightarrow & \text{H} \\ & | & \\ & \text{H} & \end{matrix} \right]^+ $$