Lewis Acid–Base Reactions: Coordination Bond Formation
Let’s examine the reaction between ammonia ($ \text{NH}_3 $) and boron trifluoride ($ \text{BF}_3 $):
- The Players:
- $ \text{NH}_3 $: A Lewis base with a lone pair of electrons on nitrogen.
- $ \text{BF}_3 $: A Lewis acid because boron is electron-deficient (it has an incomplete octet).
- The Interaction:
- The lone pair on nitrogen in $ \text{NH}_3 $ is donated to the empty $ 2p_z $ orbital on boron in $ \text{BF}_3 $.
- This forms a coordination bond, resulting in $ \text{F}_3\text{B} \to \text{NH}_3 $, where the arrow indicates the direction of electron donation.
Representing the Reaction with Lewis Structures
Here’s how to represent this reaction step-by-step:
- Draw the Lewis structure for $ \text{BF}_3 $, showing boron with three single bonds to fluorine and an empty orbital.
- Draw the Lewis structure for $ \text{NH}_3 $, showing the lone pair on nitrogen.
- Use an arrow to show the lone pair from $ \text{NH}_3 $ moving toward the boron atom in $ \text{BF}_3 $.
- Draw the resulting structure, $ \text{F}_3\text{B} \to \text{NH}_3 $, with the arrow representing the coordination bond.
