Chapter 9: Problem 31
Consider the reaction $$ \mathrm{BF}_{3}+\mathrm{NH}_{3} \longrightarrow \mathrm{F}_{3} \mathrm{~B}-\mathrm{NH}_{3} $$
Short Answer
Step by step solution
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Electron Pair Donor
- Ammonia \( \mathrm{NH}_3 \) has a stable molecular structure with a lone pair of electrons.
- The ability of nitrogen to donate this pair differentiates ammonia as a Lewis base.
- Donating its electron pair allows ammonia to bond with electron-deficient atoms like boron in \( \mathrm{BF}_3 \).
Electron Pair Acceptor
The reason \( \mathrm{BF}_3 \) accepts electrons lies in the boron atom's electronic structure:
- Boron has three valence electrons and forms three covalent bonds with fluorine in \( \mathrm{BF}_3 \), leaving it with an incomplete octet.
- Due to its electron deficiency, boron eagerly accepts a lone pair of electrons from a donor like \( \mathrm{NH}_3 \).
Coordinate Covalent Bond
This process can be broken down into the following steps:
- The lone pair on nitrogen provides the two electrons that enter a shared electron environment with boron.
- The donating atom, nitrogen, effectively becomes part of the new molecule \( \mathrm{F}_3\mathrm{B}-\mathrm{NH}_3 \), without changing its oxidation state.
- This type of bond is significant because it stabilizes molecules and complexes with deficient atoms, leading to the formation of stable compounds.
Incomplete Octet
Here are key aspects of the incomplete octet:
- Boron, in \( \mathrm{BF}_3 \), possesses only six electrons through its bonds with fluorine, making it electron-deficient.
- An incomplete octet makes boron a perfect electron pair acceptor, as it seeks additional electrons to achieve stability.