Starting with the orbital diagram of a boron atom, describe the steps needed to construct hybrid orbitals appropriate to describe the bonding in BF3.
The Correct Answer and Explanation is :
To describe the bonding in boron trifluoride (BF₃) using hybrid orbitals, we follow these steps:
Step 1: Draw the Orbital Diagram for a Boron Atom
A boron atom (B, atomic number 5) has the electron configuration:
[
1s^2 2s^2 2p^1
]
Its valence shell consists of the 2s and 2p orbitals:
- 2s orbital is fully occupied.
- 2p orbitals contain one electron in 2p₁, with the other two 2p orbitals empty.
Step 2: Promote an Electron to Maximize Bonding
Boron needs to form three bonds with fluorine, but it has only one unpaired electron in the 2p orbital. To create more bonding sites:
- One 2s electron is promoted to an empty 2p orbital, resulting in the configuration:
[
2s^1 2p^2
]
Step 3: Hybridization to Form sp² Orbitals
- The one 2s and two 2p orbitals hybridize to form three sp² hybrid orbitals.
- These three sp² hybrid orbitals are equivalent in energy and arrange themselves 120° apart to minimize electron repulsion (trigonal planar geometry).
- The remaining 2p orbital remains unhybridized (it is empty in BF₃ but crucial for other boron compounds).
Step 4: Formation of σ Bonds
- Each sp² orbital overlaps with the 2p orbital of fluorine to form three σ bonds, resulting in a BF₃ molecule with trigonal planar geometry.
Conclusion
The sp² hybridization of boron explains BF₃’s 120° bond angles and planar structure. Since boron has an incomplete octet, BF₃ is an electron-deficient molecule, making it highly reactive.
Now, I will generate an image illustrating the orbital diagram and hybridization process.
Here is the educational diagram illustrating the orbital hybridization of boron in boron trifluoride (BF₃). It includes the atomic orbital diagram of boron, the electron promotion step, the formation of sp² hybrid orbitals, and the final trigonal planar structure of BF₃. Let me know if you need any modifications!
