Why is BF3 sp2 Hybridized and Not sp3? (Stepwise Explanation & Diagrams)
The Hybridization of BF3 reveals why boron trifluoride forms a flat, symmetric, and nonpolar molecule. Understanding orbital mixing, bond angles, and shape is crucial for cracking JEE Main Chemistry. This topic connects to molecular geometry, polarity, and hybridization rules that feature heavily in JEE problems.
BF3 (boron trifluoride) is a small inorganic molecule where boron sits at the center, surrounded symmetrically by three fluorine atoms. The bonding and properties of BF3 can be explained by examining its hybridisation, molecular geometry, and electron configuration in a JEE-centric way. Hybridisation is key to explaining not just the shape, but also why boron breaks the octet rule in this compound.
Molecular Geometry and Hybridization of BF3
Start with the boron atom: its ground state electronic configuration is 1s22s22p1. For bonding in Hybridization of BF3, boron promotes an electron and mixes its 2s and two 2p orbitals, forming three equivalent sp2 hybrid orbitals. Each of these overlaps with a p orbital from fluorine to create a sigma (σ) bond, leading to trigonal planar geometry. This geometric arrangement ensures maximum separation of electron pairs, minimizing repulsion as per VSEPR theory.
All B–F bonds in BF3 are identical and separated by 120°. No lone pair on boron means the electronic and molecular shapes are both trigonal planar. The formula for steric number, SN = number of sigma bonds + number of lone pairs, gives SN = 3 for BF3. This clearly matches sp2 hybridization. You can extend this logic quickly to related group 13 molecules for comparison in exams.
Bond Angles, Polarity and Lewis Structure of BF3
The Lewis structure of BF3 shows boron at the center, bonded to three fluorine atoms. Each fluorine atom completes its octet, but boron has only six electrons around it, violating the usual octet rule. This is a classic JEE trap—remember, boron in BF3 is stable with an incomplete octet due to extensive resonance and the lack of additional incoming electron density. The molecule remains neutral, with no formal charge on any atom.
Polarity-wise, although each B–F bond is highly polar (F is much more electronegative), the symmetric triangular arrangement causes the bond dipoles to cancel. As a result, BF3 is nonpolar, which often appears in MCQ statements. In contrast, NH3 (ammonia) or H2O are polar due to their shapes.
Hybridization of BF3 vs BF4-, NH3, CH4, and BH3
| Molecule | Central Atom | Steric Number (SN) | Hybridization | Shape | Bond Angle (°) |
|---|---|---|---|---|---|
| BF3 | Boron | 3 | sp2 | Trigonal planar | 120 |
| BF4- | Boron | 4 | sp3 | Tetrahedral | 109.5 |
| NH3 | Nitrogen | 4 | sp3 | Trigonal pyramidal | 107 |
| CH4 | Carbon | 4 | sp3 | Tetrahedral | 109.5 |
| BH3 | Boron | 3 | sp2 | Trigonal planar | 120 |
Comparing sp2 (as in BF3 and BH3) and sp3 (as in BF4-, NH3, CH4) hybridization helps resolve frequent JEE questions. Remember, adding a fourth group (e.g., forming BF4-) changes boron's hybridization to sp3 and the geometry to tetrahedral. If the central atom gets an extra lone pair, angles shrink (as seen in NH3).
Stepwise Method to Determine Hybridization of BF3
- Count the number of atoms bonded to the central atom (boron): 3 for BF3.
- Count number of lone pairs on boron: 0.
- Steric number = 3 + 0 = 3.
- Steric number of 3 means sp2 hybridization (s + 2p = 3 orbitals mixed).
- Therefore, geometry is trigonal planar with 120° bond angles.
Quick calculation using the steric number is a valuable shortcut for JEE MCQs. Apply this routine to rapidly identify hybridization in exam settings.
Exceptions and Resonance in Hybridization of BF3
A common misconception is that every main group atom must complete its octet. In BF3, boron is stable with only six valence electrons due to effective delocalization (resonance) of electron density from fluorine. Resonance structures distribute negative charge, making the molecule less reactive than expected for an electron-deficient species.
- Avoid assigning extra lone pairs to boron—doing so changes the hybridization incorrectly.
- Recognize that resonance does not alter the sp2 hybridization in this system.
Applications and JEE Problem Trends on BF3 Hybridization
The hybridization of BF3 frequently appears in theory-based and calculation MCQs, especially those comparing shapes and bond angles of group 13 or 14 compounds. It is also a common reference in questions about Lewis acid behaviour, electron-deficient molecules, and exceptions to the octet rule.
- BF3 is an important industrial Lewis acid, used to initiate polymerizations and in organic synthesis.
- Shape identification is essential for logic puzzles in hybridization, e.g., “Which of these is trigonal planar/scenario of nonpolar planar molecules?”
- Hybridization can be connected with isoelectronic species, quick comparisons (BF3 vs. BF4-), and resonance explanation traps.
Revision Table: Key Facts on Hybridization of BF3
| Property | Value/Description |
|---|---|
| Central atom | Boron (B) |
| Steric number (SN) | 3 (3 σ bonds, 0 lone pairs) |
| Type of hybridization | sp2 (trigonal planar) |
| Bond angle | 120° |
| Polarity | Nonpolar |
| Lewis structure summary | Boron bound to 3 F, 6 valence e- on B |
| Octet rule | Violated (only 6 e- on B, but stable) |
For more in-depth practice and explanations on group 13 elements or hybridization logic, explore Vedantu's other JEE Chemistry pages. Related topics such as Hybridization of BCl3 and Hybridization of CH4 will reinforce your conceptual clarity. See also detailed discussion on VSEPR and trends in Chemical Bonding and Molecular Structure. Use the revision notes and comparison charts to master hybridization differences and tackle JEE MCQs on first glance.
- Practice structural logic with Lewis Dot Structures and geometry tables for easy recall.
- Learn fast identification tricks at Vedantu Hybridization page and boost exam speed.
- Compare s, sp, sp2, and sp3 compounds at sp vs sp2 vs sp3 comparison for quick contrast.
- Study reaction examples and principle exceptions detailed at p-block Elements for broader context.
Solid command of the Hybridization of BF3 topic ensures you can decode shapes, polarities, and vibrational modes across main group molecules. Master these fundamentals with Vedantu for strong JEE Main Chemistry performance.
FAQs on Hybridization of BF3: Concept, Shape & Exam Guide
1. What is the hybridization of BF3?
BF3 exhibits sp2 hybridization. In this molecule, the boron atom uses one s and two p orbitals to create three sp2 hybrid orbitals, which overlap with three fluorine atoms to form sigma bonds. This hybridization leads to a trigonal planar geometry with a 120° bond angle. This concept is crucial for JEE Main and NEET exams.
2. Why is BF3 not sp3 hybridized?
BF3 is not sp3 hybridized because boron forms bonds with only three atoms and has no lone pairs in this molecule. The steric number for boron in BF₃ is 3 (three bonding pairs, zero lone pairs), so only three orbitals mix (one s and two p), resulting in sp2 hybridization. Sp3 hybridization requires a steric number of 4 as seen in BF4- or CH4.
3. What is the shape and bond angle of BF3?
BF3 has a trigonal planar molecular geometry with a bond angle of 120°. This flat triangular shape arises from sp2 hybridization of boron and equal repulsion among three electron domains around boron.
4. How does the hybridization of BF3 differ from that of BF4-?
BF3 is sp2 hybridized, while BF4- is sp3 hybridized. This is because BF3 has three bonding pairs and no lone pairs on boron (steric number 3), resulting in trigonal planar geometry, whereas BF4- has four bonding pairs (steric number 4), leading to a tetrahedral shape.
5. Why is BH3 also considered sp2 hybridized?
BH3 is also sp2 hybridized because boron forms three sigma bonds with hydrogen atoms, with no lone pairs left on boron. This gives a trigonal planar structure, similar to BF3, making BH3 sp2 hybridized as well.
6. Does BF3 violate the octet rule for boron? Why is it stable?
BF3 violates the octet rule because boron has only six valence electrons in this compound. However, BF3 is stable due to delocalization of electrons (resonance), the high electronegativity of fluorine, and the lack of energetically favorable reactions to complete boron's octet under standard conditions.
7. Are the B–F bonds in BF3 all equivalent? Why?
Yes, all B–F bonds in BF3 are equivalent. This equivalency arises from resonance delocalization, where the electron density is shared equally among all three boron-fluorine bonds, supporting the molecule’s trigonal planar structure.
8. Can resonance affect the hybridization or geometry of BF3?
Resonance in BF3 creates partial double bond character in the B–F bonds, but hybridization remains sp2 and geometry stays trigonal planar. The resonance only influences bond lengths and stability, not the basic hybridization model.
9. How do you determine the hybridization of BF3 step by step?
Hybridization of BF3 is determined by the steric number method:
- Count valence shell electron pairs (3 bonding pairs for B, 0 lone pairs)
- Steric number = 3
- A steric number of 3 means sp2 hybridization
- Resulting geometry: trigonal planar
10. Why is the bond angle in BF3 exactly 120 degrees?
The bond angle in BF3 is 120° because the three electron domains around boron repel equally, distributing themselves at maximum separation in a plane. This is characteristic of trigonal planar geometry caused by sp2 hybridization.
