What is the Hybridization and Geometry of XeF4?
Understanding the Hybridization of XeF4 is essential for JEE Main Chemistry, as it reveals how xenon forms chemical bonds despite being a noble gas. Xenon tetrafluoride (XeF4) showcases expanded octet bonding, resulting in a unique square planar structure and specific bond angles. Recognising its hybridization type, electron arrangement, and the role of lone pairs enables students to predict molecular shapes and approach application-based problems confidently.
What is XeF4 and Why is its Hybridization Unique?
XeF4—xenon tetrafluoride—is a stable noble gas compound where a central Xe atom is bonded to four fluorine atoms. Unlike most main-group elements, xenon can expand its octet due to accessible d orbitals, forming stable molecules beyond the traditional eight electrons. Because of this, understanding XeF4 aids in mastering core topics such as expanded octet, square planar hybridization chemistry and predicting geometries of hypervalent molecules.
Electronic Structure and sp3d2 Hybridization in XeF4
The electronic configuration of xenon is [Kr] 4d105s25p6. In XeF4, xenon promotes electrons from the 5p to empty 5d orbitals, allowing bonding with four F atoms via orbital mixing. For predicting hybridization, count the number of σ-bonds (Xe–F) and lone pairs on Xe:
- Xenon provides four valence electrons paired with four from fluorine atoms for bonding.
- Two lone pairs remain on the xenon atom after bonding with four fluorine atoms.
- Total: 4 σ-bonds + 2 lone pairs = 6 electron domains.
According to the formula, number of electron domains = type of hybridization. With 6 domains, the hybridization is sp3d2 (octahedral orbital set). This involves one s, three p, and two d orbitals, forming six equivalent sp3d2 hybrid orbitals. The two lone pairs occupy axial positions, while the bonding pairs remain in a plane—resulting in the observed geometry.
Stepwise Lewis Structure and VSEPR Analysis
Draw the Lewis structure for XeF4:
- Write Xe at the center, surrounded by four F atoms.
- Distribute eight valence electrons from Xe; assign one single bond for each Xe–F.
- There are two remaining electron pairs (lone pairs) on Xe after bond formation.
- Apply VSEPR theory: Six electron pairs (4 bond pairs + 2 lone pairs) create an octahedral arrangement.
- Lone pairs prefer positions opposite each other, minimizing repulsion.
This yields a square planar shape: four fluorine atoms at the corners of a square, lone pairs above and below the plane.
Bond Angles and Square Planar Molecular Shape
In XeF4, the bond angles between two adjacent Xe–F bonds are exactly 90°, as dictated by the square planar geometry. Key factors:
- Two lone pairs occupy axial positions; no lone pairs in the equatorial (planar) positions.
- All four F atoms are in the same plane, equally spaced at 90°.
- Electron pair geometry is octahedral, but molecular shape is square planar due to lone pair placement.
| Parameter | XeF4 |
|---|---|
| Hybridization | sp3d2 |
| Electron domain geometry | Octahedral |
| Molecular shape | Square Planar |
| Lone pairs on Xe | 2 |
| Bond pairs | 4 |
| F–Xe–F bond angle | 90° |
Comparing XeF2, XeF4, XeF6 and SF4: Hybridization, Shape, Electron Domains
Understanding trends in xenon and related fluorides helps avoid confusion during exams. Compare these molecules for a clearer picture:
| Molecule | Hybridization | Electron Domains | Lone Pairs on Central Atom | Shape |
|---|---|---|---|---|
| XeF2 | sp3d | 5 | 3 | Linear |
| XeF4 | sp3d2 | 6 | 2 | Square Planar |
| XeF6 | sp3d3 | 7 | 1 | Distorted octahedral |
| SF4 | sp3d | 5 | 1 | See-saw |
Notice that hybridization and electron geometry are linked to the number of bonding domains and lone pairs—this is crucial for answer accuracy in JEE Main MCQs.
Key Mistakes and Smart Tips for XeF4 Hybridization
- Do not confuse the square planar shape (molecular geometry) with octahedral (electron domain geometry)—the lone pairs change the real shape.
- Always count both lone pairs and bond pairs to get the correct hybridization (6 = sp3d2 for XeF4).
- Remember: Lone pairs in XeF4 lie opposite each other to maximise stability.
- The value "4 + 2 = 6" is vital—don't miss lone pairs when calculating.
- Fluorine is very electronegative, influencing the shape but not the total domain count.
- For MCQs, quickly recall: XeF4=sp3d2=square planar=90°.
- Practice drawing Lewis structures; this reduces geometry confusion in higher p-block element questions.
- Some textbooks use d2sp3—for JEE, treat sp3d2 and d2sp3 as equivalent.
Quick JEE Practice on XeF4 Hybridization
- State the hybridization and shape of XeF4. (Answer: sp3d2, square planar)
- How many lone pairs are on xenon in XeF4? (Answer: 2)
- Which geometry is predicted by VSEPR for XeF4's electron pairs? (Answer: Octahedral)
- Why is the bond angle in XeF4 90° and not 109.5°?
- List hybridization and shape of XeF2 and XeF6.
Mastering such questions makes you quicker and more accurate during the exam.
Further Learning: Related XeF4 Topics for JEE Main
- Revise p-block elements trends—especially for noble gases and hypervalent molecules.
- Review SF4 hybridization contrasts to spot differences in similar compounds.
- Explore types of hybridization for fast recognition in MCQs.
- Strengthen chemical bonding basics for a strong JEE Main foundation.
- Practice on p-block practice questions for exam-level application.
- Reference square planar complexes for more shape examples.
- Use mock tests for chemical bonding to reinforce learning.
Learning the Hybridization of XeF4 builds your ability to solve advanced chemical bonding questions efficiently and with confidence. For more JEE-level clarity, keep exploring Vedantu Chemistry and strengthen your problem-solving skills with plenty of targeted examples and practice sets.
FAQs on Hybridization of XeF4: Concept, Structure & Exam Guide
1. What is the hybridization of XeF4?
XeF4 (xenon tetrafluoride) exhibits sp3d2 hybridization. This means the xenon atom uses one s-orbital, three p-orbitals, and two d-orbitals to form six hybrid orbitals.
Key points:
- 6 electron pairs arrange around xenon (including lone pairs and bonding pairs)
- Results in an octahedral electron geometry
- The molecular shape is square planar due to two lone pairs occupying axial positions
- Common in competitive exams like JEE, NEET and CBSE boards
2. Is XeF4 sp3d2 hybridized or sp3d?
XeF4 is sp3d2 hybridized, not sp3d. This is because:
- Xenon forms six electron domains in XeF4 (four bonds + two lone pairs), needing six hybrid orbitals
- This requires combination of one s, three p, and two d orbitals (= sp3d2)
- sp3d hybridization occurs with 5 electron pairs (example: XeF2)
3. What is the molecular geometry of XeF4?
The molecular geometry of XeF4 is square planar.
Explanation:
- The electron domain arrangement is octahedral
- But only 4 of 6 positions are occupied by fluorine atoms (bond pairs)
- 2 lone pairs occupy the remaining positions and stay opposite each other, minimizing repulsion
- This results in a flat, square-shaped molecule with fluorine at the four corners
4. How many lone pairs are there on xenon in XeF4?
In XeF4, the xenon atom has two lone pairs of electrons.
- Xenon starts with 8 valence electrons
- Each of four F atoms forms a single bond (using 4 electrons)
- The remaining 4 electrons (2 pairs) are nonbonding
5. What is the bond angle in XeF4?
The molecule XeF4 has a bond angle of 90º between adjacent fluorine atoms in the molecular plane.
- Square planar geometry ensures all F–Xe–F angles are exactly 90°
- The presence of lone pairs does not split these angles in this arrangement
6. What is the Lewis structure of XeF4?
The Lewis structure of XeF4 shows:
- Xenon (Xe) at the center, surrounded by four single bonds to fluorine (F) atoms
- Two lone pairs on Xe (opposite sides, in axial positions, not participating in bonding)
- Each fluorine atom has 3 lone pairs
7. Why does xenon, a noble gas, form compounds with fluorine?
Though xenon is a noble gas and typically inert, it can form compounds like XeF4 because:
- Fluorine is highly electronegative and reactive
- Xenon’s large atomic size and low ionization energy enable it to promote electrons and share them, especially under controlled conditions
- This allows expanded octet bonding, creating stable noble gas compounds
8. Can all six sp3d2 hybrid orbitals in XeF4 form bonds?
No, in XeF4, only four of the six sp3d2 hybrid orbitals form sigma bonds with fluorine atoms.
- The remaining two orbitals contain nonbonding lone pairs on xenon
- This is why the molecular shape is square planar — the lone pairs occupy positions to reduce repulsion
9. What is the difference between electronic geometry and molecular shape in XeF4?
In XeF4:
- Electronic geometry: Octahedral (all 6 electron domains considered, including bonds and lone pairs)
- Molecular shape: Square planar (only bonded atoms considered; two lone pairs are ignored in shape determination, but affect final structure)
10. How do lone pairs affect the shape of XeF4?
The two lone pairs on xenon in XeF4 force the molecule into a square planar shape.
- Lone pairs occupy opposite axial positions to minimize electron pair repulsion
- This leaves the four fluorine atoms in a single plane
- Without lone pairs, the arrangement would be octahedral, but lone pair–bond pair repulsion reduces symmetry
11. Why isn’t XeF4 tetrahedral?
XeF4 is not tetrahedral because:
- It has six electron domains (4 bonds + 2 lone pairs), not four
- Tetrahedral shape is seen with only four bonding pairs and no lone pairs (example: CH4)
- Lone pairs change its molecular geometry to square planar
12. How does the hybridization of XeF4 compare to XeF2 and XeF6?
XeF2: sp3d hybridization (5 electron domains, linear shape).
XeF4: sp3d2 hybridization (6 electron domains, square planar shape).
XeF6: sp3d3 hybridization (7 electron domains, distorted octahedral shape).
This comparison helps understand trends across xenon fluorides in competitive exams.
