Draw the Structures of XeF4, XeF6, and XeOF4 | Class 12 Chemistry

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Draw the Structures of XeF4, XeF6, and XeOF4 | Class 12 Chemistry Ch 7 Q 7.15

Understanding the structures of xenon compounds is fascinating because it challenges the old belief that noble gases are completely inert. These three compounds—XeF4, XeF6, and XeOF4—showcase how xenon can form stable compounds with different geometries based on VSEPR theory and hybridization.

📝

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📌 Quick Answer

XeF4 has square planar geometry (sp3d2 hybridization, 2 lone pairs). XeF6 has distorted octahedral geometry (sp3d3 hybridization, 1 lone pair). XeOF4 has square pyramidal geometry (sp3d2 hybridization, 1 lone pair with oxygen at apex). All three demonstrate xenon’s ability to expand its octet using vacant d-orbitals.

🔬 Understanding Xenon Compounds

Before we dive into the structures, it’s important to understand why xenon forms compounds at all. Xenon is a noble gas with a complete octet, but it has vacant d-orbitals in its valence shell. When xenon reacts with highly electronegative elements like fluorine and oxygen, electrons from filled p-orbitals can be promoted to these vacant d-orbitals, allowing xenon to form bonds. This is called expansion of octet or hypervalency.

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The geometry of these compounds is determined by VSEPR (Valence Shell Electron Pair Repulsion) theory, which states that electron pairs around a central atom arrange themselves to minimize repulsion. Both bonding pairs and lone pairs must be considered when determining molecular shape.

📐 Structure 1: XeF4 (Xenon Tetrafluoride)

⚛️ Molecular Details

Hybridization:
sp3d2
Geometry:
Square Planar
Lone Pairs:
2 (axial positions)
VSEPR:
AX4E2

🎨 Visual Structure

pasted image 
Top View (Square Planar):

                    F
                    |
                    |
         F -------- Xe -------- F
                    |
                    |
                    F

    All four F atoms lie in the same plane
    Bond angles: F-Xe-F = 90° and 180°


Side View (showing lone pairs):

              ⚫ (Lone pair above)
                    |
                    |
         F -------- Xe -------- F
                    |
                    |
              ⚫ (Lone pair below)

    Two lone pairs occupy axial positions
    (above and below the square plane)

📊 Hybridization Process

Ground state of Xe: [Kr] 5s2 5p6

Excited state: Two electrons from 5p are promoted to 5d orbitals → [Kr] 5s2 5p4 5d2

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Hybridization: One 5s + three 5p + two 5d orbitals mix to form six sp3d2 hybrid orbitals arranged octahedrally.

Orbital usage: Four orbitals form sigma bonds with four F atoms (in equatorial plane), and two orbitals contain lone pairs (in axial positions).

💡 Key Point: The two lone pairs occupy opposite positions (180° apart) to minimize lone pair-lone pair repulsion, resulting in a square planar molecular geometry.

📐 Structure 2: XeF6 (Xenon Hexafluoride)

⚛️ Molecular Details

Hybridization:
sp3d3
Geometry:
Distorted Octahedral
Lone Pairs:
1 (stereochemically active)
VSEPR:
AX6E1

🎨 Visual Structure

Distorted Octahedral Structure:

                    F
                    |
                    | (slightly bent)
         F -------- Xe -------- F
                   /|\
                  / | \
                 /  |  \
                F   F   F

    The lone pair causes distortion
    Bond angles deviate from perfect 90°


3D Representation:

              F (axial)
              ↑
              |
    F ← Xe → F  (equatorial, distorted)
         ↙ ↓ ↘
        F  F  F

    ⚫ Lone pair occupies space
       (causes asymmetry and distortion)

📊 Hybridization Process

Ground state of Xe: [Kr] 5s2 5p6

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Excited state: Electrons promoted → [Kr] 5s2 5p3 5d3

Hybridization: One 5s + three 5p + three 5d orbitals mix to form seven sp3d3 hybrid orbitals arranged in pentagonal bipyramidal geometry.

Orbital usage: Six orbitals form sigma bonds with six F atoms, and one orbital contains the lone pair.

💡 Key Point: The single lone pair is stereochemically active and causes distortion from perfect octahedral geometry. This makes XeF6 a fluxional molecule that rapidly changes shape.

📐 Structure 3: XeOF4 (Xenon Oxytetrafluoride)

⚛️ Molecular Details

Hybridization:
sp3d2
Geometry:
Square Pyramidal
Lone Pairs:
1 (below square plane)
VSEPR:
AX5E1

🎨 Visual Structure

Square Pyramidal Structure:

                    O (apex)
                    ║
                    ║ (double bond)
                    Xe
                   /|\\
                  / | \\
                 /  |  \\
                F   F   F
                    |
                    F

    Oxygen at apex (top of pyramid)
    Four F atoms form square base
    Lone pair below the base


Top View (looking down from O):

                    F
                    |
                    |
         F -------- Xe -------- F
                    |
                    |
                    F

         ⚫ Lone pair (below plane, not visible)


Side View:

              O (at top)
              ║
              ║
    F ------- Xe ------- F
              |
              F
              |
         ⚫ (Lone pair at bottom)

📊 Hybridization Process

Ground state of Xe: [Kr] 5s2 5p6

Excited state: [Kr] 5s2 5p4 5d2

Hybridization: One 5s + three 5p + two 5d orbitals mix to form six sp3d2 hybrid orbitals arranged octahedrally.

Orbital usage: One orbital forms a double bond with oxygen (at apex), four orbitals form sigma bonds with four F atoms (square base), and one orbital contains the lone pair (opposite to oxygen).

💡 Key Point: The oxygen atom occupies the axial position opposite to the lone pair. The Xe=O bond is stronger and shorter than Xe-F bonds, giving the molecule its characteristic square pyramidal shape.

📊 Comparison Table: All Three Structures

Property XeF4 XeF6 XeOF4
Hybridization sp3d2 sp3d3 sp3d2
Molecular Geometry Square Planar Distorted Octahedral Square Pyramidal
VSEPR Notation AX4E2 AX6E1 AX5E1
Number of Lone Pairs 2 1 1
Bond Angles 90°, 180° ~90° (distorted) ~90°
Polarity Non-polar Polar Polar
Physical State Colorless crystals Pale yellow solid Colorless liquid

📝 How to Write This Answer in Your Exam (5 Marks)

🎯 Marking Scheme Breakdown

✅ Part (a) XeF4: 1.5 marks

  • Structure diagram with labels (0.5 mark)
  • Hybridization and geometry (0.5 mark)
  • Lone pair positions (0.5 mark)

✅ Part (b) XeF6: 1.5 marks

  • Structure diagram with labels (0.5 mark)
  • Hybridization and geometry (0.5 mark)
  • Mention of distortion (0.5 mark)

✅ Part (c) XeOF4: 2 marks

  • Structure diagram with labels (0.75 mark)
  • Hybridization and geometry (0.75 mark)
  • Position of oxygen and lone pair (0.5 mark)

✍️ Model Answer Format

(a) XeF4 (Xenon Tetrafluoride):

Hybridization: sp3d2
Geometry: Square planar
Lone pairs: 2 (in axial positions above and below the plane)

[Draw the structure showing Xe at center with 4 F atoms in square planar arrangement and indicate 2 lone pairs in axial positions]

(b) XeF6 (Xenon Hexafluoride):

Hybridization: sp3d3
Geometry: Distorted octahedral
Lone pairs: 1 (causes distortion from perfect octahedral geometry)

[Draw the structure showing Xe at center with 6 F atoms in distorted octahedral arrangement and indicate 1 lone pair causing the distortion]

(c) XeOF4 (Xenon Oxytetrafluoride):

Hybridization: sp3d2
Geometry: Square pyramidal
Lone pairs: 1 (below the square base, opposite to oxygen)
Special feature: Oxygen occupies the apex position with a double bond to Xe

[Draw the structure showing Xe at center, O at apex with double bond, 4 F atoms forming square base, and indicate 1 lone pair below]

⏱️ Time Management Tip: Allocate 8-10 minutes for this 5-mark question. Draw neat, labeled structures and write hybridization/geometry clearly for each compound.

✅ Detailed Step-by-Step Solution

🔍 Approach to Drawing Xenon Compound Structures

1
Determine the valence electrons

Xenon has 8 valence electrons. Count the electrons needed for bonding with F and O atoms.

2
Find the hybridization

Use the formula: Hybridization = ½(V + M – C + A), where V = valence electrons of central atom, M = monovalent atoms, C = cationic charge, A = anionic charge.

3
Apply VSEPR theory

Count bonding pairs and lone pairs. Use VSEPR notation (AXnEm) to determine molecular geometry.

4
Draw the structure

Place the central atom (Xe) and arrange bonded atoms and lone pairs according to the geometry. Label all atoms and indicate lone pairs clearly.

📐 Detailed Calculations

For XeF4:

Hybridization number = ½(8 + 4 – 0 + 0) = ½(12) = 6
This means 6 hybrid orbitals → sp3d2 hybridization
Bonding pairs = 4, Lone pairs = 2
VSEPR: AX4E2 → Square planar geometry

For XeF6:

Hybridization number = ½(8 + 6 – 0 + 0) = ½(14) = 7
This means 7 hybrid orbitals → sp3d3 hybridization
Bonding pairs = 6, Lone pairs = 1
VSEPR: AX6E1 → Distorted octahedral geometry

For XeOF4:

Hybridization number = ½(8 + 4 – 0 + 0) = ½(12) = 6
(Oxygen is divalent, counted as part of bonding arrangement)
This means 6 hybrid orbitals → sp3d2 hybridization
Bonding pairs = 5 (1 double bond with O + 4 single bonds with F), Lone pairs = 1
VSEPR: AX5E1 → Square pyramidal geometry

❓ Frequently Asked Questions (FAQs)

❓ Why is XeF4 non-polar despite having polar Xe-F bonds?

XeF4 is non-polar because of its symmetrical square planar geometry. Although each Xe-F bond is polar (due to electronegativity difference), the four bonds are arranged symmetrically at 90° and 180° angles. The dipole moments of opposite bonds cancel each other out completely, resulting in zero net dipole moment. This makes the molecule non-polar overall.

❓ Why does XeF6 have distorted octahedral geometry instead of perfect octahedral?

XeF6 has one lone pair of electrons that is stereochemically active (occupies space). In a perfect octahedral geometry, all six positions are equivalent. However, the lone pair repels the bonding pairs more strongly than bonding pairs repel each other (lone pair-bond pair repulsion > bond pair-bond pair repulsion). This causes the six F atoms to be pushed away from the lone pair, resulting in a distorted octahedral structure. The molecule is also fluxional, meaning it rapidly changes between different distorted forms.

❓ How is XeOF4 different from XeF4 in terms of structure?

Both have sp3d2 hybridization, but their geometries differ. XeF4 has 4 bonding pairs and 2 lone pairs (AX4E2), giving square planar geometry with all atoms in one plane. XeOF4 has 5 bonding pairs (including the double bond with O) and 1 lone pair (AX5E1), giving square pyramidal geometry. The oxygen atom occupies the apex position with a double bond, while four F atoms form the square base. The lone pair is positioned opposite to oxygen, below the base.

❓ Can xenon form compounds with elements other than fluorine and oxygen?

Yes, but xenon compounds are primarily formed with highly electronegative elements like fluorine and oxygen because they can effectively pull electrons from xenon’s filled orbitals. Xenon can also form compounds with chlorine (like XeCl2) and in complex ions with other elements. However, fluorine is the most common partner because it’s the most electronegative element, making xenon-fluorine bonds relatively stable. Compounds like XeO3, XeO4, and various xenon-fluoride-oxide combinations are also known.

❓ Why do we need to consider lone pairs when determining molecular geometry?

Lone pairs occupy space around the central atom just like bonding pairs, but they exert stronger repulsion because they are held closer to the nucleus and are not shared between two atoms. According to VSEPR theory, the order of repulsion is: lone pair-lone pair > lone pair-bond pair > bond pair-bond pair. This means lone pairs push bonded atoms away, affecting bond angles and overall molecular shape. For example, in XeF4, the two lone pairs occupy axial positions to minimize repulsion, forcing the four F atoms into a square planar arrangement rather than a tetrahedral one.

Dr. Irfan Mansuri

✍️ Written by Dr. Irfan Mansuri

Senior Science Educator & Researcher

Independent Educational Consultant • India

Dr. Irfan Mansuri brings 25 years of extensive experience in science education, specializing in Physics and Chemistry. With a doctoral degree and a passion for making science engaging, he has mentored countless students to achieve excellence in their board examinations. His teaching methodology emphasizes conceptual clarity, practical applications, and exam-oriented strategies that help students not just memorize, but truly understand scientific principles.

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