QQuestionChemistry
QuestionChemistry
Draw the Lewis structure for the molecule XeF₄. Include all lone pairs of electrons. If resonance structures are needed to describe the Lewis structure, draw all possible resonance structures.
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Step 1:I'll solve this problem step by step, following the specified LaTeX formatting guidelines:
Step 2:: Determine the total number of valence electrons
- Total valence electrons: $$8 + (4 \times 7) = 8 + 28 = 36$$ valence electrons
- Xe (Xenon) is in Group 18 and has 8 valence electrons - F (Fluorine) has 7 valence electrons
Step 3:: Arrange atoms and form single bonds
- Remaining electrons: $$36 - 8 = 28$$ electrons
- Place Xe in the center - Attach four F atoms to Xe with single bonds
Step 4:: Distribute remaining electrons as lone pairs
- Place 3 lone pairs on each F atom ($$3 \times 4 = 12$$ electrons)
- Remaining electrons: 28 - 12 = 16 electrons - Place the remaining 16 electrons as lone pairs on the central Xe atom
Step 5:: Lewis structure details
- Central Xe atom has 2 lone pairs - Each F atom has 3 lone pairs - All atoms have a complete octet Final Lewis Structure Representation: - Four F atoms bonded to central Xe - 3 lone pairs on each F atom - 2 lone pairs on central Xe atom
Final Answer
The Lewis structure for XeF₄ shows a square planar molecular geometry with the Xe atom in the center, four F atoms symmetrically arranged around it, with additional lone pairs as described above.
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