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QuestionChemistry

Is XeF^4 polar or non-polar?
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Step 1:
I'll solve this step-by-step using molecular geometry and electronegativity principles:

Step 2:
: Determine the Lewis Structure

- Total electron count: $$8 + (4 \times 7) = 36$$ valence electrons
- Xenon (Xe) is the central atom - Four fluorine (F) atoms are bonded to Xe - Xe has 8 valence electrons - Each F contributes 7 valence electrons - Xe forms single bonds with each F atom - Xe has two lone pairs of electrons

Step 3:
: Identify Molecular Geometry

- Electron geometry is octahedral due to 6 electron domains (4 bonds + 2 lone pairs) - Molecular geometry is square planar - All F atoms are arranged symmetrically around Xe in a square

Step 4:
: Analyze Bond Polarity

- Fluorine ($$4.0$$) is much more electronegative than Xenon ($$2.6$$)
- Xe-F bonds are polar due to electronegativity difference - Individual Xe-F bonds have dipole moments

Step 5:
: Determine Overall Molecular Polarity

- Despite polar bonds, the square planar geometry results in dipole moment cancellation - Symmetrical arrangement means net dipole moment is zero

Final Answer

XeF^4 is NON-POLAR.