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Q.Explain the structures of a) XeF6 and b) XeOF4
Telangana TsbieTelangana Board of Intermediate Education 2023Subjective· 4mImportance★★★★★
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Start your 14-day free trial to unlock the full solution →XeF6 (7 electron pairs: 6 bp + 1 lp) is a distorted octahedron; XeOF4 (6 electron pairs: 5 bp + 1 lp) is square pyramidal.
(a) XeF6:
- Xenon (group 18) has 8 valence electrons. It forms 6 bonds to 6 fluorine atoms, using 6 electrons; the remaining 2 electrons form 1 lone pair.
- Total electron pairs around Xe = 6 bond pairs + 1 lone pair = 7.
- According to VSEPR, seven electron pairs give a distorted octahedral geometry (a distorted/capped octahedron). The lone pair occupies one position and distorts the shape, so XeF6 is NOT a regular octahedron; its structure is described as a distorted octahedron (fluxional in the gas phase). Hybridisation: sp3d3.
(b) XeOF4:
- Xenon has 8 valence electrons. It forms 4 single bonds to 4 F atoms and 1 double bond to O. Counting: 4 (Xe-F bond pairs) + 1 (Xe=O counted as one bonding region) = 5 bonding regions, plus 1 lone pair remaining = 6 electron pairs around Xe. …
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