[XeO₆]⁴⁻ is octahedral whereas XeF₆ is a disordered one, because · ABC26IN0317
Subject: Inorganic Chemistry · Chapter: Chemical Bonding · Topic: Ionic Lattice Born Haber · Exam: GATE 2005-2021 · Marks: 2 · Difficulty: Easy
$\mathrm{[XeO_6]^{4-}}$ is octahedral whereas $\mathrm{XeF_6}$ is a disordered one, because
(a)fluorine is more electronegative than oxygen
(b)Xe has a lone-pair in $\mathrm{XeF}_{6}$
(c)$\mathrm{XeF}$ is neutrla whereas $\mathrm{[XeO_6]^{4-}}$ is anionic.
(d)Xe-F bond has more ionic character
Answer
Answer: B ✓ checked by 4AB · confidence high
Explanation
XeF₆ has a stereochemically active lone pair on Xe (seven electron pairs), so it is a distorted octahedron, unlike [XeO₆]⁴⁻ (b).
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