The correct geometry and hybridization for XeF4 are

NEET 2016 Phase 2 · ChemistryPrevious Year Question

The correct geometry and hybridization for $XeF X_{4}$ are

Answer: (A) octahedral, $s p^{3} d^{2}$ . \textbf{Answer:} (A) octahedral, $s p^{3} d^{2}$ \textbf{Solution:} Xe in $XeF X_{4}$ has 6 electron pairs (4 bond pairs + 2 lone pairs) giving $s p^{3} d^{2}$ hybridization and an octahedral electron-pair geometry; the molecular shape is square planar.

A.octahedral, $s p^{3} d^{2}$ ✓
B.trigonal bipyramidal, $s p^{3} d$
C.planar triangle, $s p^{3} d^{3}$
D.square planar, $s p^{3} d^{2}$

Correct Answer

(A) octahedral, $s p^{3} d^{2}$

Solution & Explanation

\textbf{Answer:} (A) octahedral, $s p^{3} d^{2}$ \textbf{Solution:} Xe in $XeF X_{4}$ has 6 electron pairs (4 bond pairs + 2 lone pairs) giving $s p^{3} d^{2}$ hybridization and an octahedral electron-pair geometry; the molecular shape is square planar. The official key marks the electron-pair geometry (octahedral).

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The correct geometry and hybridization for XeFX4​ are

Solution & Explanation

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The correct geometry and hybridization for $XeF X_{4}$ are