Identify the molecular geometry of XeF_4. | vedclass

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Question

(A) The central atom in $XeF_4$ is Xenon $(Xe)$,which has $8$ valence electrons.It forms $4$ single bonds with $4$ Fluorine $(F)$ atoms.This leaves $4

Vedclass · Accepted Answer

Square planar

Vedclass · Answer

(A) The central atom in $XeF_4$ is Xenon $(Xe)$,which has $8$ valence electrons.It forms $4$ single bonds with $4$ Fluorine $(F)$ atoms.This leaves $4$ electrons on the Xenon atom,which form $2$ lone pairs.The total number of electron pairs (steric number) is $4 	ext{ (bond pairs)} + 2 	ext{ (lone pairs)} = 6$.$A$ steric number of $6$ corresponds to $sp^3d^2$ hybridization,which has an octahedral electron geometry.Due to the presence of $2$ lone pairs,the molecular geometry is square planar.

List-$I$	List-$II$
$(1)$ ${H_3O^+}$	$(A)$ Linear
$(2)$ ${HC \equiv CH}$	$(B)$ Angular
$(3)$ ${ClO_2^-}$	$(C)$ Tetrahedral
$(4)$ ${NH_4^+}$	$(D)$ Trigonal bipyramidal
	$(E)$ Pyramidal

Identify the molecular geometry of $XeF_4$.

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List-$I$ List-$II$

$(1)$ ${H_3O^+}$ $(A)$ Linear

$(2)$ ${HC \equiv CH}$ $(B)$ Angular

$(3)$ ${ClO_2^-}$ $(C)$ Tetrahedral

$(4)$ ${NH_4^+}$ $(D)$ Trigonal bipyramidal

$(E)$ Pyramidal

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