The hybridisation and structure of XeF_6 is : | vedclass

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(B) $Xe$ has $8$ valence electrons.In $XeF_6$,$6$ electrons are involved in bonding with $6$ fluorine atoms,and $2$ electrons remain as a lone pair.St

Vedclass · Accepted Answer

$sp^3d^3,$ distorted octahedral

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(B) $Xe$ has $8$ valence electrons.In $XeF_6$,$6$ electrons are involved in bonding with $6$ fluorine atoms,and $2$ electrons remain as a lone pair.Steric number = (Number of bond pairs) + (Number of lone pairs) = $6 + 1 = 7$.$A$ steric number of $7$ corresponds to $sp^3d^3$ hybridisation.Due to the presence of one lone pair,the geometry is distorted octahedral.

List-$I$	List-$II$
$(A) \ XeF_4$	$(1) \ \text{Distorted octahedral}$
$(B) \ XeF_6$	$(2) \ \text{Tetrahedral}$
$(C) \ XeO_3$	$(3) \ \text{Square planar}$
$(D) \ XeO_4$	$(4) \ \text{Pyramidal}$

The hybridisation and structure of $XeF_6$ is :

Similar Questions

Which of the following fluorides of Xenon does not exist?

Match List-$I$ with List-$II$ and select the correct answer using the codes given below the lists:

List-$I$ List-$II$

$(A) \ XeF_4$ $(1) \ \text{Distorted octahedral}$

$(B) \ XeF_6$ $(2) \ \text{Tetrahedral}$

$(C) \ XeO_3$ $(3) \ \text{Square planar}$

$(D) \ XeO_4$ $(4) \ \text{Pyramidal}$

$Xe + 2F_2 \xrightarrow{573 \ K, \ 60.70 \ bar} A$
What is $A$?

Among the fluorides below,the one which does not exist is

Why are the elements of Group $18$ known as noble gases?

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