The shapes/structures of [XeF_5]^- and XeO_3F | vedclass

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Question

(A) For $[XeF_5]^-$,the central atom $Xe$ has $8$ valence electrons. Adding $5$ electrons from $F$ atoms and $1$ electron for the negative charge give

Vedclass · Accepted Answer

pentagonal planar and trigonal bipyramidal

Vedclass · Answer

(A) For $[XeF_5]^-$,the central atom $Xe$ has $8$ valence electrons. Adding $5$ electrons from $F$ atoms and $1$ electron for the negative charge gives $14$ electrons,which corresponds to $7$ electron pairs. The hybridization is $sp^3d^3$. Due to the presence of $2$ lone pairs,the shape is pentagonal planar.For $XeO_3F_2$,the central atom $Xe$ has $8$ valence electrons. $3$ oxygen atoms form double bonds (using $6$ electrons) and $2$ fluorine atoms form single bonds (using $2$ electrons). This results in $5$ bonding pairs and $0$ lone pairs. The hybridization is $sp^3d$,and the shape is trigonal bipyramidal.

The shapes/structures of $[XeF_5]^-$ and $XeO_3F_2$ respectively,are:

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