SbF_5 reacts with XeF_4 and XeF_6 to form ion | vedclass

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(B) For $[XeF_3^+]$: The central atom $Xe$ has $8$ valence electrons. It forms $3$ bonds with $F$ atoms and has $1$ positive charge,leaving $8 - 3 - 1

Vedclass · Accepted Answer

Bent $T$-shape,square pyramidal

Vedclass · Answer

(B) For $[XeF_3^+]$: The central atom $Xe$ has $8$ valence electrons. It forms $3$ bonds with $F$ atoms and has $1$ positive charge,leaving $8 - 3 - 1 = 4$ electrons,which form $2$ lone pairs. The total number of electron pairs is $3 + 2 = 5$,corresponding to $sp^3d$ hybridization. With $3$ bond pairs and $2$ lone pairs,the geometry is trigonal bipyramidal,and the shape is bent $T$-shape.For $[XeF_5^+]$: The central atom $Xe$ has $8$ valence electrons. It forms $5$ bonds with $F$ atoms and has $1$ positive charge,leaving $8 - 5 - 1 = 2$ electrons,which form $1$ lone pair. The total number of electron pairs is $5 + 1 = 6$,corresponding to $sp^3d^2$ hybridization. With $5$ bond pairs and $1$ lone pair,the geometry is octahedral,and the shape is square pyramidal.

$SbF_5$ reacts with $XeF_4$ and $XeF_6$ to form ionic compounds $[XeF_3^+][SbF_6^-]$ and $[XeF_5^+][SbF_6^-]$. Determine the molecular shape of $[XeF_3^+]$ ion and $[XeF_5^+]$ ion respectively.

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