Question

$Sb{F}_5$ reacts with $Xe{F}_4$ and $Xe{F}_6$ to form ionic compounds $\left[Xe{F}_3^{+}\right]\left[Sb{F}_6{]}^{-}and\right[Xe{F}_5^{+}\left]\right[Sb{F}_6^{-}].$ The geometry of $Xe{F}_3^{+}$ ion and $Xe{F}_5^{+}$ ion, respectively, is :

• A. sqare pyramidal, T-shaped
• B. bent T-shaped, square pyramidal
• C. see-saw, square pyramidal
• D. square pyramidal, see-saw

Answer 1

The correct option is B bent T-shaped, square pyramidal

The geometry of $Xe{F}_3^{+}$ ion is bent T-shaped.
The central $Xe$ atom has $2$ lone pairs and $3$ bond pairs of electrons. The electron pair geometry is trigonal bipyramidal.
The geometry of $Xe{F}_5^{+}$ ion is square pyramidal.
The central $Xe$ atom has $1$ lone pair and $5$ bond pairs of electrons. The electron pair geometry is octahedral.