Question

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

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

Answer 1

Answer: (B)

bent T-shaped, square pyramidal

$H=\frac{1}{2}[V+M-C+A]$
where,
H= Number of orbitals involved in hybridization
V=Valence electrons of central atom
M- Number of monovalent atoms linked to central atom
C= Charge of cation
A= Charge of anion
Now consider,
$Xe{F}_3^{+}$
$H=\frac{1}{2}[8+3-1]=5$ $\Rightarrow s{p}^{3}d$ hybridized state with 2 lone pairs.
Hence, it is T shaped.
Now consider,
$Xe{F}_5^{+}$
$H=\frac{1}{2}[8+5-1]=6$ $\Rightarrow s{p}^3{d}^2$ hybridized state with 1 lone pair.
It is square pyramidal in shape.
Hence, option B is the right answer.