Question

$Sb{F}_5$ reacts with $Xe{F}_4$ to form an adduct. The shapes of cation and anion in the adduct are respectively

• A. Square planar, trigonal bipyramidal
• B. T-shaped, octahedral
• C. Square pyramidal, octahedral
• D. Square planar, octahedral

Answer 1

Answer: (B)

${\text{XeF}}_{\text{4}}\text{+}{\text{SbF}}_{\text{5}}\to {\text{[XeF}}_{\text{3}}\text{]}\text{+}{\text{[SbF}}_{\text{6}}{\text{]}}^{-}\to \underset{\underset{\text{T-Shaped}}{{\text{sp}}^{\text{3}}\text{d}}}{{\text{[XeF}}_{\text{3}}{\text{]}}^{\text{+}}}\text{+}\underset{\underset{\text{octahedral}}{{\text{sp}}^{\text{3}}{\text{d}}^{\text{2}}}}{{\text{[SbF}}_{\text{6}}{\text{]}}^{-}}$

As $[Xe{F}_3{]}^{+}$ has 3 bond pair and 2 lone pair of electron so it has $s{p}^{3}d$ hybridisation and T-shape.

$[Sb{F}_6{]}^{-}$ has 6 bond pair electrons so it is $s{p}^3{d}^2$ hybridise and octahedral in shape.