Question

What is the hybridization and geometry of the compound $XeO{F}_4$ ?

• A. $s{p}^3{d}^2$ and octahedral
• B. $s{p}^{3}d$ and square pyramidal
• C. $s{p}^{3}d$ and trigonal bipyramidal
• D. $s{p}^3{d}^2$ and square pyramidal

Answer 1

Answer: (D)

Number of hybrid orbitals $(X)$
$=\frac{1}{2}(V.E.+M.A.-c+a)$
For, $XeO{F}_4$
$V.E.=8;M.A.$ (monovalent atoms) $=4$
$(X)=\frac{1}{2}(8+4)\Rightarrow 6$
i.e., $s{p}^3{d}^2$ hybridisation