Question

The geometry and hybridisation of $Xe$ in $Xe{F}_4$ is:

• A. see-saw, $s{p}^{3}d$
• B. octahedral, $s{p}^3{d}^2$
• C. pIanar, $s{p}^3{d}^2$
• D. T-shaped, $s{p}^3{d}^3$

Answer 1

The correct option is C pIanar, $s{p}^3{d}^2$

The valence shell of $Xe$ contains two electrons in $5s$ orbital and six electrons in $5p$ orbital. In the $5^{th}$ orbit there is a place for $d$ orbital and $f$ orbital both of these are empty. Of the $5s$ and $5p$ electrons in the outermost orbit of $Xe$, two of the $5p$ orbital electrons get excited to the vacant $5d$ orbital resulting in 4 unpaired hybridized electrons, 2 in $5p$ and 2 in $5d$ orbitals.
And the shape is square planar.