Question

Geometry of the Xenon oxyfluoride (${\mathrm{XeOF}}_4$) is________

Answer 1

The geometry of Xenon oxyfluoride (${\mathrm{XeOF}}_4$)

1. For determining the geometry we first need to draw the lewis dot structure of ${\mathrm{XeOF}}_4$
2. First knowing the valence electrons present in the chemical is necessary in order to draw the Lewis dot structure.
3. In ${\mathrm{XeOF}}_4$ Valence electrons$=\mathrm{Xe}+\mathrm{O}+4\times \mathrm{F}$
4. Electronic configuration of $\mathrm{Xe}$ with atomic number$=54$ is$1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^{6}3{\mathrm{d}}^{10}4{\mathrm{s}}^{2}4{\mathrm{p}}^{6}4{\mathrm{d}}^{10}5{\mathrm{s}}^{2}5{\mathrm{p}}^6$, valence electron$=8$
5. Electronic configuration of oxygen with atomic number$=16$ is $1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^4$, valence electron$=6$
6. Electronic configuration of chlorine with atomic number$=17$ is $1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^5$, valence electron$=7$
7. Total valence electron$=\mathrm{Xe}+\mathrm{O}+4\times \mathrm{F}\Rightarrow 8+6+4\times 7\Rightarrow 42$
8. Therefore, these $42$ valence electrons must be distributed among the elements such that each one can attain its octet. In the given compound, Xenon is an exception because it can expand its octet.
9. 
10. The steric number can be used to predict hybridization. Steric number = sigma bond number + lone pair number
11. There are five sigma bonds and one lone pair present in ${\mathrm{XeOF}}_4$, therefore the steric number is$5+1=6$ , and the hybridization will be${\mathrm{sp}}^3{\mathrm{d}}^2$.
12. 
13. The hybrid orbitals will be created by the center atom using a combination of its s, p, and as many of its d orbitals as are required. In order to form${\mathrm{XeOF}}_4$, Xenon must have its s orbital, all three of its p-orbitals, and two of its d-orbitals. As a result, Xenon will be in the hybridization state ${\mathrm{sp}}^3{\mathrm{d}}^2$ and its shape will be square pyramidal.

Therefore, Xenon oxyfluoride ($\left(XeO{F}_4\right)$ molecular geometry is square pyramidal.