Question

Using VBT theory describe the shape of ${\text{Xe}}{{\text{O}}_{\text{4}}}$.

Answer 1

Hint: VBT is the valence bond theory. The theory defines the electronic structures of atoms or molecules. The valence bond theory explains that the electrons fill the atomic orbitals. Also, valence bond theory explains that the nucleus of one molecule is attracted to the electrons of another molecule.

Complete step by step answer:
The postulates of valence bond theory are as follows:
1.When two valence half-filled orbitals belonging to two different atoms overlap each other covalent bonds are formed. Overlapping increases the electron density in the area between the two bonding atoms. Thus, the stability of the molecule increases.
2.The atom can form multiple bonds with other atoms due to the presence of many unpaired electrons. The paired electrons do not participate in bonding.
3.The covalent bonds formed are directional and parallel to the overlapping atomic orbitals.
4.The pi bonds are formed by sidewise overlapping of the atomic orbitals and the sigma bonds are formed by the axial overlapping of the atomic orbitals.
We will now describe the shape of ${\text{Xe}}{{\text{O}}_{\text{4}}}$.
-The central atom is xenon. Four oxygen atoms are bonded to the xenon atom. Thus, the coordination number of xenon is four.
-Thus, the shape of ${\text{Xe}}{{\text{O}}_{\text{4}}}$ can be tetrahedral or square planar.
-In the formation of ${\text{Xe}}{{\text{O}}_{\text{4}}}$, xenon undergoes $s{p^3}$ hybridisation in its fourth excited state. Xenon has eight unpaired electrons in its fourth excited state.
-One s-orbital and three p-orbitals overlap to form the $s{p^3}$ hybrid orbitals. It has no unpaired electrons.
-Thus, the shape of ${\text{Xe}}{{\text{O}}_{\text{4}}}$ is tetrahedral. ${\text{Xe}}{{\text{O}}_{\text{4}}}$ has four sigma bonds and four pi bonds. The bond angle is ${\text{10}}{{\text{9}}^ \circ }$.

Note:
The coordination number of xenon is four. If the hybridisation was $ds{p^2}$ then the shape of ${\text{Xe}}{{\text{O}}_{\text{4}}}$ would have been square planar. But as the hybridisation is $s{p^3}$ the shape is tetrahedral.