Question

The structure of $\text{Xe}{{\text{F}}_{\text{4}}}$ is
A) Planar
B) Tetrahedral
C) Square Planar
D) Pyramidal

Answer 1

Hint: To solve the given question, we first need to know the hybridization of xenon tetrafluoride or $XeF_4$ with this information we can proceed to find out the hybridization of the central atom and similarly find the structure of $\text{Xe}{{\text{F}}_{\text{4}}}$.

Complete step by step answer:
The central atom in Xe has 4 bond pairs of electrons and two lone pairs of electrons. It undergoes $\text{s}{{\text{p}}^{\text{3}}}{{\text{d}}^{\text{2}}}$hybridization which results in octahedral electron geometry. This also results in the square planar molecular geometry.
The hybridization of $\text{Xe}{{\text{F}}_{\text{4}}}$ molecule.

The two lone pairs that are present on the opposite corners of an octahedron. This is because of the fact that it will minimize the repulsion between them.
Here is the diagram of the structure of $\text{Xe}{{\text{F}}_{\text{4}}}$

Hence, the correct answer is Option C.

Note: We can define hybridization as the idea of atomic orbitals to choose from newly hybridized orbitals. This, in turn, influences the molecular geometry and bonding properties of a molecule. Hybridization also defines the expansion of valence bond theory and this concept illustrates various compounds with $\text{s}{{\text{p}}^{\text{3}}}$, $\text{s}{{\text{p}}^{\text{2}}}$ and $\text{sp}$ kinds of hybridization. To define the hybridization of a particular molecule, we need to first clarify the spin direction of each electron in the molecule and similarly, we need to arrange the electrons in various energy shells such as s, p, d, f. After we attain a stable hybridization of a molecule, we can determine its pi ($\text{ }\!\!\pi\!\!\text{ }$) and sigma ($\text{ }\!\!\sigma\!\!\text{ }$) bonds and hence we can determine the structure that the molecule possesses.