Question

Which of the following has a square pyramidal shape?
(A) $XeO{F_4}$
(B) $Xe{O_3}{F_2}$
(C) $XeO{F_2}$
(D) $Xe{O_2}{F_2}$

Answer 1

Hint : Hybridization: Mixing of orbitals with different energy levels to form new hybrid orbitals with similar energies to form chemical bonds is known as hybridization. The concept of hybridization can help to determine the shape and geometry of a molecule.

Complete Step By Step Answer:
Valence shell electron pair repulsion (VSEPR) Theory:
If the central atom of the molecule is surrounded by non-bonded pairs or bond pairs of electrons, then the shape of the molecule is expected to have a different shape than the expected geometry of the molecule.
Steric number- Number of atoms bonded to a central atom with a single bond added with the number of non-bonded pairs of electrons on the central atom is known as its steric number. It is an application of VSEPR theory to find hybridization, geometry and shape of the molecule.
Properties of given molecules according to VSEPR theory are as follows:
$XeO{F_4}$ -
Number of non-bonded pair of electrons on xenon $= 8$
Number of single bonded atoms $= 4$
Steric Number $= \dfrac{{8 + 4}}{2} \Rightarrow 6$
Hybridization $= s{p^3}{d^2}$
Therefore, the geometry of the molecule is octahedral but due to the presence of one lone pair its shape is square pyramidal.
$Xe{O_3}{F_2}$ -
Number of non-bonded pair of electrons on xenon $= 8$
Number of single bonded atoms $= 2$
Steric Number $= \dfrac{{8 + 2}}{2} \Rightarrow 5$
Hybridization $= s{p^3}d$
Therefore, the geometry of the molecule is trigonal bipyramidal and there is no lone pair present on xenon, so the shape of the molecule is the same as that of its geometry.
$XeO{F_2}$ -
Number of non-bonded pair of electrons on xenon $= 8$
Number of single bonded atoms $= 2$
Steric Number $= \dfrac{{8 + 2}}{2} \Rightarrow 5$
Hybridization $= s{p^3}d$
Therefore, the geometry of the molecule is trigonal bipyramidal and there are two non-bonded pairs present on xenon, therefore the molecule is of T-shaped structure.
$Xe{O_2}{F_2}$ -
Number of non-bonded pair of electrons on xenon $= 8$
Number of single bonded atoms $= 2$
Steric Number $= \dfrac{{8 + 2}}{2} \Rightarrow 5$
Hybridization $= s{p^3}d$
Therefore, the geometry of the molecule is trigonal bipyramidal and there is one lone pair present on xenon, therefore the molecule is of V-shaped structure.
Hence $XeO{F_4}$ molecule has a square pyramidal shape.
So, option (A) is the correct answer.

Note :
Shape and geometries of the molecules may vary due to the presence of non-bonded pair or bond pairs of electrons due to repulsion between electrons. Variation in bond length of the molecules is also observed due to these factors.