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Question: The correct geometry and hybridization for \(Xe{{F}_{4}}\) are: (a)- Octahedral, \(s{{p}^{3}}{{d}^...

The correct geometry and hybridization for XeF4Xe{{F}_{4}} are:
(a)- Octahedral, sp3d2s{{p}^{3}}{{d}^{2}}
(b)- Trigonal bipyramidal, sp3ds{{p}^{3}}d
(c)- Planar triangle, sp3d3s{{p}^{3}}{{d}^{3}}
(d)- Square planar, sp3d2s{{p}^{3}}{{d}^{2}}

Explanation

Solution

The shape XeF4Xe{{F}_{4}} is based on the number of atoms joined to the central atom. Not only atoms, but lone pairs are also considered for the shape. For finding the shape of the compound the number of a total number of electron pairs and number of lone pairs should be calculated with the help of valence electron, a number of bonds, etc. and hybridization can be calculated by the number of electron pairs.

Complete step by step answer:
Both the VSEPR theory and the concept of hybridization are applied to predict the molecular geometries of xenon compounds.
According to the VSEPR theory, the shape of the molecule is predicted by the total number of electron pairs (lone pairs + bond pairs) in the valence shell of the central Xe atom.
To calculate the total number of electron pairs:
valence electrons of central atom + number of bonded atoms2\dfrac{\text{valence electrons of central atom + number of bonded atoms}}{\text{2}}
With the above formula: 8+42=6\dfrac{8+4}{2}=6
Hence, there are 6 electron pairs.
Since there are 4 fluorine atoms joined to xenon. So, there will be a 4bond pair of electrons.
Now for calculating the number of lone pairs in the compound: -
Total number of electron pairs –number of bond pairs.
Lone pairs= 6−4=26-4=2
Hence, in the compound, there are 2 lone pairs.
Depending on the number of Xe−FXe-F covalent bonds to be formed, the requisite number of electrons of the 5p−orbital5p-orbital valence shell of Xe gets unpaired and promoted to the vacant 5p−orbital5p-orbital followed by hybridization.
Since there are 6 electron pairs, the hybridization of the compound will be sp3d2s{{p}^{3}}{{d}^{2}}.
So, the hybridization is sp3d2s{{p}^{3}}{{d}^{2}} and it has 2 lone pairs, the shape of XeF4Xe{{F}_{4}} is square planar.

Hence, the correct answer is an option (d)- Square planar, sp3d2s{{p}^{3}}{{d}^{2}}

Note: Whenever you are drawing the compound structure the number of lone pairs should also be considered. In this example also there are 4 fluorine atoms with xenon, so you could get confused between tetrahedral and square planar shape.