Question

Which among the following are having ${{s}}{{{p}}^3}{{d}}$ hybridization of the central atom?
A. ${{Xe}}{{{F}}_4}$
B. ${{Xe}}{{{O}}_2}{{{F}}_2}$
C. ${{ClO}}_3^ -$
D. ${{Br}}{{{F}}_3}$

Answer 1

Hybridization of a molecule can be determined using VSEPR theory, i.e. valence shell electron pair repulsion theory. The main principle of this theory is that the electrons repel each other. So the atoms in a molecule get separated from each other to avoid the repulsion.

Complete answer:
Let’s find the molecule having ${{s}}{{{p}}^3}{{d}}$ hybridization by finding the hybridization of each of the molecules. Hybridization can be calculated using steric numbers. The formula of steric number is given below:
Steric number $=$ $\dfrac{1}{2}$ (number of valence electrons on central atom $+$ number of atoms bonded to the central atom $-$ positive charge on molecule $+$ negative charge on the molecule)
A. Valence electrons of ${{Xe}}$ is $8$ . There are four ${{F}}$ atoms. There are no positive or negative charges on the molecule.
Thus the steric number $= \dfrac{1}{2}\left( {8 + 4 - 0 + 0} \right) = 6$
Molecules having steric number $6$ have hybridization ${{s}}{{{p}}^3}{{{d}}^2}$ or ${{{d}}^2}{{s}}{{{p}}^3}$.
B. Similarly let’s calculate the steric number of ${{Xe}}{{{O}}_2}{{{F}}_2}$. Valence electrons of ${{Xe}}$ is $8$. There are only two monovalent atoms bonded to the central atom, i.e. two fluorine atoms. Oxygen is a divalent atom.
Thus the steric number $= \dfrac{1}{2}\left( {8 + 2 - 0 + 0} \right) = 5$
Molecules having steric number $5$ have hybridization ${{s}}{{{p}}^3}{{d}}$.
C. Here, the central atom is chlorine. The valence electrons of chlorine is $7$. There are no monovalent atoms, but there is an anionic charge.
Thus the steric number $= \dfrac{1}{2}\left( {7 + 0 - 0 + 1} \right) = 4$
Molecules having steric number $4$ have hybridization ${{s}}{{{p}}^3}$.
D. In ${{Br}}{{{F}}_3}$, bromine is the central atom having $7$ valence electrons and three monovalent atoms.
Thus the steric number $= \dfrac{1}{2}\left( {7 + 3 - 0 + 0} \right) = 5$
Molecules having steric number $5$ have hybridization ${{s}}{{{p}}^3}{{d}}$.
Thus the molecules having ${{s}}{{{p}}^3}{{d}}$ hybridization are ${{Xe}}{{{O}}_2}{{{F}}_2}$ and ${{Br}}{{{F}}_3}$.

Hence the correct options are B and D.

Note:
The shape of ${{Xe}}{{{O}}_2}{{{F}}_2}$ may be trigonal bipyramidal or see-saw. The bond angles are ${91^ \circ },{105^ \circ },{174^ \circ }$. In the molecule ${{Br}}{{{F}}_3}$, the bond angle is ${86^ \circ }$ and the shape is either trigonal bi pyramidal or octahedral geometry. These assumptions are made using VSEPR theory. The exact information has been introduced in several other theories.