
Atomic orbitals involved in the hybridization of the $S{F_6}$ molecule:
A. $3s,3{p_x},3{p_y},3{p_z},3{d_{{z^2}}},3{d_{{x^2} - {y^2}}}$
B. $3s,3{p_x},3{p_y},3{p_z},3{d_{{x^2}}}$
C. $4s,4{p_x},4{p_y},4{p_z},3{d_{{x^2} - {y^2}}},{d_{xy}}$
D. $3s,3{p_x},3{p_y},3{d_{xy}},3{d_{yz}},3{d_{xz}}$
Answer
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Hint: The atom sulfur has six valence electrons of which all are attached to fluorine atoms. One can draw the electronic configuration of a sulfur atom and decide which orbitals are involved in hybridization.
Complete step by step answer:
-First of all we will learn how to calculate the hybridization of the $S{F_6}$molecule. The molecule of $S{F_6}$ the central atom is sulfur which has six valence electrons in it.
-All the six valences of the sulfur atom are attached to the fluorine atom which means there are six bond pairs and zero lone pairs. Hence, the coordination number in the molecule $S{F_6}$ is six.
-The coordination number six gives us the hybridization as $s{p^3}{d^2}$ which means there is one electron of s-orbital, three electrons of p-orbital, and two electrons of d-orbital that are involved in the hybridization of the $S{F_6}$ molecule.
-Now let us write the electronic configuration of the $S{F_6}$ molecule of which electrons are involved in the hybridization,
$3{s^1}3{p^3}3{d^2}$
This shows that the two electrons from the p-orbital got promoted to d-orbital which involve in the hybridization of the molecule.
-When the electrons from the p-orbital get promoted to d-orbital the electrons get filled first in the ${d_{{z^2}}}$ and ${d_{{x^2} - {y^2}}}$. The orbitals in the p-orbital which are involved in hybridization
-Therefore, we can say that the orbitals which are involved in the hybridization of the $S{F_6}$ molecule are $3s,3{p_x},3{p_y},3{p_z},3{d_{{z^2}}},3{d_{{x^2} - {y^2}}}$
So, the correct answer is “Option A”.
Note:
The coordination number of the $S{F_6}$ molecule is six hence, the geometry of the $S{F_6}$ molecule is octahedral in which the six $s{p^2}{d^2}$ hybrid orbitals are projected towards the six corners of a regular octahedron.
Complete step by step answer:
-First of all we will learn how to calculate the hybridization of the $S{F_6}$molecule. The molecule of $S{F_6}$ the central atom is sulfur which has six valence electrons in it.
-All the six valences of the sulfur atom are attached to the fluorine atom which means there are six bond pairs and zero lone pairs. Hence, the coordination number in the molecule $S{F_6}$ is six.
-The coordination number six gives us the hybridization as $s{p^3}{d^2}$ which means there is one electron of s-orbital, three electrons of p-orbital, and two electrons of d-orbital that are involved in the hybridization of the $S{F_6}$ molecule.
-Now let us write the electronic configuration of the $S{F_6}$ molecule of which electrons are involved in the hybridization,
$3{s^1}3{p^3}3{d^2}$
This shows that the two electrons from the p-orbital got promoted to d-orbital which involve in the hybridization of the molecule.
-When the electrons from the p-orbital get promoted to d-orbital the electrons get filled first in the ${d_{{z^2}}}$ and ${d_{{x^2} - {y^2}}}$. The orbitals in the p-orbital which are involved in hybridization
-Therefore, we can say that the orbitals which are involved in the hybridization of the $S{F_6}$ molecule are $3s,3{p_x},3{p_y},3{p_z},3{d_{{z^2}}},3{d_{{x^2} - {y^2}}}$
So, the correct answer is “Option A”.
Note:
The coordination number of the $S{F_6}$ molecule is six hence, the geometry of the $S{F_6}$ molecule is octahedral in which the six $s{p^2}{d^2}$ hybrid orbitals are projected towards the six corners of a regular octahedron.
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