To determine the atomic orbitals involved in the hybridization of ${\mathrm{SF}}_6$ molecule, we first need to find the hybridization of the central sulfur atom.

Step 1: Determine the steric number of the central atom (Sulfur).
The steric number (SN) is calculated as: SN = (Number of sigma bonds) + (Number of lone pairs).
In ${\mathrm{SF}}_6$, Sulfur (S) is the central atom and is bonded to six Fluorine (F) atoms. Sulfur is in Group 16, so it has 6 valence electrons. Each Fluorine atom forms one single bond with Sulfur. Thus, there are 6 sigma bonds and 0 lone pairs on Sulfur.
SN = 6 + 0 = 6.

Step 2: Determine the hybridization based on the steric number.
A steric number of 6 corresponds to ${\mathrm{sp}}^3{\mathrm{d}}^2$ hybridization.

Step 3: Identify the atomic orbitals involved in ${\mathrm{sp}}^3{\mathrm{d}}^2$ hybridization.
${\mathrm{sp}}^3{\mathrm{d}}^2$ hybridization involves one s orbital, three p orbitals, and two d orbitals. For elements in the third period and beyond, the s and p orbitals are from the same principal quantum number (e.g., 3s, 3p) and the d orbitals are also from the same principal quantum number (e.g., 3d) or sometimes from the next lower principal quantum number if the energy difference is small. In the case of Sulfur, the valence shell is the 3rd shell, so the orbitals involved are 3s, 3p, and 3d.

The specific d orbitals involved in ${\mathrm{sp}}^3{\mathrm{d}}^2$ hybridization, which leads to an octahedral geometry, are ${\mathrm{d}}_{z^2}$ and ${\mathrm{d}}_{x^2-y^2}$. These two d orbitals have lobes directed along the axes, which are suitable for forming bonds in an octahedral arrangement.

Therefore, the atomic orbitals involved are $3s,3p_x,3p_y,3p_z,3d_{z^2},3d_{x^2-y^2}$.

Option Analysis:

• A) $3s,3p_x,3p_y,3p_z,3d_{z^2},3d_{x^2-y^2}$: This option correctly lists one s, three p, and the two specific d orbitals (${\mathrm{d}}_{z^2}$ and ${\mathrm{d}}_{x^2-y^2}$) that participate in ${\mathrm{sp}}^3{\mathrm{d}}^2$ hybridization for a third-period element like Sulfur. This is the correct set of orbitals.
• B) $3s,3p_x,3p_y,3p_z,3d_{z^2}$: This set corresponds to ${\mathrm{sp}}^3\mathrm{d}$ hybridization (steric number 5), which would result in trigonal bipyramidal geometry, not octahedral.
• C) $4s,4p_x,4p_y,4p_z,3d_{x^2-y^2},d_{xy}$: This option incorrectly uses 4s and 4p orbitals for Sulfur (which is in the 3rd period) and also lists an incorrect set of d orbitals for ${\mathrm{sp}}^3{\mathrm{d}}^2$ hybridization.
• D) $3s,3p_x,3p_y,3d_{xy},3d_{yz},3d_{xz}$: This option lists an incorrect combination of p and d orbitals, and the d orbitals listed (${\mathrm{d}}_{xy},{\mathrm{d}}_{yz},{\mathrm{d}}_{xz}$) are not the ones typically involved in ${\mathrm{sp}}^3{\mathrm{d}}^2$ hybridization for octahedral geometry, as their lobes lie between the axes.

Correct Answer: (A)