Step 1: Understanding the Question:
We need to determine the type of atomic orbital hybridization for the central sulfur (S) atom in the sulfur hexafluoride (\(SF_6\)) molecule.
Step 2: Key Formula or Approach:
Hybridization can be determined quickly using the steric number (SN) method.
\[
\text{Steric Number (SN)} = (\text{Number of atoms bonded to central atom}) + (\text{Number of lone pairs on central atom})
\]
The hybridization is then determined by the SN value:
- SN = 2 \(\rightarrow\) sp
- SN = 3 \(\rightarrow\) sp\(^2\)
- SN = 4 \(\rightarrow\) sp\(^3\)
- SN = 5 \(\rightarrow\) sp\(^3\)d
- SN = 6 \(\rightarrow\) sp\(^3\)d\(^2\)
Step 3: Detailed Explanation:
(i) Identify the central atom and its valence electrons:
The central atom is Sulfur (S). It is in Group 16 of the periodic table, so it has 6 valence electrons.
(ii) Count bonded atoms and lone pairs:
- In \(SF_6\), the sulfur atom is bonded to six Fluorine (F) atoms. So, the number of bonded atoms is 6.
- Each single bond uses one of sulfur's valence electrons. Since there are 6 bonds, all 6 valence electrons of sulfur are used for bonding.
- Number of lone pairs on S = \(\frac{1}{2}(\text{Valence e}^- - \text{Bonding e}^-) = \frac{1}{2}(6 - 6) = 0\).
(iii) Calculate the Steric Number (SN):
\[
\text{SN} = (\text{Number of bonded atoms}) + (\text{Number of lone pairs}) = 6 + 0 = 6
\]
(iv) Determine the hybridization:
A steric number of 6 corresponds to \(sp^3d^2\) hybridization. This leads to an octahedral geometry.
Step 4: Final Answer:
The hybridization of the central sulfur atom in \(SF_6\) is \(sp^3d^2\).