Step 1: Understanding the Concept:
Orbital hybridization is a valence bond theory concept where atomic valence orbitals blend together to create an equal number of identical hybridized orbitals. These new hybrid orbitals rearrange themselves in space to minimize electron-pair repulsion, determining the molecule's geometric shape.
Step 2: Key Formula or Approach:
The hybridization state of a central atom can be determined by calculating its Steric Number ($Z$):
$$ Z = (\text{Number of } \sigma \text{ bonds attached to central atom}) + (\text{Number of lone pairs on central atom}) $$
Once computed, the value of the Steric Number points directly to a specific hybridization state:
- $Z = 2 \implies \text{sp}$ hybridization (Linear geometry)
- $Z = 3 \implies \text{sp}^2$ hybridization (Trigonal Planar geometry)
- $Z = 4 \implies \text{sp}^3$ hybridization (Tetrahedral geometry)
Step 3: Detailed Explanation:
Let's apply the steric number calculation to the central carbon atom in a methane ($\text{CH}_4$) molecule:
1. Count Valence Electrons: Carbon belongs to Group 14 and possesses $4$ valence electrons.
2. Identify Bonds: In $\text{CH}_4$, the carbon atom shares its $4$ valence electrons to form $4$ single covalent bonds with $4$ separate hydrogen atoms. Each single bond represents exactly $1\sigma$ bond. Therefore, the number of $\sigma$ bonds is $4$.
3. Check for Lone Pairs: Since all $4$ of carbon's valence electrons are fully utilized in bonding, there are no unshared electrons remaining on the central atom. The number of lone pairs is $0$.
Now, calculate the final Steric Number ($Z$):
$$ Z = 4 \text{ ($\sigma$ bonds)} + 0 \text{ (lone pairs)} = 4 $$
A steric number of $4$ requires blending one $s$ orbital and three $p$ orbitals to form four identical $\text{sp}^3$ hybrid orbitals. These orbitals point toward the corners of a regular tetrahedron with ideal bond angles of $109.5^\circ$. This matches option (C).
Step 4: Final Answer:
The hybridization of the carbon atom in $\text{CH}_4$ is $\text{sp}^3$.