Concept:
Hybridization describes the mixing of atomic orbitals to form new hybrid orbitals. Different atoms in a molecule can exhibit different hybridizations.
Step 1: Benzene (A).
All carbon atoms in benzene are sp\textsuperscript{2} hybridized due to planar ring structure.
Thus, no sp\textsuperscript{3} present.
Step 2: Methanol (B).
Methanol (CH\textsubscript{3}OH) contains only sp\textsuperscript{3} hybridized atoms.
Thus, no sp\textsuperscript{2} present.
Step 3: Ethene (C).
Ethene contains only sp\textsuperscript{2} hybridized carbon atoms.
Thus, no sp\textsuperscript{3} present.
Step 4: Acetone (D).
Acetone (CH\textsubscript{3}-CO-CH\textsubscript{3}) contains:
• Carbonyl carbon (C=O) → sp\textsuperscript{2}
• Methyl carbons (CH\textsubscript{3}) → sp\textsuperscript{3}
Thus, both types of hybridization are present.
Step 5: Final evaluation.
Only acetone shows both sp\textsuperscript{2} and sp\textsuperscript{3} hybridization.
Final Conclusion:
Hence, the correct answer is option (4).