Benzene, C6H6, has the structure shown below. Considering the observation that benzene is only spari — Bonding Chemistry Question

STEPS:

1. Analyze the polarity of the two molecules:
* Water ($\text{H}_2\text{O}$): A highly polar molecule with a permanent dipole due to its bent geometry and the significant electronegativity difference between oxygen and hydrogen.
* Benzene ($\text{C}_6\text{H}_6$): A flat, highly symmetric hexagonal ring of carbon atoms with hydrogen atoms pointing outward. Because carbon and hydrogen have very similar electronegativities, and because the hexagonal ring is perfectly symmetrical, all individual bond dipoles cancel out, making benzene a nonpolar molecule.
2. Determine the interactions between a polar and a nonpolar molecule: When a polar water molecule (with a permanent dipole) approaches a nonpolar benzene molecule, the electric field of the water molecule distorts the symmetric electron cloud of the benzene molecule. This temporarily induces a weak dipole in the benzene ring. The electrostatic attraction between water's permanent dipole and benzene's newly formed temporary dipole is called a dipole-induced dipole interaction.
3. Identify other universal intermolecular forces: London dispersion forces (LDFs) arise from instantaneous fluctuations in electron density and are present between *all* molecules, whether they are polar or nonpolar. Thus, LDFs also contribute to the weak attraction between water and benzene.
4. Relate these forces to solubility: Because the dipole-induced dipole and London dispersion attractions between water and benzene are much weaker than the highly favorable hydrogen bonds water molecules must break to make room for benzene, benzene is only sparingly soluble in water. This points directly to Option D.

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WHY_OTHERS_WRONG:

• Option A is incorrect: Although benzene is nonpolar, it is a common misconception that no intermolecular forces exist between polar and nonpolar substances. Weak electrostatic attractions, specifically dipole-induced dipole and London dispersion forces, still exist between them.
• Option B is incorrect: Hydrogen bonding requires a hydrogen atom to be covalently bonded to a highly electronegative atom ($\text{N}$, $\text{O}$, or $\text{F}$). In benzene, the hydrogen atoms are covalently bonded to carbon. Because carbon is not highly electronegative, these $\text{C}-\text{H}$ groups cannot form hydrogen bonds with water.
• Option C is incorrect: "Hydrophobic" describes the macroscopic behavior where nonpolar substances do not mix well with water because the water-water hydrogen bonds are much stronger than any water-solute attractions. It does not mean there is a physical "net repulsion" pushing the molecules apart; there is still a weak attractive force between them.