Concept:Heat of vaporization reflects the energy required to overcome intermolecular forces. The molecular geometry dictates the overall polarity of these tri-atomic molecules.
Formula:$$ \Delta H_{\text{vap}} \propto \text{Strength of IMF} $$
Solution:- \( \text{CO}_2 \) has a linear molecular geometry (\( \text{O=C=O} \)). Its individual bond dipoles cancel out perfectly, resulting in a net dipole moment of zero. It only experiences weak London dispersion forces.
- \( \text{SO}_2 \) has a bent angular geometry due to a lone pair on the central Sulfur atom. Its bond dipoles do not cancel, giving it a net dipole moment (polar molecule).
- Because \( \text{SO}_2 \) is polar, it exhibits stronger dipole-dipole interactions, thereby requiring significantly more energy (higher heat of vaporization) to turn into gas.
Why other options are incorrect:Size difference and acidity do not primarily account for the drastic difference in intermolecular force strength compared to the fundamental difference in molecular polarity.
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