Two chlorinated methanes are compared at

CHEMISTRY · Multi-part · AP · Chemistry · Properties of Substances and Mixtures · Properties of Substances and Mixtures

Problem

Two chlorinated methanes are compared at 298 K298\,\text{K}. • Compound A: carbon tetrachloride (CCl4_4); molar mass =153.8 g mol−1= 153.8\,\text{g mol}^{-1}; dipole moment ≈0 D\approx 0\,\text{D}; isotropic molecular polarizability α=10.2 A˚3\alpha = 10.2\,\text{\AA}^3. • Compound B: dichloromethane (CH2_2Cl2_2); molar mass =84.9 g mol−1= 84.9\,\text{g mol}^{-1}; dipole moment =1.60 D= 1.60\,\text{D}; isotropic molecular polarizability α=5.4 A˚3\alpha = 5.4\,\text{\AA}^3. Assume that, to a first approximation, London dispersion interaction energy between two identical molecules is proportional to their polarizability α\alpha, whereas the maximum dipole–dipole interaction energy scales with the square of the dipole moment μ2\mu^2. Using only the information provided, address the following. (a) Calculate the ratio of the estimated London dispersion interaction strength for a pair of CCl4_4 molecules to that for a pair of CH2_2Cl2_2 molecules. (b) For CH2_2Cl2_2, calculate μ2\mu^2 (in D2^2) and compare it to the value for CCl4_4. What does this comparison imply about the presence or absence of a dipole–dipole term in the two substances? (c) Based on your results in (a) and (b), predict which compound is expected to have the higher normal boiling point. Provide a concise justification that refers explicitly to both intermolecular force types.

Related theory

Browse the problem library