Question:
Answer the following questions.
A. Why is NaI(s) very soluble in water, whereas I2(s) has very low solubility?
B. Why is the boiling point of CCl4 much lower than the boiling point of CBr4? (Hint: Br is a much larger atom than Cl)
C. Arrange these solids in order of boiling point, lowest first: CH3OH, C2H5OH, O2, NaCl.
D. Which of the following substances involve hydrogen bonding: HBr, H2O, HCl, HF, NH4, AsH4?
Answers:
A. NaI is an ionic solid, and strong ion-dipole interactions between the ions and water will dissolve the NaI; I2 is a non-polar molecule, so polar water will not readily dissolve it.
B. CBr4 is a larger molecule (because Br is a larger atom) than CCl4, leading to stronger dispersion forces and a higher boiling point.
C. O2, CH3OH, C2H5OH, NaCl (because O2 has only dispersion attractions; CH3OH is a small, hydrogen-bonded molecule; C2H5OH is a larger hydrogen-bonded molecule with lower attractive forces; and NaCl is an ionic liquid with very high ion-ion attractive forces).
D. H2), HF, and NH4 involve hydrogen bonding (only hydrogens hooked to an O, N, or F create hydrogen bonding).