Question #13275

1 Answer
Jun 25, 2017

Answer:

My predicted order is
#"H"_2 < "H"_2"S < HCl < CCl"_4 < "H"_2"O" < "NaBr < MgF"_2 < "Ca"_3("PO"_4)_2#.

Explanation:

#"NaBr, MgF"_2# and #"Ca"_3("PO"_4)_2# are all ionic compounds with strong ionic attractions.

The attractions are smallest in #"NaBr"# (only +1 ions), stronger in #"MgF"_2# ( +1 and +2 ions) and strongest in #"Ca"_3("PO"_4)_2# (+2 and +3 ions).

The remaining compounds are all covalent.

#"H"_2# — Lowest boiling point because it has low molecular mass and weak London dispersion forces of attraction.

#"H"_2"S"# — Higher molecular mass and higher London dispersion forces. The #"S-H"# bonds are almost nonpolar, so the dipole-dipole forces are quite weak.

#"HCl"# — Polar bond, so stronger dipole-dipole forces.

#"CCl"_4# — Nonpolar molecule, but more atoms and high molecular mass. ∴ Strong London dispersion forces.

#"H"_2"O"# — Strong hydrogen bonds between the molecules.

Thus, the predicted order of boiling points is

#"H"_2 < "H"_2"S < HCl < CCl"_4 < "H"_2"O" < "NaBr < MgF"_2 < "Ca"_3("PO"_4)_2#