Question #13275

1 Answer
Jun 25, 2017

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