Question

What are the strongest attractive forces that must be overcome to (a) boil silicon hydride \(\mathrm{SiH}_{4} ?\) (b) vaporize calcium chloride? (c) dissolve \(\mathrm{Cl}_{2}\) in carbon tetrachloride, \(\mathrm{CCl}_{4} ?\) (d) melt iodine?

Short answer

a) Boil silicon hydride SiH4 b) Vaporize calcium chloride c) Dissolve Cl2 in carbon tetrachloride CCl4 d) Melt iodine

Step-by-step solution

a) Boil silicon hydride SiH4

To boil silicon hydride (SiH4), the strongest attractive forces to overcome are the van der Waals forces (or London dispersion forces) between SiH4 molecules since SiH4 is a nonpolar covalent compound. The stronger these forces, the higher the boiling point it will have.

b) Vaporize calcium chloride

To vaporize calcium chloride (CaCl2), the strongest attractive forces to overcome are the ionic bonds between the positively charged calcium ions (Ca2+) and negatively charged chloride ions (Cl-) in the crystal lattice, as CaCl2 is an ionic compound. The strength of the ionic bond depends on the charges and sizes of the ions involved; the higher the charge and smaller the size, the stronger the bond.

c) Dissolve Cl2 in carbon tetrachloride CCl4

To dissolve Cl2 (chlorine) in carbon tetrachloride (CCl4), the attractive forces to overcome are London dispersion forces present between the nonpolar chlorine molecules and nonpolar CCl4 molecules. Cl2 and CCl4 have similar polarities, which makes it easier for Cl2 to dissolve in CCl4. However, these van der Waals forces between molecules must still be overcome for the dissolving process to occur.

d) Melt iodine

To melt iodine (I2), the strongest attractive forces to overcome are the London dispersion forces between iodine molecules in the solid-state. Iodine is a nonpolar covalent compound with a relatively high molecular weight, which results in stronger van der Waals forces and, thus, a higher melting point compared to other Group 17 elements (except for astatine).