Question

In the following pairs of binary compounds determine which one is a molecular substance and which one is an ionic substance. Use the appropriate naming convention (for ionic or molecular substances) to assign a name to each compound: (a) \(\mathrm{TiCl}_{4}\) and \(\mathrm{CaF}_{2}\), (b) \(\mathrm{ClF}_{3}\) and \(\mathrm{VF}_{3}\), (c) \(\mathrm{SbCl}_{5}\) and \(\mathrm{AlF}_{3}\).

Short answer

In short, we identified the type of compounds and assigned appropriate names as follows: (a) \(\mathrm{TiCl}_{4}\): Titanium(IV) Chloride (ionic) and \(\mathrm{CaF}_{2}\): Calcium Fluoride (ionic) (b) \(\mathrm{ClF}_{3}\): Chlorine Trifluoride (molecular) and \(\mathrm{VF}_{3}\): Vanadium(III) Fluoride (ionic) (c) \(\mathrm{SbCl}_{5}\): Antimony Pentachloride (molecular) and \(\mathrm{AlF}_{3}\): Aluminum Fluoride (ionic)

Step-by-step solution

Identify the elements involved in each compound

For \(\mathrm{TiCl}_{4}\), we have Titanium (Ti) and Chlorine (Cl). For \(\mathrm{CaF}_{2}\), we have Calcium (Ca) and Fluorine (F).

Determine the type of compound

Referring to the periodic table, we can see that Titanium (Ti) and Calcium (Ca) are metals. Chlorine (Cl) and Fluorine (F) are non-metals. Therefore, both compounds consist of a metal and non-metal combination, which means both substances are ionic compounds.

Assign names using the ionic naming convention

Using the ionic naming convention, we have: \(\mathrm{TiCl}_{4}\): Titanium(IV) Chloride \(\mathrm{CaF}_{2}\): Calcium Fluoride (b) \(\mathrm{ClF}_{3}\) and \(\mathrm{VF}_{3}\)

Identify the elements involved in each compound

For \(\mathrm{ClF}_{3}\), we have Chlorine (Cl) and Fluorine (F). For \(\mathrm{VF}_{3}\), we have Vanadium (V) and Fluorine (F).

Determine the type of compound

Chlorine (Cl) and Fluorine (F) are both non-metals, so \(\mathrm{ClF}_{3}\) is a molecular substance. Vanadium (V) is a metal, and Fluorine (F) is a non-metal, so \(\mathrm{VF}_{3}\) is an ionic substance.

Assign names using the appropriate naming conventions

Using the molecular naming convention for \(\mathrm{ClF}_{3}\) and the ionic naming convention for \(\mathrm{VF}_{3}\), we have: \(\mathrm{ClF}_{3}\): Chlorine Trifluoride \(\mathrm{VF}_{3}\): Vanadium(III) Fluoride (c) \(\mathrm{SbCl}_{5}\) and \(\mathrm{AlF}_{3}\)

Identify the elements involved in each compound

For \(\mathrm{SbCl}_{5}\), we have Antimony (Sb) and Chlorine (Cl). For \(\mathrm{AlF}_{3}\), we have Aluminum (Al) and Fluorine (F).

Determine the type of compound

Antimony (Sb) is a metalloid, while Chlorine (Cl) is a non-metal. Considering the properties of metalloids, for this case, we can treat Antimony (Sb) as a non-metal, thus \(\mathrm{SbCl}_{5}\) is a molecular substance. Aluminum (Al) is a metal, and Fluorine (F) is a non-metal, so \(\mathrm{AlF}_{3}\) is an ionic substance.

Assign names using the appropriate naming conventions

Using the molecular naming convention for \(\mathrm{SbCl}_{5}\) and the ionic naming convention for \(\mathrm{AlF}_{3}\), we have: \(\mathrm{SbCl}_{5}\): Antimony Pentachloride \(\mathrm{AlF}_{3}\): Aluminum Fluoride