Problem 46 - Chemistry: The Central Science

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 \(\mathrm{AlF}_{3}\).

Step-by-Step Solution

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 1: 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).

Step 2: 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.

Step 3: 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 \(\mathrm{VF}_{3}\)

Step 4: 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).

Step 5: 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.

Step 6: 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 \(\mathrm{AlF}_{3}\)

Step 7: 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).

Step 8: 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.

Step 9: 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