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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

Expert verified
(a) \(\mathrm{TiCl}_{4}\) is an ionic substance named Titanium(IV) Chloride, and \(\mathrm{CaF}_{2}\) is also an ionic substance named Calcium Fluoride. (b) \(\mathrm{ClF}_{3}\) is a molecular substance named Chlorine Trifluoride, and \(\mathrm{VF}_{3}\) is an ionic substance named Vanadium(III) Fluoride. (c) \(\mathrm{SbCl}_{5}\) is a molecular substance named Antimony Pentachloride, and \(\mathrm{AlF}_{3}\) is an ionic substance named Aluminum Fluoride.

Step by step solution

01

Identify the type of bond in each compound

In \(\mathrm{TiCl}_{4}\), titanium (Ti) is a metal, and chlorine (Cl) is a non-metal, which indicates an ionic bond. In \(\mathrm{CaF}_{2}\), calcium (Ca) is a metal while fluorine (F) is a non-metal, therefore an ionic bond is present in this compound as well.
02

Name the compounds using the appropriate naming conventions

Since both compounds are ionic, we will use the ionic naming conventions. \(\mathrm{TiCl}_{4}\) is named as Titanium(IV) Chloride. \(\mathrm{CaF}_{2}\) is named as Calcium Fluoride. (b) \(\mathrm{ClF}_{3}\) and \(\mathrm{VF}_{3}\)
03

Identify the type of bond in each compound

In \(\mathrm{ClF}_{3}\), both chlorine (Cl) and fluorine (F) are non-metals, which indicates it's more likely to have a covalent bond making it a molecular substance. In \(\mathrm{VF}_{3}\), vanadium (V) is a metal and fluorine (F) is a non-metal, which indicates an ionic bond.
04

Name the compounds using the appropriate naming conventions

For the molecular compound \(\mathrm{ClF}_{3}\), we use the molecular naming convention: Chlorine Trifluoride. For the ionic compound \(\mathrm{VF}_{3}\), we use the ionic naming convention: Vanadium(III) Fluoride. (c) \(\mathrm{SbCl}_{5}\) and \(\mathrm{AlF}_{3}\)
05

Identify the type of bond in each compound

In \(\mathrm{SbCl}_{5}\), antimony (Sb) is a metalloid, and chlorine (Cl) is a non-metal, which indicates it's more likely to have a covalent bond making it a molecular substance. In \(\mathrm{AlF}_{3}\), aluminum (Al) is a metal, and fluorine (F) is a non-metal, which indicates an ionic bond.
06

Name the compounds using the appropriate naming conventions

For the molecular compound \(\mathrm{SbCl}_{5}\), we use the molecular naming convention: Antimony Pentachloride. For the ionic compound \(\mathrm{AlF}_{3}\), we use the ionic naming convention: Aluminum Fluoride.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ionic Compounds
Ionic compounds are formed when a metal bonds with a non-metal. This type of bonding involves the transfer of electrons. Generally, the metal donates electrons to the non-metal. In the process, the metal becomes a positively charged ion called a cation, while the non-metal becomes a negatively charged ion called an anion. These opposite charges attract, holding the ions together in an ionic bond.
These compounds usually have high melting and boiling points because the ionic bonds are very strong. They often form crystalline structures and can conduct electricity when dissolved in water.
In the exercise, compounds like \(\mathrm{CaF}_{2}\) and \(\mathrm{AlF}_{3}\) are examples of ionic compounds. Calcium, aluminum, and the non-metal fluorine, form strong ionic bonds due to the transfer of electrons.
Molecular Compounds
Molecular compounds, also known as covalent compounds, are formed when two or more non-metal atoms share electrons. This sharing results in a covalent bond. Unlike ionic bonds, there isn't a transfer of electrons. Instead, the atoms attain stable electron configurations by sharing electrons.
Molecular compounds tend to have lower melting and boiling points compared to ionic compounds. They might not conduct electricity, as they don't form ions in solutions.
Consider the compound \(\mathrm{ClF}_{3}\) from the exercise. Chlorine and fluorine are both non-metals, and they share electrons, making \(\mathrm{ClF}_{3}\) a molecular compound. The naming reflects how many atoms of each are present, for example, Chlorine Trifluoride, indicating one chlorine and three fluorine atoms.
Naming Conventions
The naming conventions for ionic and molecular compounds are different due to their distinct bond types.
  • Ionic Compounds: Usually named by stating the cation first and then the anion. If the metal can have different charges (transition metals), the specific charge is indicated by a Roman numeral in parentheses. For example, \(\mathrm{VF}_{3}\) is called Vanadium(III) Fluoride.
  • Molecular Compounds: These use prefixes to denote the numbers of atoms. Mono- is often omitted for the first element. Examples include mono-, di-, tri-, tetra-, etc. For example, \(\mathrm{ClF}_{3}\) is named Chlorine Trifluoride, reflecting the presence of three fluorine atoms.
These systematic naming patterns help universally identify compounds by their chemical composition.

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Most popular questions from this chapter

The following three Lewis structures can be drawn for \(\mathrm{N}_{2} \mathrm{O}:\) \(\mathrm{N} \equiv \mathrm{N}-\ddot{\mathrm{O}}: \longleftrightarrow: \mathrm{N}^{*}-\mathrm{N} \equiv \mathrm{O}: \longleftrightarrow: \ddot{\mathrm{N}}=\mathrm{N}=\ddot{\mathrm{O}}:\) (a) Using formal charges, which of these three resonance forms is likely to be the most important? (b) The \(\mathrm{N}-\mathrm{N}\) bond length in \(\mathrm{N}_{2} \mathrm{O}\) is \(1.12 \mathrm{~A}\), slightly longer than a typical \(\mathrm{N} \equiv \mathrm{N}\) bond; and the \(\mathrm{N}-\mathrm{O}\) bond length is \(1.19 \AA\), slightly shorter than a typical \(\mathrm{N}=\mathrm{O}\) bond. (See Table \(8.5 .\) ) Rationalize these observations in terms of the resonance structures shown previously and your conclusion for part (a).

Consider the following statement: "For some molecules and ions, a Lewis structure that satisfies the octet rule does not lead to the lowest formal charges, and a Lewis structure that leads to the lowest formal charges does not satisfy the octet rule." Illustrate this statement using the hydrogen sulfite ion, \(\mathrm{HSO}_{3}^{-}\), as an example (the \(\mathrm{H}\) atom is bonded to one of the \(\mathrm{O}\) atoms).

Which ionic compound is expected to form from combining the following pairs of elements: (a) barium and fluorine, (b) cesium and chlorine, (c) lithium and nitrogen, (d) aluminum and oxygen?

(a) Triazine, \(\mathrm{C}_{3} \mathrm{H}_{3} \mathrm{~N}_{3}\), is like benzene except that in triazine every other \(\mathrm{C}-\mathrm{H}\) group is replaced by a nitrogen atom. Draw the Lewis structure(s) for the triazine molecule. (b) Estimate the carbon-nitrogen bond distances in the ring.

Using Lewis symbols and Lewis structures, diagram the formation of \(\mathrm{SiCl}_{4}\) from \(\mathrm{Si}\) and \(\mathrm{Cl}\) atoms.

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