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Chapter 8: Problem 45

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) \(\mathrm{SiF}_{4}\) and \(\mathrm{LaF}_{3}\), (b) \(\mathrm{FeCl}_{2}\) and \(\mathrm{ReCl}_{6}\), (c) \(\mathrm{PbCl}_{4}\) and \(\mathrm{RbCl}\).

Short Answer

Expert verified
(a) \(\mathrm{SiF}_{4}\) is a molecular compound named Silicon tetrafluoride, and \(\mathrm{LaF}_{3}\) is an ionic compound named Lanthanum fluoride. (b) \(\mathrm{FeCl}_{2}\) is an ionic compound named Iron(II) chloride, and \(\mathrm{ReCl}_{6}\) is named Rhenium hexachloride. (c) \(\mathrm{PbCl}_{4}\) is an ionic compound named Lead(IV) chloride, and \(\mathrm{RbCl}\) is an ionic compound named Rubidium chloride.

Step by step solution

01

Identify the type of compound (ionic or molecular)

First, we need to identify whether the given compounds are ionic or molecular based on the elements involved. (a) \(\mathrm{SiF}_{4}\): Si (Silicon) is a metalloid and F (Fluorine) is a non-metal. Since it is a compound of a metalloid and a non-metal, it is a molecular compound. \(\mathrm{LaF}_{3}\): La (Lanthanum) is a metal and F (Fluorine) is a non-metal. Since it is a compound of a metal and a non-metal, it is an ionic compound. (b) \(\mathrm{FeCl}_{2}\): Fe (Iron) is a metal and Cl (Chlorine) is a non-metal. Since it is a compound of a metal and a non-metal, it is an ionic compound. \(\mathrm{ReCl}_{6}\): Re (Rhenium) is a metal and Cl (Chlorine) is a non-metal. Since it is a compound of a metal and a non-metal, it is an ionic compound. (c) \(\mathrm{PbCl}_{4}\): Pb (Lead) is a metal and Cl (Chlorine) is a non-metal. Since it is a compound of a metal and a non-metal, it is an ionic compound. \(\mathrm{RbCl}\): Rb (Rubidium) is a metal and Cl (Chlorine) is a non-metal. Since it is a compound of a metal and a non-metal, it is an ionic compound. Now that we have identified the type of each compound, we can use the appropriate naming convention for each.
02

Name the compounds using appropriate naming conventions

(a) For molecular compound \(\mathrm{SiF}_{4}\), we can use the following naming convention: element(prefix for the number of atoms) + second element(prefix for the number of atoms). In this case, the name is Silicon tetrafluoride. For ionic compound \(\mathrm{LaF}_{3}\), we can use the following naming convention: cation (metal) + anion (non-metal with -ide suffix). In this case, the name is Lanthanum fluoride. (b) For ionic compound \(\mathrm{FeCl}_{2}\), we can use the naming convention: cation with the roman numeral indicating its charge + anion with -ide suffix. In this case, the name is Iron(II) chloride. For ionic compound \(\mathrm{ReCl}_{6}\), the name is Rhenium hexachloride using molecular compound naming since it's a coordination complex. (c) For ionic compound \(\mathrm{PbCl}_{4}\), we can use the naming convention: cation with the roman numeral indicating its charge + anion with -ide suffix. In this case, the name is Lead(IV) chloride. For ionic compound \(\mathrm{RbCl}\), we can use the following naming convention: cation (metal) + anion (non-metal with -ide suffix). In this case, the name is Rubidium chloride.

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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 metals and non-metals combine. These compounds result from the transfer of electrons, which creates ions. Metals tend to lose electrons and become positively charged ions, while non-metals gain electrons and become negatively charged ions. This electrostatic attraction between the positive and negative ions forms an ionic bond.

Understanding ionic compounds involves recognizing their composition. A metal and a non-metal together usually indicate an ionic compound. For example, LaF e, FeCl e, PbCl, and RbCl are all ionic compounds. In each case, you have a metal (Lanthanum, Iron, Lead, or Rubidium) and a non-metal (Fluorine or Chlorine).

In terms of nomenclature, ionic compounds follow a specific naming convention:
  • First, name the cation (metal).
  • Then name the anion (non-metal) by changing the ending to '-ide'.
  • For metals that can have multiple charges, use Roman numerals to specify the charge.
For instance, FeCl e is named Iron(II) chloride, indicating that iron is in the +2 oxidation state.
Molecular Compounds
Molecular compounds differ from ionic compounds as they consist of non-metals. In these compounds, atoms share electrons through covalent bonds. This happens because non-metals have similar electronegativities, making it energetically favorable for them to share electrons instead of transferring them.

In molecular compounds, like SiF e, the sharing of electrons results in the formation of discrete molecules. Silicon (Si) and Fluorine (F) are both non-metals, which leads to the formation of a molecular compound.

When naming molecular compounds, we apply different rules compared to ionic compounds. The naming involves the use of prefixes to indicate the number of atoms involved. For SiF e, you would name it as Silicon tetrafluoride:
  • The first element keeps its name (Silicon).
  • For the second element, use a prefix based on the number of atoms (tetra for four), followed by the element name with an '-ide' suffix (fluoride).
Naming Conventions
Naming chemical compounds correctly is crucial for clear communication in science. Different types of compounds have distinct naming conventions to avoid confusion. Whether you are dealing with ionic or molecular compounds, knowing the rules is essential.

Ionic Compounds Naming

The naming begins with the cation (typically a metal), followed by the anion (usually a non-metal). Here are the basic rules:
  • Name the metal (cation) first.
  • For anions, take the root of the element's name and add '-ide'.
  • For metals that can have different charges (like transition metals), use Roman numerals to show the charge (e.g., Iron(II) chloride for FeCl e).

Molecular Compounds Naming

Molecular naming uses prefixes to denote the number of atoms. This makes it clear how many of each type of atom is in the compound.
  • Use prefixes like mono-, di-, tri-, tetra-, penta-, etc.
  • The first element鈥檚 full name is used.
  • The second element uses a prefix and ends in '-ide'.
For example, CO e is carbon dioxide, not carbon(IV) oxide, highlighting its molecular structure. Knowing and applying these rules can significantly help in understanding chemical notation and communication.

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Most popular questions from this chapter

A major challenge in implementing the "hydrogen economy" is finding a safe, lightweight, and compact way of storing hydrogen for use as a fuel. The hydrides of light metals are attractive for hydrogen storage because they can store a high weight percentage of hydrogen in a small volume. For example, \(\mathrm{NaAlH}_{4}\) can release \(5.6 \%\) of its mass as \(\mathrm{H}_{2}\) upon decomposing to \(\mathrm{NaH}(s), \mathrm{Al}(s)\), and \(\mathrm{H}_{2}(g) . \mathrm{NaAlH}_{4}\) possesses both covalent bonds, which hold polyatomic anions together, and ionic bonds. (a) Write a balanced equation for the decomposition of \(\mathrm{NaAlH}_{4}\). (b) Which element in \(\mathrm{NaAlH}_{4}\) is the most electronegative? Which one is the least electronegative? (c) Based on electronegativity differences, predict the identity of the polyatomic anion. Draw a Lewis structure for this ion. (d) What is the formal charge on hydrogen in the polyatomic ion?

(a) Construct a Lewis structure for hydrogen peroxide, \(\mathrm{H}_{2} \mathrm{O}_{2}\), in which each atom achieves an octet of electrons. (b) How many bonding electrons are between the two oxygen atoms? (c) Do you expect the \(\mathrm{O}-\mathrm{O}\) bond in \(\mathrm{H}_{2} \mathrm{O}_{2}\) to be longer or shorter than the \(\mathrm{O}-\mathrm{O}\) bond in \(\mathrm{O}_{2}\) ? Explain.

One scale for electronegativity is based on the concept that the electronegativity of any atom is proportional to the ionization energy of the atom minus its electron affinity: electronegativity \(=k(I-E A)\), where \(k\) is a proportionality constant. (a) How does this definition explain why the electronegativity of \(\mathrm{F}\) is greater than that of \(\mathrm{Cl}\) even though \(\mathrm{Cl}\) has the greater electron affinity? (b) Why are both ionization energy and electron affinity relevant to the notion of electronegativity? (c) By using data in Chapter 7 , determine the value of \(k\) that would lead to an electronegativity of \(4.0\) for \(F\) under this definition. (d) Use your result from part (c) to determine the electronegativities of \(\mathrm{Cl}\) and \(\mathrm{O}\) using this scale. (e) Another scale for electronegativity defines electronegativity as the average of an atom's first ionization energy and its electron affinity. Using this scale, calculate the electronegativities for the halogens, and scale them so fluorine has an electronegativity of 4.0. On this scale, what is Br's electronegativity?

Which of the following bonds are polar? (a) B-F, (b) \(\mathrm{Cl}-\mathrm{Cl}\), (c) Se-O, (d) H-I. Which is the more electronegative atom in each polar bond?

Draw the dominant Lewis structures for these chlorine-oxygen molecules/ions: \(\mathrm{ClO}, \mathrm{ClO}^{-}, \mathrm{ClO}_{2}^{-}, \mathrm{ClO}_{3}^{-}, \mathrm{ClO}_{4}^{-} .\)Which of these do not obey the octet rule?
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