Problem 45
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: \((\mathbf{a}) \mathrm{SiF}_{4}\) and \(\mathrm{LaF}_{3},(\mathbf{b}) \mathrm{FeCl}_{2}\) and \(\mathrm{ReCl}_{6},(\mathbf{c}) \mathrm{PbCl}_{4}\) and RbCl.
Step-by-Step Solution
Verified Answer
(a) SiF
_{4}: Silicon tetrafluoride (molecular), LaF
_{3}: Lanthanum fluoride (ionic); (b) FeCl
_{2}: Iron(II) chloride (ionic), ReCl
_{6}: Rhenium(VI) chloride (ionic); (c) PbCl
_{4}: Lead(IV) chloride (ionic), RbCl: Rubidium chloride (ionic).
1Step 1: Determine Compound Type for Part (a)
SiF
_{4}
is a compound formed by silicon (a non-metal) and fluorine (a non-metal). This makes SiF
_{4}
a molecular substance. LaF
_{3}
, on the other hand, is comprised of lanthanum (a metal) and fluorine (a non-metal), classifying it as an ionic compound.
2Step 2: Name the Compounds for Part (a)
The molecular compound SiF
_{4}
is named using prefixes: **Silicon tetrafluoride**. The ionic compound LaF
_{3}
is named by the metal followed by the non-metal with an -ide suffix: **Lanthanum fluoride**.
3Step 3: Determine Compound Type for Part (b)
FeCl
_{2}
is a compound of iron (a metal) and chlorine (a non-metal), which is an ionic compound. ReCl
_{6}
is a compound with rhenium (a metal) and chlorine (a non-metal), also making it an ionic compound.
4Step 4: Name the Compounds for Part (b)
For ionic compounds with transition metals, specify the metal's charge. FeCl
_{2}
is named **Iron(II) chloride** because iron has a +2 charge here. ReCl
_{6}
is named **Rhenium(VI) chloride** because the charge on rhenium is +6.
5Step 5: Determine Compound Type for Part (c)
PbCl
_{4}
is composed of lead (a metal) and chlorine (a non-metal), making it an ionic compound. RbCl is made from rubidium (a metal) and chlorine (a non-metal), classifying it as another ionic compound.
6Step 6: Name the Compounds for Part (c)
PbCl
_{4}
is named using the charge of the lead: **Lead(IV) chloride**. RbCl is named simply as **Rubidium chloride**.
Key Concepts
Ionic CompoundsMolecular CompoundsNaming ConventionsBinary Compounds
Ionic Compounds
Ionic compounds are formed when metals and non-metals chemically combine by the transfer of electrons. Metals lose electrons to form positively charged ions, while non-metals gain these electrons to become negatively charged ions.
These ions attract each other due to opposite charges, creating ionic bonds.
These ions attract each other due to opposite charges, creating ionic bonds.
- They typically consist of a metal and a non-metal.
- Usually solid at room temperature.
- Have high melting and boiling points due to strong ionic bonds.
- Conduct electricity when dissolved in water or melted due to free-moving ions.
- Start with the name of the metal.
- Followed by the non-metal, altering its ending to "-ide".
- If the metal can have multiple charges, specify the charge with Roman numerals in parentheses.
Molecular Compounds
Molecular compounds, also known as covalent compounds, form when two or more non-metals share electrons. Unlike ionic compounds, there is no transfer of electrons.
This sharing allows each atom to achieve a stable electron configuration.
This sharing allows each atom to achieve a stable electron configuration.
- Typically consist of only non-metals.
- Can be gases, liquids, or solids at room temperature.
- Have relatively low melting and boiling points compared to ionic compounds.
- Do not conduct electricity because they have no charged particles.
- "Mono-" is sometimes omitted for the first element.
- "Di-", "tri-", "tetra-", and so on indicate two, three, four, etc.
- Change the ending of the last element to "-ide".
Naming Conventions
Naming conventions in chemistry help identify and communicate about compounds effectively.
Depending on whether a compound is ionic or molecular, different rules apply for naming them.
Depending on whether a compound is ionic or molecular, different rules apply for naming them.
- Ionic Compounds: Name the metal first, then the non-metal with its ending changed to "-ide". For transition metals, indicate the charge in Roman numerals.
- Molecular Compounds: Use prefixes to indicate the number of atoms. Only the second element's name is altered to "-ide", and "mono-" is often dropped for the first element.
- For polyatomic ions, such as sulfate (SO₄²⁻) and nitrate (NO₃⁻), use the ion name without modification.
Binary Compounds
Binary compounds are composed of exactly two different elements.
They can be either ionic or molecular depending on the nature of the elements involved.
An ionic binary compound consists of a metal and a non-metal, whereas a binary molecular compound comprises two non-metals.
They can be either ionic or molecular depending on the nature of the elements involved.
An ionic binary compound consists of a metal and a non-metal, whereas a binary molecular compound comprises two non-metals.
- Ionic Binary Compounds: Example: NaCl (sodium chloride) where Na is a metal and Cl is a non-metal.
- Molecular Binary Compounds: Example: CO₂ (carbon dioxide) composed of two non-metals, carbon and oxygen.
- Recognize the compound type first and apply respective naming rules.
- For ionic, use the metal name first followed by the non-metal with "-ide".
- For molecular, use appropriate prefixes for the number of atoms and alter the second element to end in "-ide".
Other exercises in this chapter
Problem 42
Arrange the bonds in each of the following sets in order of increasing polarity: \((\mathbf{a}) \mathrm{C}-\mathrm{F}, \mathrm{O}-\mathrm{F}, \mathrm{Be}-\mathr
View solution Problem 44
The iodine monobromide molecule, IBr, has a bond length of \(249 \mathrm{pm}\) and a dipole moment of \(1.21 \mathrm{D}\). (a) Which atom of the molecule is exp
View solution Problem 46
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 convent
View solution Problem 47
Draw Lewis structures for the following: \((\mathbf{a}) \mathrm{CH}_{2} \mathrm{Cl}_{2},(\mathbf{b}) \mathrm{ClCN},\) (c) AsF \(_{5},\) (d) \(\mathrm{CH}_{2} \m
View solution