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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 \(\mathrm{CaF}_{2},(\mathbf{b}) \mathrm{ClF}_{3}\) and \(\mathrm{VF}_{3},(\mathbf{c}) \mathrm{SbCl}_{5}\) and \(\mathrm{AlF}_{3} .\)

Short Answer

Expert verified
(a) TiCl₄ is an ionic substance named titanium tetrachloride, and CaF₂ is an ionic substance named calcium fluoride. (b) ClF₃ is a molecular substance named chlorine trifluoride, and VF₃ is an ionic substance named vanadium trifluoride. (c) SbCl₅ is a molecular substance named antimony pentachloride, and AlF₃ is an ionic substance named aluminum fluoride.

Step by step solution

01

(a) Identify the type of substance for TiCl4 and CaF2

In TiCl₄, we have titanium (Ti), which is a metal, and chlorine (Cl), a non-metal. This indicates that TiCl₄ is an ionic substance. In CaF₂, we have calcium (Ca), a metal, and fluorine (F), a non-metal, making it an ionic substance as well.
02

(a) Name TiCl4 and CaF2

For ionic substances, we use the metal name followed by the non-metal name with its ending changed to "-ide." TiCl₄: Titanium tetrachloride CaF₂: Calcium fluoride
03

(b) Identify the type of substance for ClF3 and VF3

In ClF₃, we have chlorine (Cl) and fluorine (F), both non-metals, making it a molecular substance. In VF₃, we have vanadium (V), a metal, and fluorine (F), a non-metal, making it an ionic substance.
04

(b) Name ClF3 and VF3

For molecular substances, we use prefixes to indicate the number of atoms of each element. For ionic substances, we use the metal name followed by the non-metal name with its ending changed to "-ide." ClF₃: Chlorine trifluoride VF₃: Vanadium trifluoride
05

(c) Identify the type of substance for SbCl5 and AlF3

In SbCl₅, we have antimony (Sb), a metalloid, and chlorine (Cl), a non-metal. Considering metalloids have properties of both metals and non-metals but mostly behave as non-metals in compounds, SbCl₅ would be a molecular substance. In AlF₃, we have aluminum (Al), a metal, and fluorine (F), a non-metal, making it an ionic substance.
06

(c) Name SbCl5 and AlF3

For molecular substances, we use prefixes to indicate the number of atoms of each element. For ionic substances, we use the metal name followed by the non-metal name with its ending changed to "-ide." SbCl₅: Antimony pentachloride AlF₃: Aluminum fluoride

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Key Concepts

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

Molecular Compounds
Molecular compounds are made up of non-metals that bond together by sharing electrons. This sharing of electrons is known as covalent bonding. The molecules that form from these bonds are distinct entities with specific arrangements of atoms.
An interesting aspect of molecular compounds is their naming convention, which involves using prefixes to indicate the number of atoms of each element present in the compound.
  • "Mono-" for one (often omitted for the first element)
  • "Di-" for two
  • "Tri-" for three
  • "Tetra-" for four, and so on
For instance, in chlorine trifluoride (\( \text{ClF}_3 \)), the prefix "tri-" indicates that three fluorine atoms are bonded to one chlorine atom. Understanding these prefixes can greatly help in identifying the composition of a molecular compound.
Ionic Compounds
Ionic compounds are formed through the transfer of electrons from one atom to another, usually between a metal and a non-metal. This transfer creates ions: cations (positively charged ions) and anions (negatively charged ions). The attraction between these oppositely charged ions results in an ionic bond.
In ionic compounds, these ions form a repeating three-dimensional lattice structure rather than individual molecules. The naming of ionic compounds follows a straightforward pattern:
  • The metal name is stated first.
  • The non-metal is mentioned second with its ending changed to "-ide."
For example, in calcium fluoride (\( \text{CaF}_2 \)), "calcium" refers to the metal calcium, and "fluoride" indicates fluorine ions. Recognizing this pattern makes understanding and naming ionic compounds a simpler task.
Naming Chemical Compounds
The naming of chemical compounds is a systematic process to ensure clear and consistent communication in chemistry. It varies depending on whether the compound is ionic or molecular. For ionic compounds, the name begins with the cation (metal) followed by the anion (non-metal), with the anion's suffix changing to "-ide." This pattern works well for simple ionic compounds, like aluminum fluoride (\( \text{AlF}_3 \)).
Naming molecular compounds, however, involves prefixes that denote the number of each type of atom. This method highlights the distinct nature of covalent bonding, where specific atom counts within molecules are important. For instance, antimony pentachloride (\( \text{SbCl}_5 \)) reflects the five chlorine atoms bonded to one antimony atom.
To accurately communicate chemical compositions, mastering these naming conventions is essential for students and professionals alike. It ensures the understanding of what elements are present and in what proportion, which is vital for studying chemical reactions and properties.

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Most popular questions from this chapter

(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.

We can define average bond enthalpies and bond lengths for ionic bonds, just like we have for covalent bonds. Which ionic bond is predicted to be stronger, \(\mathrm{Na}-\mathrm{Cl}\) or \(\mathrm{Ca}-\mathrm{O}\) ?

The compound chloral hydrate, known in detective stories as knockout drops, is composed of \(14.52 \% \mathrm{C}, 1.83 \% \mathrm{H}\) , \(64.30 \% \mathrm{Cl},\) and 13.35\(\% \mathrm{O}\) by mass, and has a molar mass of 165.4 \(\mathrm{g} / \mathrm{mol}\) . (a) What is the empirical formula of this substance? (b) What is the molecular formula of this substance? (c) Draw the Lewis structure of the molecule, assuming that the Cl atoms bond to a single \(C\) atom and that there are a \(C-C\) bond and two \(C-O\) bonds in the compound.

A carbene is a compound that has a carbon bonded to two atoms and a lone pair remaining on the carbon. Many carbenes are very reactive. (a) Draw the Lewis structure for the simplest carbene, \(\mathrm{H}_{2} \mathrm{C}\) . ( b) Predict the length of the carbon-carbon bond you would expect if two \(\mathrm{H}_{2} \mathrm{C}\) molecules reacted with each other by a combination reaction.

You and a partner are asked to complete a lab entitled "Oxides of Ruthenium" that is scheduled to extend over two lab periods. The first lab, which is to be completed by your partner, is devoted to carrying out compositional analysis. In the second lab, you are to determine melting points. Upon going to lab you find two unlabeled vials, one containing a soft yellow substance and the other a black powder. You also find the following notes in your partner's notebook Compound \(1 : 76.0 \%\) Ru and 24.0\(\%\) O (by mass), Compound \(2 : 61.2 \%\) Ru and 38.8\(\%\) O (by mass). (a) What is the empirical formula for Compound 1\(?\) (b) What is the empirical formula for Compound 2\(?\) Upon determining the melting points of these two compounds, you find that the yellow compound melts at \(25^{\circ} \mathrm{C}\) , while the black powder does not melt up to the maximum temperature of your apparatus, \(1200^{\circ} \mathrm{C}\) . (c) What is the identity of the yellow compound? (d) What is the identity of the black compound? (e) Which compound is molecular? (f) Which compound is ionic?

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