Problem 62
Question
Predict the chemical formulas of the compounds formed by the following pairs of ions: (a) \(\mathrm{Cr}^{3+}\) and \(\mathrm{Br}^{-},\) (b) \(\mathrm{Fe}^{3+}\) and \(\mathrm{O}^{2-}\) , \((\mathbf{c}) \mathrm{Hg}_{2}^{2+}\) and \(\mathrm{CO}_{3}^{2-},(\mathbf{d}) \mathrm{Ca}^{2+}\) and \(\mathrm{ClO}_{3}^{-},(\mathbf{e}) \mathrm{NH}_{4}^{+}\) and \(\mathrm{PO}_{4}^{3-}\).
Step-by-Step Solution
Verified Answer
The chemical formulas of the compounds formed by the given pairs of ions are: (a) \(\mathrm{CrBr}_{3}\), (b) \(\mathrm{Fe}_{2}\mathrm{O}_{3}\), (c) \(\mathrm{Hg}_{2}\mathrm{CO}_{3}\), (d) \(\mathrm{Ca}(\mathrm{ClO}_{3})_{2}\), (e) \(\mathrm{(NH}_{4}\mathrm{)}_{3}\mathrm{PO}_{4}\).
1Step 1: Determine the chemical formula for pair (a) \(\mathrm{Cr}^{3+}\) and \(\mathrm{Br}^{-}\)
To determine the chemical formula for the pair \(\mathrm{Cr}^{3+}\) and \(\mathrm{Br}^{-}\), we will apply the criss-cross method:
1. Swap the charges of the ions:
\(\mathrm{Cr}^{3+}\) becomes \(\mathrm{Cr}_{1}\)
\(\mathrm{Br}^{-}\) becomes \(\mathrm{Br}_{3}\)
2. The resulting compound is: \(\mathrm{CrBr}_{3}\)
2Step 2: Determine the chemical formula for pair (b) \(\mathrm{Fe}^{3+}\) and \(\mathrm{O}^{2-}\)
To determine the chemical formula for the pair \(\mathrm{Fe}^{3+}\) and \(\mathrm{O}^{2-}\), we will apply the criss-cross method:
1. Swap the charges of the ions:
\(\mathrm{Fe}^{3+}\) becomes \(\mathrm{Fe}_{2}\)
\(\mathrm{O}^{2-}\) becomes \(\mathrm{O}_{3}\)
2. The resulting compound is: \(\mathrm{Fe}_{2}\mathrm{O}_{3}\)
3Step 3: Determine the chemical formula for pair (c) \(\mathrm{Hg}_{2}^{2+}\) and \(\mathrm{CO}_{3}^{2-}\)
To determine the chemical formula for the pair \(\mathrm{Hg}_{2}^{2+}\) and \(\mathrm{CO}_{3}^{2-}\), we will apply the criss-cross method:
1. Swap the charges of the ions:
\(\mathrm{Hg}_{2}^{2+}\) becomes \(\mathrm{Hg}_{2}_{1}\)
\(\mathrm{CO}_{3}^{2-}\) becomes \(\mathrm{CO}_{3}_{1}\)
2. The resulting compound is: \(\mathrm{Hg}_{2}\mathrm{CO}_{3}\)
4Step 4: Determine the chemical formula for pair (d) \(\mathrm{Ca}^{2+}\) and \(\mathrm{ClO}_{3}^{-}\)
To determine the chemical formula for the pair \(\mathrm{Ca}^{2+}\) and \(\mathrm{ClO}_{3}^{-}\), we will apply the criss-cross method:
1. Swap the charges of the ions:
\(\mathrm{Ca}^{2+}\) becomes \(\mathrm{Ca}_{1}\)
\(\mathrm{ClO}_{3}^{-}\) becomes \(\mathrm{ClO}_{3}_{2}\)
2. The resulting compound is: \(\mathrm{Ca}(\mathrm{ClO}_{3})_{2}\)
5Step 5: Determine the chemical formula for pair (e) \(\mathrm{NH}_{4}^{+}\) and \(\mathrm{PO}_{4}^{3-}\)
To determine the chemical formula for the pair \(\mathrm{NH}_{4}^{+}\) and \(\mathrm{PO}_{4}^{3-}\), we will apply the criss-cross method:
1. Swap the charges of the ions:
\(\mathrm{NH}_{4}^{+}\) becomes \(\mathrm{NH}_{4}_{3}\)
\(\mathrm{PO}_{4}^{3-}\) becomes \(\mathrm{PO}_{4}_{1}\)
2. The resulting compound is: \(\mathrm{(NH}_{4}\mathrm{)}_{3}\mathrm{PO}_{4}\)
Key Concepts
Criss-Cross MethodIonic CompoundsChemical Nomenclature
Criss-Cross Method
Understanding the criss-cross method is essential for predicting chemical formulas, particularly for ionic compounds. This straightforward technique involves using the charges of ions to determine the subscripts for each ion in the final formula.
The process starts by identifying the charges of the cation (positively charged ion) and the anion (negatively charged ion) involved in forming the ionic compound. Once the charges are known, the absolute values of these charges are swapped, or 'criss-crossed', to become the subscripts of the opposing ion. Subscripts refer to the number of atoms or ions of a particular element present in the compound. It is important to reduce the subscripts to their lowest whole number terms, if possible, to achieve the empirical formula—the simplest ratio of atoms of each element.
For example, in the compound formed from \(\mathrm{Al}^{3+}\) and \(\mathrm{O}^{2-}\), the charge on aluminum becomes the subscript for oxygen, resulting in \(\mathrm{O}_3\). Similarly, the charge on the oxide ion becomes the subscript for aluminum, leading to \(\mathrm{Al}_2\). However, to simplify, these subscripts are reduced from \(\mathrm{Al}_2\mathrm{O}_3\) to \(\mathrm{AlO}_{1.5}\), which is not correct because subscripts must be whole numbers. Therefore, we double both subscripts to get the correct formula \(\mathrm{Al}_2\mathrm{O}_3\), which is aluminum oxide's empirical formula.
The process starts by identifying the charges of the cation (positively charged ion) and the anion (negatively charged ion) involved in forming the ionic compound. Once the charges are known, the absolute values of these charges are swapped, or 'criss-crossed', to become the subscripts of the opposing ion. Subscripts refer to the number of atoms or ions of a particular element present in the compound. It is important to reduce the subscripts to their lowest whole number terms, if possible, to achieve the empirical formula—the simplest ratio of atoms of each element.
For example, in the compound formed from \(\mathrm{Al}^{3+}\) and \(\mathrm{O}^{2-}\), the charge on aluminum becomes the subscript for oxygen, resulting in \(\mathrm{O}_3\). Similarly, the charge on the oxide ion becomes the subscript for aluminum, leading to \(\mathrm{Al}_2\). However, to simplify, these subscripts are reduced from \(\mathrm{Al}_2\mathrm{O}_3\) to \(\mathrm{AlO}_{1.5}\), which is not correct because subscripts must be whole numbers. Therefore, we double both subscripts to get the correct formula \(\mathrm{Al}_2\mathrm{O}_3\), which is aluminum oxide's empirical formula.
Ionic Compounds
Ionic compounds are chemical compounds composed of ions held together by electrostatic forces termed ionic bonding. They usually consist of a metal cation and a non-metal anion. In an ionic bond, the metal loses electrons to become a positively charged cation, whereas the non-metal gains those electrons to become a negatively charged anion.
Ionic compounds have distinct properties: they typically form crystalline solids, have high melting and boiling points, can conduct electricity when melted or dissolved in water, and are generally soluble in water.
When writing the formulas of ionic compounds, it's essential to ensure that the compound is neutral, with the total positive charge equaling the total negative charge. That's why using the criss-cross method is an effective way to achieve charge balance in the formula. For instance, when combining \(\mathrm{Na}^{+}\) with \(\mathrm{Cl}^{-}\), we get \(\mathrm{NaCl}\), where each ion only needs one of the other to balance the charges, resulting in common table salt.
Ionic compounds have distinct properties: they typically form crystalline solids, have high melting and boiling points, can conduct electricity when melted or dissolved in water, and are generally soluble in water.
When writing the formulas of ionic compounds, it's essential to ensure that the compound is neutral, with the total positive charge equaling the total negative charge. That's why using the criss-cross method is an effective way to achieve charge balance in the formula. For instance, when combining \(\mathrm{Na}^{+}\) with \(\mathrm{Cl}^{-}\), we get \(\mathrm{NaCl}\), where each ion only needs one of the other to balance the charges, resulting in common table salt.
Chemical Nomenclature
Chemical nomenclature is the systematic naming of chemical compounds based on their composition and structure, allowing chemists to communicate unambiguously about chemical substances. The International Union of Pure and Applied Chemistry (IUPAC) sets the rules for naming compounds.
For ionic compounds, the name consists of the cation's name followed by the anion's name. If the cation is a metal with a fixed charge, like sodium or magnesium, its name doesn't change when it forms an ion. However, for metals that can form ions with different charges, such as iron, the charge is indicated in Roman numerals within parentheses right after the metal's name, like \(\mathrm{Iron(II)}\) for \(\mathrm{Fe}^{2+}\) or \(\mathrm{Iron(III)}\) for \(\mathrm{Fe}^{3+}\).
The anion's name is derived from the element's name, typically by ending with '-ide', such as chloride for \(\mathrm{Cl}^{-}\), or by using polyatomic ion names like sulfate for \(\mathrm{SO}_{4}^{2-}\). Complex nomenclature arises when dealing with polyatomic ions, but learning common polyatomic names simplifies the process.
For ionic compounds, the name consists of the cation's name followed by the anion's name. If the cation is a metal with a fixed charge, like sodium or magnesium, its name doesn't change when it forms an ion. However, for metals that can form ions with different charges, such as iron, the charge is indicated in Roman numerals within parentheses right after the metal's name, like \(\mathrm{Iron(II)}\) for \(\mathrm{Fe}^{2+}\) or \(\mathrm{Iron(III)}\) for \(\mathrm{Fe}^{3+}\).
The anion's name is derived from the element's name, typically by ending with '-ide', such as chloride for \(\mathrm{Cl}^{-}\), or by using polyatomic ion names like sulfate for \(\mathrm{SO}_{4}^{2-}\). Complex nomenclature arises when dealing with polyatomic ions, but learning common polyatomic names simplifies the process.
Naming Practice
To name \(\mathrm{Ca}^{2+}\) and \(\mathrm{SO}_{4}^{2-}\) compound, we start with the cation, calcium, and then add the anion, sulfate, resulting in calcium sulfate. The compound's simplicity, \(\mathrm{CaSO}_{4}\), is due to the 2+ and 2- charges directly balancing each other.Other exercises in this chapter
Problem 60
The most common charge associated with scandium in its compounds is \(3+.\) Indicate the chemical formulas you would expect for compounds formed between scandiu
View solution Problem 61
Predict the chemical formula for the ionic compound formed by (a) \(\mathrm{Ca}^{2+}\) and \(\mathrm{Br}^{-},(\mathbf{b}) \mathrm{K}^{+}\) and \(\mathrm{CO}_{3}
View solution Problem 65
Predict whether each of the following compounds is molecular or ionic: (a) \(\mathrm{B}_{2} \mathrm{H}_{6},(\mathbf{b}) \mathrm{CH}_{3} \mathrm{OH},(\mathbf{c})
View solution Problem 66
Which of the following are ionic, and which are molecular? (a) \(\mathrm{PF}_{5},(\mathbf{b}) \mathrm{NaI},(\mathbf{c}) \mathrm{SCl}_{2},(\mathbf{d}) \mathrm{Ca
View solution