Problem 62
Question
Predict the chemical formulas of the compounds formed by the following pairs of ions: (a) \(\mathrm{Cr}^{3+}\) and \(\mathrm{CN}^{-}\), (b) \(\mathrm{Mn}^{2+}\) and \(\mathrm{ClO}_{4}^{-}\), (c) \(\mathrm{Na}^{+}\) and \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\), (d) \(\mathrm{Cd}^{2+}\) and \(\mathrm{CO}_{3}^{2-}\), (e) \(\mathrm{Ti}^{4+}\) and \(\mathrm{O}^{2-}\).
Step-by-Step Solution
Verified Answer
The chemical formulas of the compounds formed by the given pairs of ions are:
(a) \(\mathrm{Cr(CN)}_{3}\)
(b) \(\mathrm{Mn(ClO}_{4})_{2}\)
(c) \(\mathrm{Na}_{2}\mathrm{Cr}_{2}\mathrm{O}_{7}\)
(d) \(\mathrm{CdCO}_{3}\)
(e) \(\mathrm{TiO}_{2}\)
1Step 1: (a) \(\mathrm{Cr}^{3+}\) and \(\mathrm{CN}^{-}\)
To form a compound with a neutral charge between \(\mathrm{Cr}^{3+}\) and \(\mathrm{CN}^{-}\), we need to find the lowest common multiple (LCM) of the charges 3 and 1. The LCM is 3. This means that we need three \(\mathrm{CN}^{-}\) ions to balance one \(\mathrm{Cr}^{3+}\) ion. The chemical formula for the compound is \(\mathrm{Cr(CN)}_{3}\).
2Step 2: (b) \(\mathrm{Mn}^{2+}\) and \(\mathrm{ClO}_{4}^{-}\)
In the case of \(\mathrm{Mn}^{2+}\) and \(\mathrm{ClO}_{4}^{-}\), the LCM of 2 and 1 is 2. This means that we need two \(\mathrm{ClO}_{4}^{-}\) ions to balance one \(\mathrm{Mn}^{2+}\) ion. The chemical formula for the compound is \(\mathrm{Mn(ClO}_{4})_{2}\).
3Step 3: (c) \(\mathrm{Na}^{+}\) and \(\mathrm{Cr}_{2}\mathrm{O}_{7}^{2-}\)
For the ions \(\mathrm{Na}^{+}\) and \(\mathrm{Cr}_{2}\mathrm{O}_{7}^{2-}\), the LCM of 1 and 2 is 2. This means that we need two \(\mathrm{Na}^{+}\) ions to balance one \(\mathrm{Cr}_{2}\mathrm{O}_{7}^{2-}\) ion. The chemical formula for the compound is \(\mathrm{Na}_{2}\mathrm{Cr}_{2}\mathrm{O}_{7}\).
4Step 4: (d) \(\mathrm{Cd}^{2+}\) and \(\mathrm{CO}_{3}^{2-}\)
In the case of \(\mathrm{Cd}^{2+}\) and \(\mathrm{CO}_{3}^{2-}\), the charges of the ions are equal and opposite, which means that we need one of each ion to balance each other out. The chemical formula for the compound is \(\mathrm{CdCO}_{3}\).
5Step 5: (e) \(\mathrm{Ti}^{4+}\) and \(\mathrm{O}^{2-}\)
Finally, for the ions \(\mathrm{Ti}^{4+}\) and \(\mathrm{O}^{2-}\), the LCM of 4 and 2 is 4. This means that we need two \(\mathrm{O}^{2-}\) ions to balance one \(\mathrm{Ti}^{4+}\) ion. The chemical formula for the compound is \(\mathrm{TiO}_{2}\).
Key Concepts
Ionic CompoundsCharge BalancingCompound FormationNeutral ChargeIon Pairing
Ionic Compounds
Ionic compounds are formed when positive and negative ions attract each other and create a stable substance with a balanced arrangement. In these compounds, ions are held together by ionic bonds, which are formed through the transfer of electrons from one atom to another.
The process usually involves a metal, which loses electrons and becomes a positive ion, and a nonmetal, or a group of nonmetals, which gains those electrons to become negative ions. This electron exchange leads to the creation of a crystal lattice structure that gives ionic compounds distinct properties such as high melting and boiling points.
For example, common table salt (NaCl) is an ionic compound formed from sodium ions (Na⁺) and chloride ions (Cl⁻). Understanding ionic compounds is crucial for predicting the chemical formulas that arise when these ions combine.
The process usually involves a metal, which loses electrons and becomes a positive ion, and a nonmetal, or a group of nonmetals, which gains those electrons to become negative ions. This electron exchange leads to the creation of a crystal lattice structure that gives ionic compounds distinct properties such as high melting and boiling points.
For example, common table salt (NaCl) is an ionic compound formed from sodium ions (Na⁺) and chloride ions (Cl⁻). Understanding ionic compounds is crucial for predicting the chemical formulas that arise when these ions combine.
Charge Balancing
Charge balancing is a key concept when predicting chemical formulas for ionic compounds. It ensures that the total positive charge from the cations (positive ions) equals the total negative charge from the anions (negative ions), resulting in a compound with a neutral charge.
To balance charges, you determine the number of each ion needed to achieve neutrality by finding the lowest common multiple (LCM) of the ions' charges. For instance, if you have ions like \(\mathrm{Cr}^{3+}\) and \(\mathrm{CN}^{-}\), you find the LCM of 3 and 1, which is 3. This means three \(\mathrm{CN}^{-}\) ions are needed to balance one \(\mathrm{Cr}^{3+}\) ion, leading to the formula \(\mathrm{Cr(CN)}_{3}\).
This method ensures ionic compounds are electrically neutral, a necessary condition for compound stability.
To balance charges, you determine the number of each ion needed to achieve neutrality by finding the lowest common multiple (LCM) of the ions' charges. For instance, if you have ions like \(\mathrm{Cr}^{3+}\) and \(\mathrm{CN}^{-}\), you find the LCM of 3 and 1, which is 3. This means three \(\mathrm{CN}^{-}\) ions are needed to balance one \(\mathrm{Cr}^{3+}\) ion, leading to the formula \(\mathrm{Cr(CN)}_{3}\).
This method ensures ionic compounds are electrically neutral, a necessary condition for compound stability.
Compound Formation
Compound formation involves the precise combination of ions to form a neutral compound. It starts by writing the symbol for the cation first, followed by the anion. You then use subscripts to indicate the amount of each ion needed to balance the charge.
Consider \(\mathrm{Mn}^{2+}\) and \(\mathrm{ClO}_{4}^{-}\) ions, where the LCM of 2 (from \(\mathrm{Mn}^{2+}\)) and 1 (from \(\mathrm{ClO}_{4}^{-}\)) is 2. Two \(\mathrm{ClO}_{4}^{-}\) ions per \(\mathrm{Mn}^{2+}\) ion are needed, producing the formula \(\mathrm{Mn(ClO}_{4})_{2}\).
This systematic approach helps prevent errors by ensuring each compound formed has the correct stoichiometry, leading to a balanced and neutral chemical formula.
Consider \(\mathrm{Mn}^{2+}\) and \(\mathrm{ClO}_{4}^{-}\) ions, where the LCM of 2 (from \(\mathrm{Mn}^{2+}\)) and 1 (from \(\mathrm{ClO}_{4}^{-}\)) is 2. Two \(\mathrm{ClO}_{4}^{-}\) ions per \(\mathrm{Mn}^{2+}\) ion are needed, producing the formula \(\mathrm{Mn(ClO}_{4})_{2}\).
This systematic approach helps prevent errors by ensuring each compound formed has the correct stoichiometry, leading to a balanced and neutral chemical formula.
Neutral Charge
Achieving a neutral charge is fundamental for the stability of ionic compounds. A compound's neutrality ensures that all positive and negative charges are balanced. This means the total respective charges from the ions must be exactly opposite in magnitude.
An example is the formation of \(\mathrm{CdCO}_{3}\), where both the cadmium ion (\(\mathrm{Cd}^{2+}\)) and the carbonate ion (\(\mathrm{CO}_{3}^{2-}\)) naturally balance each other out. Their charges are equal and opposite, resulting in no additional ions being needed.
This principle of neutrality is why ionic compounds are so stable, as any imbalance would cause the compound to react further to regain stability.
An example is the formation of \(\mathrm{CdCO}_{3}\), where both the cadmium ion (\(\mathrm{Cd}^{2+}\)) and the carbonate ion (\(\mathrm{CO}_{3}^{2-}\)) naturally balance each other out. Their charges are equal and opposite, resulting in no additional ions being needed.
This principle of neutrality is why ionic compounds are so stable, as any imbalance would cause the compound to react further to regain stability.
Ion Pairing
Ion pairing involves the interaction and linkage of cations and anions in specific ratios to form compounds. The simplest form of ion pairing is the one-to-one pairing seen in basic compounds, but more complex interactions often occur.
For instance, in the case of \(\mathrm{Ti}^{4+}\) and \(\mathrm{O}^{2-}\), two oxygen ions are required to pair properly with one titanium ion to create a stable compound \(\mathrm{TiO}_{2}\).
It's crucial to understand ion pairing to predict how elements will combine based on their ionic charges. This knowledge helps chemists and students determine the correct formulas for compounds and understand their structure and properties.
For instance, in the case of \(\mathrm{Ti}^{4+}\) and \(\mathrm{O}^{2-}\), two oxygen ions are required to pair properly with one titanium ion to create a stable compound \(\mathrm{TiO}_{2}\).
It's crucial to understand ion pairing to predict how elements will combine based on their ionic charges. This knowledge helps chemists and students determine the correct formulas for compounds and understand their structure and properties.
Other exercises in this chapter
Problem 60
The most common charge associated with selenium is \(2-.\) Indicate the chemical formulas you would expect for compounds formed between selenium and (a) barium,
View solution Problem 61
Predict the chemical formulas of the ionic compound formed by (a) \(\mathrm{Fe}^{3+}\) and \(\mathrm{OH}^{-}\), (b) \(\mathrm{Cs}^{+}\) and \(\mathrm{NO}_{3}^{-
View solution Problem 64
Complete the table by filling in the formula for the ionic compound formed by each pair of cations and anions, as shown for the first pair. $$ \begin{array}{c|c
View solution Problem 65
Predict whether each of the following compounds is molecular or ionic: \((\mathbf{a}) \mathrm{HClO}_{4}(\mathbf{b}) \mathrm{CH}_{3} \mathrm{OCH}_{3}(\mathbf{c})
View solution