Question

For the following pairs of ions, use the concept that a chemical compound must have a net charge of zero to predict the simplest compound that the ions are most likely to form. a. ${\mathrm{Co}}^{2+}$ and ${\mathrm{O}}^{2-}$ b. $C{o}^{3+}$ and $S^{2-}$ c. $A{l}^{3+}$ and $C{l}^{-}$ d. ${\mathrm{Ba}}^{2+}$ and ${\mathrm{N}}^{3-}$ e. $C{a}^{2+}$ and $C^{4-}$ f. $K^{+}$ and $N^{3-}$

Short answer

a. CoO b. Co2S3 c. AlCl3 d. Ba3N2 e. Ca2C f. K3N

Step-by-step solution

a. Cobalt (II) ion and oxide ion

Given the ions: ${\mathrm{Co}}^{2+}$ and ${\mathrm{O}}^{2-}$. The absolute values of their charges are 2 and 2, and the lowest common multiple (LCM) of these numbers is 2. Thus, for the charges to cancel each other out, we need a 1:1 ratio of ions: ${\mathrm{Co}}^{2+}\cdot 1+{\mathrm{O}}^{2-}\cdot 1=0$ So, the simplest compound is CoO.

b. Cobalt (III) ion and sulfide ion

Given the ions: $C{o}^{3+}$ and $S^{2-}$. The absolute values of their charges are 3 and 2, respectively, and the LCM of these numbers is 6. Thus, to cancel their charges, we need a ratio of 2:3: $C{o}^{3+}\cdot 2+S^{2-}\cdot 3=0$ So, the simplest compound is Co2S3.

c. Aluminum ion and chloride ion

Given the ions: $A{l}^{3+}$ and $C{l}^{-}$. The absolute values of their charges are 3 and 1, respectively, and the LCM of these numbers is 3. Thus, to cancel their charges, we need a ratio of 1:3: $A{l}^{3+}\cdot 1+C{l}^{-}\cdot 3=0$ So, the simplest compound is AlCl3.

d. Barium ion and nitride ion

Given the ions: ${\mathrm{Ba}}^{2+}$ and ${\mathrm{N}}^{3-}$. The absolute values of their charges are 2 and 3, respectively, and the LCM of these numbers is 6. Thus, to cancel their charges, we need a ratio of 3:2: ${\mathrm{Ba}}^{2+}\cdot 3+{\mathrm{N}}^{3-}\cdot 2=0$ So, the simplest compound is Ba3N2.

e. Calcium ion and carbide ion

Given the ions: $C{a}^{2+}$ and $C^{4-}$. The absolute values of their charges are 2 and 4, respectively, and the LCM of these numbers is 4. Thus, to cancel their charges, we need a ratio of 2:1: $C{a}^{2+}\cdot 2+C^{4-}\cdot 1=0$ So, the simplest compound is Ca2C.

f. Potassium ion and nitride ion

Given the ions: $K^{+}$ and $N^{3-}$. The absolute values of their charges are 1 and 3, respectively, and the LCM of these numbers is 3. Thus, to cancel their charges, we need a ratio of 3:1: $K^{+}\cdot 3+N^{3-}\cdot 1=0$ So, the simplest compound is K3N.

Key concepts

Charge Balance

An exciting concept in ionic compounds is charge balance. Charge balance means that, for neutrality, the positive and negative charges in a compound must cancel each other out. This is similar to balancing a seesaw; both sides need to have equal weight, or in this case, equal charge, to stay level. For instance, with the pair a. ${\mathrm{Co}}^{2+}$ and ${\mathrm{O}}^{2-}$, each ion has a charge of $2+$ and $2-$, respectively. A 1:1 ratio ensures the total charge equals zero, leading to the formation of $\mathrm{CoO}$. For more complex pairs like ${\mathrm{Co}}^{3+}$ and ${\mathrm{S}}^{2-}$, the least common multiple (LCM) helps us figure out the combination required to make their charges add up to zero, resulting in ${\mathrm{Co}}_2{\mathrm{S}}_3$. Understanding charge balance helps in predicting the formula of the compound that will form.

Cation

A cation is a positively charged ion, often formed when an atom loses electrons. Imagine a hydrogen atom losing its only electron; it transforms into a positively charged cation, written as ${\mathrm{H}}^{+}$. Cations are commonly metals or other groups that can easily lose electrons. In the ionic compounds discussed, ${\mathrm{Co}}^{2+}$, ${\mathrm{Co}}^{3+}$, and ${\mathrm{Ba}}^{2+}$ are all examples of cations. These metals like cobalt and barium lose electrons, gaining a positive charge in the process. The charge on a cation is shown as a superscript next to the element symbol, such as ${\mathrm{Al}}^{3+}$. Recognizing cations and their charge is crucial for determining the formula of ionic compounds.

Anion

Anions are negatively charged ions, formed when an atom gains extra electrons. Anions are like the other half of a magnet, finding balance by accepting electrons. When chlorine gains an electron, it becomes ${\mathrm{Cl}}^{-}$, an anion. In our examples, ${\mathrm{O}}^{2-}$, ${\mathrm{S}}^{2-}$, and ${\mathrm{N}}^{3-}$ are anions, derived from non-metal elements like oxygen and sulfur. Each non-metal atom gains enough electrons to fill its outer shell, hence achieving stability by becoming fully charged. Anions are as important as cations, as they balance out the positive charge in an ionic compound, ensuring it remains electrically neutral. Spotting anions helps us accurately write the empirical formulas of ionic compounds.

Ionic Bonding

Ionic bonding is like a game of tag where each atom either hands off or grabs electrons. It involves the transfer of electrons from one atom to another, forming a bond. Consider it as atoms exchanging gifts - the metal atom gives electrons to become a cation, while the non-metal receives them to form an anion. When ${\mathrm{K}}^{+}$ interacts with ${\mathrm{N}}^{3-}$, potassium loses one electron becoming ${\mathrm{K}}^{+}$, and nitrogen gains three, forming ${\mathrm{N}}^{3-}$. As a result, three potassium atoms are needed to bond with one nitrogen atom, ultimately forming the compound ${\mathrm{K}}_3\mathrm{N}$. This "give and take" creates a solid bond, resulting in a stable compound. Ionic bonds are the glue holding these atoms together, balancing out charges and forming the basis of ionic compounds.

Empirical Formula

The empirical formula is the simplest ratio of ions in a compound. It shows the smallest whole-number ratio between the ions, reflecting the compound’s basic composition without the clutter. In other words, it's the simplest way to capture the essence of a compound. Taking ${\mathrm{Ca}}^{2+}$ and ${\mathrm{C}}^{4-}$ as an example, the most straightforward ratio that balances their charges is 2:1, forming ${\mathrm{Ca}}_2\mathrm{C}$. Even though atoms might bond in more complex ways, the empirical formula provides the most reduced, swift summary of the ionic connection.Understanding and determining the empirical formula allows us to write the correct and simplest representation for a compound, revealing the essential bonds that maintain the neutrality and identity of ionic compounds.