Question

fill in the name and empirical formula of each ionic compound that could be formed from the ions in this table:
some ionic compounds
cation \tanion \tempirical formula \tname of compound
ca²⁺ \tp³⁻ \t□ \t□
ca²⁺ \ts²⁻ \t□ \t□
ca²⁺ \tn³⁻ \t□ \t□
ca²⁺ \to²⁻ \t□ \t□

Explanation:

To solve for the empirical formulas and names of the ionic compounds formed from $\ce{Ca^{2+}}$ and the given anions, we use the criss - cross method (where the magnitude of the charge of one ion becomes the subscript of the other ion) and follow the naming rules for ionic compounds (name the cation first, then the anion with an -ide suffix for monatomic anions).

For $\ce{Ca^{2+}}$ and $\ce{P^{3-}}$:

Step 1: Determine the subscripts

The charge of $\ce{Ca^{2+}}$ is $+ 2$ and the charge of $\ce{P^{3-}}$ is $-3$. Using the criss - cross method, the subscript of $\ce{Ca}$ will be $3$ (the magnitude of the charge of $\ce{P^{3-}}$) and the subscript of $\ce{P}$ will be $2$ (the magnitude of the charge of $\ce{Ca^{2+}}$). So the empirical formula is $\ce{Ca_3P_2}$.

Step 2: Name the compound

The cation is calcium ($\ce{Ca^{2+}}$) and the anion is phosphide ($\ce{P^{3-}}$). So the name of the compound is calcium phosphide.

For $\ce{Ca^{2+}}$ and $\ce{S^{2-}}$:

Step 1: Determine the subscripts

The charge of $\ce{Ca^{2+}}$ is $+ 2$ and the charge of $\ce{S^{2-}}$ is $-2$. When we use the criss - cross method, the subscripts will simplify. The ratio of $\ce{Ca}$ to $\ce{S}$ is $1:1$ (since $2$ and $2$ have a greatest common divisor of $2$, and $\frac{2}{2}=1$ for both). So the empirical formula is $\ce{CaS}$.

Step 2: Name the compound

The cation is calcium ($\ce{Ca^{2+}}$) and the anion is sulfide ($\ce{S^{2-}}$). So the name of the compound is calcium sulfide.

For $\ce{Ca^{2+}}$ and $\ce{N^{3-}}$:

Step 1: Determine the subscripts

The charge of $\ce{Ca^{2+}}$ is $+ 2$ and the charge of $\ce{N^{3-}}$ is $-3$. Using the criss - cross method, the subscript of $\ce{Ca}$ is $3$ (magnitude of charge of $\ce{N^{3-}}$) and the subscript of $\ce{N}$ is $2$ (magnitude of charge of $\ce{Ca^{2+}}$). So the empirical formula is $\ce{Ca_3N_2}$.

Step 2: Name the compound

The cation is calcium ($\ce{Ca^{2+}}$) and the anion is nitride ($\ce{N^{3-}}$). So the name of the compound is calcium nitride.

For $\ce{Ca^{2+}}$ and $\ce{O^{2-}}$:

Step 1: Determine the subscripts

The charge of $\ce{Ca^{2+}}$ is $+ 2$ and the charge of $\ce{O^{2-}}$ is $-2$. Using the criss - cross method, the subscripts simplify to $1:1$ (since the greatest common divisor of $2$ and $2$ is $2$, and $\frac{2}{2} = 1$ for both). So the empirical formula is $\ce{CaO}$.

Step 2: Name the compound

The cation is calcium ($\ce{Ca^{2+}}$) and the anion is oxide ($\ce{O^{2-}}$). So the name of the compound is calcium oxide.

Filling the table:

cation	anion	empirical formula	name of compound
$\ce{Ca^{2+}}$	$\ce{S^{2-}}$	$\ce{CaS}$	calcium sulfide
$\ce{Ca^{2+}}$	$\ce{N^{3-}}$	$\ce{Ca_3N_2}$	calcium nitride
$\ce{Ca^{2+}}$	$\ce{O^{2-}}$	$\ce{CaO}$	calcium oxide

Answer:

To solve for the empirical formulas and names of the ionic compounds formed from $\ce{Ca^{2+}}$ and the given anions, we use the criss - cross method (where the magnitude of the charge of one ion becomes the subscript of the other ion) and follow the naming rules for ionic compounds (name the cation first, then the anion with an -ide suffix for monatomic anions).

For $\ce{Ca^{2+}}$ and $\ce{P^{3-}}$:

Step 1: Determine the subscripts

The charge of $\ce{Ca^{2+}}$ is $+ 2$ and the charge of $\ce{P^{3-}}$ is $-3$. Using the criss - cross method, the subscript of $\ce{Ca}$ will be $3$ (the magnitude of the charge of $\ce{P^{3-}}$) and the subscript of $\ce{P}$ will be $2$ (the magnitude of the charge of $\ce{Ca^{2+}}$). So the empirical formula is $\ce{Ca_3P_2}$.

Step 2: Name the compound

The cation is calcium ($\ce{Ca^{2+}}$) and the anion is phosphide ($\ce{P^{3-}}$). So the name of the compound is calcium phosphide.

For $\ce{Ca^{2+}}$ and $\ce{S^{2-}}$:

Step 1: Determine the subscripts

The charge of $\ce{Ca^{2+}}$ is $+ 2$ and the charge of $\ce{S^{2-}}$ is $-2$. When we use the criss - cross method, the subscripts will simplify. The ratio of $\ce{Ca}$ to $\ce{S}$ is $1:1$ (since $2$ and $2$ have a greatest common divisor of $2$, and $\frac{2}{2}=1$ for both). So the empirical formula is $\ce{CaS}$.

Step 2: Name the compound

The cation is calcium ($\ce{Ca^{2+}}$) and the anion is sulfide ($\ce{S^{2-}}$). So the name of the compound is calcium sulfide.

For $\ce{Ca^{2+}}$ and $\ce{N^{3-}}$:

Step 1: Determine the subscripts

The charge of $\ce{Ca^{2+}}$ is $+ 2$ and the charge of $\ce{N^{3-}}$ is $-3$. Using the criss - cross method, the subscript of $\ce{Ca}$ is $3$ (magnitude of charge of $\ce{N^{3-}}$) and the subscript of $\ce{N}$ is $2$ (magnitude of charge of $\ce{Ca^{2+}}$). So the empirical formula is $\ce{Ca_3N_2}$.

Step 2: Name the compound

The cation is calcium ($\ce{Ca^{2+}}$) and the anion is nitride ($\ce{N^{3-}}$). So the name of the compound is calcium nitride.

For $\ce{Ca^{2+}}$ and $\ce{O^{2-}}$:

Step 1: Determine the subscripts

The charge of $\ce{Ca^{2+}}$ is $+ 2$ and the charge of $\ce{O^{2-}}$ is $-2$. Using the criss - cross method, the subscripts simplify to $1:1$ (since the greatest common divisor of $2$ and $2$ is $2$, and $\frac{2}{2} = 1$ for both). So the empirical formula is $\ce{CaO}$.

Step 2: Name the compound

The cation is calcium ($\ce{Ca^{2+}}$) and the anion is oxide ($\ce{O^{2-}}$). So the name of the compound is calcium oxide.

Filling the table:

cation	anion	empirical formula	name of compound
$\ce{Ca^{2+}}$	$\ce{S^{2-}}$	$\ce{CaS}$	calcium sulfide
$\ce{Ca^{2+}}$	$\ce{N^{3-}}$	$\ce{Ca_3N_2}$	calcium nitride
$\ce{Ca^{2+}}$	$\ce{O^{2-}}$	$\ce{CaO}$	calcium oxide