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 anion empirical formula name of compound
ca²⁺ p³⁻ □ □
ca²⁺ s²⁻ □ □
ca²⁺ n³⁻ □ □
ca²⁺ o²⁻ □ □

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 	anion 	empirical formula 	name of compound
ca²⁺ 	p³⁻ 	□ 	□
ca²⁺ 	s²⁻ 	□ 	□
ca²⁺ 	n³⁻ 	□ 	□
ca²⁺ 	o²⁻ 	□ 	□

Accepted Answer

# Explanation:
To determine the empirical formula of an ionic compound, we use the criss - cross method. The charge of the cation becomes the subscript of the anion and vice - versa. Then we simplify the subscripts to the smallest whole - number ratio. For naming ionic compounds, we use the name of the cation (calcium in this case, as it has a fixed charge) followed by the name of the anion (with the appropriate suffix, like - ide for monatomic anions).

## Step 1: For $\ce{Ca^{2+}}$ and $\ce{P^{3-}}$
- 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 anion's charge) and the subscript of $\ce{P}$ will be 2 (the magnitude of the cation's charge). So the empirical formula is $\ce{Ca_{3}P_{2}}$.
- The name of the compound is calcium phosphide (since the anion is $\ce{P^{3-}}$, named phosphide).

## Step 2: For $\ce{Ca^{2+}}$ and $\ce{S^{2-}}$
- The charge of $\ce{Ca^{2+}}$ is $+ 2$ and the charge of $\ce{S^{2-}}$ is $-2$. Using the criss - cross method, the subscript of $\ce{Ca}$ will be 2 and the subscript of $\ce{S}$ will be 2. But we can simplify this ratio by dividing both subscripts by 2. So the empirical formula is $\ce{CaS}$.
- The name of the compound is calcium sulfide (the anion $\ce{S^{2-}}$ is named sulfide).

## Step 3: For $\ce{Ca^{2+}}$ and $\ce{N^{3-}}$
- 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 $\ce{Ca}$ will be 3 and the subscript of $\ce{N}$ will be 2. So the empirical formula is $\ce{Ca_{3}N_{2}}$.
- The name of the compound is calcium nitride (the anion $\ce{N^{3-}}$ is named nitride).

## Step 4: For $\ce{Ca^{2+}}$ and $\ce{O^{2-}}$
- The charge of $\ce{Ca^{2+}}$ is $+ 2$ and the charge of $\ce{O^{2-}}$ is $-2$. Using the criss - cross method, the subscript of $\ce{Ca}$ will be 2 and the subscript of $\ce{O}$ will be 2. We simplify this ratio by dividing both subscripts by 2. So the empirical formula is $\ce{CaO}$.
- The name of the compound is calcium oxide (the anion $\ce{O^{2-}}$ is named oxide).

# Answer:

| cation | anion | empirical formula | name of compound |
| ---- | ---- | ---- | ---- |
| $\ce{Ca^{2+}}$ | $\ce{P^{3-}}$ | $\ce{Ca_{3}P_{2}}$ | calcium phosphide |
| $\ce{Ca^{2+}}$ | $\ce{S^{2-}}$ | $\ce{CaS}$ | calcium sulfide |
| $\ce{Ca^{2+}}$ | $\ce{N^{3-}}$ | $\ce{Ca_{3}N_{2}}$ | calcium nitride |
| $\ce{Ca^{2+}}$ | $\ce{O^{2-}}$ | $\ce{CaO}$ | calcium oxide |

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QUESTION IMAGE

Question

Explanation:

Step 1: For $\ce{Ca^{2+}}$ and $\ce{P^{3-}}$

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

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

Step 4: For $\ce{Ca^{2+}}$ and $\ce{O^{2-}}$

Answer:

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