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
ba²⁺ | o²⁻ | □ | □
ba²⁺ | p³⁻ | □ | □
ba²⁺ | s²⁻ | □ | □
ba²⁺ | n³⁻ | □ | □

Explanation:

To solve for the empirical formulas and names of the ionic compounds formed from \( \text{Ba}^{2+} \) 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 (metal name + non - metal name with suffix - ide, and for polyatomic ions, use their specific names; here all anions are monatomic non - metal ions).

For \( \text{Ba}^{2+} \) and \( \text{O}^{2 - } \)

Step 1: Determine the empirical formula

The charge of \( \text{Ba}^{2+} \) is \( + 2 \) and the charge of \( \text{O}^{2 - } \) is \( - 2 \). Using the criss - cross method, we take the magnitude of the charge of each ion as the subscript of the other. So the formula is \( \text{Ba}_2\text{O}_2 \), but we simplify it by dividing by the greatest common divisor (which is 2) to get \( \text{BaO} \).

Step 2: Determine the name of the compound

The cation is barium (\( \text{Ba}^{2+} \)) and the anion is oxide (\( \text{O}^{2 - } \)). So the name of the compound is barium oxide.

For \( \text{Ba}^{2+} \) and \( \text{P}^{3 - } \)

Step 1: Determine the empirical formula

The charge of \( \text{Ba}^{2+} \) is \( + 2 \) and the charge of \( \text{P}^{3 - } \) is \( - 3 \). Using the criss - cross method, the subscript of \( \text{Ba} \) will be 3 (the magnitude of the charge of \( \text{P}^{3 - } \)) and the subscript of \( \text{P} \) will be 2 (the magnitude of the charge of \( \text{Ba}^{2+} \)). So the empirical formula is \( \text{Ba}_3\text{P}_2 \).

Step 2: Determine the name of the compound

The cation is barium (\( \text{Ba}^{2+} \)) and the anion is phosphide (\( \text{P}^{3 - } \)). So the name of the compound is barium phosphide.

For \( \text{Ba}^{2+} \) and \( \text{S}^{2 - } \)

Step 1: Determine the empirical formula

The charge of \( \text{Ba}^{2+} \) is \( + 2 \) and the charge of \( \text{S}^{2 - } \) is \( - 2 \). Using the criss - cross method, we get \( \text{Ba}_2\text{S}_2 \), and simplifying by dividing by 2 (the greatest common divisor), we obtain \( \text{BaS} \).

Step 2: Determine the name of the compound

The cation is barium (\( \text{Ba}^{2+} \)) and the anion is sulfide (\( \text{S}^{2 - } \)). So the name of the compound is barium sulfide.

For \( \text{Ba}^{2+} \) and \( \text{N}^{3 - } \)

Step 1: Determine the empirical formula

The charge of \( \text{Ba}^{2+} \) is \( + 2 \) and the charge of \( \text{N}^{3 - } \) is \( - 3 \). Using the criss - cross method, the subscript of \( \text{Ba} \) is 3 (the magnitude of the charge of \( \text{N}^{3 - } \)) and the subscript of \( \text{N} \) is 2 (the magnitude of the charge of \( \text{Ba}^{2+} \)). So the empirical formula is \( \text{Ba}_3\text{N}_2 \).

Step 2: Determine the name of the compound

The cation is barium (\( \text{Ba}^{2+} \)) and the anion is nitride (\( \text{N}^{3 - } \)). So the name of the compound is barium nitride.

Filling the table:

cation	anion	empirical formula	name of compound
\( \text{Ba}^{2+} \)	\( \text{P}^{3 - } \)	\( \text{Ba}_3\text{P}_2 \)	barium phosphide
\( \text{Ba}^{2+} \)	\( \text{S}^{2 - } \)	\( \text{BaS} \)	barium sulfide
\( \text{Ba}^{2+} \)	\( \text{N}^{3 - } \)	\( \text{Ba}_3\text{N}_2 \)	barium nitride

Answer:

To solve for the empirical formulas and names of the ionic compounds formed from \( \text{Ba}^{2+} \) 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 (metal name + non - metal name with suffix - ide, and for polyatomic ions, use their specific names; here all anions are monatomic non - metal ions).

For \( \text{Ba}^{2+} \) and \( \text{O}^{2 - } \)

Step 1: Determine the empirical formula

The charge of \( \text{Ba}^{2+} \) is \( + 2 \) and the charge of \( \text{O}^{2 - } \) is \( - 2 \). Using the criss - cross method, we take the magnitude of the charge of each ion as the subscript of the other. So the formula is \( \text{Ba}_2\text{O}_2 \), but we simplify it by dividing by the greatest common divisor (which is 2) to get \( \text{BaO} \).

Step 2: Determine the name of the compound

The cation is barium (\( \text{Ba}^{2+} \)) and the anion is oxide (\( \text{O}^{2 - } \)). So the name of the compound is barium oxide.

For \( \text{Ba}^{2+} \) and \( \text{P}^{3 - } \)

Step 1: Determine the empirical formula

The charge of \( \text{Ba}^{2+} \) is \( + 2 \) and the charge of \( \text{P}^{3 - } \) is \( - 3 \). Using the criss - cross method, the subscript of \( \text{Ba} \) will be 3 (the magnitude of the charge of \( \text{P}^{3 - } \)) and the subscript of \( \text{P} \) will be 2 (the magnitude of the charge of \( \text{Ba}^{2+} \)). So the empirical formula is \( \text{Ba}_3\text{P}_2 \).

Step 2: Determine the name of the compound

The cation is barium (\( \text{Ba}^{2+} \)) and the anion is phosphide (\( \text{P}^{3 - } \)). So the name of the compound is barium phosphide.

For \( \text{Ba}^{2+} \) and \( \text{S}^{2 - } \)

Step 1: Determine the empirical formula

The charge of \( \text{Ba}^{2+} \) is \( + 2 \) and the charge of \( \text{S}^{2 - } \) is \( - 2 \). Using the criss - cross method, we get \( \text{Ba}_2\text{S}_2 \), and simplifying by dividing by 2 (the greatest common divisor), we obtain \( \text{BaS} \).

Step 2: Determine the name of the compound

The cation is barium (\( \text{Ba}^{2+} \)) and the anion is sulfide (\( \text{S}^{2 - } \)). So the name of the compound is barium sulfide.

For \( \text{Ba}^{2+} \) and \( \text{N}^{3 - } \)

Step 1: Determine the empirical formula

The charge of \( \text{Ba}^{2+} \) is \( + 2 \) and the charge of \( \text{N}^{3 - } \) is \( - 3 \). Using the criss - cross method, the subscript of \( \text{Ba} \) is 3 (the magnitude of the charge of \( \text{N}^{3 - } \)) and the subscript of \( \text{N} \) is 2 (the magnitude of the charge of \( \text{Ba}^{2+} \)). So the empirical formula is \( \text{Ba}_3\text{N}_2 \).

Step 2: Determine the name of the compound

The cation is barium (\( \text{Ba}^{2+} \)) and the anion is nitride (\( \text{N}^{3 - } \)). So the name of the compound is barium nitride.

Filling the table:

cation	anion	empirical formula	name of compound
\( \text{Ba}^{2+} \)	\( \text{P}^{3 - } \)	\( \text{Ba}_3\text{P}_2 \)	barium phosphide
\( \text{Ba}^{2+} \)	\( \text{S}^{2 - } \)	\( \text{BaS} \)	barium sulfide
\( \text{Ba}^{2+} \)	\( \text{N}^{3 - } \)	\( \text{Ba}_3\text{N}_2 \)	barium nitride