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
fe³⁺ | i⁻ | |
fe²⁺ | i⁻ | |
rb⁺ | i⁻ | |
ba²⁺ | i⁻ | |

Explanation:

To solve for the empirical formulas and names of the ionic compounds, we use the principle of charge balance (the total positive charge equals the total negative charge in an ionic compound) and naming conventions for ionic compounds (metal name + non - metal name with suffix -ide, and for transition metals, include the charge in Roman numerals).

For $\boldsymbol{Fe^{3+}}$ and $\boldsymbol{I^-}$:

Step 1: Determine the ratio of ions

Let the number of $Fe^{3+}$ ions be $x$ and the number of $I^-$ ions be $y$. The charge balance equation is $3x + (- 1)y=0$. To balance the charges, we find that when $x = 1$ and $y = 3$ (because $3\times(+3)+3\times(- 1)=0$), the formula is $FeI_3$.

Step 2: Name the compound

Iron is a transition metal with a +3 charge. So the name is Iron(III) iodide.

For $\boldsymbol{Fe^{2+}}$ and $\boldsymbol{I^-}$:

Step 1: Determine the ratio of ions

Using the charge balance equation $2x+(-1)y = 0$. When $x = 1$ and $y = 2$ (since $1\times(+2)+2\times(-1)=0$), the formula is $FeI_2$.

Step 2: Name the compound

Iron has a +2 charge here. So the name is Iron(II) iodide.

For $\boldsymbol{Rb^+}$ and $\boldsymbol{I^-}$:

Step 1: Determine the ratio of ions

The charge balance equation is $1\times(+1)+1\times(-1)=0$. So the ratio of $Rb^+$ to $I^-$ is $1:1$, and the formula is $RbI$.

Step 2: Name the compound

Rubidium is an alkali metal. The name is Rubidium iodide.

For $\boldsymbol{Ba^{2+}}$ and $\boldsymbol{I^-}$:

Step 1: Determine the ratio of ions

Using the charge balance equation $2x+(-1)y = 0$. When $x = 1$ and $y = 2$ (because $1\times(+2)+2\times(-1)=0$), the formula is $BaI_2$.

Step 2: Name the compound

Barium is an alkaline earth metal. The name is Barium iodide.

Filling in the table:

Cation	Anion	Empirical Formula	Name of Compound
$Fe^{2+}$	$I^-$	$FeI_2$	Iron(II) iodide
$Rb^+$	$I^-$	$RbI$	Rubidium iodide
$Ba^{2+}$	$I^-$	$BaI_2$	Barium iodide

Final Answers (for each row):

• For $Fe^{3+}$ and $I^-$: Empirical formula: $\boldsymbol{FeI_3}$, Name: $\boldsymbol{Iron(III) \ iodide}$
• For $Fe^{2+}$ and $I^-$: Empirical formula: $\boldsymbol{FeI_2}$, Name: $\boldsymbol{Iron(II) \ iodide}$
• For $Rb^+$ and $I^-$: Empirical formula: $\boldsymbol{RbI}$, Name: $\boldsymbol{Rubidium \ iodide}$
• For $Ba^{2+}$ and $I^-$: Empirical formula: $\boldsymbol{BaI_2}$, Name: $\boldsymbol{Barium \ iodide}$

Answer:

To solve for the empirical formulas and names of the ionic compounds, we use the principle of charge balance (the total positive charge equals the total negative charge in an ionic compound) and naming conventions for ionic compounds (metal name + non - metal name with suffix -ide, and for transition metals, include the charge in Roman numerals).

For $\boldsymbol{Fe^{3+}}$ and $\boldsymbol{I^-}$:

Step 1: Determine the ratio of ions

Let the number of $Fe^{3+}$ ions be $x$ and the number of $I^-$ ions be $y$. The charge balance equation is $3x + (- 1)y=0$. To balance the charges, we find that when $x = 1$ and $y = 3$ (because $3\times(+3)+3\times(- 1)=0$), the formula is $FeI_3$.

Step 2: Name the compound

Iron is a transition metal with a +3 charge. So the name is Iron(III) iodide.

For $\boldsymbol{Fe^{2+}}$ and $\boldsymbol{I^-}$:

Step 1: Determine the ratio of ions

Using the charge balance equation $2x+(-1)y = 0$. When $x = 1$ and $y = 2$ (since $1\times(+2)+2\times(-1)=0$), the formula is $FeI_2$.

Step 2: Name the compound

Iron has a +2 charge here. So the name is Iron(II) iodide.

For $\boldsymbol{Rb^+}$ and $\boldsymbol{I^-}$:

Step 1: Determine the ratio of ions

The charge balance equation is $1\times(+1)+1\times(-1)=0$. So the ratio of $Rb^+$ to $I^-$ is $1:1$, and the formula is $RbI$.

Step 2: Name the compound

Rubidium is an alkali metal. The name is Rubidium iodide.

For $\boldsymbol{Ba^{2+}}$ and $\boldsymbol{I^-}$:

Step 1: Determine the ratio of ions

Using the charge balance equation $2x+(-1)y = 0$. When $x = 1$ and $y = 2$ (because $1\times(+2)+2\times(-1)=0$), the formula is $BaI_2$.

Step 2: Name the compound

Barium is an alkaline earth metal. The name is Barium iodide.

Filling in the table:

Cation	Anion	Empirical Formula	Name of Compound
$Fe^{2+}$	$I^-$	$FeI_2$	Iron(II) iodide
$Rb^+$	$I^-$	$RbI$	Rubidium iodide
$Ba^{2+}$	$I^-$	$BaI_2$	Barium iodide

Final Answers (for each row):

• For $Fe^{3+}$ and $I^-$: Empirical formula: $\boldsymbol{FeI_3}$, Name: $\boldsymbol{Iron(III) \ iodide}$
• For $Fe^{2+}$ and $I^-$: Empirical formula: $\boldsymbol{FeI_2}$, Name: $\boldsymbol{Iron(II) \ iodide}$
• For $Rb^+$ and $I^-$: Empirical formula: $\boldsymbol{RbI}$, Name: $\boldsymbol{Rubidium \ iodide}$
• For $Ba^{2+}$ and $I^-$: Empirical formula: $\boldsymbol{BaI_2}$, Name: $\boldsymbol{Barium \ iodide}$