Answer :
Sure! To write the empirical formulas for ionic compounds formed from the given ions [tex]\(CN^-\)[/tex], [tex]\(Fe^{2+}\)[/tex], [tex]\(CrO_4^{2-}\)[/tex], and [tex]\(NH_4^+\)[/tex], we need to combine these ions to form neutral compounds. Let's determine the possible combinations step-by-step:
### 1. Combination of [tex]\(Fe^{2+}\)[/tex] and [tex]\(CN^-\)[/tex]:
The iron ion ([tex]\(Fe^{2+}\)[/tex]) has a charge of [tex]\(2+\)[/tex], while the cyanide ion ([tex]\(CN^{-}\)[/tex]) has a charge of [tex]\(1-\)[/tex]. To form a neutral compound, we need two cyanide ions to balance the charge of one iron ion.
- Charges: [tex]\(Fe^{2+}\)[/tex] and [tex]\(2 \times CN^{-}\)[/tex]
- Formula: [tex]\(Fe(CN)_2\)[/tex]
The empirical formula is [tex]\( \text{Fe(CN)}_2 \)[/tex].
### 2. Combination of [tex]\(NH_4^+\)[/tex] and [tex]\(CrO_4^{2-}\)[/tex]:
The ammonium ion ([tex]\(NH_4^+\)[/tex]) has a charge of [tex]\(1+\)[/tex], while the chromate ion ([tex]\(CrO_4^{2-}\)[/tex]) has a charge of [tex]\(2-\)[/tex]. To balance the charges, we need two ammonium ions for every one chromate ion.
- Charges: [tex]\(2 \times NH_4^+\)[/tex] and [tex]\(CrO_4^{2-}\)[/tex]
- Formula: [tex]\((NH_4)_2CrO_4\)[/tex]
The empirical formula is [tex]\((\text{NH}_4)_2\text{CrO}_4\)[/tex].
### 3. Combination of [tex]\(Fe^{2+}\)[/tex] and [tex]\(CrO_4^{2-}\)[/tex]:
Both the iron ion ([tex]\(Fe^{2+}\)[/tex]) and the chromate ion ([tex]\(CrO_4^{2-}\)[/tex]) have charges of [tex]\(2+\)[/tex] and [tex]\(2-\)[/tex] respectively. They can combine in a 1:1 ratio to form a neutral compound.
- Charges: [tex]\(Fe^{2+}\)[/tex] and [tex]\(CrO_4^{2-}\)[/tex]
- Formula: [tex]\(FeCrO_4\)[/tex]
The empirical formula is [tex]\( \text{FeCrO}_4 \)[/tex].
### 4. Combination of [tex]\(NH_4^+\)[/tex] and [tex]\(CN^-\)[/tex]:
The ammonium ion ([tex]\(NH_4^+\)[/tex]) has a charge of [tex]\(1+\)[/tex], and the cyanide ion ([tex]\(CN^{-}\)[/tex]) has a charge of [tex]\(1-\)[/tex]. These ions can combine in a 1:1 ratio to form a neutral compound.
- Charges: [tex]\(NH_4^+\)[/tex] and [tex]\(CN^-\)[/tex]
- Formula: [tex]\(NH_4CN\)[/tex]
The empirical formula is [tex]\( \text{NH}_4\text{CN} \)[/tex].
In summary, the empirical formulas for the four ionic compounds that could be formed from the given ions are:
1. [tex]\( \text{Fe(CN)}_2 \)[/tex]
2. [tex]\( (\text{NH}_4)_2\text{CrO}_4 \)[/tex]
3. [tex]\( \text{FeCrO}_4 \)[/tex]
4. [tex]\( \text{NH}_4\text{CN} \)[/tex]
### 1. Combination of [tex]\(Fe^{2+}\)[/tex] and [tex]\(CN^-\)[/tex]:
The iron ion ([tex]\(Fe^{2+}\)[/tex]) has a charge of [tex]\(2+\)[/tex], while the cyanide ion ([tex]\(CN^{-}\)[/tex]) has a charge of [tex]\(1-\)[/tex]. To form a neutral compound, we need two cyanide ions to balance the charge of one iron ion.
- Charges: [tex]\(Fe^{2+}\)[/tex] and [tex]\(2 \times CN^{-}\)[/tex]
- Formula: [tex]\(Fe(CN)_2\)[/tex]
The empirical formula is [tex]\( \text{Fe(CN)}_2 \)[/tex].
### 2. Combination of [tex]\(NH_4^+\)[/tex] and [tex]\(CrO_4^{2-}\)[/tex]:
The ammonium ion ([tex]\(NH_4^+\)[/tex]) has a charge of [tex]\(1+\)[/tex], while the chromate ion ([tex]\(CrO_4^{2-}\)[/tex]) has a charge of [tex]\(2-\)[/tex]. To balance the charges, we need two ammonium ions for every one chromate ion.
- Charges: [tex]\(2 \times NH_4^+\)[/tex] and [tex]\(CrO_4^{2-}\)[/tex]
- Formula: [tex]\((NH_4)_2CrO_4\)[/tex]
The empirical formula is [tex]\((\text{NH}_4)_2\text{CrO}_4\)[/tex].
### 3. Combination of [tex]\(Fe^{2+}\)[/tex] and [tex]\(CrO_4^{2-}\)[/tex]:
Both the iron ion ([tex]\(Fe^{2+}\)[/tex]) and the chromate ion ([tex]\(CrO_4^{2-}\)[/tex]) have charges of [tex]\(2+\)[/tex] and [tex]\(2-\)[/tex] respectively. They can combine in a 1:1 ratio to form a neutral compound.
- Charges: [tex]\(Fe^{2+}\)[/tex] and [tex]\(CrO_4^{2-}\)[/tex]
- Formula: [tex]\(FeCrO_4\)[/tex]
The empirical formula is [tex]\( \text{FeCrO}_4 \)[/tex].
### 4. Combination of [tex]\(NH_4^+\)[/tex] and [tex]\(CN^-\)[/tex]:
The ammonium ion ([tex]\(NH_4^+\)[/tex]) has a charge of [tex]\(1+\)[/tex], and the cyanide ion ([tex]\(CN^{-}\)[/tex]) has a charge of [tex]\(1-\)[/tex]. These ions can combine in a 1:1 ratio to form a neutral compound.
- Charges: [tex]\(NH_4^+\)[/tex] and [tex]\(CN^-\)[/tex]
- Formula: [tex]\(NH_4CN\)[/tex]
The empirical formula is [tex]\( \text{NH}_4\text{CN} \)[/tex].
In summary, the empirical formulas for the four ionic compounds that could be formed from the given ions are:
1. [tex]\( \text{Fe(CN)}_2 \)[/tex]
2. [tex]\( (\text{NH}_4)_2\text{CrO}_4 \)[/tex]
3. [tex]\( \text{FeCrO}_4 \)[/tex]
4. [tex]\( \text{NH}_4\text{CN} \)[/tex]
