Answer :
Let's find the empirical formulas for the ionic compounds formed from the given ions: [tex]\( \text{Pb}^{4+} \)[/tex], [tex]\( \text{CN}^{-} \)[/tex], [tex]\( \text{CO}_3^{2-} \)[/tex], and [tex]\( \text{Fe}^{2+} \)[/tex].
### Compound 1: [tex]\( \text{Pb}^{4+} \)[/tex] and [tex]\( \text{CN}^{-} \)[/tex]
1. [tex]\( \text{Pb}^{4+} \)[/tex] has a charge of [tex]\( +4 \)[/tex].
2. [tex]\( \text{CN}^{-} \)[/tex] has a charge of [tex]\( -1 \)[/tex].
3. To balance the charges, we need 4 [tex]\( \text{CN}^{-} \)[/tex] ions to balance the single [tex]\( \text{Pb}^{4+} \)[/tex] ion.
[tex]\[ \text{Pb}^{4+} + 4(\text{CN}^{-}) \rightarrow \text{Pb(CN)}_4 \][/tex]
So, the empirical formula for this compound is [tex]\( \text{Pb(CN)}_4 \)[/tex].
### Compound 2: [tex]\( \text{Pb}^{4+} \)[/tex] and [tex]\( \text{CO}_3^{2-} \)[/tex]
1. [tex]\( \text{Pb}^{4+} \)[/tex] has a charge of [tex]\( +4 \)[/tex].
2. [tex]\( \text{CO}_3^{2-} \)[/tex] has a charge of [tex]\( -2 \)[/tex].
3. To balance the charges, we need 2 [tex]\( \text{CO}_3^{2-} \)[/tex] ions to balance the single [tex]\( \text{Pb}^{4+} \)[/tex] ion.
[tex]\[ \text{Pb}^{4+} + 2(\text{CO}_3^{2-}) \rightarrow \text{Pb(CO}_3\text{)}_2 \][/tex]
So, the empirical formula for this compound is [tex]\( \text{Pb(CO}_3\text{)}_2 \)[/tex].
### Compound 3: [tex]\( \text{Fe}^{2+} \)[/tex] and [tex]\( \text{CN}^{-} \)[/tex]
1. [tex]\( \text{Fe}^{2+} \)[/tex] has a charge of [tex]\( +2 \)[/tex].
2. [tex]\( \text{CN}^{-} \)[/tex] has a charge of [tex]\( -1 \)[/tex].
3. To balance the charges, we need 2 [tex]\( \text{CN}^{-} \)[/tex] ions to balance the single [tex]\( \text{Fe}^{2+} \)[/tex] ion.
[tex]\[ \text{Fe}^{2+} + 2(\text{CN}^{-}) \rightarrow \text{Fe(CN)}_2 \][/tex]
So, the empirical formula for this compound is [tex]\( \text{Fe(CN)}_2 \)[/tex].
### Compound 4: [tex]\( \text{Fe}^{2+} \)[/tex] and [tex]\( \text{CO}_3^{2-} \)[/tex]
1. [tex]\( \text{Fe}^{2+} \)[/tex] has a charge of [tex]\( +2 \)[/tex].
2. [tex]\( \text{CO}_3^{2-} \)[/tex] has a charge of [tex]\( -2 \)[/tex].
3. To balance the charges, 1 [tex]\( \text{Fe}^{2+} \)[/tex] ion matches perfectly with 1 [tex]\( \text{CO}_3^{2-} \)[/tex] ion.
[tex]\[ \text{Fe}^{2+} + \text{CO}_3^{2-} \rightarrow \text{FeCO}_3 \][/tex]
So, the empirical formula for this compound is [tex]\( \text{FeCO}_3 \)[/tex].
Putting it all together, the empirical formulas for the four ionic compounds are:
1. [tex]\( \text{Pb(CN)}_4 \)[/tex]
2. [tex]\( \text{Pb(CO}_3\text{)}_2 \)[/tex]
3. [tex]\( \text{Fe(CN)}_2 \)[/tex]
4. [tex]\( \text{FeCO}_3 \)[/tex]
### Compound 1: [tex]\( \text{Pb}^{4+} \)[/tex] and [tex]\( \text{CN}^{-} \)[/tex]
1. [tex]\( \text{Pb}^{4+} \)[/tex] has a charge of [tex]\( +4 \)[/tex].
2. [tex]\( \text{CN}^{-} \)[/tex] has a charge of [tex]\( -1 \)[/tex].
3. To balance the charges, we need 4 [tex]\( \text{CN}^{-} \)[/tex] ions to balance the single [tex]\( \text{Pb}^{4+} \)[/tex] ion.
[tex]\[ \text{Pb}^{4+} + 4(\text{CN}^{-}) \rightarrow \text{Pb(CN)}_4 \][/tex]
So, the empirical formula for this compound is [tex]\( \text{Pb(CN)}_4 \)[/tex].
### Compound 2: [tex]\( \text{Pb}^{4+} \)[/tex] and [tex]\( \text{CO}_3^{2-} \)[/tex]
1. [tex]\( \text{Pb}^{4+} \)[/tex] has a charge of [tex]\( +4 \)[/tex].
2. [tex]\( \text{CO}_3^{2-} \)[/tex] has a charge of [tex]\( -2 \)[/tex].
3. To balance the charges, we need 2 [tex]\( \text{CO}_3^{2-} \)[/tex] ions to balance the single [tex]\( \text{Pb}^{4+} \)[/tex] ion.
[tex]\[ \text{Pb}^{4+} + 2(\text{CO}_3^{2-}) \rightarrow \text{Pb(CO}_3\text{)}_2 \][/tex]
So, the empirical formula for this compound is [tex]\( \text{Pb(CO}_3\text{)}_2 \)[/tex].
### Compound 3: [tex]\( \text{Fe}^{2+} \)[/tex] and [tex]\( \text{CN}^{-} \)[/tex]
1. [tex]\( \text{Fe}^{2+} \)[/tex] has a charge of [tex]\( +2 \)[/tex].
2. [tex]\( \text{CN}^{-} \)[/tex] has a charge of [tex]\( -1 \)[/tex].
3. To balance the charges, we need 2 [tex]\( \text{CN}^{-} \)[/tex] ions to balance the single [tex]\( \text{Fe}^{2+} \)[/tex] ion.
[tex]\[ \text{Fe}^{2+} + 2(\text{CN}^{-}) \rightarrow \text{Fe(CN)}_2 \][/tex]
So, the empirical formula for this compound is [tex]\( \text{Fe(CN)}_2 \)[/tex].
### Compound 4: [tex]\( \text{Fe}^{2+} \)[/tex] and [tex]\( \text{CO}_3^{2-} \)[/tex]
1. [tex]\( \text{Fe}^{2+} \)[/tex] has a charge of [tex]\( +2 \)[/tex].
2. [tex]\( \text{CO}_3^{2-} \)[/tex] has a charge of [tex]\( -2 \)[/tex].
3. To balance the charges, 1 [tex]\( \text{Fe}^{2+} \)[/tex] ion matches perfectly with 1 [tex]\( \text{CO}_3^{2-} \)[/tex] ion.
[tex]\[ \text{Fe}^{2+} + \text{CO}_3^{2-} \rightarrow \text{FeCO}_3 \][/tex]
So, the empirical formula for this compound is [tex]\( \text{FeCO}_3 \)[/tex].
Putting it all together, the empirical formulas for the four ionic compounds are:
1. [tex]\( \text{Pb(CN)}_4 \)[/tex]
2. [tex]\( \text{Pb(CO}_3\text{)}_2 \)[/tex]
3. [tex]\( \text{Fe(CN)}_2 \)[/tex]
4. [tex]\( \text{FeCO}_3 \)[/tex]
