Answer :
Certainly! Let's write out the formulas for the compounds formed between the provided pairs of ions. To do this, we need to balance the charges of the ions such that the resulting compound is electrically neutral.
### a. Aluminum (Al) and Bromine (Br)
Aluminum typically has a charge of [tex]\( +3 \)[/tex] (Al[tex]\(^{3+}\)[/tex]), and bromine has a charge of [tex]\( -1 \)[/tex] (Br[tex]\(^-\)[/tex]). To balance the charges, we need three bromine ions to balance one aluminum ion:
[tex]\[ \text{AlBr}_3 \][/tex]
### b. Sodium (Na) and Oxygen (O)
Sodium typically has a charge of [tex]\( +1 \)[/tex] (Na[tex]\(^+\)[/tex]), and oxygen has a charge of [tex]\( -2 \)[/tex] (O[tex]\(^{2-}\)[/tex]). To balance the charges, we need two sodium ions to balance one oxygen ion:
[tex]\[ \text{Na}_2\text{O} \][/tex]
### c. Magnesium (Mg) and Iodine (I)
Magnesium typically has a charge of [tex]\( +2 \)[/tex] (Mg[tex]\(^{2+}\)[/tex]), and iodine has a charge of [tex]\( -1 \)[/tex] (I[tex]\(^-\)[/tex]). To balance the charges, we need two iodine ions to balance one magnesium ion:
[tex]\[ \text{MgI}_2 \][/tex]
### d. [tex]\( \text{Pb}^{2+} \)[/tex] and [tex]\( \text{O}^{2-} \)[/tex]
Lead(II), denoted as [tex]\( \text{Pb}^{2+} \)[/tex], and oxygen ion [tex]\( \text{O}^{2-} \)[/tex] have equal and opposite charges that balance in a 1:1 ratio:
[tex]\[ \text{PbO} \][/tex]
### e. [tex]\( \text{Sn}^{2+} \)[/tex] and [tex]\( \text{I}^{-} \)[/tex]
Tin(II), denoted as [tex]\( \text{Sn}^{2+} \)[/tex], and iodine ion [tex]\( \text{I}^{-} \)[/tex]. To balance the charges, we need two iodine ions to balance one tin ion:
[tex]\[ \text{SnI}_2 \][/tex]
### f. [tex]\( \text{Fe}^{3+} \)[/tex] and [tex]\( \text{S}^{2-} \)[/tex]
Iron(III), denoted as [tex]\( \text{Fe}^{3+} \)[/tex], and sulfur ion [tex]\( \text{S}^{2-} \)[/tex]. To balance the charges, we use the least common multiple of 3 and 2, which is 6. Hence, we need two Fe[tex]\(^{3+}\)[/tex] ions and three S[tex]\(^{2-}\)[/tex] ions:
[tex]\[ \text{Fe}_2\text{S}_3 \][/tex]
### g. [tex]\( \text{Cu}^{2+} \)[/tex] and [tex]\( \text{NO}_3^{-} \)[/tex]
Copper(II), denoted as [tex]\( \text{Cu}^{2+} \)[/tex], and nitrate ion [tex]\( \text{NO}_3^{-} \)[/tex]. To balance the charges, we need two nitrate ions to balance one copper ion:
[tex]\[ \text{Cu}(\text{NO}_3)_2 \][/tex]
### h. [tex]\( \text{NH}_4^{+} \)[/tex] and [tex]\( \text{SO}_4^{2-} \)[/tex]
Ammonium ion [tex]\( \text{NH}_4^{+} \)[/tex] and sulfate ion [tex]\( \text{SO}_4^{2-} \)[/tex]. To balance the charges, we need two ammonium ions to balance one sulfate ion:
[tex]\[ (\text{NH}_4)_2\text{SO}_4 \][/tex]
To summarize, the formulas for the compounds are:
a. [tex]\( \text{AlBr}_3 \)[/tex]
b. [tex]\( \text{Na}_2\text{O} \)[/tex]
c. [tex]\( \text{MgI}_2 \)[/tex]
d. [tex]\( \text{PbO} \)[/tex]
e. [tex]\( \text{SnI}_2 \)[/tex]
f. [tex]\( \text{Fe}_2\text{S}_3 \)[/tex]
g. [tex]\( \text{Cu}(\text{NO}_3)_2 \)[/tex]
h. [tex]\( (\text{NH}_4)_2\text{SO}_4 \)[/tex]
### a. Aluminum (Al) and Bromine (Br)
Aluminum typically has a charge of [tex]\( +3 \)[/tex] (Al[tex]\(^{3+}\)[/tex]), and bromine has a charge of [tex]\( -1 \)[/tex] (Br[tex]\(^-\)[/tex]). To balance the charges, we need three bromine ions to balance one aluminum ion:
[tex]\[ \text{AlBr}_3 \][/tex]
### b. Sodium (Na) and Oxygen (O)
Sodium typically has a charge of [tex]\( +1 \)[/tex] (Na[tex]\(^+\)[/tex]), and oxygen has a charge of [tex]\( -2 \)[/tex] (O[tex]\(^{2-}\)[/tex]). To balance the charges, we need two sodium ions to balance one oxygen ion:
[tex]\[ \text{Na}_2\text{O} \][/tex]
### c. Magnesium (Mg) and Iodine (I)
Magnesium typically has a charge of [tex]\( +2 \)[/tex] (Mg[tex]\(^{2+}\)[/tex]), and iodine has a charge of [tex]\( -1 \)[/tex] (I[tex]\(^-\)[/tex]). To balance the charges, we need two iodine ions to balance one magnesium ion:
[tex]\[ \text{MgI}_2 \][/tex]
### d. [tex]\( \text{Pb}^{2+} \)[/tex] and [tex]\( \text{O}^{2-} \)[/tex]
Lead(II), denoted as [tex]\( \text{Pb}^{2+} \)[/tex], and oxygen ion [tex]\( \text{O}^{2-} \)[/tex] have equal and opposite charges that balance in a 1:1 ratio:
[tex]\[ \text{PbO} \][/tex]
### e. [tex]\( \text{Sn}^{2+} \)[/tex] and [tex]\( \text{I}^{-} \)[/tex]
Tin(II), denoted as [tex]\( \text{Sn}^{2+} \)[/tex], and iodine ion [tex]\( \text{I}^{-} \)[/tex]. To balance the charges, we need two iodine ions to balance one tin ion:
[tex]\[ \text{SnI}_2 \][/tex]
### f. [tex]\( \text{Fe}^{3+} \)[/tex] and [tex]\( \text{S}^{2-} \)[/tex]
Iron(III), denoted as [tex]\( \text{Fe}^{3+} \)[/tex], and sulfur ion [tex]\( \text{S}^{2-} \)[/tex]. To balance the charges, we use the least common multiple of 3 and 2, which is 6. Hence, we need two Fe[tex]\(^{3+}\)[/tex] ions and three S[tex]\(^{2-}\)[/tex] ions:
[tex]\[ \text{Fe}_2\text{S}_3 \][/tex]
### g. [tex]\( \text{Cu}^{2+} \)[/tex] and [tex]\( \text{NO}_3^{-} \)[/tex]
Copper(II), denoted as [tex]\( \text{Cu}^{2+} \)[/tex], and nitrate ion [tex]\( \text{NO}_3^{-} \)[/tex]. To balance the charges, we need two nitrate ions to balance one copper ion:
[tex]\[ \text{Cu}(\text{NO}_3)_2 \][/tex]
### h. [tex]\( \text{NH}_4^{+} \)[/tex] and [tex]\( \text{SO}_4^{2-} \)[/tex]
Ammonium ion [tex]\( \text{NH}_4^{+} \)[/tex] and sulfate ion [tex]\( \text{SO}_4^{2-} \)[/tex]. To balance the charges, we need two ammonium ions to balance one sulfate ion:
[tex]\[ (\text{NH}_4)_2\text{SO}_4 \][/tex]
To summarize, the formulas for the compounds are:
a. [tex]\( \text{AlBr}_3 \)[/tex]
b. [tex]\( \text{Na}_2\text{O} \)[/tex]
c. [tex]\( \text{MgI}_2 \)[/tex]
d. [tex]\( \text{PbO} \)[/tex]
e. [tex]\( \text{SnI}_2 \)[/tex]
f. [tex]\( \text{Fe}_2\text{S}_3 \)[/tex]
g. [tex]\( \text{Cu}(\text{NO}_3)_2 \)[/tex]
h. [tex]\( (\text{NH}_4)_2\text{SO}_4 \)[/tex]
