Question

1. write the chemical formula. as always, the number of atoms is written as a subscript after the element symbol. you do not need to write the charges in the final formula.

example

name	cation	anion	chemical formula

practice exercises
write the chemical formula for each of the following compounds.

1. antimony (v) iodide
2. lead (ii) oxide
3. tin (iv) phosphide
4. germanium (iv) fluoride
5. gallium (iii) carbide
6. copper (i) bromide
7. platinum (iv) silicide
8. cobalt (ii) chloride
9. rhenium (vi) hydride
10. tantalum (v) selenide

rules for writing formulas for ternary compounds with fixed charges

• ternary ionic compounds are composed of ions of more than two different elements - one of which is a metal cation, and the other a polyatomic nonmetal anion. the cation, anion, or both is a polyatomic ion.
• the polyatomic ion is a single ion, and cannot be divided into parts.

1. identify the cation part of the compound by either looking it up on the polyatomic ions table. the first word of the compound name usually indicates the cation.

Explanation:

Step1: Determine cations and anions

For each compound, identify the cation and its charge from the name (the Roman - numeral indicates the charge of the metal cation), and the anion and its charge.

Step2: Balance charges

Use the criss - cross method to balance the charges between the cation and the anion to get the chemical formula.

1. Antimony (V) iodide: Cation is $Sb^{5 + }$, anion is $I^{-}$. To balance charges, we need 5 iodide ions for 1 antimony ion. Chemical formula is $SbI_{5}$.
2. Lead (II) oxide: Cation is $Pb^{2+}$, anion is $O^{2 - }$. Chemical formula is $PbO$.
3. Tin (IV) phosphide: Cation is $Sn^{4+}$, anion is $P^{3 - }$. To balance charges, we find the least - common multiple of 4 and 3 which is 12. So we need 3 tin ions and 4 phosphide ions. Chemical formula is $Sn_{3}P_{4}$.
4. Germanium (IV) fluoride: Cation is $Ge^{4+}$, anion is $F^{-}$. We need 4 fluoride ions for 1 germanium ion. Chemical formula is $GeF_{4}$.
5. Gallium (III) carbide: Cation is $Ga^{3+}$, anion is $C^{4 - }$. The least - common multiple of 3 and 4 is 12. So we need 4 gallium ions and 3 carbide ions. Chemical formula is $Ga_{4}C_{3}$.
6. Copper (I) bromide: Cation is $Cu^{+}$, anion is $Br^{-}$. Chemical formula is $CuBr$.
7. Platinum (IV) silicide: Cation is $Pt^{4+}$, anion is $Si^{4 - }$. Chemical formula is $PtSi$.
8. Cobalt (II) chloride: Cation is $Co^{2+}$, anion is $Cl^{-}$. We need 2 chloride ions for 1 cobalt ion. Chemical formula is $CoCl_{2}$.
9. Rhenium (VI) hydride: Cation is $Re^{6+}$, anion is $H^{-}$. We need 6 hydride ions for 1 rhenium ion. Chemical formula is $ReH_{6}$.
10. Tantalum (V) selenide: Cation is $Ta^{5+}$, anion is $Se^{2 - }$. The least - common multiple of 5 and 2 is 10. So we need 2 tantalum ions and 5 selenide ions. Chemical formula is $Ta_{2}Se_{5}$.

Answer:

1. $SbI_{5}$
2. $PbO$
3. $Sn_{3}P_{4}$
4. $GeF_{4}$
5. $Ga_{4}C_{3}$
6. $CuBr$
7. $PtSi$
8. $CoCl_{2}$
9. $ReH_{6}$
10. $Ta_{2}Se_{5}$