## Chemistry: An Atoms-Focused Approach

(a) Ruthenium (II) sulfide This is a compound involving a transition metal, so we have to be aware of which valence to use for the metal. Let us look at sulfur first to be able to determine what the charge on the metal would be. Sulfur has a valence of 2-. Since ruthenium (Ru) is in a 1:1 relationship with sulfur in this compound, ruthenium must have a valence of 2+. We will use the full name of the cation (ruthenium), follow it with the Roman numeral II in parentheses to denote the cation's valence, and then add the anion (sulfur), changing its ending to $-ide$. (b) Palladium (II) chloride This is a compound involving a transition metal, so we have to be aware of which valence to use for the metal. Let us look at the anion first to be able to determine what the charge on the metal would be. Chlorine has a valence of 1-. From the chemical formula, we see that we need two chlorine atoms for every one palladium atom. We see that palladium must have a valence of 2+. We will use the full name of the cation (palladium), follow it with the Roman numeral II in parentheses to denote the cation's valence, and then add the anion (chlorine), changing its ending to $-ide$. (c) Silver oxide This is a compound involving a transition metal; however, since we are dealing with silver, we don't need to worry about the valence because silver typically has one valence. Silver has a valence of 1+ whereas oxygen has a valence of 2-. We will use the full name of the cation (silver) and then add the anion (oxygen), changing the ending to $-ide$. (d) Tungsten (VI) oxide This is a compound involving a transition metal, so we have to be aware of which valence to use for the metal. Let us look at the anion first to be able to determine what the charge on the metal would be. Oxygen has a valence of 2-. From the chemical formula, we see that we need one tungsten atom for every three oxygen atoms. The ratio of tungsten to oxygen in this compound is 1:3. We see that tungsten must have a valence of 6+. We will use the full name of the cation (tungsten), follow it with the Roman numeral VI in parentheses to denote the cation's valence, and then add the anion (oxygen), changing the ending to $-ide$. (e) Platinum (IV) oxide This is a compound involving a transition metal, so we have to be aware of which valence to use for the metal. Let us look at the anion first to be able to determine what the charge on the metal would be. Oxygen has a valence of 2-. From the chemical formula, we see that we need one platinum atom for every two oxygen atoms. The ratio of platinum to oxygen in this compound is 1:2. We see that platinum must have a valence of 4+. We will use the full name of the cation (platinum), follow it with the Roman numeral IV in parentheses to denote the cation's valence, and then add the anion (oxygen), changing the ending to $-ide$.