# Chapter 4 - Chemical Bonding - Questions and Problems - Page 179: 51

(a) Manganese (II) sulfide (b) Vanadium (II) nitride (c) Chromium (III) sulfate (d) Cobalt (II) nitrate (e) Iron (III) oxide

#### Work Step by Step

(a) Manganese (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 manganese (Mn) is in a 1:1 relationship with sulfur in this compound, manganese must have a valence of 2+. We will use the full name of the cation (manganese), 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) Vanadium (II) nitride 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. Nitrogen has a valence of 3-. From the chemical formula, we see that we need three vanadium atoms for every two nitrogen atoms. We see that vanadium must have a valence of 2+. We will use the full name of the cation (vanadium), follow it with the Roman numeral II in parentheses to denote the cation's valence, and then add the anion (nitrogen), changing its ending to $-ide$. (c) Chromium (III) sulfate 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. The oxoanion sulfate has a valence of 2-. From the chemical formula, we see that we need two chromium atoms for every three sulfate ions. We see that chromium must have a valence of 3+. We will use the full name of the cation (chromium), follow it with the Roman numeral III in parentheses to denote the cation's valence, and then add the anion (sulfate), keeping the ending as-is. (d) Cobalt (II) nitrate 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. The oxoanion nitrate has a valence of 1-. From the chemical formula, we see that we need one chromium atom for every two nitrate ions. We see that chromium must have a valence of 2+. We will use the full name of the cation (chromium), follow it with the Roman numeral II in parentheses to denote the cation's valence, and then add the anion (nitrate), keeping the ending as-is. (e) Iron (III) 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 two iron atoms for every three oxygen atoms. We see that iron must have a valence of 3+. We will use the full name of the cation (iron), follow it with the Roman numeral III in parentheses to denote the cation's valence, and then add the anion (oxygen), changing the ending to $-ide$.

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