Name the compounds in parts a-d and write the formulas for the compounds in parts e-h. a. \(\mathrm{Hg}_{2} \mathrm{O}\) b. \(\mathrm{FeBr}_{3}\) c. \(\operatorname{CoS}\) d. \(\mathrm{TiCl}_{4}\) e. tin(II) nitride f. cobalt(III) iodide g. mercury(II) oxide h. chromium(VI) sulfide

To name compounds a-d, we will use the rules for naming ionic compounds and identify the cation and anion. a. \(\mathrm{Hg}_{2} \mathrm{O}\) This compound is composed of mercury (Hg) and oxygen (O). Since mercury is a transition metal with variable oxidation states, we have to determine its oxidation number. Oxygen has a fixed oxidation state with the charge of -2, and there are two mercury atoms in the compound, each with a charge of +1. Therefore, the compound's name is mercury(I) oxide. b. \(\mathrm{FeBr}_{3}\) This compound is composed of iron (Fe) and bromine (Br). Since iron is another transition metal, its oxidation number must be determined. Bromine has a fixed oxidation state with the charge of -1, and there are three bromine atoms in the compound, meaning the iron atom has a total charge of +3. The compound's name is iron(III) bromide. c. \(\operatorname{CoS}\) This compound is composed of cobalt (Co) and sulfur (S). Cobalt is a transition metal with variable oxidation states, so we need to find its oxidation number. Sulfur has a fixed oxidation state of -2, which means that cobalt has an oxidation number of +2. The compound's name is cobalt(II) sulfide. d. \(\mathrm{TiCl}_{4}\) This compound is composed of titanium (Ti) and chlorine (Cl). Since titanium is a transition metal with variable oxidation states, we need to determine its oxidation number. Chlorine has a fixed oxidation state of -1, and there are four chlorine atoms in the compound, which means the titanium atom has a total charge of +4. The compound's name is titanium(IV) chloride.

To write the chemical formulas for compounds e-h, we need to identify the cations and anions and use their charges to determine the correct chemical formula. e. tin(II) nitride Tin(II) has a 2+ charge represented by Sn, while nitride has a 3- charge represented by N. To determine the correct formula, we need to balance the charges. The chemical formula will be Sn₃N₂. f. cobalt(III) iodide Cobalt(III) has a 3+ charge represented by Co, and iodide has a 1- charge represented by I. To balance the charges, the chemical formula will be CoI₃. g. mercury(II) oxide Mercury(II) has a 2+ charge represented by Hg, and oxide has a 2- charge represented by O. The charges are balanced, so the chemical formula will be HgO. h. chromium(VI) sulfide Chromium(VI) has a 6+ charge represented by Cr, and sulfide has a 2- charge represented by S. To balance the charges, the chemical formula will be Cr₂S₆.