Problem 64
Question
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
Step-by-Step Solution
Verified Answer
a. mercury(I) oxide
b. iron(III) bromide
c. cobalt(II) sulfide
d. titanium(IV) chloride
e. Sn₃N₂
f. CoI₃
g. HgO
h. Cr₂S₆
1Step 1: Naming Compounds given the Chemical Formulas
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.
2Step 2: Writing Chemical Formulas given the Names
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.
Other exercises in this chapter
Problem 62
For each of the following atomic numbers, use the periodic table to write the formula (including the charge) for the simple ion that the element is most likely
View solution Problem 63
Name the compounds in parts a-d and write the formulas for the compounds in parts e-h. a. \(\mathrm{NaBr}\) b. \(\mathrm{Rb}_{2} \mathrm{O}\) c. \(\mathrm{CaS}\
View solution Problem 65
Name each of the following compounds: a. \(\mathrm{CsF}\) b. \(\mathrm{Li}_{3} \mathrm{~N}\) c. \(\mathrm{Ag}_{2} \mathrm{~S}\) d. \(\mathrm{MnO}_{2}\) e. \(\ma
View solution Problem 66
Write the formula for each of the following compounds: a. zinc chloride d. aluminum sulfide b. \(\operatorname{tin}(\mathrm{IV})\) fluoride e. mercury(I) seleni
View solution