Problem 65
Question
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. \(\mathrm{TiO}_{2}\) f. \(\mathrm{Sr}_{3} \mathrm{P}_{2}\)
Step-by-Step Solution
Verified Answer
a. Cesium Fluoride
b. Lithium Nitride
c. Silver Sulfide
d. Manganese (IV) Oxide
e. Titanium (IV) Oxide
f. Strontium Phosphide
1Step 1: Compound a: CsF
In CsF, Cs is Cesium, a group 1 metal with a +1 charge, and F is the halide Fluorine, with a -1 charge. The compound name is Cesium Fluoride.
2Step 2: Compound b: Li3N
In Li3N, Li is Lithium, a group 1 metal with a +1 charge, and N is Nitrogen, a group 15 nonmetal with a -3 charge. The compound name is Lithium Nitride.
3Step 3: Compound c: Ag2S
In Ag2S, Ag is Silver, a transition metal that has a +1 charge, and S is Sulfur, a group 16 nonmetal with a -2 charge. The compound name is Silver Sulfide.
4Step 4: Compound d: MnO2
In MnO2, Mn is Manganese, a transition metal with a +4 charge, and O is Oxygen, a group 16 nonmetal with a -2 charge. The compound name is Manganese (IV) Oxide. The Roman numeral (IV) is required to indicate the oxidation state of the transition metal Manganese.
5Step 5: Compound e: TiO2
In TiO2, Ti is Titanium, a transition metal with a +4 charge, and O is Oxygen, a group 16 nonmetal with a -2 charge. The compound name is Titanium (IV) Oxide. The Roman numeral (IV) indicates the oxidation state of the transition metal Titanium.
6Step 6: Compound f: Sr3P2
In Sr3P2, Sr is Strontium, a group 2 alkaline earth metal with a +2 charge, and P is Phosphorus, a group 15 nonmetal with a -3 charge. The compound name is Strontium Phosphide.
Key Concepts
Ionic CompoundsTransition MetalsOxidation StatesChemical Formulas
Ionic Compounds
Ionic compounds are formed when metals and non-metals exchange electrons. A metal typically loses one or more electrons to become positively charged, while a non-metal gains those electrons to become negatively charged. The resulting positive and negative ions attract each other and form an ionic bond. This type of bond is strong due to the electrostatic forces between the charged ions. Common characteristics of ionic compounds include high melting and boiling points, and they usually crystallize into solid forms.
When naming ionic compounds, the metal's name is written first, followed by the non-metal with an -ide suffix. For example, in the compound \(\mathrm{CsF}\), Cesium is the metal, and fluorine is changed to fluoride. This means the compound is named Cesium Fluoride.
When naming ionic compounds, the metal's name is written first, followed by the non-metal with an -ide suffix. For example, in the compound \(\mathrm{CsF}\), Cesium is the metal, and fluorine is changed to fluoride. This means the compound is named Cesium Fluoride.
Transition Metals
Transition metals are a special group of elements found in the middle of the periodic table. These metals have unique properties that include forming various oxidation states. This means they can lose different numbers of electrons when forming compounds. Because of this variability, naming compounds with transition metals requires more detail.
For instance, in \(\mathrm{MnO_2}\), Manganese is a transition metal. To correctly name the compound, you must specify Manganese's oxidation state using Roman numerals. Hence, \(\mathrm{MnO_2}\) is named Manganese (IV) Oxide, indicating a +4 charge.
For instance, in \(\mathrm{MnO_2}\), Manganese is a transition metal. To correctly name the compound, you must specify Manganese's oxidation state using Roman numerals. Hence, \(\mathrm{MnO_2}\) is named Manganese (IV) Oxide, indicating a +4 charge.
Oxidation States
Oxidation states represent the charge of an atom in a compound. Each atom has a preferred oxidation state, but transition metals can have several. This affects how compounds are named. Understanding oxidation states helps in determining the correct formula and name for a compound.
To find the oxidation state, use the rule that the sum of oxidation states in a neutral compound is zero. For \(\mathrm{MnO_2}\), Oxygen always has an oxidation state of -2. Since there are two Oxygen atoms, they contribute -4 overall. Consequently, the Manganese must have a +4 state to balance this, giving Manganese (IV) Oxide as its name.
To find the oxidation state, use the rule that the sum of oxidation states in a neutral compound is zero. For \(\mathrm{MnO_2}\), Oxygen always has an oxidation state of -2. Since there are two Oxygen atoms, they contribute -4 overall. Consequently, the Manganese must have a +4 state to balance this, giving Manganese (IV) Oxide as its name.
Chemical Formulas
Chemical formulas represent the number and type of atoms in a compound. They are a shorthand way of describing which elements are present and in what ratio. In ionic compounds, the total positive and negative charges in the formula must equal zero, ensuring electrical neutrality.
Writing chemical formulas involves identifying the correct number of each type of atom. For example, \(\mathrm{Li_3N}\) has 3 Lithium atoms each with a +1 charge, balancing the single Nitrogen's -3 charge, resulting in the name Lithium Nitride. This balance ensures compounds are stable and is essential for correctly writing and naming chemical formulas.
Writing chemical formulas involves identifying the correct number of each type of atom. For example, \(\mathrm{Li_3N}\) has 3 Lithium atoms each with a +1 charge, balancing the single Nitrogen's -3 charge, resulting in the name Lithium Nitride. This balance ensures compounds are stable and is essential for correctly writing and naming chemical formulas.
Other exercises in this chapter
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