Question

Chrome Chromium, a transition metal used in chrome plating, forms both the \(\mathrm{Cr}^{2+}\) and \(\mathrm{Cr}^{3+}\) ions. Write the formulas for the ionic compounds formed when each of these ions react with fluorine and oxygen ions.

Short answer

When chromium ions react with fluorine and oxygen ions, the formulas for the ionic compounds formed are: - \(\mathrm{Cr}^{2+}\) and \(\mathrm{F}^{-}\): \(\mathrm{CrF_2}\) - \(\mathrm{Cr}^{2+}\) and \(\mathrm{O}^{2-}\): \(\mathrm{CrO}\) - \(\mathrm{Cr}^{3+}\) and \(\mathrm{F}^{-}\): \(\mathrm{CrF_3}\) - \(\mathrm{Cr}^{3+}\) and \(\mathrm{O}^{2-}\): \(\mathrm{Cr_2O_3}\)

Step-by-step solution

Determine the charges on the fluorine and oxygen ions.

Fluorine is a halogen, belonging to Group 17 in the periodic table. Halogens form anions with a charge of -1 by gaining one electron, so the fluorine ion has a charge of \(\mathrm{F}^{-}\). Oxygen is a chalcogen, belonging to Group 16 in the periodic table. Chalcogens form anions with a charge of -2 by gaining two electrons, so the oxygen ion has a charge of \(\mathrm{O}^{2-}\).

Combine ions to form compounds so that charges are neutralized.

We will now combine the ions in such a way that the charges are neutralized: 1. For \(\mathrm{Cr}^{2+}\) ion: a. With \(\mathrm{F}^{-}\) ion: Since the charges are +2 and -1, we will need 2 fluorine ions for each chromium(II) ion to neutralize the charges. Thus, the formula for this compound is \(\mathrm{CrF_2}\). b. With \(\mathrm{O}^{2-}\) ion: As the charges on both ions are +2 and -2, we only need one oxygen ion for each chromium(II) ion to neutralize the charges. Therefore, the formula for this compound is \(\mathrm{CrO}\). 2. For \(\mathrm{Cr}^{3+}\) ion: a. With \(\mathrm{F}^{-}\) ion: For the +3 and -1 charges to be neutralized, we need 3 fluorine ions for each chromium(III) ion. Hence, the formula for this compound is \(\mathrm{CrF_3}\). b. With \(\mathrm{O}^{2-}\) ion: In order to neutralize the +3 and -2 charges, we'll need two chromium(III) ions and three oxygen ions. The formula for this compound is \(\mathrm{Cr_2O_3}\). In conclusion, the formulas for the ionic compounds formed when chromium ions react with fluorine and oxygen ions are as follows: - \(\mathrm{Cr}^{2+}\) and \(\mathrm{F}^{-}: \;\mathrm{CrF_2}\) - \(\mathrm{Cr}^{2+}\) and \(\mathrm{O}^{2-}: \;\mathrm{CrO}\) - \(\mathrm{Cr}^{3+}\) and \(\mathrm{F}^{-}: \;\mathrm{CrF_3}\) - \(\mathrm{Cr}^{3+}\) and \(\mathrm{O}^{2-}: \;\mathrm{Cr_2O_3}\)

Key concepts

Transition Metals

Transition metals are elements found in the center of the periodic table, specifically in groups 3 to 12. They are characterized by their ability to form compounds in multiple oxidation states, which is a result of the filling of their d-orbital electrons. For instance, chromium (Cr), a widely known transition metal, can form a variety of ions, such as \(\mathrm{Cr}^{2+}\) and \(\mathrm{Cr}^{3+}\), each with a different charge. This variety underpins the diverse chemical behaviors and applications of transition metals, such as chromium being used for chrome plating due to its corrosion resistance properties.

Learning how transition metals combine with other elements to form ionic compounds is fundamental for understanding many aspects of chemistry, from industrial processes to biological systems. As seen in the example with chromium, determining the correct ionic compound formulas requires balancing the charges between the metal ions and the non-metal ions they bond with.

Chemical Nomenclature

Chemical nomenclature is a system used for naming chemical compounds in a standardized manner. This allows scientists and students alike to communicate clearly and efficiently about chemical substances. In the context of ionic compounds, the metal part of the compound (usually a cation) is named first, followed by the non-metal part (usually an anion). For transition metals that can have more than one possible charge, the oxidation state of the metal ion is indicated using Roman numerals in parentheses after the metal name.

For example, the compound \(\mathrm{CrF}_2\) is named chromium(II) fluoride, indicating that chromium has an oxidation state of +2 in this compound. Proper use of chemical nomenclature is crucial in chemistry to avoid misunderstandings and ensure precise communication of compound structures and compositions.

Oxidation States

Oxidation states, often referred to as oxidation numbers, are a way of indicating the hypothetical charge that an atom would have if all the compound’s bonds were completely ionic. This concept is particularly important in ionic compounds involving transition metals because these metals can have multiple oxidation states. For example, chromium exhibits +2 and +3 oxidation states as shown by the ions \(\mathrm{Cr}^{2+}\) and \(\mathrm{Cr}^{3+}\).

Understanding oxidation states helps predict the formulas of compounds formed when ions combine. The goal is to achieve a neutral compound, so the sum of the oxidation states of all atoms in the compound must be zero. This principle is essential when writing formulas, especially for compounds with polyatomic ions or with metals having multiple oxidation states, like the compounds \(\mathrm{CrO}\) and \(\mathrm{Cr}_2\mathrm{O}_3\), where chromium has different oxidation states.

Polyatomic Ions

Polyatomic ions are charged species composed of two or more atoms covalently bonded, and they behave as a single ion. These ions are prevalent in various types of chemical compounds. A clear understanding of polyatomic ions is vital, particularly when they interact with transition metals to form ionic compounds. For example, while chromium can bond with single-element anions like \(\mathrm{F}^{-}\) and \(\mathrm{O}^{2-}\), it can also combine with polyatomic ions to create compounds such as chromates or dichromates.

Remembering common polyatomic ions and their charges, like sulfate \(\mathrm{SO}_4^{2-}\) or nitrate \(\mathrm{NO}_3^{-}\), can greatly ease the process of writing chemical formulas. It's crucial to consider the overall charge balance within the ionic compound, which involves a comprehension of how the charges of polyatomic ions contribute to the neutrality of the compound.