Question

Determine the oxidation number for the indicated element in each of the following compounds: (a) Ti in \(\mathrm{TiO}_{2}\) (b) Sn in \(\mathrm{SnCl}_{3}^{-}\), (c) \(\mathrm{C}\) in \(\mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-}\), (d) \(\mathrm{N}\) in \(\mathrm{N}_{2} \mathrm{H}_{4}\), (e) \(\mathrm{N}\) in \(\mathrm{HNO}_{2}\), (f) \(\mathrm{Cr}\) in \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\)

Short answer

The oxidation numbers for each element in the given compounds are as follows: (a) Ti in \(\mathrm{TiO}_{2}\) is +4, (b) Sn in \(\mathrm{SnCl}_{3}^{-}\) is +2, (c) C in \(\mathrm{C}_{2}\mathrm{O}_{4}^{2-}\) is +3, (d) N in \(\mathrm{N}_{2}\mathrm{H}_{4}\) is -2, (e) N in \(\mathrm{HNO}_{2}\) is +3, and (f) Cr in \(\mathrm{Cr}_{2}\mathrm{O}_{7}^{2-}\) is +6.

Step-by-step solution

Rule reminder

Remember that the oxidation number of an element represents its charge when all its bonds are treated as ionic bonds. Here are some general rules for oxidation numbers: 1. The oxidation number of an element in its elemental form is zero. 2. The oxidation number of a monatomic ion is equal to its charge. 3. The sum of the oxidation numbers of all atoms in a neutral compound is zero; for a polyatomic ion, it equals the charge of the ion. 4. The oxidation number of oxygen is typically -2, except in peroxides, where it is -1, or in compounds with fluorine, where it can be positive. 5. The oxidation number of hydrogen is +1 when bonded to a non-metal and -1 when bonded to a metal.

(a) Ti in \(\mathrm{TiO}_{2}\)

In \(\mathrm{TiO}_{2}\), oxygen has an oxidation number of -2. Since there are two oxygen atoms in the compound, their total charge is -4. As the compound is neutral, the oxidation number of Ti must be +4 to balance the overall charge. Therefore, the oxidation number for Ti in \(\mathrm{TiO}_{2}\) is +4.

(b) Sn in \(\mathrm{SnCl}_{3}^{-}\)

In \(\mathrm{SnCl}_{3}^{-}\), chlorine has an oxidation number of -1. Since there are three chlorine atoms in the ion, their total charge is -3. To balance the -1 charge of the polyatomic ion, the oxidation number of Sn must be +2. Therefore, the oxidation number for Sn in \(\mathrm{SnCl}_{3}^{-}\) is +2.

(c) C in \(\mathrm{C}_{2}\mathrm{O}_{4}^{2-}\)

In \(\mathrm{C}_{2}\mathrm{O}_{4}^{2-}\) (oxalate ion), oxygen has an oxidation number of -2. As there are four oxygen atoms, their total charge is -8. To balance the -2 charge of the polyatomic ion, the total charge of the two carbon atoms must be +6. Since these two carbon atoms are identical, each carbon atom has an oxidation number of +3. Therefore, the oxidation number for C in \(\mathrm{C}_{2}\mathrm{O}_{4}^{2-}\) is +3.

(d) N in \(\mathrm{N}_{2}\mathrm{H}_{4}\)

In \(\mathrm{N}_{2}\mathrm{H}_{4}\) (hydrazine), hydrogen has an oxidation number of +1. Since there are four hydrogen atoms in the compound, their total charge is +4. As the compound is neutral, the total charge of the two nitrogen atoms must be -4. Since these two nitrogen atoms are identical, each nitrogen atom has an oxidation number of -2. Therefore, the oxidation number for N in \(\mathrm{N}_{2}\mathrm{H}_{4}\) is -2.

(e) N in \(\mathrm{HNO}_{2}\)

In \(\mathrm{HNO}_{2}\) (nitrous acid), hydrogen has an oxidation number of +1 and oxygen has an oxidation number of -2. Since there are two oxygen atoms in the compound, their total charge is -4. To balance the overall charge, the oxidation number of nitrogen must be +3. Therefore, the oxidation number for N in \(\mathrm{HNO}_{2}\) is +3.

(f) Cr in \(\mathrm{Cr}_{2}\mathrm{O}_{7}^{2-}\)

In \(\mathrm{Cr}_{2}\mathrm{O}_{7}^{2-}\) (dichromate ion), oxygen has an oxidation number of -2. As there are seven oxygen atoms, their total charge is -14. To balance the -2 charge of the polyatomic ion, the total charge of the two chromium atoms must be +12. Since these two chromium atoms are identical, each chromium atom has an oxidation number of +6. Therefore, the oxidation number for Cr in \(\mathrm{Cr}_{2}\mathrm{O}_{7}^{2-}\) is +6.

Key concepts

Understanding Redox Chemistry

Redox chemistry, short for reduction-oxidation chemistry, is a fundamental concept in chemistry that deals with the transfer of electrons between species in a chemical reaction. This exchange of electrons leads to changes in the oxidation states of the atoms involved, with one species being oxidized (losing electrons) and another being reduced (gaining electrons).

In simple terms, redox reactions are all about the flow of electrons. If you think of electrons as currency, redox chemistry can be understood as a transaction where one party (the reducing agent) gives up electrons, and another party (the oxidizing agent) receives them. This electron 'currency' allows atoms and molecules to engage in chemical reactions and form new substances with different properties.

To make redox chemistry easier to understand, scientists use oxidation numbers—a helpful bookkeeping tool—to keep track of electron flow. When solving problems, like determining the oxidation number for elements in chemical compounds, we apply a set of rules that serve as a guide to predict the outcome of redox reactions.

Chemical Compounds and Their Significance

Chemical compounds are substances made up of two or more elements that are chemically bonded together. These bonds can be ionic (where electrons are transferred from one atom to another) or covalent (where electrons are shared between atoms). Each compound has a unique composition, represented by a chemical formula, and possesses specific physical and chemical properties.

The nature of the bonds within a compound directly influences the oxidation states of the constituent elements. Understanding the structure of chemical compounds, including the number of atoms of each element and their arrangement, is crucial when determining oxidation numbers. For example, knowing that there are two oxygen atoms in \( \mathrm{TiO}_{2} \) or seven in \( \mathrm{Cr}_{2}\mathrm{O}_{7}^{2-} \) helps us calculate the oxidation numbers for titanium and chromium, respectively.

Grasping the concept of chemical compounds and their structures is essential not just for solving textbook problems, but also for predicting the behavior of substances in various chemical reactions, including redox processes.

Deciphering Oxidation States

Oxidation states, also known as oxidation numbers, play a crucial role in the study of redox chemistry. They are assigned to atoms to describe the degree of oxidation or reduction relative to the elemental state of the atom. While the term 'state' might seem abstract, think of oxidation states as a way to keep score of the electrons during a game of redox reactions.

The rules for determining oxidation states are straightforward, yet require careful application. For example, oxygen typically has an oxidation number of -2, except in peroxides or when it's bonded to fluorine. Knowing this, we can tackle problems like finding the oxidation number of different elements within a variety of compounds. Let's consider \( \mathrm{SnCl}_{3}^{-} \)—knowing that chlorine usually has an oxidation state of -1 allows us to figure out that the oxidation number for tin (Sn) must be +2 to balance the charge of the polyatomic ion.

Understanding and applying oxidation states in exercises helps us not only to balance redox reactions but also to comprehend the nuances of electron transfer, giving us deeper insights into the mechanisms behind the chemical changes we observe.