Question

Obtain the oxidation number for the element noted in each of the following. (a) \(\mathrm{Sn}\) in \(\mathrm{SnO}_{2}\) (b) \(\mathrm{Ta}\) in \(\mathrm{Ta}_{2} \mathrm{O}_{3}\) (c) Br in \(\mathrm{KBrO}_{4}\) (d) Os in \(\mathrm{K}_{2} \mathrm{O}_{\mathrm{s}} \mathrm{O}_{4}\)

Short answer

Sn is +4, Ta is +3, Br is +7, Os is +6.

Step-by-step solution

Oxidation Numbers in SnO2

To find the oxidation number of Sn in SnO2, start with the knowledge that the oxidation number of O is typically -2. Therefore, since there are two O atoms, the total contribution from oxygen is \(-2 \times 2 = -4\). Let \(x\) be the oxidation number of Sn. The compound is neutral, so \(x - 4 = 0\). Solving this gives \(x = +4\). So, the oxidation number of Sn in \(\mathrm{SnO}_{2}\) is +4.

Oxidation Numbers in Ta2O3

In \(\mathrm{Ta}_{2}\mathrm{O}_{3}\), start by recognizing the oxidation number of O is -2. With three O atoms, the oxygen contributes \(-2 \times 3 = -6\). Let \(y\) be the oxidation number of Ta. Since there are two Ta atoms, the total contribution from Ta is \(2y\). The compound is neutral, so \(2y - 6 = 0\). Solving for \(y\), we find \(2y = 6\), so \(y = +3\). The oxidation number of Ta in \(\mathrm{Ta}_{2}\mathrm{O}_{3}\) is +3.

Oxidation Numbers in KBrO4

For \(\mathrm{KBrO}_{4}\), note that K has an oxidation number of +1, and each O is -2. There are four O atoms, making its total contribution \(-2 \times 4 = -8\). Let \(z\) be the oxidation number of Br. The entire compound is neutral, so \(1 + z - 8 = 0\). Solving this gives \(z = +7\). The oxidation number of Br in \(\mathrm{KBrO}_{4}\) is +7.

Oxidation Numbers in K2OsO4

In \(\mathrm{K}_{2} \mathrm{Os} \mathrm{O}_{4}\), K has an oxidation number of +1, and with two K atoms, its contribution is \(+1 \times 2 = +2\). Each O is -2, and there are four O atoms, totaling \(-2 \times 4 = -8\). Let \(w\) be the oxidation number of Os. Thus, the equation is \(2 + w - 8 = 0\). Solving for \(w\) results in \(w = +6\). Hence, the oxidation number of Os in \(\mathrm{K}_{2} \mathrm{Os} \mathrm{O}_{4}\) is +6.

Key concepts

Redox Reactions

Redox reactions are a fascinating type of chemical reaction where two key processes happen simultaneously: oxidation and reduction. In a redox reaction, one substance donates electrons and gets oxidized, while another substance accepts those electrons and gets reduced. These reactions are vital because they are central to many real-life processes, such as burning, respiration, and even corrosion.

The principle of redox reactions revolves around the exchange of electrons. To understand this better, think of electrons as the currency of chemical energy. When one atom loses electrons (oxidized), its oxidation state increases. Conversely, when another atom gains those electrons (reduced), its oxidation state decreases. Hence the term "redox," which combines the concepts of reduction and oxidation.

• Oxidation: Loss of electrons, increase in oxidation state.
• Reduction: Gain of electrons, decrease in oxidation state.

By analyzing changes in the oxidation numbers of various atoms in a reaction, we can identify which atoms are oxidized and which are reduced. This insight helps in understanding the flow of electrons which is crucial for delineating the reaction mechanism.

Chemical Compounds

Chemical compounds are made of molecules containing at least two different elements combined in fixed ratios. Each compound has a unique structure and a distinct set of chemical properties, making them the basic building blocks of chemistry. The chemical formula often provides insights into the compound's makeup and its possible chemical behaviors.

Taking \[\mathrm{SnO}_{2}, \, \mathrm{Ta}_{2} \mathrm{O}_{3}, \, \mathrm{KBrO}_{4}, \, \text{and} \, \mathrm{K}_{2} \mathrm{OsO}_{4}\]as examples, these formulas disclose the number and types of atoms in each compound. Here's a breakdown of these compounds:

• \(\mathrm{SnO}_{2}\): Composed of tin (Sn) and oxygen (O). Tin oxide, in this case, is used in various industrial applications.
• \(\mathrm{Ta}_{2} \mathrm{O}_{3}\): Made of tantalum (Ta) and oxygen, known as tantalum oxide, primarily used in electronics.
• \(\mathrm{KBrO}_{4}\): Contains potassium (K), bromine (Br), and oxygen, an example of a perbromate used in chemical synthesis.
• \(\mathrm{K}_{2} \mathrm{OsO}_{4}\): Composed of potassium (K), osmium (Os), and oxygen, and is an osmium compound often used in various catalytic processes.

These formulas help chemists understand not just what elements are present, but also guide predictions about each compound's reactivity and role in redox reactions.

Oxidation States

Oxidation states, also called oxidation numbers, are a way to track how electrons are distributed in a molecule. They are assigned to atoms within molecules to indicate the degree of oxidation for an atom in a chemical compound. These numbers assist in understanding electron transfer processes by showing the loss or gain of electrons by atoms, which is crucial in redox reactions.

Here's a simplistic explanation of how they are assigned: Generally, certain rules are followed, such as oxygen usually having an oxidation state of -2 and hydrogen typically being +1. Based on these rules, each element in a compound is assigned an oxidation state to confirm that the compound is electrically neutral overall.

• For example, in \(\mathrm{SnO}_{2}\), oxygen is \(-2\), so tin \(\mathrm{Sn}\) must be +4 to balance and make the compound neutral.
• In \(\mathrm{Ta}_{2} \mathrm{O}_{3}\), each oxygen is \(-2\), leading to each tantalum \(\mathrm{Ta}\) being +3.
• In \(\mathrm{KBrO}_{4}\), with potassium \(+1\) and oxygen \(-2\), bromine \(\mathrm{Br}\) ends up with an oxidation state of +7.
• Lastly, for \(\mathrm{K}_{2} \mathrm{OsO}_{4}\), each potassium is \(+1\) and each oxygen \(-2\), resulting in osmium \(\mathrm{Os}\) having an oxidation state of +6.

This assignment of oxidation numbers helps chemists predict how compounds will react, especially in redox processes where changes in oxidation states are paramount.