Question

Determine the oxidation number for the indicated element in each of the following substances: (a) \(\mathrm{S}\) in \(\mathrm{SO}_{2},\) (b) \(\mathrm{C}\) in \(\mathrm{COCl}_{2}\), (c) \(\mathrm{Mn}\) in \(\mathrm{KMnO}_{4}\) (d) \(\mathrm{Br}\) in \(\mathrm{HBrO}\) (e) As in As \(_{4}\), (f) \(\mathrm{O}\) in \(\mathrm{K}_{2} \mathrm{O}_{2}\)

Short answer

(a) The oxidation number of $\mathrm{S}$ in $\mathrm{SO}_{2}$ is +4. (b) The oxidation number of $\mathrm{C}$ in $\mathrm{COCl}_{2}$ is +4. (c) The oxidation number of $\mathrm{Mn}$ in $\mathrm{KMnO}_{4}$ is +7. (d) The oxidation number of $\mathrm{Br}$ in $\mathrm{HBrO}$ is +1. (e) The oxidation number of As in As$_{4}$ is 0. (f) The oxidation number of $\mathrm{O}$ in $\mathrm{K}_{2} \mathrm{O}_{2}$ is -1.

Step-by-step solution

(a) Determine the oxidation number of S in SO2

To find the oxidation number of S in SO2, we need to consider the oxidation numbers of the other elements in the compound. Oxygen generally has an oxidation number of -2. Thus, in SO2, there are two oxygen atoms, each with an oxidation number of -2, contributing a total of -4. Therefore, in SO2, the oxidation number of S must be +4 to balance out the contribution from the oxygen atoms.

(b) Determine the oxidation number of C in COCl2

For COCl2, let's first focus on the Cl atoms. Chlorine generally has an oxidation number of -1. Since there are two Cl atoms, they have a combined oxidation number of -2. Oxygen has an oxidation number of -2 as well. So, to balance the oxidation numbers, the C atom must have an oxidation number of +4.

(c) Determine the oxidation number of Mn in KMnO4

In KMnO4, we can see that K is a Group 1 metal, which generally has an oxidation number of +1. Oxygen is -2, and there are four oxygen atoms, with a total oxidation number of -8. To achieve an overall charge of zero, Mn must have an oxidation number of +7.

(d) Determine the oxidation number of Br in HBrO

In HBrO, we already know that the oxidation number of H is +1 and the oxidation number of O is -2. To balance the oxidation state and achieve overall charge of zero, the oxidation number of Br in HBrO must be +1.

(e) Determine the oxidation number of As in As4

In a pure elemental form, such as As in As4, the oxidation number is always zero. Hence, the oxidation number of As in As4 is 0.

(f) Determine the oxidation number of O in K2O2

In K2O2, there are two K atoms, each with an oxidation number of +1, contributing a total oxidation number of +2. Peroxide, O2^2-, generally has an oxidation number of -1 for each oxygen atom, making the total for two O atoms -2. As a result, the oxidation number of O in K2O2 is -1.

Key concepts

Chemical Compounds

Chemical compounds are substances formed when two or more different elements bond together in a fixed ratio. These bonds can be ionic, where electrons are transferred between atoms, or covalent, where electrons are shared. Understanding chemical compounds is essential because their properties and behavior are determined by the type and arrangement of atoms within them. In the context of this exercise, we see compounds like \( \mathrm{SO}_2 \) and \( \mathrm{K}_2 \mathrm{O}_2 \) which involve elements such as sulfur, oxygen, and potassium. Knowing how these elements interact to form different chemical structures helps us determine things like oxidation numbers, which reflect the electron distribution in these compounds.

Oxidation States

Oxidation states (or oxidation numbers) are indicators of the degree of oxidation of an atom in a chemical compound. These numbers are essential for balancing chemical equations and understanding redox (reduction-oxidation) reactions. They often vary based on the element's environment within a compound. For example:

• In \( \mathrm{SO}_2 \), sulfur has an oxidation number of +4.
• Carbon in \( \mathrm{COCl}_2 \) holds an oxidation number of +4.
• Manganese in \( \mathrm{KMnO}_4 \) carries an oxidation number of +7.

Understanding these states involves identifying the typical oxidation numbers of common elements:

• Oxygen usually has an oxidation number of -2, except in peroxides where it is -1.
• Hydrogen is typically +1, unless it's bonded to metals where it can be -1.
• Alkali metals like potassium always have a +1 oxidation state.

This exercise demonstrates how oxidation numbers are used to balance charges within compounds, ensuring electrons are correctly accounted for.

Chemical Reactions

Chemical reactions involve the transformation of chemical substances through the breaking and formation of bonds. During this process, atoms can experience changes in their oxidation states, reflecting the redistribution of electrons.For example, redox reactions are specific types of chemical reactions where the oxidation states of atoms change due to the transfer of electrons. A classic example from the exercise is the manganese in \( \mathrm{KMnO}_4 \) which can undergo redox reactions where its oxidation state plays a critical role.Understanding chemical reactions includes:

• Balancing equations to ensure the same number of each type of atom on both sides.
• Predicting products based on reactants' properties.
• Using oxidation numbers to identify which atoms are oxidized and reduced.

Chemical reactions are fundamental to chemistry because they explain how substances interact, transform, and transfer energy, all of which are vital for scientific fields ranging from biology to material science.