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 \(\mathrm{COCl}_{2},(\mathrm{c}) \mathrm{Mn}\) in \(\mathrm{MnO}_{4}^{-}\), (d) \(\mathrm{Br}\) in \(\mathrm{HBrO}\), (e) \(\mathrm{As}\) in \(\mathrm{As}_{4}\), (f) \(\mathrm{O}\) in \(\mathrm{K}_{2} \mathrm{O}_{2}\).

Short answer

The oxidation numbers of the indicated elements in each compound are: (a) S in \(\mathrm{SO}_{2}\): +4 (b) C in \(\mathrm{COCl}_{2}\): +4 (c) Mn in \(\mathrm{MnO}_{4}^{-}\): +7 (d) Br in \(\mathrm{HBrO}\): +1 (e) As in \(\mathrm{As}_{4}\): 0 (f) O in \(\mathrm{K}_{2}\mathrm{O}_{2}\): -1

Step-by-step solution

Determine known oxidation numbers and unknown oxidation numbers

For each given compound, determine known oxidation numbers of some elements and identify the element whose oxidation number needs to be found. We can use general rules such as: - Oxidation number of hydrogen is usually +1 - Oxidation number of oxygen is usually -2 - Elements in their elemental state have an oxidation number of zero - Alkali metals (group 1) have an oxidation number of +1 - Alkaline earth metals (group 2) have an oxidation number of +2

Write equations to find the unknown oxidation number

Use the sum of the oxidation numbers in the compound being equal to its overall charge to write the equation to find the unknown oxidation number. (a) In \(\mathrm{SO}_{2}\): Oxidation number of \(\mathrm{S}\) + 2(Oxidation number of \(\mathrm{O}\)) = 0 (b) In \(\mathrm{COCl}_{2}\): Oxidation number of \(\mathrm{C}\) + (Oxidation number of \(\mathrm{O}\)) + 2(Oxidation number of \(\mathrm{Cl}\)) = 0 (c) In \(\mathrm{MnO}_{4}^{-}\): Oxidation number of \(\mathrm{Mn}\) + 4(Oxidation number of \(\mathrm{O}\)) = -1 (d) In \(\mathrm{HBrO}\): Oxidation number of \(\mathrm{H}\) + Oxidation number of \(\mathrm{Br}\) + Oxidation number of \(\mathrm{O}\) = 0 (e) In \(\mathrm{As}_{4}\): 4(Oxidation number of \(\mathrm{As}\)) = 0 (f) In \(\mathrm{K}_{2}\mathrm{O}_{2}\): 2(Oxidation number of \(\mathrm{K}\)) + 2(Oxidation number of \(\mathrm{O}\)) = 0

Solve equations to find unknown oxidation numbers

Use previously known oxidation numbers of other elements and solve the equations to find the unknown oxidation numbers. (a) Oxidation number of \(\mathrm{S}\) + 2(-2) = 0 Oxidation number of \(\mathrm{S}\) = +4 (b) Oxidation number of \(\mathrm{C}\) + (-2) + 2(-1) = 0 Oxidation number of \(\mathrm{C}\) = +4 (c) Oxidation number of \(\mathrm{Mn}\) + 4(-2) = -1 Oxidation number of \(\mathrm{Mn}\) = +7 (d) +1 + Oxidation number of \(\mathrm{Br}\) - 2 = 0 Oxidation number of \(\mathrm{Br}\) = +1 (e) 4(Oxidation number of \(\mathrm{As}\)) = 0 Oxidation number of \(\mathrm{As}\) = 0 (f) 2(+1) + 2(Oxidation number of \(\mathrm{O}\)) = 0 Oxidation number of \(\mathrm{O}\) = -1 The oxidation numbers of the indicated elements in each compound are: (a) S in \(\mathrm{SO}_{2}\): +4 (b) C in \(\mathrm{COCl}_{2}\): +4 (c) Mn in \(\mathrm{MnO}_{4}^{-}\): +7 (d) Br in \(\mathrm{HBrO}\): +1 (e) As in \(\mathrm{As}_{4}\): 0 (f) O in \(\mathrm{K}_{2}\mathrm{O}_{2}\): -1

Key concepts

Redox Reactions

Understanding redox reactions is crucial for comprehending various aspects of chemistry, including how substances interact and change.

A redox reaction, short for reduction-oxidation reaction, involves the transfer of electrons between two species. It’s a simultaneous process where one substance gets oxidized by losing electrons, and another gets reduced by gaining electrons. This fundamental concept explains why certain reactions produce energy, such as those in batteries, or why metals corrode over time.

Examining Electron Transfer

In the case of our exercise example with \(\mathrm{MnO}_{4}^{-}\), manganese (\(\mathrm{Mn}\)) is being oxidized as its oxidation state increases from a lower number to a higher one when forming the compound. On the other side of that reaction, somewhere in the chemical reaction mechanism, reduction is taking place as another species is gaining those electrons lost by manganese.

It’s essential to identify the changes in oxidation states to understand the role each element plays in a redox reaction. This ability can help predict the outcomes of reactions and is vital both in industrial applications and in understanding biological processes like respiration.

Chemical Compounds

Chemical compounds are substances formed when two or more elements combine in fixed ratios, resulting in a product with different properties than the constituent elements.

For example, when we look at \(\mathrm{SO}_{2}\), sulfur dioxide, it is a compound of sulfur (\(\mathrm{S}\)) and oxygen (\(\mathrm{O}\)), with distinct properties from either sulfur or oxygen alone. Each element within a compound can exhibit different oxidation states, which are essential for dictating the nature of chemical bonds that hold the compound together.

Understanding Compound Formation

In forming chemical compounds, elements follow certain rules, such as achieving a stable electron configuration. This often leads to the transfer or sharing of electrons, creating ionic or covalent bonds, respectively. In our examples where we calculated the oxidation numbers, these numbers also provide insight into the nature of the bonding and the compound's overall structure. Through practicing with exercises such as determining oxidation numbers, students gain a better understanding of the principles underlying compound formation and stability.

Oxidation States

Determining oxidation states of elements in compounds is a fundamental skill in chemistry, as it gives insights into the electronic structure and possible reactions a compound might undergo.

Oxidation states, often called oxidation numbers, are the hypothetical charges that an atom would have if all bonds to atoms of different elements were completely ionic. Although few bonds are purely ionic, this concept is a useful tool for understanding redox reactions.

Applying Oxidation State Rules

When solving for oxidation states, we apply a set of rules, like the ones mentioned in our exercise's solution. These include the fact that the oxidation state of oxygen is typically -2 and hydrogen is generally +1, with variations occurring in certain situations. For instance, in \(\mathrm{K}_{2} \mathrm{O}_{2}\), peroxide, oxygen has an unusual oxidation state of -1.

It is crucial to note that oxidation states do not necessarily represent real charges on atoms in molecules or complexes but are rather a bookkeeping device to help chemists keep track of electron movement in chemical reactions, particularly redox reactions.