Question

Oxidation number of iodine in \(\mathrm{IO}_{3}^{-}, 1 \mathrm{O}_{4}^{-}, \mathrm{Kl}\) and \(\mathrm{I}_{2}\) respectively is (a) \(-2,-5,-1,0\) (b) \(+5,+7,-1,0\) (c) \(+2,+5,+1,0\) (d) \(-1,+1,0,+1\)

Short answer

+5 for \text{IO}_3^{-}, +7 for \text{IO}_4^{-}, -1 for \text{KI}, and 0 for \text{I}_2. Hence, the correct answer is (b) +5, +7, -1, 0.

Step-by-step solution

Determining oxidation number of iodine in \text{IO}_3^{-}

Oxygen typically has an oxidation number of -2. Since there are 3 oxygens, this contributes -6 to the oxidation state. The molecule has a -1 charge. To find the oxidation number of iodine, set up the equation: oxidation number of iodine (I) + 3(-2) = -1. Solve for I: I - 6 = -1, so I = +5.

Determining oxidation number of iodine in \text{IO}_4^{-}

Following the same logic, with 4 oxygens, this contributes -8 to the oxidation state, and the overall charge is -1. The equation is: I + 4(-2) = -1. Solving for I gives: I - 8 = -1, so I = +7.

Determining oxidation number of iodine in \text{KI}

In ionic compounds like KI (potassium iodide), potassium (K) has an oxidation number of +1. Because the compound is neutral, the iodine must balance it with an oxidation number of -1.

Determining oxidation number of iodine in \text{I}_2

In the diatomic molecule I2, the iodine atoms are in their elemental form, and the oxidation number for any element in its standard state is 0. So, the oxidation number of iodine in I2 is 0.

Key concepts

Oxidation State

Understanding the concept of oxidation state is crucial for analyzing how elements interact in chemical reactions. It is a theoretical charge that an atom would have if the compound was composed of ions. This concept helps us keep track of electron movement during chemical reactions, especially redox reactions. For example, determining the oxidation number of iodine in various compounds requires us to consider its bonding and the oxidation states of other atoms, like oxygen, which typically has an oxidation state of -2. When iodine pairs with three oxygens in \(\mathrm{IO}_{3}^{-}\), its oxidation state is calculated by adding up the oxidation states of all atoms and equating them to the overall charge of the molecule.

In our exercise, we set up an equation where the oxidation number of iodine plus the total oxidation number of oxygens equals the charge of the ion. Solving this, we uncovered that iodine’s oxidation state in \(\mathrm{IO}_{3}^{-}\) is +5. A similar approach is used for \(\mathrm{IO}_{4}^{-}\), where iodine has an oxidation state of +7. These exercises improve a student’s ability to apply the rules for calculating oxidation numbers, an essential skill in understanding redox reactions.

Redox Reactions

Redox reactions are the bread and butter of understanding how oxidation states function within chemical reactions. These reactions involve the transfer of electrons between two species, leading to a change in oxidation numbers. In these processes, one species will lose electrons and become oxidized, and another will gain electrons and become reduced.

Considering our problem with iodine compounds, knowing the oxidation state of iodine can help us predict its behavior in redox reactions. For instance, iodine in \(\mathrm{KI}\) has an oxidation state of -1, but when it becomes \(\mathrm{I}_{2}\), its oxidation state is 0. This change implies that a redox process might have occurred. Students can trace the flow of electrons and identify the oxidizing and reducing agents by understanding these concepts.

Chemical Bonding

Chemical bonding is the foundation for the formation of compounds and dictates the arrangements of atoms within them. The oxidation numbers are closely tied to the type of chemical bonds formed, whether they are ionic or covalent. In ionic bonds, electrons are transferred, resulting in full charges on the ions, as seen in potassium iodide \(\mathrm{KI}\), where potassium gives up an electron to iodine, forming ions with oxidation numbers of +1 and -1, respectively.

Hence, when discussing iodine's oxidation numbers in various states, we have to take into account the nature of its bonding with other elements. In the provided exercises, we analyzed compounds where iodine forms different chemical bonds with oxygen and potassium, showcasing the versatility and significance of chemical bonding in determining oxidation states.

Periodicity

Periodicity refers to the recurring trends that are observed in the properties of elements, including their oxidation states, as we move across the periodic table. These trends are the result of variations in the atomic structure, particularly the valence electrons. The periodic table is organized in such a way that elements with similar properties fall under the same group, and therefore tend to exhibit similar oxidation numbers in compounds.

For example, understanding that halogens like iodine often have negative oxidation states when paired with metals can be explained by periodic trends. In our examples with iodine compounds, iodine has various oxidation states, showcasing how its position on the periodic table allows it to form multiple stable oxidation states, reflecting the versatility of its chemical reactivity. This periodicity concept aids students in predicting the type of bonding and possible reactions of elements by looking at their position in the periodic table.