Question

Mssign oxidation numbers to each element in (a) \(\mathrm{HIO}_{3}\) (b) \(\mathrm{NaMnO}_{4}\) (c) \(\mathrm{SnO}_{2}\) (d) NOF (e) \(\mathrm{NaO}_{2}\)

Short answer

Answer: The oxidation numbers of the elements in the given compounds are as follows: - HIO3: H (+1), I (+5), and O (-2) - NaMnO4: Na (+1), Mn (+7), and O (-2) - SnO2: Sn (+4) and O (-2) - NOF: N (+3), O (-2), and F (-1) - NaO2: Na (+1) and O (-1/2)

Step-by-step solution

HIO3: Assigning Oxidation Numbers for Hydrogen, Iodine, and Oxygen

1. Hydrogen generally has an oxidation number of +1. 2. Oxygen generally has an oxidation number of -2. 3. Knowing that the overall charge of the compound is 0, we will determine the oxidation number of Iodine. \(+1*x -2*3+y=0\), x being the amount of H-atoms and y being the amount of I-atoms. In this compound: x=1 and y=1 Hence, the oxidation number of Iodine is +5. So, the oxidation numbers of the elements are: H (+1), I (+5), and O (-2).

NaMnO4: Assigning Oxidation Numbers for Sodium, Manganese, and Oxygen

1. Sodium generally has an oxidation number of +1. 2. Oxygen generally has an oxidation number of -2. 3. Knowing that the overall charge of the compound is 0, we will determine the oxidation number of Manganese. \(+1-2*4+y=0\), y being the oxidation number for Mn. Hence, the oxidation number of Manganese is +7. So, the oxidation numbers of the elements are: Na (+1), Mn (+7), and O (-2).

SnO2: Assigning Oxidation Numbers for Tin and Oxygen

1. Oxygen generally has an oxidation number of -2. 2. Knowing that the overall charge of the compound is 0, we will determine the oxidation number of Tin. \(y-2*2=0\), y being the oxidation number for Sn. Hence, the oxidation number of Tin is +4. So, the oxidation numbers of the elements are: Sn (+4) and O (-2).

NOF: Assigning Oxidation Numbers for Nitrogen, Oxygen, and Fluorine

1. Oxygen generally has an oxidation number of -2. 2. Fluorine generally has an oxidation number of -1. 3. Knowing that the overall charge of the compound is 0, we will determine the oxidation number of Nitrogen. \(y-1-2=0\), y being the oxidation number for N. Hence, the oxidation number of Nitrogen is +3. So, the oxidation numbers of the elements are: N (+3), O (-2), and F (-1).

NaO2: Assigning Oxidation Numbers for Sodium and Oxygen

1. Sodium generally has an oxidation number of +1. 2. Knowing that the overall charge of the compound is 0, we will determine the oxidation number of Oxygen. \(+1+y*2=0\), y being oxidation number for O. Hence, the oxidation number of Oxygen is -1/2 (in this case, oxygen is in a peroxide form). So, the oxidation numbers of the elements are: Na (+1) and O (-1/2).

Key concepts

Understanding Chemical Compounds

Chemical compounds are substances made from two or more different elements that are chemically bonded together. Each compound has unique properties that differ from its constituent elements. Understanding how these compounds are formed and summarized is crucial in chemistry. By looking at the chemical formula, you can determine:

• The types of atoms involved
• The number of each atom
• The arrangement of these atoms in the compound

The chemical formula tells us a great deal about the compound's makeup. For instance, in \(\mathrm{HIO}_3\), we see one hydrogen, one iodine, and three oxygen atoms. This specific arrangement dictates the compound's characteristics.
Always recognize that a compound’s name and formula offer insights into the chemical nature of the substance, including potential reactivity and properties. Understanding oxidation numbers helps in figuring out how electrons are distributed among atoms in a compound.

Explaining Ionic Charges

Ionic charges are crucial for understanding the behavior and reactivity of chemical compounds. When atoms bond to form compounds, electrons are transferred or shared to attain stability, resulting in ions if the electrons are transferred.

• Positive ions (cations) are formed when electrons are lost.
• Negative ions (anions) are formed when electrons are gained.

Each element has a typical oxidation state that usually corresponds to its ionic charge. This is especially helpful when examining how compounds interact, as shown in sodium (\(\mathrm{Na}\)), which commonly forms a +1 cation.
In neutral compounds, the total charge must balance to zero, which means positive and negative charges from different ions need to offset each other. For example, in \(\mathrm{NaMnO}_4\), sodium (\(+1\)) and oxygen (\(-2\)) balance with manganese to make the compound neutral.

Redox Reactions and Oxidation Numbers

Redox reactions, short for reduction-oxidation reactions, involve the transfer of electrons between substances. These reactions are vital in both biological processes and industrial applications. The core idea is that:

• Oxidation involves losing electrons and increasing an oxidation number.
• Reduction involves gaining electrons and decreasing an oxidation number.

Oxidation numbers help track the transfer of electrons in these reactions. For example, in \(\mathrm{SnO}_2\), the oxidation number of tin (\(+4\)) shows that tin has lost four electrons. This concept helps in balancing chemical equations and understanding the flow of energy in reactions.
Recognizing oxidation and reduction as processes that often occur simultaneously can deepen your understanding of chemistry, enhancing both theoretical knowledge and practical skills in analyzing reactions.