91Ó°ÊÓ

Understanding oxidation states helps us identify how electrons are transferred in chemical reactions. There are several general rules to keep in mind when determining oxidation numbers:

• The oxidation number of any elemental form is always zero. For example, in \( ext{H}_2\), the oxidation number of hydrogen is zero.
• The oxidation number of a monoatomic ion is the same as its charge. For example, \( ext{Na} ^{+} \) has an oxidation number of +1.
• Oxygen has an oxidation number of -2 in most compounds, except in peroxides where it's -1.
• Hydrogen’s oxidation number is +1 when bound to nonmetals, such as in water (\( ext{H}_2 ext{O} \)).
• The sum of oxidation numbers in a neutral compound is zero. In a polyatomic ion, it is equal to the charge of the ion.

These guidelines are foundational tools for chemists to understand the electron dynamics in compounds.

Chemical compounds are substances formed from two or more elements. Understanding their composition is vital to determining oxidation numbers.
A compound’s nature—ionic or covalent—affects how you identify oxidation states. In ionic compounds, metal ions always maintain their positive charges while their nonmetal counterparts pick up negative charges. For example, in \( ext{NaCl} \), sodium has a +1 oxidation state while chlorine has -1.
Covalent compounds, with shared electrons between atoms, can exhibit variable oxidation states depending on the molecule's structure. Take \( ext{H}_2 ext{O} \); here, oxygen carries most of the electrons and shows -2 oxidation, whereas hydrogen's shared electrons indicate a +1 oxidation state.
Knowing whether a compound is ionic or covalent gets you a step closer to understanding the compound's properties and interactions.

Ions are atoms or groups of atoms that carry an electric charge due to the loss or gain of electrons. Identifying ions is a crucial step in calculating oxidation numbers in compounds. For example, consider \( ext{Hg}_{2}( ext{BrO}_{3})_{2} \), which has \( ext{Hg}_2^{2+} \) and \( ext{BrO}_3^- \).
Recognizing ions involves recognizing the common charges and chemistries of elements:

• Cations usually arise from metals; for instance, \( ext{Cr}^{3+} \) in \( ext{Cr}_{2}( ext{SO}_{4})_{3} \). Here, chromium appears as a cation with a charge of +3.
• Anions often involve nonmetals; for example, \( ext{OH}^- \) in \( ext{Pb}( ext{OH})_{2} \), where hydroxide is a familiar anion with a -1 charge.

Accurately identifying ions simplifies assigning oxidation states to different atoms in compounds.

To determine oxidation numbers in the exercise, follow a step-by-step approach. We will break down the process using practical examples from specified compounds.
Take compound 'a': \( ext{Hg}_{2}( ext{BrO}_{3})_{2} \). This contains mercury ions \( ext{Hg}_2^{2+} \) and bromate ions \( ext{BrO}_3^- \). First, assign +1 to each mercury because their sum should equal the ion's +2 overall charge.
Next, look at the bromate ion. Oxygen is usually -2, so solving \( 3 imes (-2) + ext{x} = -1 \) gives bromine an oxidation number of +5.
Continue similarly for compounds 'b', 'c', and 'd', always respecting the oxidation state rules and ion charges identified. By systematically addressing each component, you ensure accuracy in resolving each compound’s individual oxidation numbers.