Question

Determine the oxidation number of each element in the following ions or compounds. (a) \(\mathrm{BrO}_{3}^{-}\) (b) \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\) (c) \(\mathrm{F}^{-}\) (d) \(\mathrm{CaH}_{2}\) (e) \(\mathrm{H}_{4} \mathrm{Si} \mathrm{O}_{4}\) (f) \(\mathrm{HSO}_{4}^{-}\)

Short answer

(a) Br: +5, O: -2; (b) C: +3, O: -2; (c) F: -1; (d) Ca: +2, H: -1; (e) H: +1, Si: +4, O: -2; (f) H: +1, S: +6, O: -2.

Step-by-step solution

Determine Oxidation Number of Each Element in \( \mathrm{BrO}_{3}^{-} \)

For \( \mathrm{BrO}_{3}^{-} \), we assign the oxidation number of \( \mathrm{O} \) as \(-2\). Let \( x \) be the oxidation number of \( \mathrm{Br} \). The total charge of the ion is \(-1\), so we write the equation: \( x + 3(-2) = -1 \). Solving for \( x \), we find \( x = +5 \). Thus, \( \mathrm{Br} \) has an oxidation number of \(+5\), and \( \mathrm{O} \) is \(-2\).

Determine Oxidation Number of Each Element in \( \mathrm{C}_{2} \mathrm{O}_{4}^{2-} \)

The oxidation number of \( \mathrm{O} \) is \(-2\). Let \( y \) be the oxidation number of \( \mathrm{C} \). The total charge is \(-2\), so we write: \( 2y + 4(-2) = -2 \). Solving for \( y \), we get \( 2y - 8 = -2 \) or \( 2y = 6 \), leading to \( y = +3 \). Thus, \( \mathrm{C} \) is \(+3\) and \( \mathrm{O} \) is \(-2\).

Determine Oxidation Number of \( \mathrm{F}^{-} \)

Since \( \mathrm{F}^{-} \) is a monatomic ion, its oxidation number is equal to its charge, which is \(-1\).

Determine Oxidation Number of Each Element in \( \mathrm{CaH}_{2} \)

In \( \mathrm{CaH}_{2} \), calcium \( \mathrm{Ca} \) is an element in the second group of the periodic table, commonly having an oxidation number of \(+2\). Hydrogen in metal hydrides such as \( \mathrm{CaH}_{2} \) has an oxidation number of \(-1\).

Determine Oxidation Number of Each Element in \( \mathrm{H}_{4} \mathrm{SiO}_{4} \)

In \( \mathrm{H}_{4} \mathrm{SiO}_{4} \), hydrogen has an oxidation number of \(+1\) and oxygen \(-2\). Let the oxidation number of silicon be \( z \). The compound is neutral, so we write: \( 4(+1) + z + 4(-2) = 0 \), which simplifies to \( 4 + z - 8 = 0 \). Solving gives \( z = +4 \). Therefore, \( \mathrm{H} \) is \(+1\), \( \mathrm{Si} \) is \(+4\), and \( \mathrm{O} \) is \(-2\).

Determine Oxidation Number of Each Element in \( \mathrm{HSO}_{4}^{-} \)

For \( \mathrm{HSO}_{4}^{-} \), the oxidation number of \( \mathrm{H} \) is \(+1\) and \( \mathrm{O} \) is \(-2\). Let \( a \) be the oxidation number of sulfur \( \mathrm{S} \). The total charge is \(-1\), so write: \( +1 + a + 4(-2) = -1 \), which simplifies to \( 1 + a - 8 = -1 \), giving \( a = +6 \). Thus, \( \mathrm{H} \) is \(+1\), \( \mathrm{S} \) is \(+6\), and \( \mathrm{O} \) is \(-2\).

Key concepts

Ionic Compounds

Ionic compounds are chemical compounds made up of ions that are bonded together by ionic bonds. These compounds typically form between metals and non-metals. When elements come together to form ionic compounds, they do so because of the electrostatic attraction between oppositely charged ions. For example:

• Metal atoms, such as sodium ( Na ), tend to lose electrons, forming positively charged ions known as cations.
• Non-metal atoms, like chlorine ( Cl ), typically gain these electrons to become negatively charged ions called anions.

This transfer of electrons allows each atom to achieve a stable electron configuration, often described as having a full outer shell of electrons.
In such compounds, the overall charge is neutral because the positive and negative charges balance each other out. The classic example of an ionic compound is sodium chloride ( NaCl ), better known as table salt.

Oxidation State Determination

The oxidation state, or oxidation number, is a concept that helps chemists keep track of electrons in chemical reactions. Determining the oxidation state of an element in a molecule or ion involves using some basic rules:

• For an atom in its elemental form, the oxidation number is zero. For instance, in O elements such as O ₂ or N ₂, each atom has an oxidation number of zero.
• In a monatomic ion, the oxidation number equals the ion's charge. For example, in Na⁺ , the oxidation number of sodium is +1 .
• The sum of oxidation numbers for all atoms in a neutral compound is zero, and in a polyatomic ion, it equals the ion's charge.

To find the oxidation number of a specific element, we can use these rules to set up and solve equations based on the charges and known oxidation states of other elements in the compound. This systematic approach is very handy in complex molecules and ions, such as the polyatomic ion BrO₃⁻ , where balancing the equation involves calculations like x + 3(-2) = -1 to find the value of x .

Chemical Reactions

Chemical reactions involve the transformation of reactants into products and involve changes in the chemical makeup of the substances involved. These reactions are fundamental to understanding chemistry as they allow us to see how different compounds interact. During a chemical reaction, atoms in the reactants are rearranged to form new substances. Reactions are classified into various types, such as:

• Synthesis reactions, where simpler substances combine to form a complex compound.
• Decomposition reactions, where a compound breaks into simpler elements or compounds.
• Single replacement reactions, where an element in a compound is replaced by another element.
• Double replacement reactions, where two compounds exchange elements to form two new compounds.

For students, learning these classifications helps in predicting the outcomes of reactions and understanding the changes at a molecular level.

Oxidation-Reduction Reactions

Oxidation-reduction reactions, commonly known as redox reactions, are a category of chemical reactions involving the transfer of electrons between two substances. They are vital in energy production, corrosion, and many biological processes. In these reactions:

• Oxidation refers to the loss of electrons, resulting in an increase in oxidation state.
• Reduction means gaining electrons, which decreases the oxidation state.

During a redox reaction, one substance is oxidized while the other is reduced. This transfer of electrons is crucial for processes like combustion, respiration, and even in the workings of batteries.
Identifying redox reactions involves pinpointing the changes in oxidation numbers of the reactants. For example, understanding that when Na metal oxidizes to Na⁺ , it loses an electron, making it a classic oxidation process. Simultaneously, any associated reduction must balance the equation by accepting those electrons. This complementary nature of oxidation and reduction in reactions offers an essential perspective on chemical processes.