Question

Assign oxidation numbers to the atoms in each substance. a) \(\mathrm{P}_{4}\) b) \(\mathrm{SO}_{3}\) c) \(\mathrm{SO}_{3}^{2-}\) d) \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{3}\right) 2\)

Short answer

a) P: 0, b) S: +6, c) S: +4, d) Ca: +2, P: +3, O: -2.

Step-by-step solution

Understanding Elements in Their Elemental Form

For substance (a) \( \mathrm{P}_{4} \): Elements in their elemental form have an oxidation number of 0. Since phosphorus is in a molecular form of \( \mathrm{P}_{4} \), each phosphorus atom has an oxidation number of 0.

Assigning Oxidation Numbers Based on Known Oxidation States

For substance (b) \( \mathrm{SO}_{3} \):1. Oxygen usually has an oxidation number of \(-2\).2. Let the oxidation number of sulfur be \( x \).3. The sum of the oxidation numbers in a neutral compound must be zero:- \( x + 3(-2) = 0 \).- Solve for \( x \): \( x = 6 \).4. The oxidation number of sulfur in \( \mathrm{SO}_{3} \) is \(+6\).

Calculating Oxidation Numbers in a Polyatomic Ion

For substance (c) \( \mathrm{SO}_{3}^{2-} \):1. Oxygen has an oxidation number of \(-2\).2. Let the oxidation number of sulfur be \( y \).3. The sum of the oxidation numbers equals the charge of the ion:- \( y + 3(-2) = -2 \).- Solve for \( y \): \( y = 4 \).4. The oxidation number of sulfur in \( \mathrm{SO}_{3}^{2-} \) is \(+4\).

Assigning Oxidation Numbers in Ionic Compounds

For substance (d) \( \mathrm{Ca}_{3}\left(\mathrm{PO}_{3}\right)_2 \):1. Calcium typically has an oxidation number of \(+2\) as it is an alkaline earth metal.2. For the phosphate ion \( \mathrm{PO}_{3}^{3-} \): - Let the oxidation number of phosphorus be \( z \). - Oxygen is \(-2\), so: - \( z + 3(-2) = -3 \). - Solve for \( z \): \( z = 3 \).3. The oxidation number of phosphorus in \( \mathrm{PO}_{3}^{3-} \) is \(+3\). 4. The compound is neutral, so calculation checks with: - \( 3(+2) + 2(z + 3 \times (-2)) = 0 \).5. This verifies the oxidation numbers: Ca is \(+2\), P is \(+3\), and O is \(-2\).

Key concepts

Chemical Compounds

Chemical compounds are substances formed by the chemical union of two or more elements. These elements combine in fixed ratios and are held together by chemical bonds. Chemical compounds can be broadly classified into two types: ionic compounds and molecular compounds.

• Ionic Compounds: These are formed when metals and non-metals combine and involve a transfer of electrons from one atom to another. A classic example is sodium chloride (NaCl), where sodium transfers an electron to chlorine.
• Molecular Compounds: Also known as covalent compounds, they are typically formed between non-metal elements where electrons are shared. Water (\( ext{H}_2 ext{O}\)) is an example of a molecular compound.

Identifying the nature of a compound—whether ionic or molecular—helps to predict its physical and chemical properties. In general, ionic compounds tend to have high melting points and conduct electricity when molten. Molecular compounds usually have lower melting and boiling points and do not conduct electricity.

Oxidation States

The oxidation state, also known as the oxidation number, reflects the degree of oxidation (loss of electrons) of an atom in a chemical compound. This concept is crucial in understanding redox reactions and balancing chemical equations.
Assigning oxidation numbers involves using a set of rules:

• The oxidation state of an atom in its elemental form is always 0. Examples include \(O_2\), \(N_2\), or \(P_4\).
• In a simple monoatomic ion, the oxidation number equals the charge of the ion. For example, Na⁺ is \(+1\).
• Oxygen is typically assigned an oxidation number of \(-2\) except in peroxides or superoxides.
• Hydrogen has an oxidation number of \(+1\) when bonded with non-metals and \(-1\) when bonded with metals.

Understanding oxidation states is fundamental to predicting the outcome of chemical reactions and determining the nature of the chemical changes taking place.

Ionic and Molecular Compounds

Ionic and molecular compounds differ by the type of bonding and the properties they exhibit. Exploring these differences can clarify why compounds behave the way they do.

• Ionic Compounds: Their structure forms a crystalline lattice, giving these compounds their characteristic hard and brittle nature. Ionic bonds result from the electrostatic attraction between ions of opposite charges.
• Molecular Compounds: These compounds typically exist as discrete molecules. They are often gases or liquids at room temperature, such as \( ext{CO}_2\) or \( ext{H}_2 ext{O}\), due to the weaker intermolecular forces compared to ionic bonds.

These differences in structure and bonding influence not only physical properties but also thermal and electrical conductivity characteristics of substances.

Balancing Redox Reactions

Balancing redox reactions (reduction-oxidation reactions) involves ensuring that the number of electrons lost in oxidation equals the number of electrons gained in reduction. This ensures that charge and mass are conserved in the reactions, aligning with the law of conservation of mass.
To balance redox reactions, follow these steps:

• Identify the Oxidation and Reduction Half-Reactions: Determine which atoms experience a change in oxidation state and separate the reaction into two half-reactions.
• Balance the Atoms and Charges: First balance all the atoms except for hydrogen and oxygen, then balance the charges by adding electrons where necessary.
• Combine the Half-Reactions: Multiply the half-reactions by appropriate coefficients to equalize the number of electrons transferred, then add them together.
• Verify the Balance: Ensure both mass and charge are balanced in the final reaction.

Mastering the technique of balancing redox reactions is essential for understanding the underlying electron transfer processes in chemical reactions.