## Oxidation Numbers: A Comprehensive Guide This guide explains how to determine oxidation numbers (states) of elements in various chemical species. Understanding oxidation numbers is crucial for balancing redox reactions and predicting chemical behavior. ### Rule 1: Pure Elements The oxidation state of any pure element is always zero. * **Example:** The oxidation number of Zn (zinc metal) is 0. This applies to all elements in their elemental form, such as O₂ (oxygen gas), F₂ (fluorine gas), and P₄ (white phosphorus). ### Rule 2: Monatomic Ions For monatomic ions, the oxidation number is equal to the charge of the ion. * **Example:** The oxidation number of Zn²⁺ is +2, and the oxidation number of Fe³⁺ is +3. * **Example (Polyatomic Ion):** Hg₂²⁺ (mercury(I) ion). Each mercury atom has an oxidation state of +1 because 2(+1) = +2. We solve for the individual oxidation state by dividing the total charge by the number of atoms. * **Example (Polyatomic Ion):** O₂²⁻ (peroxide ion). Each oxygen atom has an oxidation state of -1 because 2(-1) = -2. * **Example (Polyatomic Ion):** O₂⁻ (superoxide ion). Each oxygen atom has an oxidation state of -1/2 because 2(-1/2) = -1. ### Rule 3: Common Oxidation States in Compounds These rules are guidelines; exceptions exist, especially with transition metals. * **Fluorine (F):** Always -1 in compounds. It's the most electronegative element. * **Oxygen (O):** Usually -2, except: * In peroxides (e.g., H₂O₂), it's -1. * In superoxides (e.g., KO₂), it's -1/2. * When bonded to fluorine (e.g., OF₂), it's positive. * **Hydrogen (H):** * +1 when bonded to nonmetals. * -1 when bonded to metals (e.g., NaH). Forgetting the exceptions to the rules for oxygen and hydrogen is a common mistake. ### Working with Compounds The sum of oxidation numbers in a neutral compound is always zero. In a polyatomic ion, the sum equals the charge of the ion. Find the oxidation states of Mg and Cl in MgCl₂. * We know Cl is usually -1. * The equation is: Mg + 2Cl = 0 * Mg + 2(-1) = 0 * Mg = +2 Find the oxidation states of Al and F in AlF₃. * F is always -1. * The equation is: Al + 3F = 0 * Al + 3(-1) = 0 * Al = +3 Find the oxidation states of V and O in V₂O₅. * O is usually -2. * The equation is: 2V + 5O = 0 * 2V + 5(-2) = 0 * 2V = +10 * V = +5 ### Polyatomic Ions Find the oxidation state of S in SO₄²⁻ (sulfate). * O is usually -2. * The equation is: S + 4O = -2 * S + 4(-2) = -2 * S = +6 Find the oxidation state of P in PO₄³⁻ (phosphate). * O is usually -2. * The equation is: P + 4O = -3 * P + 4(-2) = -3 * P = +5 Find the oxidation states of N in NO₃⁻ (nitrate) and Cl in ClO₄⁻ (perchlorate). * For NO₃⁻: N + 3(-2) = -1; N = +5 * For ClO₄⁻: Cl + 4(-2) = -1; Cl = +7 ### Electronegativity and Oxidation Numbers Electronegativity increases across a period and up a group on the periodic table. The more electronegative element in a bond tends to have a negative oxidation number. What are the oxidation states of O and F in OF₂? * F is more electronegative than O, so F = -1. * O + 2F = 0 * O + 2(-1) = 0 * O = +2 HCl and NaH * In HCl, Cl (3.0 electronegativity) is more electronegative than H (2.1), so Cl = -1 and H = +1. * In NaH, H (2.1) is more electronegative than Na (approximately 0.9), so H = -1 and Na = +1. BH₃ * H (2.1) is slightly more electronegative than B (2.0), so H = -1 and B = +3. H₂S and SO₂ * In H₂S, S is more electronegative than H, so S = -2 and H = +1. * In SO₂, O is more electronegative than S, so O = -2 and S = +4. NH₃ and NO₂ * In NH₃, N is more electronegative than H, so N = -3 and H = +1. * In NO₂, O is more electronegative than N, so O = -2 and N = +4. CH₄ and CO₂ * In CH₄, C is more electronegative than H, so C = -4 and H = +1. * In CO₂, O is more electronegative than C, so O = -2 and C = +4. ### Average Oxidation States Sometimes, the average oxidation state is not a whole number. This indicates that multiple oxidation states are present. C₃H₈ * 3C + 8H = 0 * 3C + 8(+1) = 0 * C = -8/3 Fe₃O₄ * 3Fe + 4O = 0 * 3Fe + 4(-2) = 0 * Fe = +8/3 (This means some Fe atoms are +2 and others are +3). HSO₃⁻ * H + S + 3O = -1 * (+1) + S + 3(-2) = -1 * S = +4 K₂CrO₄ * 2K + Cr + 4O = 0 * 2(+1) + Cr + 4(-2) = 0 * Cr = +6 KHCO₃ * K + H + C + 3O = 0 * (+1) + (+1) + C + 3(-2) = 0 * C = +4 BrCl₃ and IBr₅ * In BrCl₃, Cl is more electronegative, so Cl = -1 and Br = +3. * In IBr₅, Br is more electronegative, so Br = -1 and I = +5. This guide provides a comprehensive overview of oxidation numbers. Remember to practice applying these rules to various compounds and ions to solidify your understanding.