Determining oxidation numbers (also called oxidation states) is an essential skill in understanding redox (reduction-oxidation) reactions, balancing chemical equations, and interpreting chemical bonding. An oxidation number is a theoretical charge assigned to an atom in a compound or ion, reflecting its degree of oxidation. Below is a comprehensive guide for determining oxidation numbers, including examples.
General Rules for Assigning Oxidation Numbers
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Elemental Form: The oxidation number of an atom in its elemental form is 0.
- Example: O₂, N₂, H₂ → Oxidation number = 0
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Monatomic Ions: The oxidation number of a monatomic ion equals the charge of the ion.
- Example: Na⁺ → Oxidation number = +1; Cl⁻ → Oxidation number = -1
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Oxygen: The oxidation number of oxygen is usually -2 in compounds. Exceptions include:
- Peroxides (e.g., H₂O₂) where it is -1.
- Superoxides (e.g., KO₂) where it is -1/2.
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Hydrogen: The oxidation number of hydrogen is +1 when bonded to nonmetals and -1 when bonded to metals.
- Example: In HCl, H = +1; in NaH, H = -1
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Fluorine: The oxidation number of fluorine in all compounds is -1.
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Alkali Metals: The oxidation number of group 1 alkali metals is always +1.
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Alkaline Earth Metals: The oxidation number of group 2 alkaline earth metals is always +2.
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Sum of Oxidation Numbers:
- In neutral compounds, the sum of the oxidation numbers equals 0.
- In polyatomic ions, the sum of the oxidation numbers equals the charge of the ion.
- Example: In sulfate ion (SO₄²⁻), the sum of oxidation states = -2.
Examples
Example 1: Determining Oxidation Numbers in a Compound
Take the compound \( \text{H}_2\text{SO}_4 \) (sulfuric acid):
- Hydrogen (H): Each H is +1, so 2 H contributes \( 2 \times +1 = +2 \).
- Oxygen (O): Each O is -2, so 4 O contributes \( 4 \times -2 = -8 \).
- Let the oxidation number of sulfur (S) be \( x \).
Setting up the equation: \[ 2 + x + (-8) = 0 \] \[ x - 6 = 0 \] So, \( x = +6 \).
- Final Oxidation States:
- H = +1
- S = +6
- O = -2
Example 2: Determining Oxidation Numbers in an Ion
Consider the ion \( \text{MnO}_4^- \) (permanganate):
- Oxygen (O): Again, each O is -2, so in 4 O, we have \( 4 \times -2 = -8 \).
- Let the oxidation number of manganese (Mn) be \( x \).
Setting up the equation: \[ x + (-8) = -1 \] \[ x - 8 = -1 \] Hence, \( x = +7 \).
- Final Oxidation States:
- Mn = +7
- O = -2
Example 3: Transition Metals
In \( \text{FeCl}_3 \):
- Chlorine (Cl): Each Cl is -1, so 3 Cl contributes \( 3 \times -1 = -3 \).
- Let the oxidation number of iron (Fe) be \( x \).
Setting up the equation: \[ x + (-3) = 0 \] \[ x = +3 \].
- Final Oxidation States:
- Fe = +3
- Cl = -1
Summary
Determining oxidation numbers requires systematic application of rules, and practice helps in mastering these concepts. Always start by assuming oxidation numbers based on the rules, and then solve equations where necessary, especially in complex molecules or ions. This can be essential for understanding electron transfer in reactions and for balancing chemical equations correctly.
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