The Periodic Table S3.1.6 Oxidation States IB DP Chemistry Study Notes - New Syllabus 2025
The periodic table: Classification of elements- IB DP Chemistry- Study Notes
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Structure 3.1.6 – Oxidation States
Structure 3.1.6 – Oxidation States
Oxidation and Reduction
In redox (reduction–oxidation) reactions, the concepts of oxidation and reduction are defined in terms of either electron transfer or changes in oxidation state.
- Oxidation
- addition of oxygen
- loss of hydrogen
- loss of electrons
- Reduction
- loss of oxygen
- addition of hydrogen
- gain of electrons
- These processes always occur together — when one species is oxidized, another is reduced.
What is an Oxidation State?
The oxidation state (or oxidation number) is a hypothetical charge that an atom would have if all bonds were ionic. It helps track electron transfer in reactions, especially redox processes.
- Oxidation states are represented by a sign (+ or –) followed by a number (e.g., +2, –1).
- On oxidation : increase in oxidation state(number).
- On reduction : decrease in oxidation state(number).
Why is the Oxidation State of an Element Zero?
For elements in their standard, uncombined form (not part of a compound or ion), the oxidation state is always zero. This is because the atom is electrically neutral — it has not lost or gained electrons.
Examples:
- \( \text{O}_2 \): Oxidation state of O = 0
- \( \text{H}_2 \): Oxidation state of H = 0
- \( \text{Na} \): Oxidation state of Na = 0
Atoms and Simple Ions
- Oxidation number = number of electrons to add/remove to make atom neutral.
- Examples:
- Na in Na = 0
- Na in Na⁺ = +1
- Cl in Cl⁻ = –1
- For simple ions, oxidation number = ionic charge:
- Na⁺, K⁺, H⁺ = +1
- Mg²⁺, Ca²⁺ = +2
- Cl⁻, Br⁻ = –1
- O²⁻, S²⁻ = –2
Important Naming Notes
- For naming compounds (especially polyatomic ions and transition metal compounds), oxidation states may be shown in Roman numerals.
- However, many traditional names are still accepted. For example:
- \( \text{NO}_3^- \) = nitrate (not nitrogen(V) oxide)
- \( \text{SO}_4^{2-} \) = sulfate (not sulfur(VI) oxide)
- \( \text{NO}_2^- \) = nitrite
- \( \text{SO}_3^{2-} \) = sulfite
What Oxidation State Tells Us
- Helps determine whether a species is oxidized or reduced
- Essential for balancing redox equations
- Supports understanding of bonding and compound types
Example
What is the oxidation state of oxygen in the following species?
- \( \text{O}_2 \)
- \( \text{H}_2\text{O} \)
- \( \text{H}_2\text{O}_2 \)
▶️ Answer/Explanation
- In \( \text{O}_2 \): Oxygen is uncombined → Oxidation state = 0
- In \( \text{H}_2\text{O} \): Oxygen = –2 (usual state in compounds)
- In \( \text{H}_2\text{O}_2 \): Oxygen = –1 (peroxide exception)
General Rules for Assigning Oxidation States
Use the following standard rules to determine oxidation states of atoms in compounds or ions:
- Uncombined elements: Oxidation state is 0. (e.g., \( \text{Cl}_2 \), \( \text{O}_2 \), \( \text{Fe} \))
- Group 1 elements (alkali metals): Always +1 in compounds.
- Group 2 elements (alkaline earth metals): Always +2 in compounds.
- Fluorine (F): Always –1 in compounds.
- Hydrogen (H): Usually +1, but –1 in metal hydrides (e.g., \( \text{NaH} \)).
- Oxygen (O): Usually –2, but:
–1 in peroxides (e.g., \( \text{H}_2\text{O}_2 \))
+2 in \( \text{F}_2\text{O} \) (fluorine is more electronegative) - Halogens: Usually –1, unless bonded to more electronegative elements.
- Monatomic ions: Oxidation number = charge of ion (e.g., \( \text{Na}^+ = +1 \), \( \text{O}^{2-} = -2 \))
- Compounds: Sum of oxidation numbers = 0.
- Polyatomic ions: Sum of oxidation numbers = charge of the ion.
Assigning Oxidation Numbers: Step-by-Step
To deduce oxidation states:
- Assign known oxidation numbers using the rules above.
- Use the total charge of the compound or ion to calculate the unknown oxidation state.
- Remember that the more electronegative atom is given the negative oxidation number.
Example
What is the oxidation state of sulfur in \( \text{SO}_4^{2-} \)?
▶️ Answer/Explanation
- Each oxygen has an oxidation number of –2.
- Total from oxygen: \( 4 \times (-2) = -8 \)
- Let sulfur = x. Total charge = –2:
- \( x + (-8) = -2 \Rightarrow x = +6 \)
- Answer: Oxidation state of S = +6
Example
Determine the oxidation state of nitrogen in \( \text{NH}_2\text{OH}\).
▶️ Answer/Explanation
- Each hydrogen = +1
- Total from H: \( 3 \times (+1) = +3 \)
- Each oxygen = –2.
- Total from O: \(-2 \)
- Let N = x. The molecule is neutral:
- \( x + 3 – 2 = 0 \Rightarrow x = -1 \)
- Answer: Oxidation state of N = –1
Example
What is the oxidation state of oxygen in \( \text{H}_2\text{O}_2 \)?
▶️ Answer/Explanation
- Each hydrogen = +1
- Total from H: \( 2 \times (+1) = +2 \)
- Let each oxygen = x. There are 2 oxygens:
- \( 2x + 2 = 0 \Rightarrow x = -1 \)
- Answer: Oxidation state of O = –1 in peroxides
Fractional Oxidation Numbers
In some compounds containing the same element in different oxidation states, the overall oxidation number may be fractional (an average).
Example
Determine the average oxidation state of sulfur in \( \text{S}_4\text{O}_6^{2-} \).
▶️ Answer/Explanation
- Oxygen = –2; \( 6 \times (-2) = -12 \)
- Let total sulfur = x
- \( x – 12 = -2 \Rightarrow x = +10 \)
- There are 4 sulfur atoms → \( 10 / 4 = +2.5 \)
- Answer: Average oxidation state of S = +2.5
Naming Transition Metal Compounds
Transition metals often have variable oxidation states, which must be shown using Roman numerals in compound names.
- \( \text{Fe}^{2+} \) → iron(II)
- \( \text{Fe}^{3+} \) → iron(III)
- \( \text{Cu}_2\text{O} \) → copper(I) oxide
- \( \text{MnO}_2 \) → manganese(IV) oxide