7.2.2 Rules for Assigning Oxidation States or Numbers

Rule 1: The oxidation state of any uncombined or free element is always zero.

Examples: Zn, Na, H₂, S₈, O₂ — oxidation state = 0

Rule 2: In simple ions, the oxidation state is the same as their charge.

• Na⁺ → oxidation state = +1
• Ca²⁺ → oxidation state = +2
• Cl⁻ → oxidation state = −1

Rule 3: In a complex ion, the sum of oxidation states of atoms is equal to the charge on the ion.

• In CO₃²⁻, the sum of oxidation states of C and 3 O atoms = −2
• In NH₄⁺, the sum of oxidation states of N and 4 H atoms = +1

Rule 4: The oxidation number of each of the atoms in a molecule or compound is counted separately and their algebraic sum is zero.

• In HCl, the sum of oxidation states of H and Cl = 0
• In CO₂, the sum of oxidation states of one C and 2 O atoms = 0

Table 7.2: Oxidation States of Some Elements in Binary Compounds

Detailed Explanation

What Is Oxidation State?

Oxidation state (or oxidation number) is the charge an atom would have if all its bonds were ionic.

• It tells you the degree of oxidation of an atom.
• It can be positive, negative, or zero.
• It helps identify which element is oxidized and which is reduced.

Simple explanation: Oxidation state is like a "score" that tells you whether an atom has lost or gained electrons.

Rule 1: Free Elements Have Oxidation State Zero

Any element in its free (uncombined) state has an oxidation state of 0.

• Zn (solid zinc) → 0
• Na (solid sodium) → 0
• H₂ (hydrogen gas) → 0
• S₈ (sulfur) → 0
• O₂ (oxygen gas) → 0

Simple explanation: If an element is alone, it has no charge — so its oxidation state is zero.

Rule 2: Simple Ions — Oxidation State = Charge

For monoatomic ions, the oxidation state is the same as the charge on the ion.

• Na⁺ → +1
• Ca²⁺ → +2
• Al³⁺ → +3
• Cl⁻ → −1
• O²⁻ → −2
• N³⁻ → −3

Simple explanation: The oxidation state of a simple ion is just its charge.

Rule 3: Complex Ions — Sum = Charge on the Ion

In a polyatomic (complex) ion, the sum of the oxidation states of all atoms equals the charge on the ion.

Example 1: Carbonate ion (CO₃²⁻)

• Let the oxidation state of C be x.
• Oxidation state of O = −2 (Rule from Table 7.2)
• Sum: x + 3(−2) = −2
• x − 6 = −2
• x = +4

So the oxidation state of C in CO₃²⁻ is +4.

Example 2: Ammonium ion (NH₄⁺)

• Let the oxidation state of N be x.
• Oxidation state of H = +1
• Sum: x + 4(+1) = +1
• x + 4 = +1
• x = −3

So the oxidation state of N in NH₄⁺ is −3.

Simple explanation: For polyatomic ions, the sum of all oxidation states must equal the overall charge.

Rule 4: Neutral Compounds — Sum = Zero

In a neutral molecule or compound, the sum of the oxidation states of all atoms is zero.

Example 1: HCl

• H = +1 (Rule from Table 7.2)
• Cl = ?
• Sum: (+1) + x = 0
• x = −1

So the oxidation state of Cl in HCl is −1.

Example 2: CO₂

• C = ?
• O = −2 each
• Sum: x + 2(−2) = 0
• x − 4 = 0
• x = +4

So the oxidation state of C in CO₂ is +4.

Simple explanation: For neutral compounds, all the oxidation states must add up to zero.

Example 7.3: Determining Oxidation Number (from the book)

Problem: A device called a Breathalyzer is used by police to test a person's breath for alcohol. It contains an acidic solution of potassium dichromate (K₂Cr₂O₇). It is a strong oxidizing agent. Determine the oxidation state of Cr in it.

Problem Solving Strategy: Use rules 1 to 4 and Table 7.2 to get as many oxidation numbers as you can. Use rule 4 to get the oxidation number that has not been assigned.

Solution:

Step 1: Oxidation number of K is +1 (Group IA). There are 2 K atoms, so overall = 2(+1) = +2.

Step 2: There are 7 oxygen atoms, so overall = 7(−2) = −14.

Step 3: Suppose the oxidation state of Cr is x. Since there are 2 Cr atoms, overall = 2x.

Step 4: The sum of oxidation numbers must be zero (Rule 4).

+2 + 2x + (-14) = 0

2x - 12 = 0

2x = 12

x = +6

Final Answer: The oxidation state of Cr in K₂Cr₂O₇ is +6.

Example 7.4: Determining Oxidation State (from the book)

Problem: Boric acid (H₃BO₃) is used in eyewash. What is the oxidation state of B in this acid?

Solution:

Step 1: There are 3 H-atoms, so overall = 3(+1) = +3.

Step 2: There are 3 O-atoms, so overall = 3(−2) = −6.

Step 3: Suppose the oxidation state of B is x.

Step 4: The total oxidation states must be zero.

+3 + x + (-6) = 0

+3 + x - 6 = 0

x - 3 = 0

x = +3

Final Answer: The oxidation state of B in H₃BO₃ is +3.

Example 7.5: Determining Oxidation Number in an Ion (from the book)

Problem: What is the oxidation number of C in the carbonate ion, CO₃²⁻?

Solution:

Step 1: Oxidation number of O = −2.

Step 2: Use Rule 3 (sum = charge on ion).

Oxidation state of one C + Oxidation state of 3 O atoms = -2

x + 3(-2) = -2

x - 6 = -2

x = -2 + 6

x = +4

Final Answer: The oxidation state of C in CO₃²⁻ is +4.

Concept Assessment Exercise 7.4 (from the book)

Determine the oxidation state of:

1. S in sulphate ion, SO₄²⁻

• O = −2 × 4 = −8
• Sum = −2
• x + (−8) = −2
• x = **+6**

2. P in phosphate ion, PO₄³⁻

• O = −2 × 4 = −8
• Sum = −3
• x + (−8) = −3
• x = **+5**

3. N in ammonium ion, NH₄⁺

• H = +1 × 4 = +4
• Sum = +1
• x + (+4) = +1
• x = **−3**

Intermediate Level

Why Oxidation States Are Useful

• Identify which element is oxidized and which is reduced.
• Balance redox equations.
• Name compounds (e.g., iron(II) chloride vs iron(III) chloride).
• Predict chemical behaviour.

Simple explanation: Oxidation states are the "bookkeeping" system for electrons in redox reactions.

Determining Oxidation States — General Strategy

• Write down what you know (using rules and Table 7.2).
• Assign a variable (x) to the unknown oxidation state.
• Use Rule 3 or Rule 4 to set up an equation.
• Solve for x.

Worked Examples

Example 1: K₂Cr₂O₇

• K = +1 × 2 = +2
• O = −2 × 7 = −14
• Cr = x × 2 = 2x
• Sum = 0
• +2 + 2x − 14 = 0
• 2x = 12
• x = **+6**

Example 2: H₃BO₃

• H = +1 × 3 = +3
• O = −2 × 3 = −6
• B = x
• Sum = 0
• +3 + x − 6 = 0
• x = **+3**

Example 3: CO₃²⁻

• O = −2 × 3 = −6
• C = x
• Sum = −2
• x − 6 = −2
• x = **+4**

Comparison of Oxidation and Reduction

Important Points

• - Oxidation state (oxidation number): number of charges an atom will have in a molecule or compound.
• - Rule 1: Free elements have oxidation state 0.
• - Rule 2: Simple ions have oxidation state = their charge.
• - Rule 3: In a complex ion, the sum of oxidation states = charge on the ion.
• - Rule 4: In a neutral compound, the sum of oxidation states = 0.
• - Table 7.2: Group IA = +1; Group IIA = +2; Group IIIA = +3; H = +1 (except metal hydrides −1); Group VIIA = −1; O = −2 (except peroxides and OF₂).
• - Examples:
• - Cr in K₂Cr₂O₇ = **+6**
• - B in H₃BO₃ = **+3**
• - C in CO₃²⁻ = **+4**
• - S in SO₄²⁻ = **+6**
• - P in PO₄³⁻ = **+5**
• - N in NH₄⁺ = **−3**

Remember

• - Oxidation state = charge an atom would have.
• - Rule 1: Free element = 0.
• - Rule 2: Simple ion = charge.
• - Rule 3: Complex ion → sum = charge on ion.
• - Rule 4: Neutral compound → sum = 0.
• - Group IA = +1; Group IIA = +2; H = +1; O = −2; Group VIIA = −1.
• - **Cr in K₂Cr₂O₇** = +6.
• - **C in CO₃²⁻** = +4.
• - **N in NH₄⁺** = −3.

Common Mistakes

• 1. Forgetting that free elements have oxidation state 0. This is Rule 1.
• 2. Thinking oxygen is always −2. It is −2 in most compounds, but −1 in peroxides and +2 in OF₂.
• 3. Thinking hydrogen is always +1. It is +1 in most compounds, but −1 in metal hydrides.
• 4. Forgetting to multiply by the number of atoms. In K₂Cr₂O₇, there are 2 K atoms and 7 O atoms.
• 5. Forgetting that the sum must equal the charge on the ion for complex ions. For neutral compounds, the sum must be zero.
• 6. Confusing oxidation state with charge for polyatomic ions. In polyatomic ions, individual atoms have oxidation states, but the overall ion has a charge.

Quick Revision

• Oxidation state (oxidation number): the number of charges an atom will have in a molecule or a compound.
• Rules for assigning oxidation states:
• Table 7.2: Group IA = +1; Group IIA = +2; Group IIIA = +3; H = +1 (except metal hydrides −1); Group VIIA = −1; O = −2 (except peroxides and OF₂).
• Examples:
• Cr in K₂Cr₂O₇ = +6
• B in H₃BO₃ = +3
• C in CO₃²⁻ = +4
• S in SO₄²⁻ = +6
• P in PO₄³⁻ = +5
• N in NH₄⁺ = −3
• Oxidation: increase in oxidation number; Reduction: decrease in oxidation number.

60-Second Revision

• Oxidation state = charge an atom would have.
• Rules: free element = 0; simple ion = charge; complex ion → sum = charge; neutral compound → sum = 0.
• **Group IA** = +1; **Group IIA** = +2; **H** = +1; **O** = −2.
• **Cr in K₂Cr₂O₇** = +6; **C in CO₃²⁻** = +4; **N in NH₄⁺** = −3.
• Warning: oxygen is not always −2 (exception: peroxides, OF₂).