Oxidation Number

1.0What Is Oxidation Number?

The oxidation number (also known as oxidation state) is a formalism that indicates the degree of oxidation (loss of electrons) of an atom in a chemical compound. It represents the hypothetical charge that an atom would have if all bonds were completely ionic. Oxidation numbers help balance redox reactions, determine electron transfer, and understand redox chemistry.
Key points:

• Oxidation number is a bookkeeping tool, not always equal to actual charge.
• It facilitates the identification of oxidizing and reducing agents.
• The concept applies across ionic and covalent compounds, making it versatile for chemical analysis.

2.0How to Find the Oxidation Number of an Atom?

Determining oxidation numbers relies on a set of well-defined rules:

• Rule 1: The oxidation number of an atom in its elemental form (e.g., O₂, H₂, N₂) is zero.
• Rule 2: For monatomic ions, the oxidation number equals the ion’s charge (e.g., Na⁺ is +1, Cl⁻ is –1).
• Rule 3: In compounds, hydrogen generally has an oxidation number of +1 (except in metal hydrides, where it is –1).
• Rule 4: Oxygen usually has an oxidation number of –2 (except in peroxides like H₂O₂ where it is –1, or in superoxides where it is –½).
• Rule 5: Fluorine always has an oxidation number of –1; other halogens typically have –1 unless bonded to oxygen or other halogens of lower electronegativity.
• Rule 6: The sum of oxidation numbers in a neutral compound is zero; in a polyatomic ion, the sum equals the ion’s charge.

These rules form the basis for systematic computation of oxidation numbers in complex compounds.

3.0Calculation of Oxidation Number

Let's walk through a systematic approach to calculate oxidation numbers:

Step 1: Identify known oxidation states using the rules (e.g., assign O as –2, H as +1, etc.).
Step 2: Let the unknown oxidation number be x.
Step 3: Write the algebraic sum equation: sum of oxidation numbers = overall charge.
Step 4: Solve for x.

Example 1: Determine the oxidation number of sulfur in ${\text{SO}}_4^{2-}$

• Oxygen: –2 × 4 = –8
• Let sulfur = x
• Sum: x + (–8) = –2 → x = +6

Example 2: Oxidation number of chromium in $\text{Cr}2\text{O}{7}^{2-}$

• Oxygen: –2 × 7 = –14
• Let each Cr = x → total for 2 Cr = 2x
• Sum: 2x + (–14) = –2 → 2x = +12 → x = +6

This method applies to a vast array of compounds and ions.

4.0Examples of Oxidation Numbers

1. Water ($H_{2}O$)
   
   H: +1
   O: -2
   Sum = 2(+1) + (-2) = 0
2. Sulphuric Acid (${\text{H}}_2{\text{SO}}_4$)
   
   H: +1
   O: -2
   S: ? → Let x = oxidation number of S
3. Potassium Permanganate (${\text{KMnO}}_4$)
   
   K: +1
   O: -2 × 4 = -8
   Mn: x
4. Carbonate Ion (${\text{CO}}_3^{2-}$)
   
   O: -2 × 3 = -6
   C: x

5.0Oxidation Number of Atoms in a Diatomic Molecule

In a diatomic molecule comprised of two identical atoms (e.g., O₂, N₂, Cl₂), each atom is in its elemental state. By rule, the oxidation number of each atom in such molecules is zero. This is because there is no net electron transfer between identical atoms in a homonuclear bond.

6.0Fractional Oxidation States

While oxidation numbers are typically integers, certain compounds exhibit fractional oxidation states due to resonance or mixed oxidation environments. Examples include:

• ( \text{Fe}3\text{O}4 ) (magnetite): has both Fe²⁺ and Fe³⁺; the average oxidation number of Fe is +8/3 (≈ +2.667).
• Transition metal complexes with mixed valence states can yield fractional averages, useful for understanding electron delocalization.

Interpretation of fractional oxidation states relies on recognizing average electron distribution rather than discrete ionic charges.

7.0Importance of Oxidation Number in Chemistry

• Redox Reactions: Oxidation numbers help identify which species undergo oxidation (increase in oxidation number) or reduction (decrease in oxidation number).
• Balancing Equations: Essential for systematic balancing of complex redox reactions, especially in acidic or basic media.
• Predicting Reactivity: Higher oxidation states often correlate with stronger oxidizing behavior.
• Electrochemistry: Oxidation numbers assist in determining half-reactions, electrode potentials, and cell voltage

8.0Why Oxidation Numbers Matter in Chemistry

Oxidation numbers are foundational for:

• Balancing redox equations.
• Determining electron flow in electrochemical cells.
• Understanding reaction mechanisms in inorganic and organic chemistry.
• Analyzing oxidation catalysts and electron transfer processes.

9.0Applications of Oxidation Numbers

1. Balancing Redox Reactions:
2. • Used in both acidic and basic solutions.
2. Electrochemistry:
3. • Determining the electron flow in galvanic and electrolytic cells.
3. Analytical Chemistry:
4. • Titration reactions using KMnO₄, K₂Cr₂O₇, etc.
4. Organic Chemistry:
5. • Identifying oxidation states in functional group transformations.
5. Industrial Chemistry:
6. • Understanding processes like Haber process, Contact process, and metal extraction.

10.0Solved Examples

Example 1

Oxidation Number of Nitrogen in Ammonium Nitrate (NH₄NO₃) 

Step 1: Separate the compound into its two ions:

• Ammonium ion → NH₄⁺
• Nitrate ion → NO₃⁻

Step 2: Determine the oxidation number of nitrogen in each ion. Let nitrogen = X.

In Ammonium ion (NH₄⁺)

Step 3: The total charge on NH₄⁺ is +1.
Hydrogen has an oxidation number of +1.

So the equation becomes:

X+4(+1)=+1

Step 4: Solving,

X+4=1⇒X=−3

So, nitrogen in NH₄⁺ has an oxidation number of −3.

In Nitrate ion (NO₃⁻)

Step 5: Oxygen has an oxidation number of −2, and the ion has an overall charge of −1.

Step 6: Write the equation:

X+3(−2)=−1X + 3(-2) = -1X+3(−2)=−1

Step 7: Solving,

 X = +5X − 6 = −1 ⇒ X = +5

So, nitrogen in NO₃⁻ has an oxidation number of +5.

Answer:

In ammonium nitrate (NH₄NO₃), nitrogen shows two different oxidation numbers: −3 and +5.

Example 2

Oxidation Number of Carbon in CN⁻ 

Let the oxidation number of carbon in the cyanide ion (CN⁻) be x.
Nitrogen has a fixed oxidation number of −3, and the ion has a total charge of −1.

So the equation becomes:

x+(−3)=−1x + (-3) = -1x+(−3)=−1

Solving:

X−3 = −1

X = −1 + 3 = +2

Answer: The oxidation number of carbon in CN⁻ is +2.