Oxidation Number: How To Find It, Assigning Rules and Exceptions

Questions based on this section of oxidation number are asked in NEET and CUET. Students must ensure that they have understood what exactly is oxidation number. The oxidation number is the charge assigned to an atom in a molecule or ion, assuming all bonds are ionic. It reflects:

• Electron Loss/Gain: Positive oxidation numbers indicate electron loss (oxidation); negative indicate electron gain (reduction).
• Redox Reactions: Changes in oxidation numbers identify the species oxidized or reduced.
• Balancing Equations: Used in the oxidation number method to balance redox reactions.

 

The oxidation states in a compound needs to add up the overall charge. For those neutral compounds, they will add up to zero. In case of ions, they will add up to the ion charge. For instance, let us consider H₂SO₄. 

• Hydrogen: +1 for each atom = +2 total
• Oxygen: -2 for each. There are 4 = -8 total
• Since the entire molecule is neutral. So (+2) + Sulphur + (-8) = 0
• Therefore Sulfur must be +6

The following rules, based on periodic properties, are used to calculate oxidation numbers:

1. Uncombined Elements: The oxidation number of an element in its free state (e.g., ${\mathrm{C}\mathrm{l}}_2,{\mathrm{O}}_2,{\mathrm{S}}_8$ ) is zero.

2. Monoatomic Ions: The oxidation number equals the ion's charge (e.g., ${\mathrm{N}\mathrm{a}}^{+}$ $=+1,{\mathrm{C}\mathrm{l}}^{-}=-1$ ).

3. Hydrogen: Usually it is +1 ; -1 in metal hydrides (e.g., NaH ).

4. Oxygen: Typically $-2;-1$ in peroxides (e.g., ${\mathrm{H}}_2{\mathrm{O}}_2$ ), $-1/2$ in superoxides (e.g., ${\mathrm{K}\mathrm{O}}_2$ ), +2 in ${\mathrm{O}\mathrm{F}}_2$ .

5. Fluorine: Always -1 due to its high electronegativity.

6. Halogens: -1 as halides (e.g., ${\mathrm{C}\mathrm{l}}^{-}$ in NaCl); positive in oxoacids or interhalogen compounds (e.g., Cl in ${\mathrm{H}\mathrm{C}\mathrm{l}\mathrm{O}}_4=+7$ ).

7. Alkali Metals: Always +1 (e.g., K in ${\mathrm{K}\mathrm{M}\mathrm{n}\mathrm{O}}_4$ ).

8. Alkaline Earth Metals: Always +2 (e.g., Ca in ${\mathrm{C}\mathrm{a}\mathrm{C}\mathrm{O}}_3$ ).

9. Sulphur: -2 as sulphide (e.g., ${\mathrm{H}}_2\mathrm{S}$ ); positive in oxoacids (e.g., ${\mathrm{S}\mathrm{O}}_4^{2-}$ ).

10. Metals: Always positive (e.g., Fe in ${\mathrm{F}\mathrm{e}}_2{\mathrm{O}}_3$ ).

11. Neutral Compounds: The sum of oxidation numbers equals zero (e.g., ${\mathrm{K}\mathrm{M}\mathrm{n}\mathrm{O}}_4$ : $+1+\mathrm{x}-8=0,\mathrm{x}=+7$ for Mn ).

12. Polyatomic Ions: The sum equals the ion's charge (e.g., ${\mathrm{C}\mathrm{O}}_3^{2-}:\mathrm{x}+3(-2)=$ for C ).

The following table summarizes the rules for oxidation numbers:

Element	Oxidation State
Group 1 metals	Always +1
Group 2 metals	Always +2
Fluorine	Always -1
Oxygen	Usually -2
Chlorine	Usually -1
Hydrogen	Usually +1

Important Related Links
CBSE Class 12 Chemistry Notes	NCERT Notes
NCERT Class 11 Chemistry Notes	NCERT Class 11 Notes

Following exceptions are there where rule for oxidation does not work. CBSE students must learn about this to solve questions in class 12th board exams:

• Peroxides: Oxygen's oxidation number is -1 (e.g., ${\mathrm{N}\mathrm{a}}_2{\mathrm{O}}_2,{\mathrm{H}}_2{\mathrm{O}}_2$ ).
• Superoxides: Oxygen is $-1/2$ (e.g., ${\mathrm{K}\mathrm{O}}_2$ ).
• OF₂: Oxygen is +2 due to fluorine's higher electronegativity.
• Fractional Oxidation Numbers: Occur in compounds like ${\mathrm{F}\mathrm{e}}_3{\mathrm{O}}_4$ (average Fe oxidation number $=+8/3$ ) due to mixed oxidation states.

NCERT Reference: Page 238 discusses exceptions like peroxides and superoxides.

Oxidation numbers are used for the following reasons. Do note that application based questions are asked in IISER and IIT JAM exams. 

• Identify Redox Processes: A change in oxidation number indicates oxidation (increase) or reduction (decrease).
• Balance Equations: The oxidation number method equates electron loss and gain.
• Determine Oxidizing/Reducing Agents:
• Highest oxidation state: Oxidizing agent (e.g., ${\mathrm{K}\mathrm{M}\mathrm{n}\mathrm{O}}_4,\mathrm{M}\mathrm{n}=+7$ ).
• Lowest oxidation state: Reducing agent (e.g., ${\mathrm{H}}_2\mathrm{S},\mathrm{S}=-2$ ).
• Intermediate state: Both (e.g., ${\mathrm{S}\mathrm{O}}_2,\mathrm{S}=+4$ ).

JEE Main students need to keep the following points in mind to solve conceptual questions based on the topic:

• Memorize oxidation number rules (NCERT, Pages 237-238).
• Practice calculating oxidation numbers in complex ions (e.g., ${\mathrm{C}\mathrm{r}}_2{\mathrm{O}}_7^{2-},{\mathrm{M}\mathrm{n}\mathrm{O}}_4^{-}$ ).
• Solve problems involving redox identification and balancing using the oxidation number method.
• Review JEE PYQs on oxidation states in compounds like ${\mathrm{K}}_2{\mathrm{C}\mathrm{r}}_2{\mathrm{O}}_7$ or ${\mathrm{H}}_2{\mathrm{O}}_2$ .
• You can practice the following questions to assess the types of questions asked in the exam:

1. Calculate the oxidation number of Cr in ${\mathrm{C}\mathrm{r}}_2{\mathrm{O}}_7^{2-}$ .

2. Determine the oxidation number of S in ${\mathrm{N}\mathrm{a}}_2{\mathrm{S}}_2{\mathrm{O}}_3$ .

3. Identify the oxidizing and reducing agents in the reaction: ${\mathrm{H}}_2\mathrm{S}+{\mathrm{S}\mathrm{O}}_2\to \mathrm{S}+$ ${\mathrm{H}}_2\mathrm{O}$ .

The following table summarizes the oxidation number rules for students to remember during board and entrance examinations:

Element/Condition	Oxidation Number
Free element (e.g., ${\mathrm{C}\mathrm{l}}_2,{\mathrm{O}}_2$ )	0
Monoatomic ion (e.g., ${\mathrm{N}\mathrm{a}}^{+},{\mathrm{C}\mathrm{l}}^{-}$ )	Charge of ion
Hydrogen	+1; -1 in hydrides
Oxygen	-2 ; -1 in peroxides, -1/2 in superoxides, +2 in ${\mathrm{O}\mathrm{F}}_2$
Fluorine	-1
Alkali metals	+1
Alkaline earth metals	+2
Neutral compound	Sum = 0
Polyatomic ion	Sum = Ion’s charge