JEE Mains Chemistry · Some Basic Concepts of Chemistry

Oxidation Number and Redox Reactions

Assigning oxidation numbers, naming the type of a redox reaction, and balancing redox equations by the half-reaction method.

All Some Basic Concepts of Chemistry notes JEE Mains Chemistry strategy

Why this matters

Twenty-four PYQs, fourteen of them multiple choice and none yet from 2026. Nine assign oxidation numbers. Eight classify a redox reaction, most often asking whether it is a disproportionation, and seven balance a half-reaction or a whole equation. The electron counts found here are the n-factors the titration page uses next.

Concept 1 of 3: Assigning oxidation numbers

An oxidation number is the charge an atom would carry if every bond went wholly to the more electronegative partner. Fix the atoms whose values are set, then solve for the one that is left.

Definition

The oxidation numbers in a species add up to its charge.
F is always $-1$ . O is $-2$ , but $-1$ in peroxides and $+2$ in $\mathrm{OF_2}$ . H is $+1$ , but $-1$ in metal hydrides.
A neutral ligand adds nothing: Fe in $\mathrm{Fe(CO)_5}$ is 0.
Examples: Mn is $+7$ in $\mathrm{KMnO_4}$ , C is $+3$ in $\mathrm{H_2C_2O_4}$ , Mo is $+6$ in $\mathrm{(NH_4)_3[PMo_{12}O_{40}]}$ .
An atom at its lowest oxidation number ( $\mathrm{N^{3-}}$ ) cannot be an oxidising agent; one at its highest cannot be a reducing agent.
The oxidation state is based on electronegativity, not on electron gain enthalpy.

Charge balance

\sum(\text{oxidation numbers})=\text{charge on the species}

Worked example

Find the oxidation number of Cr in

\mathrm{K_2Cr_2O_7}

and of S in

\mathrm{Na_2S_2O_3}

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 27 Jan 2024 · Q60Moderate

Example 1 · Some Basic Concepts of Chemistry · Oxidation Number and Redox Reactions

From the given list, the number of compounds with +4 oxidation state of Sulphur _____ .

SO_{3},H_{2}SO_{3},SOCl_{2},SF_{4},BaSO_{4},H_{2}{\text{ }S}_{2}O_{7}

A fraction is an average

S in

\mathrm{S_4O_6^{2-}}

works out to

+2.5

. That is an average: two S atoms are

+5

and two are 0. A fractional answer means unlike atoms, not an error.

Drill 8 more on assigning oxidation numbers

Concept 2 of 3: Types of redox reaction

Name the reaction by what happens to the reactants: two combine, one splits, one element pushes another out of a compound, or one element goes both up and down. That last one, disproportionation, needs an element in an in-between oxidation state.

Definition

Combination: two substances form one.
Decomposition: one substance breaks into two or more.
Displacement: an element replaces another in a compound.
Disproportionation: one element, in one oxidation state, is both oxidised and reduced.
Cannot disproportionate: $\mathrm{F_2}$ (fluorine has no positive state), an element at its highest state ( $\mathrm{MnO_4^-}$ , $\mathrm{Cr_2O_7^{2-}}$ , $\mathrm{ClO_4^-}$ ) or at its lowest ( $\mathrm{K^+}$ , $\mathrm{N^{3-}}$ , Ag).
Two states meeting in the middle (comproportionation) is not disproportionation. Nor is $\mathrm{2KMnO_4\rightarrow K_2MnO_4+MnO_2+O_2}$ , where Mn is reduced but O is oxidised.

Reaction	Type	Why
$\mathrm{2Mg+O_2\rightarrow2MgO}$	Combination	Two reactants, one product
$\mathrm{2Pb(NO_3)_2\rightarrow2PbO+4NO_2+O_2}$	Decomposition	N goes from $+5$ to $+4$ ; O goes from $-2$ to 0
$\mathrm{V_2O_5+5Ca\rightarrow2V+5CaO}$	Metal displacement	Ca pushes V out of its oxide
$\mathrm{2Na+2H_2O\rightarrow2NaOH+H_2}$	Hydrogen displacement	Na pushes H out of water
$\mathrm{2H_2O_2\rightarrow2H_2O+O_2}$	Disproportionation	O at $-1$ goes to $-2$ and to 0
$\mathrm{Cl_2+2OH^-\rightarrow Cl^-+ClO^-+H_2O}$	Disproportionation	Cl at 0 goes to $-1$ and to $+1$
$\mathrm{2MnO_4^-+3Mn^{2+}+2H_2O\rightarrow5MnO_2+4H^+}$	Comproportionation	$+7$ and $+2$ meet at $+4$ ; not a disproportionation

Disproportionation needs one element in one intermediate oxidation state going both up and down.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2022 · 26 June 2022 · Q128Moderate

Example 2 · Some Basic Concepts of Chemistry · Oxidation Number and Redox Reactions

Which one of the following is an example of disproportionation reaction?

Burning methane is 'combination' here

In the NCERT scheme the paper follows, a fuel burning in oxygen,

\mathrm{CH_4+2O_2\rightarrow CO_2+2H_2O}

, is filed as a combination reaction, and it is keyed that way.

Two manganese species are not enough

A reaction with Mn on both sides is a disproportionation only if all the Mn started in one state.

\mathrm{MnO_4^-}

with

\mathrm{Mn^{2+}}

starts from two states, so it is not one.

Drill 7 more on types of redox reaction

Concept 3 of 3: Balancing redox equations by half-reactions

Split the reaction into an oxidation and a reduction. In each, balance the main atoms, then O with water, then H with

\mathrm{H^+}

, then the charge with electrons. Scale the halves so the electrons cancel, and add.

Definition

In acid: atoms other than O and H first; O with $\mathrm{H_2O}$ ; H with $\mathrm{H^+}$ ; charge with $e^-$ .
In base: balance as in acid, then add $\mathrm{OH^-}$ to both sides to turn every $\mathrm{H^+}$ into water.
$\mathrm{MnO_4^-+8H^++5e^-\rightarrow Mn^{2+}+4H_2O}$ in acid; $\mathrm{MnO_4^-+2H_2O+3e^-\rightarrow MnO_2+4OH^-}$ in neutral or basic solution.
$\mathrm{Cr_2O_7^{2-}+14H^++6e^-\rightarrow2Cr^{3+}+7H_2O}$ .
$\mathrm{C_2O_4^{2-}\rightarrow2CO_2+2e^-}$ ; $\mathrm{Fe^{2+}\rightarrow Fe^{3+}+e^-}$ .

Electron balance

\text{electrons lost in oxidation}=\text{electrons gained in reduction}

Worked example

Balance

\mathrm{MnO_4^-+Fe^{2+}\rightarrow Mn^{2+}+Fe^{3+}}

in acid.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 13 April 2023 · Q143Moderate

Example 3 · Some Basic Concepts of Chemistry · Oxidation Number and Redox Reactions

See the following chemical reaction:

Cr_{2}O_{7}^{2 -}+XH^{+}+ 6Fe^{2 +}\rightarrow YCr^{3 +}+ 6Fe^{3 +}+ZH_{2}O

The sum of

X, Y

and

Z

is____

The medium sets the product

Permanganate takes 5 electrons in acid (to

\mathrm{Mn^{2+}}

) but 3 in neutral or basic solution (to

\mathrm{MnO_2}

). Read the medium before counting.

Check the charge, not just the atoms

An equation can balance every atom and still be wrong. Add the charges on each side; they must match.

Drill 6 more on balancing redox equations by half-reactions

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Assigning oxidation numbers
Charge balance
$\sum(\text{oxidation numbers})=\text{charge on the species}$
Balancing redox equations by half-reactions
Electron balance
$\text{electrons lost in oxidation}=\text{electrons gained in reduction}$

Reference tables (1)

Types of redox reaction7 rows

Reaction	Type	Why
$\mathrm{2Mg+O_2\rightarrow2MgO}$	Combination	Two reactants, one product
$\mathrm{2Pb(NO_3)_2\rightarrow2PbO+4NO_2+O_2}$	Decomposition	N goes from $+5$ to $+4$ ; O goes from $-2$ to 0
$\mathrm{V_2O_5+5Ca\rightarrow2V+5CaO}$	Metal displacement	Ca pushes V out of its oxide
$\mathrm{2Na+2H_2O\rightarrow2NaOH+H_2}$	Hydrogen displacement	Na pushes H out of water
$\mathrm{2H_2O_2\rightarrow2H_2O+O_2}$	Disproportionation	O at $-1$ goes to $-2$ and to 0
$\mathrm{Cl_2+2OH^-\rightarrow Cl^-+ClO^-+H_2O}$	Disproportionation	Cl at 0 goes to $-1$ and to $+1$
$\mathrm{2MnO_4^-+3Mn^{2+}+2H_2O\rightarrow5MnO_2+4H^+}$	Comproportionation	$+7$ and $+2$ meet at $+4$ ; not a disproportionation

Disproportionation needs one element in one intermediate oxidation state going both up and down.

Watch out for (5)

A fraction is an average
→ Assigning oxidation numbers
Burning methane is 'combination' here
→ Types of redox reaction
Two manganese species are not enough
→ Types of redox reaction
The medium sets the product
→ Balancing redox equations by half-reactions
Check the charge, not just the atoms
→ Balancing redox equations by half-reactions

Test yourself on Some Basic Concepts of Chemistry

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Some Basic Concepts of Chemistry chapter test Drill all 24 questions