Top 10 Most Repeated Topics In Chemistry For JEE Mains 2026 - High Scoring Areas to Crack IIT JEE

Most Repeated Chemistry Topics For JEE Main 2026

Preparing for any exam requires research, and especially for a competitive entrance exam like JEE Main, requires thorough preparation. So, our experts have collected and done thorough research from the JEE Main Latest Syllabus 2026. In the table below, we have provided you with the top 10 most repeated topics in chemistry for JEE Main 2026. This data comprises the last 10 years of JEE Main question papers of all slots.

This table contains the most scoring topics in JEE Main Chemistry. For more details, we have also provided you with a count of the total number of questions from these topics, along with the chapters that they are from.

As seen, the topic Concentration Terms happens to be the most important with the highest weighted topic. The 10th most important topic is Oxidation State, a part of both the chapters: d-Block Element/Redox Reactions. Following this list, we have more highly weighted topics that could be a game-changer for the JEE Main chemistry exam.

Now, after seeing the topics, let’s dive into the most important chapters for JEE Main Chemistry 2026. Understanding this aspect will give you an upper hand in the preparation for the exam!

Top 10 Most Important Chapters For JEE Main Chemistry 2026

We have already seen the most soring chemistry topics for JEE Main and the chapters that they are asked from. Now, let’s see which chapters have the highest number of questions from the past 10 years. The most important chapters for JEE Main Chemistry 2026 are curated by analyzing the frequency of the questions asked from the whole chapter in the last 10 years.

This table contains the chapter name on one side and the total number of questions asked from each chapter on the other side. The following listed chapters constitute the most important chapters for chemistry, with organic compounds containing oxygen being the highest weighted chapter. This list has the most important chapters ranked in order of most to least important.

Refer to JEE Main Chapter-Wise Weightage for a better understanding of important topics and help you to prepare accordingly.

Most Repeated Questions in JEE Main Chemistry 2026

In this section, we will be diving into each topic and looking at the set of the most important questions from those particular topics listed above individually. Understanding the level of these questions and practicing them will help you increase your chances of scoring better and increasing your rank in the JEE Main exam 2026. Let’s dive in!

1. Concentration Terms

Question: Some amount of dichloromethane $\left(\mathrm{CH}_2 \mathrm{Cl}_2\right.$) is added to 671.141 mL of chloroform $\left(\mathrm{CHCl}_3\right)$ to prepare $2.6 \times 10^{-3} \mathrm{M}$ solution of $\mathrm{CH}_2 \mathrm{Cl}_2(\mathrm{DCM})$. The concentration of DCM is ppm (by mass).
Given : atomic mass : $\mathrm{C}=12$
$\begin{aligned} & \mathrm{H}=1 \\ & \mathrm{Cl}=35.5\end{aligned}$
density of $\mathrm{CHCl}_3=1.49 \mathrm{~g} \mathrm{~cm}^{-3}$

Solution:

Molar mass $=12+2+71$

= 85mmoles of DCM $=671.141 \times 2.6 \times 10^{-3}$

mass of solution $=1.49 \times 671.141$

$\mathrm{PPM}=\frac{671.141 \times 2.6 \times 10^{-3} \times 85 \times 10^{-3}}{1.49 \times 671.141} \times 10^6$

$=148.322$..

Hence, the answer is (148.322).

Question: The molarity of 0.006 moles of NaCl in 100 ml solutions in -

(1) 0.6

(2) 0.06

(3) 0.006

(4) 0.066

Solution:

As we learn

Molarity -

$\begin{aligned} & \text { Molarity }=\frac{\text { Moles of solute }}{\text { Vol.of solution }(L)} \\ & \text { Molarity }=\frac{\text { Moles of solute }}{\text { Vol.of solution }(L)} \\ & M=\frac{n}{V(l)}=\frac{0.006}{0.1}=0.06\end{aligned}$

Hence, the answer is the option (2).

2. Mole Concept and Molar Mass

Question: 1 gram of a carbonate ($\mathrm{M}_2 \mathrm{CO}_3$) on treatment with excess HCl produces 0.01186 mole of $\mathrm{CO}_2$.. The molar mass of $\mathrm{M}_2 \mathrm{CO}_3$ in $\mathrm{g} \mathrm{mol}^{-1}$ is :

(1) 118.6

(2) 11.86

(3) 1186

(4) 84.3

Solution:

Given,

Mass of carbonate ($\left.\mathrm{M}_2 \mathrm{CO}_3\right)$ = 1 gram

As we have learned,

Number of Moles -

No of moles = given mass of substance/ molar mass of a substance

Given the Chemical reaction,

$\mathrm{M}_2 \mathrm{CO}_3+2 \mathrm{HCl} \rightarrow 2 \mathrm{MCl}+2 \mathrm{H}_2 \mathrm{O}+\mathrm{CO}_2$

From the balanced equation

$\frac{1}{M}=0.01186 \Rightarrow M=\frac{1}{0.01186}=84.3$

Hence, the answer is an option (4).

3. Magnetic Moment (On the Basis of VBT)

Magnetic moment explains the magnetic behavior of coordination compounds using Valence Bond Theory (VBT). Questions are mostly conceptual and numerical, moderate in difficulty.

Question: The calculated magnetic moments (spin only value) for species $\left[\mathrm{FeCl}_4\right]^{2-},\left[\mathrm{Co}\left(\mathrm{C}_2 \mathrm{O}_4\right)_3\right]^{3-}$ and $\mathrm{MnO}_4^{2-}$ respectively are :

(1) 5.82, 0 and $0 B M$

(2) $5.92,4.90$ and $0 B M$

(3) $4.90,0$ and $1.73 B M$

(4) $4.90,0$ and $2.83 B M$

Solution:

$\left[\mathrm{FeCl}_4\right]^{2-} \quad \mathrm{Fe}^{2+} \quad 3 \mathrm{~d}^6 \rightarrow 4$ unpaired electron

as Cl– in a weak field liquid.

$\mu_{\text {spin }}=\sqrt{24} 8 \mathrm{M}=4.9 \mathrm{BM}$

$\left[\mathrm{Co}\left(\mathrm{C}_2 \mathrm{O}_4\right)_3\right]^3 \quad \mathrm{Co}^{3+} \quad 3 \mathrm{~d}^6 \rightarrow$ for $\mathrm{Co}^{3+}$ with coodination no. 6

$\mathrm{C}_2 \mathrm{O}_4^{2-}$ is strong field ligand & causes pairing & hence no. unpaired electron.

$\mu_{\text {spin }}=0$

$\left[\mathrm{MnO}_4\right]^{2-} \quad \mathrm{Mn}^{+6} \quad$ it has one unpaired electron.

$\mu_{\text {spin }}=\sqrt{3} \mathrm{BM}=1.73 \mathrm{BM}$

Hence,the answer is the option(3).

4. Stoichiometry, Stoichiometric Calculations and Limiting Reagent

Question: A sample of $\mathrm{NaClO}_3$ is converted by heat to NaCl with a loss of 0.16 g of oxygen. The residue is dissolved in water and precipitated as AgCl. The mass of AgCl (in g) obtained will be:

(1) 0.35

(2) 0.41

(3) 0.48

(4) 0.54

Solution:

As we learnt in

Stoichiometry -

Stoichiometry deals with measurements of reactants and products in a chemical reaction.

- wherein

$\mathrm{aA}(\mathrm{g})+\mathrm{bB}(\mathrm{g}) \rightarrow \mathrm{cC}(\mathrm{g})+\mathrm{dD}(\mathrm{g})$

Here, ‘a’ moles of A(g) reacts with ‘b’ moles of B(g) to give ‘c’ mole of C(g) and ‘d’ moles of D(g)

$2 \mathrm{NaClO}_3 \xrightarrow{\Delta} 2 \mathrm{NaCl}+\underset{0.16 \mathrm{~g}}{3 \mathrm{O}_2}$

$\frac{n_{\mathrm{NaCl}}}{2}=\frac{n_{\mathrm{O}_2}}{3}$

$n_{\mathrm{NaCl}}=\frac{0.16}{32} \times \frac{2}{3}=\frac{1}{200} \times \frac{2}{3}=\frac{1}{300}$

$\mathrm{NaCl} \rightarrow \mathrm{AgCl}$

PoAC on Cl

$1 \times n_{\mathrm{NaCl}}=1 \times n_{\mathrm{AgCl}}$

$\frac{1}{300}=n_{A g C l}$

the weight of $A g C l=\frac{1}{300} \times[108+35.5]=\frac{1}{300} \times 143.5=0.48 g$

Hence, the answer is an option (3).

5. Applications of CFT (Crystal Field Theory)

CFT explains color, magnetism, and stability of coordination compounds. Questions are mostly conceptual and numerical, moderate in difficulty.

Question: Which of the following 3 d -metal ion will give the lowest enthalpy of hydration $\left(\Delta_{\text {hyd }} \mathrm{H}\right)$ when dissolved in water?

(1) $\mathrm{Cr}^{2+}$

(2) $\mathrm{Mn}^{2+}$

(3) $\mathrm{Fe}^{2+}$

(4) $\mathrm{Co}^{2+}$

Solution:

Water act as a weak ligand.C F S E of metal ions depend on strength of ligand.

Since the CFSE of $\mathrm{Mn}^{2+}$ is the least, $\Delta \mathrm{H}_{\mathrm{Hyd}}$ of it is also lowest.
Hence, the answer is the option (2).

6. Reduction and Oxidation Reaction

Redox reactions are key in electrochemistry and organic transformations. Questions are mostly numerical and moderate in difficulty.

Question: Experimentally reducing a functional group cannot be done by which one of the following reagents?

(1) $\mathrm{Zn} / \mathrm{H}_2 \mathrm{O}$

(2) $\mathrm{Pt}-\mathrm{C} / \mathrm{H}_2$

(3) $\mathrm{Pd}-\mathrm{C} / \mathrm{H}_2$

(4) $\mathrm{Na} / \mathrm{H}_2$

Solution:

Out of the given reagents, $\mathrm{Na} / \mathrm{H}_2$ is not used as a reducing agent.

Hence, the answer is the option (4).

7. Stereoisomerism

Stereoisomerism explains spatial arrangement of atoms in molecules, important in organic chemistry. Questions are mostly conceptual and moderate in difficulty.

Question: The total number of possible isomers for square-planar $\left[\mathrm{Pt}(\mathrm{Cl})\left(\mathrm{NO}_2\right)\left(\mathrm{NO}_3\right)(\mathrm{SCN})\right]^{2-}$ is :

(1) 8

(2) 12

(3) 16

(4) 24

Solution:

As we have learnt,

$\mathrm{NO}_2^{-}$ and $S C N^{-}$ are ambidentate ligands and each of them can attach through two different donor sites.

Now, the given square planar complex can show geometrical isomerism as well as linkage isomers.

The number of isomers possible is listed below:

Thus, the total number of Isomers is 12.

Hence, the answer is an option (2).

8. Disaccharides and Polysaccharides

This topic is important in biomolecules, especially for carbohydrates' structure and properties. Questions are mostly conceptual and numerical, moderate in difficulty.

Question: Compound A gives D-Galactose and D-Glucose on hydrolysis. The compound A is :

(1) Amylose

(2) Sucrose

(3) Maltose

(4) Lactose

Solution:

As we have learned,
The monosaccharides are involved in the formation of the given sugars.
Amylose: $\alpha-\mathrm{D}$ Glucose

Sucrose : $\alpha-$ D-Glucose $+\beta-$ D-Fructose

Maltose : $\alpha$ - D-Glucose

Lactose : $\beta$ - D-Galactose $+\beta$ - D-Glucose

Hence, the answer is the option (4).

9. First Order Reaction

First-order reactions are important in chemical kinetics. Questions are mostly numerical and moderate in difficulty.

Question: For a first-order reaction, $\mathrm{A} \rightarrow \mathrm{P}, \mathrm{t}_{\frac{1}{2}}$ , (half-life) is 10 days. The time required for $\frac{1}{4}^{\text {th }}$conversion of A (in days) is :

(ln 2=0.693, ln 3=1.1)

(1) 4.1

(2) 3.2

(3) 5

(4) 2.5

Solution:

For the first-order reaction: -

$\begin{aligned} & K=\frac{0.693}{t_{\frac{1}{2}}}=\frac{0.693}{10} \\ & K=\frac{2.303}{t} \log \frac{R_o}{R_t} \\ & \frac{0.693}{10}=\frac{2.303}{t} \log \frac{R_o \times 4}{3 R_0} \\ & \frac{0.693}{10}=\frac{2.303}{t}[\log 4-\log 3]=\frac{2.303}{t}[0.6020-0.4771] \\ & \frac{0.693}{10}=\frac{2.303}{t} \times 0.1249 \\ & t=\frac{2.303 \times 0.1249 \times 10}{0.603}=4.15 \text { days }\end{aligned}$

Hence, the correct answer is option (1)

10. Oxidation State

Question: The amphoteric oxide among $\mathrm{V}_2 \mathrm{O}_3, \mathrm{~V}_2 \mathrm{O}_4$ and $\mathrm{V}_2 \mathrm{O}_5$ upon reaction with alkali leads to formation of an oxide anion. The oxidation state of V in the oxide anion is:

(1) +3

(2) +7

(3) +5

(4) +4

Solution:

$\mathrm{V}_2 \mathrm{O}_3$ - basic
$\mathrm{V}_2 \mathrm{O}_4$ - weakly acidic or amphoteric
$\mathrm{V}_2 \mathrm{O}_5$ - amphoteric

$
\mathrm{V}_2 \mathrm{O}_5+\text { alkali } \rightarrow \mathrm{VO}_4^{3-}
$
In $\mathrm{VO}_4^{3-}$ ion, vanadium is in a +5 oxidation state.

Hence, the correct answer is option (3).

Students can try to solve the JEE Main 2026 Sample Paper. for better practice and revision.

Chemistry Study Plan for JEE Main 2026

Chemistry is divided into three branches: Physical, Organic, and Inorganic Chemistry. And it is one of the most important and scoring subjects if you are preparing for JEE Main 2026. To study chemistry in a better way, students must study each branch separately, focusing on its core concepts, formulas, and reaction mechanisms.

• It is important to learn important reactions, formulas, and periodic table trends.

• Read the NCERT textbook and solve exercises

• Revise key topics frequently and solve full-length mock tests

• Solve previous year questions

For the complete preparation strategy, follow the JEE Main 2026 Study Plan