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Top 10 Most Repeated Topics In Chemistry For JEE Mains 2026 - High Scoring Areas to Crack IIT JEE
Top 10 Most Repeated Topics In Chemistry For JEE Mains - Chemistry is often seen as the most scoring section in JEE Main, as it requires less time to solve compared to Physics and Mathematics. With the right preparation, students can secure strong marks by focusing on the areas that carry the highest weightage. Over the years, certain chapters in Chemistry have been repeated more frequently in JEE Main, making them important for every aspirant to prioritize during their revision.
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This Story also Contains
- Most Repeated Chemistry Topics For JEE Mains 2026
- Top 10 Most Important Chapters For JEE Mains Chemistry 2026
- Most Repeated Questions in JEE Mains Chemistry 2026
- Chemistry Study Plan for JEE Mains 2026
The JEE Mains chemistry is known for being the most scoring subject so far. Hence, in this article we will help you prepare for chemistry by looking at the most repeated chemistry topics for JEE mains. Not only that, we will also be delving into top 10 most repeated topics in chemistry for JEE Mains to help you get exam ready. First, let's start with understanding the exam pattern and getting familiar with the details of JEE Mains 2026.
Most Repeated Chemistry Topics For JEE Mains 2026
Preparing for any exam requires research and specially for a competitive entrance exam like JEE Mains requires thorough preparation. So, our experts have collected and done the thorough research from JEE Main Latest Syllabus 2026. In the below table we have provided you with the top 10 most repeated topics in chemistry for JEE Mains 2026. This data comprises last 10 years JEE mains question papers of all slots.
This table contains the most repeated chemistry topics for JEE Mains. For more details we have also provided you with a count of the total numbers of questions from these topics along with the chapters that they are from.
|
Chapter |
Topic |
Number of Questions |
|
Some Basic Concepts of Chemistry |
Concentration Terms |
48 |
|
Some Basic Concepts of Chemistry |
Mole Concept and Molar Mass |
48 |
|
Some Basic Concepts of Chemistry |
Stoichiometry, Stoichiometric Calculations and Limiting Reagent |
39 |
|
Organic Compounds containing Oxygen |
Reduction and Oxidation Reaction |
37 |
|
Organic Chemistry – General Organic Chemistry + Hydrocarbons |
Stereoisomerism |
34 |
|
Biomolecules |
Disaccharides and Polysaccharides |
33 |
|
Chemical Kinetics |
33 | |
|
Redox Reactions/d-Block Elements |
Oxidation State |
32 |
As seen, the topic Concentration Terms happen 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 high weighted topics which could be a game changer for the JEE mains chemistry exam.
Now, after seeing the topics, let’s dive into the most important chapters for JEE mains chemistry 2026. Understanding this aspect will give you an upper hand in the preparation for the exam!
Top 10 Most Important Chapters For JEE Mains Chemistry 2026
We have already seen the most repeated chemistry topics for jee mains 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. Most important chapters for JEE mains 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. These following listed chapters constitute the most important chapters for chemistry with organic compounds containing oxygen being the highest weighted chapter. This list has most important chapters ranked in order of most to least important.
|
Chapter |
No. of Questions (Last 10 Years) |
|
Organic Compounds containing Oxygen |
259 |
|
216 | |
|
Co-ordination Compounds |
185 |
|
188 | |
|
Redox Reaction and Electrochemistry |
173 |
|
Organic Compounds Containing Nitrogen |
172 |
|
Chemical Thermodynamics |
166 |
|
Atomic Structure |
155 |
|
Chemical Bonding and Molecular Structure |
151 |
|
149 | |
|
d - and f - BLOCK ELEMENTS |
148 |
|
Some basic concepts in chemistry |
142 |
|
136 | |
|
Some Basic Principles of Organic Chemistry |
136 |
|
120 |
Refer to JEE Main Chapter-Wise Weightage for for a better understanding of important topics and helps you to prepare accordingly.
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Most Repeated Questions in JEE Mains Chemistry 2026
In this section, we will be diving into each topic and looking at the set of 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 JEE Mains 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:
| Compound | Number of Isomers |
| $\left[\mathrm{Pt}(\mathrm{Cl})\left(\mathrm{NO}_2\right)\left(\mathrm{NO}_3\right)(\mathrm{SCN})\right]^{2-}$ | 3 |
| $\left[\mathrm{Pt}(\mathrm{Cl})(\mathrm{ONO})\left(\mathrm{NO}_3\right)(\mathrm{SCN})\right]^{2-}$ | 3 |
| $\left[\mathrm{Pt}(\mathrm{Cl})(\mathrm{ONO})\left(\mathrm{NO}_3\right)(\mathrm{NCS})\right]^{2-}$ | 3 |
| $\left[\mathrm{Pt}(\mathrm{Cl})\left(\mathrm{NO}_2\right)\left(\mathrm{NO}_3\right)(\mathrm{NCS})\right]^{2-}$ | 3 |
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Download EBookThus, 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 follow JEE Main- Top 30 Most Repeated Questions & Topics for better practice and revision. Along with that try to solve JEE Main 2026 Sample Paper.
Chemistry Study Plan for JEE Mains 2026
Chemistry is divided into three branches Physical, Organic, and Inorganic Chemistry. And it is one of the most important and scoring subject if you are preparing for JEE Mains 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 complete preparation strategy follow JEE Main 2026 complete preparation strategy
JEE Main Syllabus: Subjects & Chapters
Select your preferred subject to view the chapters
Frequently Asked Questions (FAQs)
Q: How should I prioritize Chemistry chapters for JEE Mains 2026?
A:
Start with high-weightage and repeated topics like Mole Concept, Thermodynamics, Coordination Compounds, Redox Reactions, and Hydrocarbons. Once strong in these, cover remaining chapters systematically to ensure no topic is left uncovered.
Q: Which Chemistry is easiest to score?
A:
Inorganic chemistry has the easiest syllabus and is the highest to score. It is advisable to not leave any topic of this part of chemistry if you’re planning to bag a good rank in the exam.
Q: Is NCERT enough for the preparation of JEE Mains 2026?
A:
Actually, no! JEE Mains is a national competitive exam and students will refer to a lot of other books as well for the in depth preparation of chemistry. So you must refer to 2-3 books outside NCERT as well!
Q: Which are the most repeated topics for JEE Mains so far?
A:
The most repeated topics include Concentration Terms (Solutions), Mole Concept, Redox Reactions, Stoichiometry, Coordination Compounds, Chemical Kinetics, Chemical Bonding, and Oxidation States. These topics consistently appear in JEE Mains over the years.
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Hello,
To prepare for the JEE paper 2 or the Architecture exam, you need to understand the exam pattern and syllabus clearly. Then strengthen the fundamental concepts with daily revision. After that, take a mock test and practice with the PYQ to get the exam-like experience.
I hope it will clear your query!!
JEE Main exam is a national-level entrance test for admission into top engineering colleges like NITs, IIITs, and GFTIs. It mainly tests your understanding of Physics, Chemistry, and Mathematics. To prepare well, focus on NCERT books first, then refer to standard JEE preparation books for deeper concepts and practice. Regular mock tests and solving previous year papers also help in improving speed and accuracy. I’ll be attaching some useful JEE Main preparation links from Careers360 to help you get started.
https://engineering.careers360.com/articles/best-books-for-jee-main
https://engineering.careers360.com/articles/best-study-material-for-jee-main
https://learn.careers360.com/engineering/jee-main-preparation-material/
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Generally an income certificate isn't required for the JEE Main registration, but if you want to claim the EWS quota, then you need this. You must provide the certificate, issued by a government authority, as proof of your family's income being below the specified limit for the reservation category you wish to apply under.
I hope it will clear your query!!
Yes, as JEE does accepts improvement examination scores, so you must go for it but most of the state boards have already conducted or are conducting their 2025 improvement exams. If you have already given your improvement that's fine. If you have not given improvement this year then you can take your improvement next year.
Thank You.
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Yes, you can prepare for and take the JEE Main exam after completing your intermediate (12th year) exams. This is a common path for students who want to dedicate a year to intensive preparation without the pressure of simultaneous board exams.
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