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. JEE Main 2026 registration has already started and students can register from 31 October 2025 to 27 November 2025, session 1 is scheduled from 21 to 30 January 2026.
The NTA has clarified that the JEE Mains 2026 exam dates will not be changed even if they clash with any other national and state-level exams. "The NTA will not change the allotted date or shift of the examination for the candidates," it said.
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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.
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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!
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.
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!
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).
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).
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).
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).
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).
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).
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 |
Thus, the total number of Isomers is 12.
Hence, the answer is an option (2).
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).
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)
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 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
Frequently Asked Questions (FAQs)
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.
Actually, no! JEE Mains is a national competitive exam and students should 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!
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.
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.
On Question asked by student community
Hello,
Your OBC-NCL certificate issued in June 2024 will not be accepted for JEE Main or JEE Advanced reservation. For these exams, the OBC-NCL certificate must be issued on or after 1 April 2025 . So yes, there will be an issue if you use the June 2024 certificate.
What you should do:
Apply for a new OBC-NCL certificate with an issue date of 1 April 2025 or later.
At the time of form filling, if the portal asks for the certificate and you do not have the new one yet, you can upload the self-declaration format (the exam authorities usually allow this). Later, during counselling or document verification, you must show the updated certificate.
Since your Aadhaar has mistakes in name and DOB, make sure it gets updated before the exam application and counselling. Wrong Aadhaar details can cause mismatch issues.
So, get your Aadhaar corrected and plan to get a fresh OBC-NCL certificate after 1 April 2025. This will prevent any problem during verification.
Hope it helps !
Hello Satyam,
Yes, your EWS certificate is valid.
In Bihar, the EWS certificate can be issued by the Circle Officer, BDO, SDO, or DM. So a certificate signed by the Circle Officer (Revenue Officer) is acceptable.
If the format of your certificate is the same as the Central Government EWS format, then it is valid for JEE Main registration and also for JoSAA counselling. The heading “Government of Bihar” does not create any problem.
Just remember one point:
For
JoSAA 2026
, you will need an EWS certificate issued on or after 1 April 2025. Even if your current certificate works for registration, you must update it before counselling.
Hope it helps !
Hello murali
No, your son is not eligible for OBC NCL for IIT JEE because you fall in the "creamy layer" occupational category, regardless of your current employment status or family income. Students whose family income is less than Rs. 8 lakhs annually and they are not belong to the "creamy layer".
Note -
Thank You
Hello,
You can fill the JEE Main form even if you are a private candidate
Write the
name of the school/board from where you are appearing as a private candidate
.
If your Class 12 admit card or registration slip shows a school/centre name, use that exactly.
If your board lists you as a “Private Candidate” under the board name , then write:
CBSE – Private Candidate
(or your board name – Private Candidate)
Use the pin code of the examination centre/school mentioned on your Class 12 private candidate admit card or registration details.
If your board does not give any school address and only shows the regional office address, then use the regional office address pin code given by your board.
Hope it helps !
Hello Aspirant
You should not leave the OBC-NCL certificate ID blank in the JEE Main form it can create problems later.
NTA wants the certificate details while filling the form, not just at counselling. If you can, apply for the OBC-NCL certificate immediately so you get the ID on time.
If you fail to submit the certificate during counselling, your category will shift to General. It’s safer to enter OBC-NCL only if you’re sure you’ll get the certificate before counselling.
Hope it will help you
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