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JEE Main Chemistry Formulas 2026 - The Joint Entrance Exam is one of the most competitive exam of engineering entrance exams. It is divided into three subsections (Physics, Chemistry, Maths). Chemistry is an important subject for JEE Main preparation. Hence, having a list of important formulas for JEE Main Chemistry 2025 is necessary. Every aspirant should have the short notes for JEE Mains that comprise the important formulas. These JEE important formulas for Chemistry help in solving the typical problems easily. The registration process for JEE Main 2026 is now open, and students can apply from 31 October 2025 to 27 November 2025. Session 1 of the exam will be conducted from 21 to 30 January 2026.
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JEE Mains Chemistry all Formulas PDF Download
There will be a total of 25 questions from Chemistry in JEE Main 2026, all of which have to be attempted. It is further segregated into Physical, Organic, and Inorganic sections. The maximum number of topics has been reduced from this section (chemistry) only. This makes Chemistry easier than before. However, the high-weightage topics for JEE Main remain the same.
Below are important formulas for JEE Main 2026 Chemistry that are helpful for both sessions. Candidates should solve as many Chemistry questions as they can and implement these formulas to remember them quickly. To crack the JEE Main exam with good marks, paste these formulas near the study table and memorise them.
Candidates while studying chemistry, they need to revise and practice the chemical equations and symbols. To some, chemistry is a tough subject, but when candidates practice chemical equations, revise the properties, formulas and symbols, they will have command over the subject. Candidates can check the JEE Main Chemistry formulas below
Some Basic Concepts in chemistry
Atomic Structure
Line Spectrum of Hydrogen-like atoms
$\frac{1}{\lambda}=R Z^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$
Bohr radius of nth orbit:
$\mathrm{r}_{\mathrm{n}}=0.529 \frac{\mathrm{n}^2}{\mathrm{Z}} \mathrm{~A}^0$
Velocity of electron in nth orbit:
$\mathrm{V}_{\mathrm{n}}=\left(2.18 \times 10^6\right) \frac{\mathrm{Z}}{\mathrm{n}} \mathrm{~m} / \mathrm{s}$
where Z is atomic number
Total energy of electron in nth orbit:
$\mathrm{E}_{\mathrm{n}}=-13.6 \frac{\mathrm{Z}^2}{\mathrm{n}^2} \mathrm{eV}=-2.18 \times 10^{-18} \frac{\mathrm{Z}^2}{\mathrm{n}^2} \mathrm{~J}$
where Z is atomic number
Hisenberg Uncertainity Principle: $\Delta x . \Delta P \geq \frac{h}{4 \pi}$
$\mathrm{E}_{\mathrm{n}}=-\frac{1312 \times \mathrm{Z}^2}{\mathrm{n}^2} \mathrm{~kJ} / \mathrm{mol}$
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Chemical Thermodynamics
Work done in a reversible isothermal process
$$
\begin{aligned}
& \mathrm{W}=-2.303 \mathrm{nRT} \log _{10} \frac{\mathrm{~V}_2}{\mathrm{~V}_1} \\
& \mathrm{~W}=-2.303 \mathrm{nRT} \log _{10} \frac{\mathrm{P}_1}{\mathrm{P}_2}
\end{aligned}
$$
Work done in an irreversible isothermal process
Work $=-\mathrm{P}_{\text {ext. }}\left(\mathrm{V}_2-\mathrm{V}_1\right)$
That is, Work $=-\mathrm{P} \times \Delta \mathrm{V}$
$W=\Delta E=n C_V \Delta T$
Enthalpy: $H=U+p V$
First Law of Thermodynamics: $\Delta U=q+W$
$\Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{T} \Delta(\mathrm{S})$
$\Delta G=-n F E$
Equilibrium
$\mathrm{mA}+\mathrm{nB} \rightleftharpoons \mathrm{pC}+\mathrm{qD}$ $\frac{\mathrm{K}_{\mathrm{f}}}{\mathrm{K}_{\mathrm{b}}}=\frac{[\mathrm{C}]^{\mathrm{p}}[\mathrm{D}]^{\mathrm{q}}}{[\mathrm{A}]^{\mathrm{m}}[\mathrm{B}]^{\mathrm{n}}}=\mathrm{K}_{\mathrm{c}}$
$\mathrm{pH}=-\log _{10}\left[\mathrm{H}^{+}\right]$
$\mathrm{k}_{\mathrm{w}}=\left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right]=10^{-14}$
$\mathrm{pH}=\mathrm{pK}_{\mathrm{a}}+\log _{10} \frac{[\text { Salt }]}{\text { Acid }}$
$\mathrm{pOH}=\mathrm{pK}_{\mathrm{b}}+\log _{10} \frac{[\text { Salt }]}{[\text { Base }]}$
ELECTROCHEMISTRY
$\begin{aligned}
& \frac{E_1}{E_2}=\frac{M_1}{M_2} \text { or } \frac{W_1}{W_2}=\frac{Z_1}{Z_2} \\
& E_1=\text { equivalent weight } \\
& E_2=\text { equivalent weight }
\end{aligned}$
W or M = mass deposited
$\begin{aligned} & E_{\text {cell }} \text { or } E M F=\left[E_{\text {red }}(\text { cathode })-E_{\text {red }}(\text { anode })\right] \\ & E_{\text {eell }}^{\circ} \text { or } E M F^{\circ} \\ & =\left[E_{\text {red }}^{\circ}(\text { cathode })-E_{\text {red }}^{\circ}(\text { anode })\right]\end{aligned}$
$\mathrm{E}=\mathrm{E}^{\circ}-\frac{\mathrm{RT}}{\mathrm{nF}} \ln Q$
$\mathrm{xA}+\mathrm{yB} \xrightarrow{\mathrm{ne}^{-}} \mathrm{mC}+\mathrm{nD}$
The emf can be calculated as
$\text { Ecell }=\mathrm{E}^{\circ} \text { cell }-\frac{0.059}{\mathrm{n}} \log \frac{[\mathrm{C}]^{\mathrm{m}}[\mathrm{D}]^{\mathrm{n}}}{[\mathrm{~A}]^x[\mathrm{~B}]^{\mathrm{y}}}$
$\wedge_{\mathrm{m}}=\kappa \times \frac{1000}{\mathrm{c}}$
$\wedge_{\text {eq }}=\frac{1000 \times \kappa}{\mathrm{N}}$
Solutions
Chemical kinetics
Unit of Rate Constant-
$\begin{aligned}
& \text { The differential rate expression for } \mathrm{n}^{\text {th }} \text { order reaction is as follows: } \\
& \qquad-\frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{k}(\mathrm{a}-\mathrm{x})^{\mathrm{n}} \\
& \text { or } \quad \mathrm{k}=\frac{\mathrm{dx}}{(\mathrm{a}-\mathrm{x})^{\mathrm{n}} \mathrm{dt}}=\frac{(\text { concentration })}{(\text { concentration })^{\mathrm{n}} \text { time }}=(\text { conc. })^{1-\mathrm{n}} \text { time }^{-1}
\end{aligned}$
For the first-order reaction,
$k=\frac{2.303}{t} \log \frac{[\mathrm{R}]_0}{[\mathrm{R}]}$
$t_{1 / 2}=\frac{0.693}{k}$
For any general nth order reaction it is evident that,
$\mathrm{t}_{\frac{1}{2}} \propto[\mathrm{~A}]_0^{1-\mathrm{n}}$
It is to be noted that the above formula is applicable for any general nth-order reaction except $\mathrm{n}=1$.
Arrhenius Equation: $\mathrm{k}=\mathrm{Ae}^{-\mathrm{Ea} / \mathrm{RT}}$
$\log \frac{\mathrm{K}_2}{\mathrm{~K}_1}=\frac{\mathrm{Ea}}{2.303 \mathrm{R}}\left[\frac{1}{\mathrm{~T}_1}-\frac{1}{\mathrm{~T}_2}\right]$
Coordination Compounds
$\mathrm{EAN}=Z-O+2 L$
Where:
$\mathbf{Z}=$ Atomic number of the central metal atom/ion
$\mathbf{O}=$ Oxidation state of the metal atom/ion
L = Number of ligands (or donor atoms) $\times$ number of electrons donated per ligand
Crystal Field Stabilization Energy (CFSE):
Octahedral:
$$
\mathrm{CFSE}=(-0.4 x+0.6 y) \Delta_0
$$
Tetrahedral:
$$
\mathrm{CFSE}=(-0.6 x+0.4 y) \Delta_t
$$
where $x=t_2 g$ electrons, $y=$ e_g electrons
d- & f-Block Elements
Magnetic Moment:
$\mu=\sqrt{n(n+2)} \mathrm{BM}$
Chemical Bonding and Molecular Structure
Formal Charge:
$\text { F.C. }=V-N-\frac{B}{2}$
( $\mathrm{V}=$ valence electrons, $\mathrm{N}=$ non-bonding, $\mathrm{B}=$ bonding electrons)
Bond Order (Molecular Orbital Theory):
$\text { Bond Order }=\frac{\left(N_b-N_a\right)}{2}$
Dipole Moment:
$\mu=q \times d$
( $q=$ charge,$d=$ distance between charges)
Some Basic Principles of Organic Chemistry
Application of Inductive Effect
The decreasing -I effect or increasing +I effect order is as follows:
$\begin{aligned}
& -\mathrm{NH}_3+>-\mathrm{NO}_2>-\mathrm{SO}_2 \mathrm{R}>-\mathrm{CN}>-\mathrm{SO}_3 \mathrm{H}>-\mathrm{CHO}>-\mathrm{CO}>-\mathrm{COOH}>-\mathrm{F}>-\mathrm{COCl}>-\mathrm{CONH}_2>-\mathrm{Cl}>-\mathrm{Br}>-\mathrm{I}>-\mathrm{OR}>-\mathrm{OH}>-\mathrm{NR}_2>-\mathrm{NH}_2> \\
& -\mathrm{C}_6 \mathrm{H}_5>-\mathrm{CH}=\mathrm{CH}_2>-\mathrm{H}
\end{aligned}$
Degree of Unsaturation (DU or IHD):
$\mathrm{DU}=\frac{2 C+2-H+N-X}{2}$
( $\mathrm{C}=$ carbon, $\mathrm{H}=$ hydrogen, $\mathrm{N}=$ nitrogen, $\mathrm{X}=$ halogen)
Hydrocarbons
Carboxylic Acids and Derivatives
Method of Preparation of Carboxylic Acid

Given below are some tips to help you prepare for JEE Main and score good marks in the exam:
1. First, students need to understand the Syllabus and Exam Pattern so that they can refer to the JEE Main syllabus from the official website.
2. Try to identify the important and high-weightage topics and prepare according to that.
3. Create an effective study plan according to your preparation level. Divide your preparation into monthly, weekly, and daily targets and allocate more time to difficult subjects or topics.
4. Students must focus on conceptual clarity; they must understand the logic and derivations behind every formula.
5. Try to solve questions regularly. Solve previous years' JEE Main question papers and attempt mock tests and sample papers regularly.
Students find it difficult to learn formulas for JEE Main, but with the right approach, they can remember them. Given below are some points to remember:
1. Students must try to understand why a formula works and how chemical reactions occur, and their mechanism.
2. Then break down formulas into chapters or topics.
3. To learn these formulas easily, try to make a formula notebook.
4. Sometimes students must try to make Mnemonics and short tricks, as it helps in quick revision.
5. Try to solve as many questions and revise
6. Try to use diagrams and flowcharts.
Frequently Asked Questions (FAQs)
Chemistry has three subparts (Physical, inorganic and organic). Physical chemistry has many numericals based on different formulas.
Yes, Class 12 syllabus carries more than 50% of weightage, so it is important for JEE Main Exam.
JEE Main is an exam conducted for those aspirants who want to take admission in NIITs, IITs and other engineering colleges, and it is also an eligibility test for JEE advance.
On Question asked by student community
Hello there,
Understanding and solving different question papers is one of the best practices for preparation ,especially when it comes to JEE. It gives you a proper understanding of the exam pattern, important topics to cover, and the marking scheme.
Here is the link attached from the official website of Careers360 which will provide you with the question papers of JEE exam. Hope it helps!
https://engineering.careers360.com/articles/jee-main-question-papers
thank you!
Hello aspirant,
In order to prepare for the JEE Main 2026 exam, it is strongly advised that students complete the JEE Main 2025 Question Paper. This will help them comprehend the exam's difficulty level and question kinds. The JEE Main 2026 session 1 test will be administered by NTA from January 21 to 30, while the session 2 exam will take place from April 2 to 9.
To get the previous year question papers, you can visit our site through following link:
https://engineering.careers360.com/articles/jee-main-question-papers
Thank you
With a BC-E category certificate, EWS certificate, and Telangana home-state quota, your chances of getting admission to AIML or related CSE branches in CBIT, VNR VJIET, or similar top private colleges in Telangana depend mainly on your JEE Main percentile and the TS EAMCET counselling process, because these colleges fill seats through both TS EAMCET ranks and JEE Main scores (Category-B seats).
For Category-A seats (through TS EAMCET), JEE Main percentile is not considered. Admission is purely based on your TS EAMCET rank, reservation category, gender, and home-state status. CBIT and VNR are highly competitive, especially for AIML, CSE, and IT branches, so you generally need a very strong TS EAMCET rank (usually within a few thousand for BC-E candidates, sometimes slightly relaxed for girls).
For Category-B seats (JEE Main quota), JEE Main percentile matters. Based on recent trends, to have a realistic chance:
For CBIT (AIML / CSE / IT), a JEE Main percentile of around 95+ is usually expected, even for reserved categories, because Category-B seats are limited and competition is high.
For VNR VJIET (AIML / CSE / IT), candidates with 92–95 percentile sometimes have chances, depending on availability and the specific branch.
If your percentile is below 90, getting AIML or core CSE in CBIT or VNR becomes difficult, but you may still have chances in slightly lower-demand branches (like ECE, EEE, or emerging specialisations) or in other good private colleges affiliated with JNTU or autonomous institutions.
Your BC-E and EWS certificates do help mainly in TS EAMCET counselling (Category-A seats), not significantly in Category-B (JEE Main) admissions, where merit largely dominates. Home-state status is already assumed for these colleges, so there is no extra advantage beyond eligibility.
In summary, if your JEE Main percentile is 95 or above, you can reasonably target AIML/CSE in VNR and possibly CBIT through Category-B. If it is 90–94, VNR or related branches are more realistic than CBIT. Below that, it is better to focus on TS EAMCET performance or consider other reputed Telangana colleges offering AIML. If you want, you can share your exact JEE Main percentile and TS EAMCET rank, and I can give you a more precise college- and branch-wise estimate.
Hello,
JEE Mains is a national level entrance examination conducted for admission to UG engineering course in IITs, NITs, etc. This also is a qualifying test for JEE Advanced.
Practicing with previous year question paper will give you on overview of the entire exam pattern, marking scheme, types of questions asked, etc.
Check out the official website of Careers360 for the question papers, preparation tips, etc.
https://engineering.careers360.com/articles/jee-main-question-papers
Careers360 is also conducting a free JEE Mains mock test which you can attempt. The last date for registration is 8th January, 2026.
https://learn.careers360.com/test-series-jee-main-2026-free-mock-test/
Thank you.
Hello,
The link to the mock test series is attached here. The mock test of Careers360 is completely free, and the structure and questions were prepared keeping in mind the exam of JEE Mains. The registration is ongoing. the last date of registration on 8th January.
https://learn.careers360.com/test-series-jee-main-free-mock-test/
Thank you.
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