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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)
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.
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.
On Question asked by student community
Hello Suhana
If you want to prepare for the JEE Mains exam 2026, then you need to manage your study time and try to give very much attention to your studies. JEE Mains is an entrance exam which is conducted by NTA(National Test Agency), and only science stream students give this exam to get admission in top engineering colleges to get admission in B.Tech or B.E courses.
Students give this exam after completing or appearing in the 12th class. This entrance exam is conducted at the undergraduate level. After clearing the JEE Mains exam, you become eligible to give the JEE Advanced, which helps to get admission in top IIT & NIT colleges of our country.
Here are some tips which help you at the time of your preparation-
I hope this information he;lps you. You also do your preparation on our official website, careers360.
All the best for your exam.
Hello,
Yes, you are eligible to attempt JEE Main 2026, as eligibility is based on appearing for or passing the class 12 exam in 2024, 2025, or 2026. Since you passed Class 12 in 2024, you are within the eligibility window to appear in 2026. Here in this article you will find more about the JEE Main eligibility criteria.
I hope it will clear your query!!
Hello,
If you have already passed Class 12 earlier and you are only taking the improvement exam in 2026, then you must choose “Passed” in the JEE Main 2026 form .
JEE Main always considers the first year in which you passed Class 12 as your official passing year. Improvement does not change that. So your status is not “appeared” and not “appearing.” It stays “Passed.”
You should enter:
Pass Status: Passed
Year of Passing: Your original Class 12 pass year
Marks: Your original marks (you can update later only if NTA allows it)
If you are giving Class 12 for the first time in 2026, then you must choose “Appearing.”
This is the correct and safe option for your JEE Main 2026 registration .
Hope it helps !
Hi Hriday
If your father's name is different on your Aadhar card and 10th marksheet, you need to change your father's name in the Aadhar card and match it with the spelling of the 10th marksheet, because when you take admission in college through JEE Mains score. They match your name, your father's name, and your mother's name with your 10th class marksheet, which serves as a real identity for document verification.
What college typically accepts-
I hope you will understand my point. These are some major issues which many students face during the time of admission in college.
Thank you.
Hello,
Yes, you can apply for JEE Main even if you do not have a caste certificate right now.
You only need the caste certificate if you want to claim a reserved category (SC/ST/OBC/EWS).
If you do not have it at the time of filling the form, you can apply as
General
or
General-EWS/OBC-NCL without certificate
, and later you can upload the valid certificate during counselling or document verification.
So yes, you can fill the JEE Main form without a caste certificate.
Hope it helps !
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