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.
NTA has reiterated that it will not revise the JEE Main 2026 exam date or shift in case of a clash with any other entrance or school-level examination. The agency said the slot allotment is fully computer-based and randomised. Candidates have been asked to prepare according to the schedule published in the information bulletin.
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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
Hey there,
Your OBC-NCL certificate will not be valid for the JEE Mains 2026 application as it is from June 2024. You can apply for a new one as soon as possible. For the JEE Mains, the OBC-NCL certificate should be issued on or after 1 April 2025.
And as for your Aadhar, the mistakes in your name and Date of Birth will be an issue during the document verification process. The details should match your 10th-class certificate. So you will also need to update it as soon as possible.
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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