JEE Main Chemistry Formulas 2026 - Topic wise Important Formulas

JEE Main 2026 Chemistry Formula Overview

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.

JEE Main Chemistry Important Formulas

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

JEE Main Important Formulas of Physical Chemistry

Some Basic Concepts in chemistry

1. Boyle's Law: $P_1{V}_1=P_2{V}_2$ (at constant T and n )
2. Charles's Law: $\frac{V_1}{T_1}=\frac{V_2}{T_2}($ at constant P and n$)$
3. Avogadro's Law: $\frac{V}{n}=$ constant
4. Average Atomic Mass $=\frac{\mathrm{\Sigma }(\text{Mass of Isotopes}{)}_i\times (\mathrm{\%}\text{abundance}{)}_i}{100}$
5. Mole $=\frac{\mathrm{W}}{\mathrm{M}}=\frac{(\mathrm{Wt}.\text{of substance in gm.})}{(\text{Molar mass of substance}(\mathrm{G}.\mathrm{m}.\mathrm{m}))}$
6. Mass $\mathrm{\%}$ of an element $=\frac{\text{Mass of that element in one mole of the compound}}{\text{Molar mass of the compound}}\times 100$
7. Equivalent Weight $=\frac{\text{Molecular weight}}{\mathrm{n}-\text{factor}(\mathrm{x})}$

Atomic Structure

1. Frequency $\nu =\frac{1}{\mathrm{T}}$
2. Wave number $(\overline{\nu })$ $\overline{\nu }=\frac{1}{\lambda }$
3. $E=h\nu =\frac{hc}{\lambda }$

4. Line Spectrum of Hydrogen-like atoms

   $\frac{1}{\lambda }=R{Z}^2(\frac{1}{n_1^2}-\frac{1}{n_2^2})$

5. Bohr radius of nth orbit:

   ${\mathrm{r}}_{\mathrm{n}}=0.529\frac{{\mathrm{n}}^2}{\mathrm{Z}}{\mathrm{A}}^0$

6. Velocity of electron in nth orbit:

   ${\mathrm{V}}_{\mathrm{n}}=(2.18\times {10}^6)\frac{\mathrm{Z}}{\mathrm{n}}\mathrm{m}/\mathrm{s}$

   where Z is atomic number

7. 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

8. Hisenberg Uncertainity Principle: $\mathrm{\Delta }x.\mathrm{\Delta }P\ge \frac{h}{4\pi }$

9. ${\mathrm{E}}_{\mathrm{n}}=-\frac{1312\times {\mathrm{Z}}^2}{{\mathrm{n}}^2}\mathrm{kJ}/\mathrm{mol}$

Chemical Thermodynamics

1. Expansion Work $=\mathrm{P}\times \mathrm{\Delta }\mathrm{V}=-{\mathrm{P}}_{\text{ext.}}[{\mathrm{V}}_2-{\mathrm{V}}_1]$
   $\mathrm{P}=$ external pressure And $\mathrm{\Delta }\mathrm{V}=$ increase or decrease in volume.

2. Work done in a reversible isothermal process

   $$\begin{array}{r}\mathrm{W}=-2.303\mathrm{nRT}{\log }_{10}\frac{{\mathrm{V}}_2}{{\mathrm{V}}_1}\\ \mathrm{W}=-2.303\mathrm{nRT}{\log }_{10}\frac{{\mathrm{P}}_1}{{\mathrm{P}}_2}\end{array}$$

3. Work done in an irreversible isothermal process
   Work $=-{\mathrm{P}}_{\text{ext.}}({\mathrm{V}}_2-{\mathrm{V}}_1)$
   That is, Work $=-\mathrm{P}\times \mathrm{\Delta }\mathrm{V}$

4. $W=\mathrm{\Delta }E=n{C}_V\mathrm{\Delta }T$

5. Enthalpy: $H=U+pV$

6. First Law of Thermodynamics: $\mathrm{\Delta }U=q+W$

7. $\mathrm{\Delta }\mathrm{G}=\mathrm{\Delta }\mathrm{H}-\mathrm{T}\mathrm{\Delta }(\mathrm{S})$

8. $\mathrm{\Delta }G=-nFE$

Equilibrium

1. For a reaction:

   $\mathrm{mA}+\mathrm{nB}\rightleftharpoons \mathrm{pC}+\mathrm{qD}$ $\frac{{\mathrm{K}}_{\mathrm{f}}}{{\mathrm{K}}_{\mathrm{b}}}=\frac{[\mathrm{C}{]}^{\mathrm{p}}[\mathrm{D}{]}^{\mathrm{q}}}{[\mathrm{A}{]}^{\mathrm{m}}[\mathrm{B}{]}^{\mathrm{n}}}={\mathrm{K}}_{\mathrm{c}}$

2. $\mathrm{pH}=-{\log }_{10}\left[{\mathrm{H}}^{+}\right]$

3. ${\mathrm{k}}_{\mathrm{w}}=\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{OH}}^{-}\right]={10}^{-14}$

4. $\mathrm{pH}={\mathrm{pK}}_{\mathrm{a}}+{\log }_{10}\frac{[\text{Salt}]}{\text{Acid}}$

5. $\mathrm{pOH}={\mathrm{pK}}_{\mathrm{b}}+{\log }_{10}\frac{[\text{Salt}]}{[\text{Base}]}$

ELECTROCHEMISTRY

1. $\mathrm{W}=\frac{\text{Eit}}{96500}$

2. $\begin{array}{r}\frac{E_1}{E_2}=\frac{M_1}{M_2}\text{or}\frac{W_1}{W_2}=\frac{Z_1}{Z_2}\\ E_1=\text{equivalent weight}\\ E_2=\text{equivalent weight}\end{array}$
   W or M = mass deposited

3. 
   

   $\begin{array}{r}{E}_{\text{cell}}\text{or}EMF=[E_{\text{red}}(\text{cathode})-E_{\text{red}}(\text{anode})]\\ E_{\text{eell}}^{\circ }\text{or}EM{F}^{\circ }\\ =[E_{\text{red}}^{\circ }(\text{cathode})-E_{\text{red}}^{\circ }(\text{anode})]\end{array}$

4. 
   

   $\mathrm{E}={\mathrm{E}}^{\circ }-\frac{\mathrm{RT}}{\mathrm{nF}}\ln Q$

5. 
   

   $\mathrm{xA}+\mathrm{yB}\stackrel{{\mathrm{ne}}^{-}}{\to }\mathrm{mC}+\mathrm{nD}$
   The emf can be calculated as

   $\text{Ecell}={\mathrm{E}}^{\circ }\text{cell}-\frac{0.059}{\mathrm{n}}\log \frac{[\mathrm{C}{]}^{\mathrm{m}}[\mathrm{D}{]}^{\mathrm{n}}}{[\mathrm{A}{]}^x[\mathrm{B}{]}^{\mathrm{y}}}$

6. 
   

   ${\wedge }_{\mathrm{m}}=\kappa \times \frac{1000}{\mathrm{c}}$

7. ${\wedge }_{\text{eq}}=\frac{1000\times \kappa }{\mathrm{N}}$

Solutions

1. Mass $\mathrm{\%}$ of a component $=\frac{\text{Mass of the component in the solution}}{\text{Total mass of the solution}}\times 100$
2. Volume $\mathrm{\%}$ of a component $=\frac{\text{Volume of the component}}{\text{Total volume of solution}}\times 100$
3. Mass by Volume $\mathrm{\%}$ of a component $=\frac{\text{Mass of the component}}{\text{Total volume of solution}}\times 100$
4. Parts per million $=\frac{\text{Number of parts of the component}}{\text{Total number of parts of all components of the solution}}\times {10}^6$
5. Mole fraction of a component $=\frac{\text{Number of moles of the component}}{\text{Total number of moles of all the components}}$
6. Molarity: $(M)=\frac{\text{No. of Moles of Solutes}}{\text{Volume of Solution in Liters}}$
7. Molality: $(m)=\frac{\text{No. of Moles of Solutes}}{\text{Mass of solvent in}\mathrm{kg}}$
8. $\left(P_T\right)=P_A^o{X}_A+P_B^o{X}_B$ ($\begin{array}{r}{P}_A=P_A^o{X}_A\\ P_B=P_B^o{X}_B\\ P_T=P_A+P_B\end{array}$)
9. $\mathrm{\Delta }{\mathrm{T}}_{\mathrm{b}}={\mathrm{K}}_{\mathrm{b}}\times \frac{\mathrm{w}}{\mathrm{M}}\times \frac{1000}{\mathrm{W}}$
10. $\mathrm{\Delta }{T}_f=K_f\times \frac{w}{M}\times \frac{1000}{W}$
11. $\mathrm{\Pi }=CRT$
12. $\begin{array}{r}\mathrm{i}=\frac{\text{Observed number of solute particles}}{\text{Number of particles initially taken}}\\ \mathrm{i}=\frac{\text{Observed value of colligative property}}{\text{Theoretical value of colligative property}}\end{array}$

Chemical kinetics

1. Unit of average velocity $=\frac{\text{Unit of concentration}}{\text{Unit of time}}=\frac{\text{mole}}{\text{litre second}}=$ mole litre ${-1}^{}$ second ${-1}^{}$
2. $\mathrm{aA}+\mathrm{bB}\to \mathrm{cC}+\mathrm{dD}$
   Rate w.r.t. $[\mathrm{A}]=-\frac{\mathrm{d}[\mathrm{A}]}{\mathrm{dt}}\times \frac{1}{\mathrm{a}}$
   Rate w.r.t. $[B]=-\frac{d[B]}{dt}\times \frac{1}{b}$
   Rate w.r.t. $[\mathrm{C}]=-\frac{\mathrm{d}[\mathrm{C}]}{\mathrm{dt}}\times \frac{1}{\mathrm{c}}$
   Rate w.r.t. $[\mathrm{D}]=-\frac{\mathrm{d}[\mathrm{D}]}{\mathrm{dt}}\times \frac{1}{\mathrm{d}}$
3. $\mathrm{R}\propto [\mathrm{A}{]}^{\mathrm{p}}[\mathrm{B}{]}^{\mathrm{q}}$

4. Unit of Rate Constant-

   $\begin{array}{r}\text{The differential rate expression for}{\mathrm{n}}^{\text{th}}\text{order reaction is as follows:}\\ -\frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{k}(\mathrm{a}-\mathrm{x}{)}^{\mathrm{n}}\\ \text{or}\mathrm{k}=\frac{\mathrm{dx}}{(\mathrm{a}-\mathrm{x}{)}^{\mathrm{n}}\mathrm{dt}}=\frac{(\text{concentration})}{(\text{concentration}{)}^{\mathrm{n}}\text{time}}=(\text{conc.}{)}^{1-\mathrm{n}}{\text{time}}^{-1}\end{array}$

5. For the first-order reaction,

   $k=\frac{2.303}{t}\log \frac{[\mathrm{R}{]}_0}{[\mathrm{R}]}$

6. $t_{1/2}=\frac{0.693}{k}$

7. 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 $\mathrm{n}=1$.

8. Arrhenius Equation: $\mathrm{k}={\mathrm{Ae}}^{-\mathrm{Ea}/\mathrm{RT}}$

9. $\log \frac{{\mathrm{K}}_2}{{\mathrm{K}}_1}=\frac{\mathrm{Ea}}{2.303\mathrm{R}}[\frac{1}{{\mathrm{T}}_1}-\frac{1}{{\mathrm{T}}_2}]$

JEE Main Important Formulas of Inorganic Chemistry

Coordination Compounds

1. $\mathrm{EAN}=Z-O+2L$
   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

2. 
   

   Crystal Field Stabilization Energy (CFSE):
   Octahedral:

   $$\mathrm{CFSE}=(-0.4x+0.6y){\mathrm{\Delta }}_0$$

   Tetrahedral:

   $$\mathrm{CFSE}=(-0.6x+0.4y){\mathrm{\Delta }}_t$$

   where $x=t_{2}g$ electrons, $y=$ e_g electrons

d- & f-Block Elements

1. Magnetic Moment:

   $\mu =\sqrt{n(n+2)}\mathrm{BM}$

Chemical Bonding and Molecular Structure

1. Formal Charge:

   $\text{F.C.}=V-N-\frac{B}{2}$

   ( $\mathrm{V}=$ valence electrons, $\mathrm{N}=$ non-bonding, $\mathrm{B}=$ bonding electrons)

2. Bond Order (Molecular Orbital Theory):

   $\text{Bond Order}=\frac{(N_b-N_a)}{2}$

3. Dipole Moment:

   $\mu =q\times d$

   ( $q=$ charge,$d=$ distance between charges)

JEE Main Important Formulas: Inorganic Chemistry

Some Basic Principles of Organic Chemistry

1. Application of Inductive Effect

   The decreasing -I effect or increasing +I effect order is as follows:

   $\begin{array}{r}-{\mathrm{NH}}_3+>-{\mathrm{NO}}_2>-{\mathrm{SO}}_2\mathrm{R}>-\mathrm{CN}>-{\mathrm{SO}}_3\mathrm{H}>-\mathrm{CHO}>-\mathrm{CO}>-\mathrm{COOH}>-\mathrm{F}>-\mathrm{COCl}>-{\mathrm{CONH}}_2>-\mathrm{Cl}>-\mathrm{Br}>-\mathrm{I}>-\mathrm{OR}>-\mathrm{OH}>-{\mathrm{NR}}_2>-{\mathrm{NH}}_2>\\ -{\mathrm{C}}_6{\mathrm{H}}_5>-\mathrm{CH}={\mathrm{CH}}_2>-\mathrm{H}\end{array}$

2. Degree of Unsaturation (DU or IHD):

   $\mathrm{DU}=\frac{2C+2-H+N-X}{2}$

   ( $\mathrm{C}=$ carbon, $\mathrm{H}=$ hydrogen, $\mathrm{N}=$ nitrogen, $\mathrm{X}=$ halogen)

Hydrocarbons

1. Alkanes: $C_n{H}_{2n+2}$
2. Alkenes: $C_n{H}_{2n}$
3. Alkynes: $C_n{H}_{2n-2}$

Carboxylic Acids and Derivatives

Method of Preparation of Carboxylic Acid

Preparation tips for JEE Mains

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.

How to Memorize Chemistry Formulas Effectively

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.