Most Repeated Questions in JEE Mains Chemistry 2026
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!
1. Concentration Terms
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).
2. Mole Concept and Molar Mass
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).
3. Magnetic Moment (On the Basis of VBT)
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).
4. Stoichiometry, Stoichiometric Calculations and Limiting Reagent
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).
5. Applications of CFT (Crystal Field Theory)
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).
6. Reduction and Oxidation Reaction
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).
7. Stereoisomerism
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 |
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Thus, the total number of Isomers is 12.
Hence, the answer is an option (2).
8. Disaccharides and Polysaccharides
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).
9. First Order Reaction
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)
10. Oxidation State
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.
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