जेईई मेन्स केमिस्ट्री 2026 में सबसे अधिक बार पूछे जाने वाले प्रश्न (Most Repeated Questions in JEE Mains Chemistry 2026)
इस अनुभाग में, हम प्रत्येक विषय पर गहनता से विचार करेंगे तथा ऊपर सूचीबद्ध विशेष विषयों से सबसे महत्वपूर्ण प्रश्नों के समूह पर अलग-अलग चर्चा करेंगे। इन प्रश्नों के स्तर को समझने और उनका अभ्यास करने से आपको बेहतर स्कोर करने और जेईई मेन परीक्षा 2026 में बेहतर रैंक प्राप्त करने में मदद मिलेगी। आइए शुरू करते हैं!
1. कंसंट्रेशन टर्म
Question: Some amount of dichloromethane
(CH2Cl2
) is added to 671.141 mL of chloroform
(CHCl3)
to prepare
2.6×10−3M
solution of
CH2Cl2(DCM)
. The concentration of DCM is ppm (by mass).
Given : atomic mass :
C=12
H=1Cl=35.5
density of
CHCl3=1.49 g cm−3
Solution:
Molar mass
=12+2+71
= 85mmoles of DCM
=671.141×2.6×10−3
mass of solution
=1.49×671.141
PPM=671.141×2.6×10−3×85×10−31.49×671.141×106
=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 -
Molarity = Moles of solute Vol.of solution (L) Molarity = Moles of solute Vol.of solution (L)M=nV(l)=0.0060.1=0.06
Hence, the answer is the option (2).
2. मोल अवधारणा और मोलर द्रव्यमान
Question: 1 gram of a carbonate (
M2CO3
) on treatment with excess HCl produces 0.01186 mole of
CO2
.. The molar mass of
M2CO3
in
gmol−1
is :
(1) 118.6
(2) 11.86
(3) 1186
(4) 84.3
Solution:
Given,
Mass of carbonate (
M2CO3)
= 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,
M2CO3+2HCl→2MCl+2H2O+CO2
From the balanced equation
1M=0.01186⇒M=10.01186=84.3
Hence, the answer is an option (4).
3. चुंबकीय आघूर्ण (वीबीटी के आधार पर)
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
[FeCl4]2−,[Co(C2O4)3]3−
and
MnO42−
respectively are :
(1) 5.82, 0 and
0BM
(2)
5.92,4.90
and
0BM
(3)
4.90,0
and
1.73BM
(4)
4.90,0
and
2.83BM
Solution:
[FeCl4]2−Fe2+3 d6→4
unpaired electron
as Cl– in a weak field liquid.
μspin =248M=4.9BM
[Co(C2O4)3]3Co3+3 d6→
for
Co3+
with coodination no. 6
C2O42−
is strong field ligand & causes pairing & hence no. unpaired electron.
μspin =0
[MnO4]2−Mn+6
it has one unpaired electron.
μspin =3BM=1.73BM
Hence,the answer is the option(3).
4. स्टोइकोमेट्री, स्टोइकोमेट्रिक गणना और सीमांत अभिकर्मक
Question: A sample of
NaClO3
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
aA(g)+bB(g)→cC(g)+dD(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)
2NaClO3→Δ2NaCl+3O20.16 g
nNaCl2=nO23
nNaCl=0.1632×23=1200×23=1300
NaCl→AgCl
PoAC on Cl
1×nNaCl=1×nAgCl
1300=nAgCl
the weight of
AgCl=1300×[108+35.5]=1300×143.5=0.48g
Hence, the answer is an option (3).
5. सीएफटी (क्रिस्टल फील्ड थ्योरी) के अनुप्रयोग
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
(Δhyd H)
when dissolved in water?
(1)
Cr2+
(2)
Mn2+
(3)
Fe2+
(4)
Co2+
Solution:
Water act as a weak ligand.C F S E of metal ions depend on strength of ligand.
Since the CFSE of
Mn2+
is the least,
ΔHHyd
of it is also lowest.
Hence, the answer is the option (2).
6. अपचयन और ऑक्सीकरण अभिक्रिया
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)
Zn/H2O
(2)
Pt−C/H2
(3)
Pd−C/H2
(4)
Na/H2
Solution:
Out of the given reagents,
Na/H2
is not used as a reducing agent.
Hence, the answer is the option (4).
7. स्टीरियोआइसोमेरिज्म
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
[Pt(Cl)(NO2)(NO3)(SCN)]2−
is :
(1) 8
(2) 12
(3) 16
(4) 24
Solution:
As we have learnt,
NO2−
and
SCN−
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 |
[Pt(Cl)(NO2)(NO3)(SCN)]2− | 3 |
[Pt(Cl)(ONO)(NO3)(SCN)]2− | 3 |
[Pt(Cl)(ONO)(NO3)(NCS)]2− | 3 |
[Pt(Cl)(NO2)(NO3)(NCS)]2− | 3 |
Thus, the total number of Isomers is 12.
Hence, the answer is an option (2).
8. डाइसैकेराइड और पॉलीसैकेराइड
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:
α−D
Glucose
Sucrose :
α−
D-Glucose
+β−
D-Fructose
Maltose :
α
- D-Glucose
Lactose :
β
- D-Galactose
+β
- D-Glucose
Hence, the answer is the option (4).
9. प्रथम क्रम अभिक्रिया
First-order reactions are important in chemical kinetics. Questions are mostly numerical and moderate in difficulty.
Question: For a first-order reaction,
A→P,t12
, (half-life) is 10 days. The time required for
14th
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: -
K=0.693t12=0.69310K=2.303tlogRoRt0.69310=2.303tlogRo×43R00.69310=2.303t[log4−log3]=2.303t[0.6020−0.4771]0.69310=2.303t×0.1249t=2.303×0.1249×100.603=4.15 days
Hence, the correct answer is option (1)
10. ऑक्सीकरण अवस्था
Question: The amphoteric oxide among
V2O3, V2O4
and
V2O5
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:
V2O3
- basic
V2O4
- weakly acidic or amphoteric
V2O5
- amphoteric
V2O5+ alkali →VO43−
In
VO43−
ion, vanadium is in a +5 oxidation state.
Hence, the correct answer is option (3).
बेहतर अभ्यास और रिवीजन के लिए छात्र जेईई मेन के टॉप 30 सबसे ज़्यादा बार पूछे जाने वाले प्रश्नों और विषयों को देख सकते हैं। साथ ही, जेईई मेन 2026 के सैंपल पेपर को हल करने का प्रयास करें।
