JEE Main 2026 January 24, Shift 1 Question Paper with Solutions

All 74 questions from the JEE Main 2026 (January 24, Shift 1) shift — Physics (24), Chemistry (25) and Mathematics (25) — with the correct answer and a step-by-step solution for every question.

Physics PYQs 2026 Chemistry PYQs 2026 Mathematics PYQs 2026

Physics24 questions

Q26Single correctProperties of Solids and Liquids

Density of water

4^{\circ}c

and

20^{\circ}c

are

1000\,kg/m^3

and

998\,kg/m^3

respectively the increase in internal energy of

4\,kg

of water when it is heated from

4^{\circ}c

20^{\circ}c

is ____ J. (specific heat capacity of water

=4.2\,J/kg

and

1

atmospheric pressure

=10^5\,Pa

)

(A)

\displaystyle 268799.2

(B)

\displaystyle 315826.2

(C)

\displaystyle 258700.8

(D)

\displaystyle 234699.2

Q27Single correctOptics

In a microscope of tube length

10\,cm

, two convex len's are arranged with focal length of

2\,cm

and

5\,cm

Total magnification obtained with this system for normal advistament is

(5)^k

the value of K is

(A)

\displaystyle 4

(B)

\displaystyle 5

(C)

\displaystyle 2

(D)

\displaystyle 3.5

Q28Single correctAtoms and Nuclei

Two electrons are moving in orbits of two hydrogen like atoms with speed

3\times10^5\,m/s

and

2.5\times10^5\,m/s

respectively. If the radii of these orbits are nearly same then, the possible order of energy states are ____ respectively

(A)

\displaystyle 9

and

\displaystyle 8

(B)

\displaystyle 10

and

\displaystyle 12

(C)

\displaystyle 8

and

\displaystyle 10

(D)

\displaystyle 6

and

\displaystyle 5

Q29Single correctOptics

An unpolarised light is incident at an interface of two dielectric media having refractive indices of

2

and

2\sqrt{3}

respectively to satisfy the condition that reflected and refracted rays are perpendicular to each other, the angle of incidence is ___

(A)

\displaystyle 30^{O}

(B)

\displaystyle 60^{O}

(C)

\displaystyle 45^{O}

(D)

\displaystyle 10^{O}

Q30Single correctMagnetic Effects of Current and Magnetism

Match the list -1 with list-II

List-I	List-II
A. Magnetic induction	I. $M\,L\,T^{-2}\,A^{-2}$
B. Magnetic flues	II. $M\,L^{2}\,T^{-2}\,A^{-2}$
C. Magnetic permeability	III. $M\,L^{0}\,T^{-2}\,A^{-1}$
D. Self induction	IV. $M\,L^{2}\,T^{-2}\,A^{-1}$

(A)

\displaystyle A-\text{III},\,B-IV,\,C-II,\,D-I

(B)

\displaystyle A-I,\,B-\text{III},\,C-IV,\,D-II

(C)

\displaystyle A-IV,\,B-\text{III},\,C-I,\,D-II

(D)

\displaystyle A-\text{III},\,B-IV,\,C-I,\,D-II

Q31Single correctElectromagnetic Induction and Alternating Currents

For a series LCR circuit connected with

220V

50\,Hz

a.c source as shown in the figure The power factory is

\dfrac{\alpha}{10}

the valve of

\alpha

is ___

Series LCR circuit drawn as a rectangular loop. A 220 V, 50 Hz AC source (circle with sine wave) is in the top wire. The bottom wire carries, left to right in series, an inductor L (X_L = 70 ohm), a capacitor C (X_C = 150 ohm), and a resistor R (60 ohm).

(A)

\displaystyle 10

(B)

\displaystyle 4

(C)

\displaystyle 6

(D)

\displaystyle 8

Q32Single correctElectromagnetic Waves

Match the list -1 with list-II

List-I	List-II
A. Ratio wave	I. Is produced by magnetron value
B. Micro wave	II. Due to change in vibrational modes of atoms
C. Infrared - wave	III. Due to inner shell electron moving from higher energy level to lower energy level
D. X-ray	IV. Due to rapid acceleration of chares

(A)

\displaystyle A\to IV,\,B\to I,\,C\to II,\,D\to \text{III}

(B)

\displaystyle A\to II,\,B\to IV,\,C\to \text{III},\,D\to I

(C)

\displaystyle A\to IV,\,B\to II,\,C\to I,\,D\to \text{III}

(D)

\displaystyle A\to IV,\,B\to \text{III},\,C\to I,\,D\to II

Q33Single correctOscillations and Waves

A cylindrical block of mass M and area of cross section A is floating in a liquid of density

\rho

and with x-axis vertical when depressed a little an released the block starts oscillating the period of oscillation is ___

(A)

\displaystyle 2\pi\sqrt{\dfrac{A}{Mg}}

(B)

\displaystyle 2\pi\sqrt{\dfrac{M}{\rho A g}}

(C)

\displaystyle \pi\sqrt{\dfrac{A}{Mg}}

(D)

\displaystyle \pi\sqrt{\dfrac{2M}{Ag}}

Q34Single correctAtoms and Nuclei

Given below are two statements:
Statement -I :- For all elements, grater then mass of the nucleus, greater is the binding energy per nucleon
Statement -II :-For all elements, nuclei with less binding energy per nucleon transforms to nuclei with grater bonding energy per nucleon
In the light of the above statements, choose the correct answer from the option given below

(A)

Statement -I is true but Statement -II are false

(B)

Both Statement -I and Statement -II are false

(C)

Both Statement -I and Statement -II are true

(D)

Statement -I is false but Statement -II is true

Q35Single correctElectrostatics

The electro static potential in a charged spherical region of radius r varies as

V = ar^3 + b

. where a and b are constants. The total charge in the sphere of unit radius is

\alpha\times\pi a\varepsilon_0

. The value of

\alpha

is ______. (permittivity of vacuum is

\varepsilon_0

)

(A)

\displaystyle -8

(B)

\displaystyle -9

(C)

\displaystyle -12

(D)

\displaystyle -6

Q36Single correctElectrostatics

Three charges

+2q, +3q

and

-4q

are satiated at

(0,-3a), (2a,0)

and

(-2a,0)

respectively in the x y plane. The resultant dipole moment about origin is______.

(A)

\displaystyle 2qa\left[3\hat{i}-7\hat{j}\right]

(B)

\displaystyle 2qa\left[3\hat{j}-7\hat{i}\right]

(C)

\displaystyle 2qa\left(7\hat{i}-3\hat{j}\right)

(D)

\displaystyle 2qa\left(3\hat{j}-\hat{i}\right)

Q37Single correctCurrent Electricity

Two resistor's of

100\,\Omega

each are connected in series with a

9V

battery a voltmeter of

400\,\Omega

resistance is connected two measure the voltage drop a cross one of the resistance the voltmeter reading is______V

(A)

\displaystyle 4.5

(B)

\displaystyle 4

(C)

\displaystyle 3

(D)

\displaystyle 2

Q38Single correctRotational Motion

Two masses 400g and 350g are suspended from the ends of a light string passing over a heavy pulley of radius 2 cm when released from rest the heavier mass is observed to fall 81 cm is 9s. The rotational inertia of the pulley is ___kg.

m^2

. (g=9.8 m/

s^2

)

(A)

\displaystyle 4.75\times10^{-3}

(B)

\displaystyle 1.86\times10^{-2}

(C)

\displaystyle 9.5\times10^{-3}

(D)

\displaystyle 8.3\times10^{-3}

Q39Single correctKinematics

A boy throw's a ball in to air at 4

5^{\circ}

from the horizontal to land on a roof of a building of height H. If the ball attains maximum height in 2s and land on the building in 3s after launch then the value of H is ___m (g=10 m/

s^2

)

(A)

(B)

(C)

(D)

Q40Single correctProperties of Solids and Liquids

A brass wire of length 2m and radius 1mm at 2

7^{\circ}

c is held taut between two rigid supports initially it was cooled to temperature of -4

3^{\circ}

c creating a tension T in the wire. The temperature to which the wire has to be cooled in order to increase the tension in it to 1.4T is ___

^{\circ}

(A)

-80

(B)

-65

(C)

-71

(D)

-86

Q41Single correctGravitation

Three masses 200 kg, 300 kg and 400 kg are placed at the vertices of an equilateral triangle with sides 20 m. They are rearranged on the vertices of a bigger triangle of side 25 m and with same centre. The work done in this process is ___J. (Gravitational constant

G = 6.7 \times 10^{-11} Nm^2/kg^2

)

(A)

\displaystyle 2.85\times10^{-7}

(B)

\displaystyle 4.77\times10^{-7}

(C)

\displaystyle 9.86\times10^{-6}

(D)

\displaystyle 1.74\times10^{-7}

Q42Single correctCurrent Electricity

Two resistors

2\Omega

and

3\Omega

are connected in the gap of bridge as shown in fig the null point is obtained with the contact of jockey at some point on wire XY. When an unknown resistor is connected in parallel with

3\Omega

resistor, the null point is shifted by 22.5cm towards Y. The resistance of unknown resistor is ___

\Omega

Metre-bridge circuit as a rectangular loop. The top side has two resistors in series, 2 ohm on the left and 3 ohm on the right, meeting at a central junction. From that junction a jockey contact (wavy line) runs down to the bottom wire XY, with X at the bottom-left and Y at the bottom-right.

(A)

(B)

(C)

(D)

Q43Single correctOscillations and Waves

A spring of force constant 15N/m is cut into two pieces If the ratio of their lengths is 1:3, then force constant of smaller pieces is ___N/m

(A)

(B)

(C)

(D)

Q44Single correctOptics

The exit surface of prism with Refractive index n is coated with material having Refractive index n/2 when this prism is set for minimum angle of deviation It Exactly meet's the condition of critical angle, the prism angle is ___

(A)

\displaystyle 15^{O}

(B)

\displaystyle 60^{O}

(C)

\displaystyle 45^{O}

(D)

\displaystyle 30^{O}

Q45NumericalLaws of Motion

In the given figure the blocks A, B and C weigh 4 kg , 6 kg and 8 kg respectively. The co-efficient of sliding friction between any two surfaces is 0.5. The force F required to slide the block C with constant speed is ___N. (Use g=10 m/

s^2

)

Block C rests on hatched ground. Hatched block A sits on the top-left of C. At the top-right corner of C a pulley is mounted against a fixed hatched vertical wall. A string runs horizontally from block A over the pulley and down to a hanging block. A leftward force F (vector arrow) is applied to the left face of block C.

Q46NumericalElectronic Devices

A voltage regulating circuit consisting of Zener diode, having break-down voltage of 10 V and maximum power dissipation of 0.4 W, is operated at 15 V. The approximate value of protective resistance in this circuit is ___

\Omega

Q47NumericalKinetic Theory of Gases

A gas of certain mass filled in a closed cylinder at a pressure of 3.23 kPa has temperature 50

^{0}

C. The gas is now heated to double its temperature. The modified pressure is ___Pa.

Q48NumericalMagnetic Effects of Current and Magnetism

A short bar magnet placed with its axis at 3

0^{0}

with an external field of 800 Gauss, experiences a torque of 0.016 N.m. The work done in moving it from most stable to most unstable position is

\alpha\times10^{-3}J

. The value of

\alpha

is ___

Q49NumericalProperties of Solids and Liquids

Sixty four rain drops of radius 1 mm each falling down with a terminal velocity of 10 cm/s coalesce to form a bigger drop. The terminal velocity of bigger drop is ___ cm/s.

Chemistry25 questions

Q50Single correctCoordination Compounds

Given below are two statements:
Statement-I: Hybridisation, shape and spin only magnetic moment of

K_3[Co(CO_3)_3]

sp^3d^2

, octahedral and 4.9 BM respectively.
Statement-II: Geometry, hybridisation and spin only magnetic moment values (BM) of the ions

[Ni(CN)_4]^{2-}

[\text{MnBr}_4]^{2-}

and

[\text{CoF}_6]^{2-}

respectively are square planar, tetrahedral, octahedral;

\text{dsp}^2

sp^3

sp^3d^2

and 0, 5.9, 4.9.
In light of the above statements, choose the correct answer from the options given below

(A)

S-I is true but S-II is false

(B)

Both S-I and S-II are true

(C)

S-I is false but S-II is true

(D)

Both S-I and S-II are false

SolutionAnswer: Option 2Both S-I and S-II are true

Approach:

Assign the oxidation state and d-electron configuration of each metal centre, apply the ligand field strength to fix the spin state, then deduce hybridisation, geometry and spin-only magnetic moment for every complex named in the two statements.

Step 1:Statement-I requires the configuration of cobalt in

K_3[Co(CO_3)_3]

and the configurations of the three ions of Statement-II. The target is to test the truth of each statement against valence bond theory.

\text{Targets: }K_3[Co(CO_3)_3];\ [Ni(CN)_4]^{2-};\ [MnBr_4]^{2-};\ [CoF_6]^{2-}

Step 2:In

K_3[Co(CO_3)_3]

cobalt carries a

+3

charge giving a

3d^6

configuration. Carbonate is a weak field ligand, so the

3d^6

set stays as

t_{2g}^4 e_g^2

with four unpaired electrons, forcing outer-orbital

sp^3d^2

hybridisation and an octahedral shape.

Co^{3+}:3d^6,\ n=4,\ sp^3d^2,\ \text{octahedral}

Step 3:Substituting

n=4

into the spin-only expression gives the moment quoted in Statement-I.

\mu=\sqrt{4(4+2)}=\sqrt{24}=4.9\ \text{BM}

Step 4:For

[Ni(CN)_4]^{2-}

, nickel is

Ni^{2+}

(

3d^8

); the strong field

CN^-

pairs the electrons, leaving one vacant

3d

orbital and producing

\text{dsp}^2

hybridisation, square planar geometry and zero unpaired electrons.

Ni^{2+}:3d^8,\ dsp^2,\ \text{square planar},\ \mu=0

Step 5:For

[\text{MnBr}_4]^{2-}

, manganese is

Mn^{2+}

(

3d^5

); the weak field

Br^-

leaves all five electrons unpaired, giving

sp^3

hybridisation and a tetrahedral shape.

\mu=\sqrt{5(5+2)}=\sqrt{35}=5.9\ \text{BM}

Step 6:For the cobalt fluoro ion (

Co^{3+}

3d^6

) the weak field

F^-

leaves four unpaired electrons, giving

sp^3d^2

hybridisation, octahedral geometry and

\mu=4.9

BM. Each geometry, hybridisation and moment in Statement-II is reproduced, so Statement-II is true.

\mu=\sqrt{4(4+2)}=\sqrt{24}=4.9\ \text{BM}

Final answer: Both S-I and S-II are true

Q51Single correctChemical Bonding and Molecular Structure

Given below are statements about some molecules/ions.
Identify the CORRECT statements.
A. The dipole moment value of

NF_3

is higher than that of

NH_3

.
B. The dipole moment value of

\text{BeH}_2

is zero.
C. The bond order of

O_2^{2-}

and

F_2

is same.
D. The formal charge on the central oxygen atom of ozone is -1.
E. In

NO_2

, all the three atoms satisfy the octet rule, hence it is very stable.
Choose the correct answer from the options given below:

(A)

B & C only

(B)

B, C & D only

(C)

A, C & D only

(D)

A, B, C, D & E

SolutionAnswer: Option 1B & C only

Approach:

Test each lettered statement separately using dipole-moment vector addition, molecular-orbital bond-order counting, formal-charge calculation and octet analysis, then collect the correct statements.

Step 1:Five statements A to E are to be assessed and the subset that is correct selected. Both

NH_3

and

NF_3

are pyramidal with a lone pair on nitrogen.

\text{Assess A, B, C, D, E}

Step 2:In

NH_3

the lone-pair dipole points the same way as the resultant of the three N

\to

H bond dipoles, so they add. In

NF_3

the bond dipoles point from N toward the more electronegative F, opposing the lone-pair dipole, so they subtract. Therefore

NH_3

has the larger dipole moment and statement A is false.

\mu(NH_3)>\mu(NF_3)

Step 3:

\text{BeH}_2

is linear (

AB_2

, no lone pair on Be), so its two equal and opposite bond dipoles cancel exactly, giving a net dipole moment of zero. Statement B is true.

\mu(\text{BeH}_2)=0

Step 4:

O_2^{2-}

and

F_2

each carry 18 electrons. Filling the molecular orbitals gives

N_b=10

bonding and

N_a=8

antibonding electrons in both, so each has bond order

1

and statement C is true.

\text{B.O.}=\tfrac{1}{2}(10-8)=1

Step 5:In ozone the central oxygen has one lone pair (2 electrons), one O=O double bond and one O-O single bond (6 bonding electrons total). Its valence electron count is 6, so the formal charge is

6-2-\tfrac{1}{2}(6)=+1

, not

-1

. Statement D is false.

FC=6-2-\tfrac{1}{2}(6)=+1

Step 6:

NO_2

has an odd total of 17 valence electrons, leaving an unpaired electron on nitrogen, which therefore carries seven electrons and does not complete its octet. Statement E is false.

N:\,7e^-,\ \text{odd electron}

Step 7:The true statements are B and C only.

\{B,C\}

Final answer: B & C only

Q52Single correctSome Basic Principles of Organic Chemistry

Arrange the following combinations in the decreasing order of stability

Five para-substituted benzylic carbanion species labelled I to V, each drawn as a benzene ring written as C6H4 (or C6H5) attached to a CH2 carbanion carbon bearing a lone pair (two dots) and a circled negative charge.

(A)

IV > II > I > III > V

(B)

I > IV > II > V > III

(C)

I > II > IV > V > III

(D)

IV > I > II > V > III

SolutionAnswer: Option 4IV > I > II > V > III

Approach:

Rank the five para-substituted benzylic carbanions by the electron-withdrawing or electron-donating power of the para substituent acting on the negatively charged benzylic carbon.

Step 1:The five species are para-substituted benzyl carbanions: I = p-Br, II = p-H (unsubstituted), III = p-OC

H_3

, IV = p-CHO, V = p-C

H_3

. The objective is the decreasing stability order of the carbanion.

\text{I:}p\text{-Br},\ \text{II:H},\ \text{III:}p\text{-OCH}_3,\ \text{IV:}p\text{-CHO},\ \text{V:}p\text{-CH}_3

Step 2:A carbanion bears a localised negative charge, so an electron-withdrawing group (

-I

-M

) at the para position disperses that charge and stabilises the anion, while an electron-donating group intensifies the charge and destabilises it.

\text{EWG}\Rightarrow\text{stabilise};\ \text{EDG}\Rightarrow\text{destabilise}

Step 3:The para-formyl group (IV) withdraws strongly by resonance (

-M

), delocalising the negative charge onto its oxygen, so it gives the most stable carbanion.

\text{IV }(p\text{-CHO}):\ \text{most stable}

Step 4:Para-bromine (I) withdraws by the inductive effect, stabilising more than the unsubstituted carbanion (II), placing I above II.

\text{I }(p\text{-Br}) > \text{II (H)}

Step 5:Para-methyl (V) donates by hyperconjugation and weak

+I

, destabilising the anion below the unsubstituted case, while para-methoxy (III) donates most strongly by resonance (

+M

), making III the least stable.

\text{II (H)} > \text{V }(p\text{-CH}_3) > \text{III }(p\text{-OCH}_3)

Step 6:Combining the partial orders gives the full decreasing stability sequence.

\text{IV} > \text{I} > \text{II} > \text{V} > \text{III}

Final answer: IV > I > II > V > III

Q53Single correctOrganic Compounds Containing Nitrogen

The correct stability order of the following diazonium salts is

Four benzenediazonium chloride structures drawn in a row, each a benzene ring with a diazonium group at the top and labelled A, B, C, D by para substituent: A has para-OCH3, B has para-NO2, C is unsubstituted, D has para-CN.

(A)

C > A > D > B

(B)

A > B > C > D

(C)

A > C > D > B

(D)

C > D > B > A

Q54Single correctChemical Bonding and Molecular Structure

Among the following, the correct combinations are
A.

IF_3 \rightarrow T-\text{shaped}\ (sp^3d)

IF_5 \rightarrow Square\ \text{pyramidal}\ (sp^3d^2)

IF_7 \rightarrow \text{Pentagonal}\ \text{bipyramidal}\ (sp^3d^3)

\text{ClO}_4^- \rightarrow Square\ \text{planar}\ (sp^2d)

Choose the correct answer from the options given below:

(A)

A, B and C only

(B)

A and B only

(C)

A, B, C and D

(D)

B, C and D only

SolutionAnswer: Option 1A, B and C only

Approach:

Apply VSEPR theory to each species by computing the steric number (bond pairs plus lone pairs) on the central atom, then assign hybridisation and shape and test each lettered combination.

Step 1:Four combinations of species, shape and hybridisation are to be tested. The central atom for A, B, C is iodine and for D is chlorine.

IF_3,\ IF_5,\ IF_7,\ \text{ClO}_4^-

Step 2:Iodine in

IF_3

has 7 valence electrons; three form bonds with F and the remaining four form two lone pairs, giving

SN=3+2=5

. This is

AB_3E_2

sp^3d

hybridised, T-shaped. Statement A is correct.

IF_3:SN=5,\ AB_3E_2,\ sp^3d,\ \text{T-shaped}

Step 3:Iodine in

IF_5

uses five electrons in bonds with F, leaving one lone pair, giving

SN=5+1=6

. This is

AB_5E

sp^3d^2

hybridised, square pyramidal. Statement B is correct.

IF_5:SN=6,\ AB_5E,\ sp^3d^2,\ \text{square pyramidal}

Step 4:Iodine in

IF_7

uses all seven valence electrons in bonds with F, leaving no lone pair, giving

SN=7

. This is

AB_7

sp^3d^3

hybridised, pentagonal bipyramidal. Statement C is correct.

IF_7:SN=7,\ AB_7,\ sp^3d^3,\ \text{pentagonal bipyramidal}

Step 5:In

\text{ClO}_4^-

chlorine has four bonding regions to oxygen and no lone pair, giving

SN=4

sp^3

hybridisation and a tetrahedral shape, not square planar with

sp^2d

. Statement D is false.

ClO_4^-:SN=4,\ sp^3,\ \text{tetrahedral}

Step 6:The correct combinations are A, B and C only.

\{A,B,C\}

Final answer: A, B and C only

Q55Single correctChemical Thermodynamics

\rightarrow

D is an endothermic reaction occurring in three steps (elementary).
(i)

A \rightarrow B

\Delta H_i = +ve

(ii)

B \rightarrow C

\Delta H_{ii} = -ve

(iii)

C \rightarrow D

\Delta H_{iii} = -ve

Which of the following graphs between potential energy (y-axis) vs reaction coordinate (x-axis) correctly represents the reaction profile of A

\rightarrow

(A)

Potential-energy versus reaction-coordinate curve with three humps (three activation peaks). The curve starts at a level on the left, rises and falls through three peaks, and the final product level on the right sits below the starting reactant level.

(B)

Potential-energy versus reaction-coordinate curve with three deep humps separated by deep valleys; the first peak is moderate, the middle and last peaks are taller, and the final product level ends near or slightly below the start.

(C)

Potential-energy versus reaction-coordinate curve with three humps of increasing height; the curve starts low, rises through three progressively taller peaks and ends at a product level, with the overall trend rising.

(D)

Potential-energy versus reaction-coordinate curve with three humps; the reactant starts at the lowest level, the first hump leads to an intermediate higher than the reactant, subsequent steps descend stepwise, and the final product level ends above the starting reactant level (net endothermic).

SolutionAnswer: Option 4Three-peak reaction profile in which intermediate B lies above reactant A (step i endothermic), C lies below B and D lies below C (steps ii and iii exothermic), with the final product D ending above the starting reactant A so that the overall A to D process is endothermic.

Approach:

Convert the sign of each step enthalpy into the relative energy levels of A, B, C and D, then select the three-peak profile whose level pattern matches every constraint.

Step 1:The given data are three elementary steps with

\Delta H_i>0

\Delta H_{ii}<0

\Delta H_{iii}<0

, and an overall endothermic reaction

A\to D

. The objective is the energy-level ordering of A, B, C, D.

\Delta H_i>0,\ \Delta H_{ii}<0,\ \Delta H_{iii}<0,\ \Delta H_{A\to D}>0

Step 2:Three elementary steps each pass through one transition state, so the profile must contain three activation peaks.

3\ \text{elementary steps}\Rightarrow 3\ \text{peaks}

Step 3:Since step (i) has positive

\Delta H

, the product of that step lies higher in energy than its reactant, so intermediate B is above A.

\Delta H_i=+ve\Rightarrow H_B>H_A

Step 4:Steps (ii) and (iii) have negative

\Delta H

, so each product falls below its reactant, placing C below B and D below C.

\Delta H_{ii},\Delta H_{iii}=-ve\Rightarrow H_B>H_C>H_D

Step 5:The overall reaction is endothermic, so the final product D sits above the starting reactant A. The matching profile rises to B, descends stepwise to C and D, yet ends above A.

\Delta H_{A\to D}=+ve\Rightarrow H_D>H_A

Final answer: Three-peak reaction profile in which intermediate B lies above reactant A (step i endothermic), C lies below B and D lies below C (steps ii and iii exothermic), with the final product D ending above the starting reactant A so that the overall A to D process is endothermic.

Q56Single correctOrganic Compounds Containing Halogens

Given below are two statements
Statement-I: 'C - Cl' bond is strong in

CH_2 = CH - Cl

than

CH_3 - CH_2 - Cl

Statement-II: The given optically active molecule, on hydrolysis gives a solution that can rotate the plane polarized light.
In the light of the above statements, choose the correct answer from the options given below.

A wedge-dash structure of a chiral carbon bonded to four different groups: a phenyl (Ph) group on a plain bond, a methyl (Me) group on a hashed (dashed) bond, a chlorine (Cl) on a plain horizontal bond, and an ethyl (Et) group on a bold wedge bond.

(A)

S-I is true but S-II is false

(B)

Both S-I and S-II are false

(C)

Both S-I and S-II are true

(D)

S-I is false but S-II is true

Q57Single correctOrganic Compounds Containing Halogens

Match List-I with List-II

List-I (Chloro derivative)	List-II (Example)
A. Vinyl Chloride	I. $CH_2 = CH - CH_2Cl$
B. Benzyl Chloride	II. $CH_3 - CH(Cl)CH_3$
C. Alkyl Chloride	III. $CH_2 = \text{CHCl}$
D. Allyl Chloride	IV.

(A)

A-III, B-IV, C-I, D-II

(B)

A-IV, B-I, C-III, D-II

(C)

A-I, B-II, C-IV, D-III

(D)

A-III, B-IV, C-II, D-I

Q58Single correctCoordination Compounds

Consider a mixture 'X' which is made by dissolving 0.4 mol of

[Co(NH_3)_5SO_4]Br

and 0.4 mol of

[Co(NH_3)_5Br]SO_4

in water to make 4 L of solution. When 2 L of mixture 'X' is allowed to react with excess of

\text{AgNO}_3

, it forms precipitate 'Y'. The rest 2 L of mixture 'X' reacts with excel

\text{BaCl}_2

to form precipitate 'Z'. which of the following statements is CORRECT?

(A)

0.1 mol of 'Y' is formed

(B)

0.2 mol of 'Z' is formed

(C)

0.4 mol of 'Z' is formed

(D)

'Y' is

\displaystyle \text{BaSO}_4

and 'Z' is AgBr

SolutionAnswer: Option 20.2 mol of 'Z' is formed

Approach:

Identify the free counter-ion released by each ionisation isomer, find the moles available in each 2 L portion, then apply precipitation stoichiometry to evaluate each statement.

Step 1:The 4 L mixture contains 0.4 mol

[Co(NH_3)_5SO_4]Br

and 0.4 mol

[Co(NH_3)_5Br]SO_4

. The objective is to determine the moles and identity of precipitates Y (with

\text{AgNO}_3

) and Z (with

\text{BaCl}_2

) from each 2 L half.

0.4\ \text{mol each complex in }4\ L

Step 2:In

[Co(NH_3)_5SO_4]Br

the sulphate is inside the coordination sphere, so bromide is the free counter-ion. This releases 0.4 mol

Br^-

into 4 L.

[Co(NH_3)_5SO_4]Br\to[Co(NH_3)_5SO_4]^{+}+Br^{-}

Step 3:In

[Co(NH_3)_5Br]SO_4

the bromide is coordinated, so sulphate is the free counter-ion. This releases 0.4 mol

SO_4^{2-}

into 4 L.

[Co(NH_3)_5Br]SO_4\to[Co(NH_3)_5Br]^{2+}+SO_4^{2-}

Step 4:Each 2 L portion is half of the 4 L mixture, so it contains 0.2 mol free

Br^-

and 0.2 mol free

SO_4^{2-}

2\,L:\ 0.2\ \text{mol }Br^-,\ 0.2\ \text{mol }SO_4^{2-}

Step 5:Excess

\text{AgNO}_3

precipitates the free bromide as Y = AgBr (0.2 mol). Excess

\text{BaCl}_2

precipitates the free sulphate as Z =

\text{BaSO}_4

(0.2 mol). The coordinated

SO_4

and Br stay in their spheres and do not precipitate.

Ag^++Br^-\to \text{AgBr};\ Ba^{2+}+SO_4^{2-}\to \text{BaSO}_4

Step 6:Testing the options: 0.1 mol Y is wrong (it is 0.2), 0.4 mol Z is wrong (it is 0.2), and the identities in option 4 are swapped. Only 0.2 mol of Z is correct.

\text{Z}=0.2\ \text{mol }BaSO_4

Final answer: 0.2 mol of 'Z' is formed

Q59Single correctPrinciples Related to Practical Chemistry

Consider three metal chlorides x, y and z, where x is water soluble at room temperature, y is sparingly soluble in water at room temperature and z is soluble in hot water. x, y and z re respectively

(A)

\displaystyle \text{MgCl}_2, \text{AgCl}

and

\displaystyle \text{AlCl}_3

(B)

\displaystyle \text{CuCl}_2, \text{AgCl}

and

\displaystyle \text{PbCl}_2

(C)

\displaystyle \text{AgCl}, Hg_2Cl_2

and

\displaystyle \text{PbCl}_2

(D)

\displaystyle \text{AlCl}_3, \text{PbCl}_2

and

\displaystyle \text{BaCl}_2

SolutionAnswer: Option 2

\text{CuCl}_2, \text{AgCl}

and

\text{PbCl}_2

Approach:

Match each of the three solubility descriptions (freely soluble at room temperature, sparingly soluble at room temperature, soluble only in hot water) to a metal chloride, then test which option satisfies all three simultaneously.

Step 1:Three roles are defined: x is freely soluble in cold water, y is sparingly soluble in cold water, and z dissolves only on heating. The objective is to assign each role to a chloride that uniquely fits.

x:\text{cold-soluble};\ y:\text{sparingly soluble};\ z:\text{hot-soluble}

Step 2:Copper(II) chloride is a typical freely soluble chloride in cold water, fitting role x.

x=\text{CuCl}_2

Step 3:Silver chloride has a very low solubility product (

K_{sp}\approx1.8\times10^{-10}

) and is sparingly soluble in cold water, fitting role y.

y=\text{AgCl},\ K_{sp}\approx1.8\times10^{-10}

Step 4:Lead(II) chloride is only sparingly soluble in cold water but its solubility rises strongly with temperature, so it dissolves appreciably in hot water and is recovered on cooling, fitting role z.

z=PbCl_2\ (\text{hot-soluble})

Step 5:Option 2 assigns

x=\text{CuCl}_2

y=\text{AgCl}

z=\text{PbCl}_2

, satisfying all three descriptions cleanly. Options 1 and 4 require

\text{AlCl}_3

\text{BaCl}_2

(both freely soluble in cold water) to fill the hot-water-only role, and option 3 makes AgCl the freely soluble salt, both of which conflict with standard solubility behaviour.

\{CuCl_2,\ AgCl,\ PbCl_2\}

Final answer:

\text{CuCl}_2, \text{AgCl}

and

\text{PbCl}_2

Q60Single correctOrganic Compounds Containing Oxygen

A student is given one compound among the following compounds that gives positive test with Tollen's reagent.
The compound is

Four oxygen-containing five-membered (tetrahydrofuran-type) ring structures labelled A, B, C, D. A: ring with OCH3 at the 2-position (carbon adjacent to ring oxygen). B: ring with OCH3 at the 3-position. C: ring with OH at the 2-position (adjacent to ring oxygen). D: ring with OH at the 3-position.

(A)

(B)

(C)

(D)

Q61Single correctOrganic Compounds Containing Oxygen

Consider the following two reactions A and B

Two reaction schemes. Reaction (A): a benzene ring bearing an OH group (phenol) reacts with Na over an arrow to give 'Main Product + gas (x)'. Reaction (B): a benzene ring bearing a COOH group (benzoic acid) reacts with NaHCO3 over an arrow to give 'Main Product + gas (y)'.

(A)

(B)

(C)

(D)

160

Q62Single correctCoordination Compounds

Given below are two statements;
Statement-I: The number of paramagnetic species among

[\text{CoF}_6]^{3-}

[\text{TiF}_6]^{3-}

V_2O_5

and

[Fe(CN)_6]^{3-}

is 3.
Statement-II:

K_4[Fe(CN)_6] < K_3[Fe(CN)_6] < [Fe(H_2O)_6]SO_4 \cdot H_2O < [Fe(H_2O)_6]Cl_3

is the correct order in terms of number of unpaired element(s) present in the complexes.
In the light of the above statements, choose the correct answer from the options given below

(A)

Both S-I and S-II are true

(B)

S-I is false but S-II is true

(C)

S-I is true but S-II is false

(D)

Both S-I and S-II are false

SolutionAnswer: Option 1Both S-I and S-II are true

Approach:

Determine the number of unpaired electrons in each species from the metal oxidation state, d-electron count and ligand field strength, then test the count in Statement-I and the ordering in Statement-II.

Step 1:Statement-I lists four species whose unpaired-electron counts must be found; Statement-II gives an increasing-unpaired-electron ordering of four iron complexes. The objective is to test both.

[\text{CoF}_6]^{3-},\ [\text{TiF}_6]^{3-},\ V_2O_5,\ [Fe(CN)_6]^{3-}

Step 2:

[\text{CoF}_6]^{3-}

has

Co^{3+}

(

3d^6

) with the weak field

F^-

, so it stays high spin with 4 unpaired electrons (paramagnetic).

[\text{TiF}_6]^{3-}

has

Ti^{3+}

(

3d^1

) with 1 unpaired electron (paramagnetic).

Co^{3+}:n=4;\ Ti^{3+}:n=1

Step 3:

V_2O_5

has vanadium as

V^{5+}

(

3d^0

) with zero unpaired electrons (diamagnetic).

[Fe(CN)_6]^{3-}

has

Fe^{3+}

(

3d^5

) with the strong field

CN^-

, giving the low-spin

t_{2g}^5

set with 1 unpaired electron (paramagnetic).

V^{5+}:n=0;\ Fe^{3+}(CN):n=1

Step 4:For Statement-II:

K_4[Fe(CN)_6]

has

Fe^{2+}

(

3d^6

) low spin with 0 unpaired;

K_3[Fe(CN)_6]

has

Fe^{3+}

(

3d^5

) low spin with 1 unpaired;

[Fe(H_2O)_6]SO_4\cdot H_2O

has

Fe^{2+}

(

3d^6

) high spin with 4 unpaired;

[Fe(H_2O)_6]Cl_3

has

Fe^{3+}

(

3d^5

) high spin with 5 unpaired.

0,\ 1,\ 4,\ 5

Step 5:The unpaired counts

0<1<4<5

are monotonically increasing in the order listed in Statement-II, so Statement-II is true. Both statements being true selects the corresponding option.

0<1<4<5

Final answer: Both S-I and S-II are true

Q63Single correctClassification of Elements and Periodicity in Properties

Given below are two statements:
Statement-I:

K > Mg > Al > B

is the correct order in terms of metallic character.
Statement-II: Atomic radius is always greater than the ionic radius for any element.
In the light of the above statements, the correct answer from the options given below

(A)

Both S-I and S-II are false

(B)

Both S-I and S-II are true

(C)

S-I is false but S-II is true

(D)

S-I is true but S-II is false

Q64Single correctSolutions

A solution is prepared by dissolving 0.3 g of non-volatile non-electrolyte solute 'A' of molar mass 60 g mo

l^{-1}

and 0.9 g of non-volatile non-electrolyte solute 'B' of molar mass 180 g mo

l^{-1}

in 100 mL

H_2

O at

27^\circ\text{C}

. Osmatic pressure of the solution will be [Given: R = 0.082 L atm

K^{-1}

l^{-1}

]

(A)

1.23 atm

(B)

1.47 atm

(C)

2.46 atm

(D)

0.82 atm

Q65Single correctOrganic Compounds Containing Oxygen

The hydroxy compound (X) with molar mass 122 g mo

l^{-1}

is acetylated with acetic anhydride, using a large excess of the reagent ensuring complete acetylation of all hydroxyl groups. The product obtained has a molar mass of 290 g mo

l^{-1}

. The number of hydroxyl groups present in compound (X) is:

(A)

(B)

(C)

(D)

Q66Single correctChemical Kinetics

27^\circ\text{C}

in presence of a catalyst, activation energy of a reaction is lowered by 10 kJ mo

l^{-1}

. The logarithm of ratio of

\dfrac{k(catalysed)}{k(uncatalysed)}

is …. (Consider that the frequency factor for both reactions is same)

(A)

3.482

(B)

1.741

(C)

17.41

(D)

0.1741

Q67Single correctHydrocarbons

Arrange the following alkenes in decreasing order of stability

Four skeletal alkene structures labelled I, II, III and IV. I is a tetrasubstituted alkene (four alkyl groups, no vinylic H); II is a trans (E) disubstituted alkene; III is a trisubstituted alkene; IV is a cis (Z) disubstituted alkene.

(A)

I > III > II > IV

(B)

III > II > I > IV

(C)

I > III > IV > II

(D)

III > I > II > IV

Q68Single correctSolutions

'W' g of non-volatile electrolyte solid solute of molar mass 'M' g mo

l^{-1}

when dissolved in 100 mL water, decreases vapour pressure of water from 640mm Hg to 600 mm Hg. If aqueous solution of the electrolyte boils at 375 K and

K_b

for water is 0.52 K kg mo

l^{-1}

, then the mole fraction of the electrolyte solution (

x_2

) in the solution can be expressed as ( Given: Density of water = 1 g/mL and boiling point of water = 373 K)

(A)

\displaystyle \dfrac{16}{2.6} \times \dfrac{W}{M}

(B)

\displaystyle \dfrac{1.3}{8} \times \dfrac{M}{W}

(C)

\displaystyle \dfrac{2.6}{16} \times \dfrac{M}{W}

(D)

\displaystyle \dfrac{1.3}{8} \times \dfrac{W}{M}

Q69Single correctChemical Thermodynamics

Match List-I with List-II

List-I (Isothermal process for ideal gas system)	List-II (Work done $(V_f > V_i)$ )
A. Reversible expansion	I. $w = 0$
B. Free expansion	II. $w = -\text{nRT} \ln \dfrac{V_f}{V_i}$
C. Irreversible expansion	III. $w = -P_{ex}(V_f - V_i)$
D. Irreversible compression	IV. $w = -P_{ex}(V_i - V_f)$

(A)

A-IV, B-II, C-III, D-I

(B)

A-II, B-I, C-III, D-IV

(C)

A-I, B-III, C-II, D-IV

(D)

A-IV, B-I, C-III, D-II

Q70NumericalRedox Reactions and Electrochemistry

X and Y are the number of electrons involved, respectively during the oxidation of I

^-

I_2

and

S^{2-}

to S by acidified

K_2Cr_2O_7

. The value of X + Y is ____.

Q71NumericalRedox Reactions and Electrochemistry

Electricity is passed through an acidic solution of

Cu^{2+}

till all the

Cu^{2+}

was exhausted, leading to the deposition of 300 mg of Cu metal. However, a current of 600 mA was continued to pass through the same solution for another 28 minutes by keeping the total volume of the solution fixed at 200 mL. The total volume of oxygen evolved at STP during the entire process is __mL. (nearest integer)

Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)\ E^o_{red} = +0.34\,V

O_2(g) + 4H^+ + 4e^- \rightarrow 2H_2O\ E^o_{red} = +1.23\,V

Molar mass of Cu = 63.54 g mo

l^{-1}

Molar mass of

O_2

= 32 g mo

l^{-1}

Faraday Constant = 96500 C mo

l^{-1}

Molar volume at STP = 22.4 L

Q72NumericalEquilibrium

Consider two group IV metal ions

X^{2+}

and

Y^{2+}

A solution containing 0.01 M

X^{2+}

and 0.01 M

Y^{2+}

is saturated with

H_2

S. The pH at which the metal sulphide YS will form as a precipitate is … (nearest integer)
(Given:

K_{sp}

(XS) = 1

\times

0^{-22}

25^\circ\text{C}

K_{sp}

(YS) = 4

\times

0^{-16}

25^\circ\text{C}

, [

H_2

S] = 0.1M in solution,

K_{a1}

\times

K_{a2}

(

H_2

S) = 1.0

\times

0^{-21}

, log 2 = 0.30, log 3 = 0.48, log 5 = 0.70)

Q73NumericalAtomic Structure

The hydrogen spectrum consists of several spectral lines in Lyman series (

L_1

L_2

L_3

…..;

L_1

has lowest energy among Lyman series). Similarly it consists of several spectral lines in Balmer series (

B_1

B_2

B_3

…;

B_1

has lowest energy among Balmer lines). The energy of

L_1

is x times the energy of

B_1

. The value of x is … x 1

0^{-1}

(Nearest integer)

Q74NumericalPurification and Characterisation of Organic Compounds

In Dumas method for estimation of nitrogen, 0.50 g of an organic compound gave 70 mL of nitrogen collected at 300 K and 715 mm pressure. The percentage of nitrogen in the organic compound is ….% (Aqueous tension at 300 K is 15 mm).

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