JEE Main 2026 January 23, Shift 1 Question Paper with Solutions

All 75 questions from the JEE Main 2026 (January 23, Shift 1) shift — Physics (25), 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

Physics25 questions

Q26Single correctCurrent Electricity

A Wire of uniform resistance

\lambda\,\Omega/\text{m}

is bent into a circle of radius r and another piece of wire with length

2r

is connected between points A and B (AOB) as shown in figure. The equivalent resistance between point A and B is______

\Omega

A circle of radius r centred at point O. Point A is on the left of the circle on the horizontal diameter. Point B is on the top of the circle. A horizontal line segment runs from A through the centre O (labelled r between O and B/A) and a radius r is marked from O upward to B. The straight wire path goes A to O to B (the chord/diameter AOB). Labels: B at top of circle, A at left, O at centre, r marked on the radius from O to B.

(A)

\displaystyle 2\pi\lambda r

(B)

\displaystyle \dfrac{3\pi\lambda r}{8}

(C)

\displaystyle \dfrac{6\pi\lambda r}{3\pi+16}

(D)

\displaystyle (\pi+1)2r\lambda

SolutionAnswer: Option 3

\dfrac{6\pi\lambda r}{3\pi+16}

Approach:

Givens: uniform wire resistance per unit length

\lambda

(units

\Omega/\text{m}

), circle radius r (m), straight connecting wire of length 2r. Point A sits on the left of the horizontal diameter and point B at the top of the circle, so A and B are separated by a quarter turn (90 deg). The straight path A-O-B has total length

2r

. Three conductors join A and B in parallel: the straight wire, the minor (quarter) arc, and the major (three-quarter) arc. Target: equivalent resistance between A and B.

Step 1:The major arc spans three quarters of the circumference; its length is

\tfrac{3}{4}(2\pi r)

, giving resistance

R_1

R_1 = \lambda\cdot\dfrac{3}{4}(2\pi r) = \dfrac{3\pi\lambda r}{2}

Step 2:The minor arc spans one quarter of the circumference; its length is

\tfrac{1}{4}(2\pi r)

, giving resistance

R_2

R_2 = \lambda\cdot\dfrac{1}{4}(2\pi r) = \dfrac{\pi\lambda r}{2}

Step 3:The straight diametric wire A-O-B has length

2r

, giving resistance

R_3

R_3 = \lambda\cdot 2r = 2\lambda r

Step 4:Combine the three branches in parallel and factor out

\tfrac{1}{\lambda r}

\dfrac{1}{R_{eq}} = \dfrac{2}{3\pi\lambda r} + \dfrac{2}{\pi\lambda r} + \dfrac{1}{2\lambda r} = \dfrac{1}{\lambda r}\left[\dfrac{2}{3\pi} + \dfrac{6}{3\pi} + \dfrac{1}{2}\right]

Step 5:Place the bracket over the common denominator

6\pi

and invert.

\dfrac{1}{R_{eq}} = \dfrac{1}{\lambda r}\cdot\dfrac{16 + 3\pi}{6\pi} \Rightarrow R_{eq} = \dfrac{6\pi\lambda r}{3\pi + 16}

Final answer:

\dfrac{6\pi\lambda r}{3\pi+16}\ \Omega

Q27Single correctOscillations and Waves

Two blocks with masses 100g and 200g are attached to the ends of springs A and B as shown in figure. The energy stored in A is E. The energy stored in B, when spring constants

k_A, k_B

f A and B respectively satisfy the relation

4k_A = 3k_B

Two vertical springs hang from a fixed hatched ceiling. The left spring is labelled A at the top and carries a small block labelled '100 g' at its lower end. The right spring is labelled B at the top and carries a hatched block labelled '200 g' at its lower end, which hangs lower than the 100 g block. Both blocks are drawn as small squares.

(A)

\displaystyle 4E

(B)

\displaystyle 3E

(C)

\displaystyle 2E

(D)

\displaystyle \dfrac{4}{3}E

Q28Single correctUnits and Measurements

Four persons measure the length of a rod as 20.00 cm, 19.75 cm, 17.01 cm and 18.25 cm. The relative error in the measurement of average length of the rod is:

(A)

\displaystyle 0.06

(B)

\displaystyle 0.08

(C)

\displaystyle 0.24

(D)

\displaystyle 0.18

Q29Single correctMagnetic Effects of Current and Magnetism

Given Below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R) Consider a ferromagnetic material:
Assertion (A): The individual atoms in a ferromagnetic material possess a magnetic dipole moment and interact with one another in such a way that they spontaneously align themselves forming domains.
Reason (R): At high enough temperature, the domain structure of ferromagnetic material disintegrate. Thus magnetization will disappear at high enough temperature known as Curie temperature.
In the light of the above statement choose the correct answer from the options given below:

(A)

Both (A) and (R) are true but (R) is not the correct explanation of (A)

(B)

(A) is true but (R) is false

(C)

Both (A) and (R) are true and (R) is the correct explanation of (A)

(D)

(A) is false but (R) is true

Q30Single correctWork, Energy and Power

In a perfectly inelastic collision, two spheres made of the same material with masses 15kg and 25kg moving in opposite directions with speeds of 10 m/s and 30 m/s respectively, strike each other and stick together. The rise in temperature (in C), if all the heat produced during the collision is retained by these spheres is: (specific heat of sphere material 31 cal/kg.

^{0}

C and 1 cal

=4.2

(A)

\displaystyle 1.95

(B)

\displaystyle 1.44

(C)

\displaystyle 1.75

(D)

\displaystyle 1.15

Q31Single correctRotational Motion

Two small balls with masses m and 2m are attached to both ends of a rigid rod of length d and negligible mass. If angular momentum of this system is L about an axis (A) passing through its centre of mass and perpendicular to the rod then angular velocity of the system about A is:

(A)

\displaystyle \dfrac{3}{2}\dfrac{L}{md^2}

(B)

\displaystyle \dfrac{4}{3}\dfrac{L}{md^2}

(C)

\displaystyle \dfrac{2L}{5md^2}

(D)

\displaystyle \dfrac{2L}{md^2}

Q32Single correctElectromagnetic Induction

A 20m long uniform copper wire held horizontally is allowed to fall under the gravity

(g=10\,m/s^2)

trough a uniform horizontal magnetic field of 0.5 Gauss perpendicular to the length of the wire. The induced EMF across the wire when it travels a vertical distance of 200 m is____ mV

(A)

\displaystyle 2\sqrt{10}

(B)

\displaystyle 0.2\sqrt{10}

(C)

\displaystyle 20\sqrt{10}

(D)

\displaystyle 200\sqrt{10}

Q33Single correctOptics

A thin prism with angle

5^{0}

of refractive index 1.72 is combined with another prism of refractive index 1.9 to produced dispersion without deviation. The angle of second prism is______

(A)

\displaystyle 5^{0}

(B)

\displaystyle 6^{0}

(C)

\displaystyle 4^{0}

(D)

\displaystyle 4.5^{0}

Q34Single correctElectrostatics

Two point charges 2q and q are placed at vertex A and centre of face CDEF of the cube as shown in figure. The electric flux passing through the cube is:

A cube drawn in perspective. The top face has corners labelled B (back-left), C (back-right), D (front-right near top), and the front-top-left corner is A. The charge 2q sits at vertex A (top-left-front corner). The bottom face has corners H (front-left), G (back-left-bottom), E (front-right-bottom), F (right). A small dot labelled q is placed at the centre of the right face CDEF (the face with corners C, D, E, F), located roughly mid-height on the right side. Labels around the cube: B and C on top-back edge, D top-front-right, A top-front-left with '2q' just below it, H bottom-front-left, G bottom interior-left, E bottom-front-right, F right side; q marked at centre of face CDEF.

(A)

\displaystyle \dfrac{3q}{2\,\epsilon_0}

(B)

\displaystyle \dfrac{3q}{\epsilon_0}

(C)

\displaystyle \dfrac{3q}{4\,\epsilon_0}

(D)

\displaystyle \dfrac{q}{\epsilon_0}

Q35Single correctUnits and Measurements

In a screw gauge, the zero of the circular scale lies 3 divisions above the horizontal pitch line when their metallic studs are brought in contact. Using this instrument thickness of a sheet is measure If pitch scale reading is 1mm and the circular scale reading is 51 then the correct thickness of the sheet is______ mm. [Assume least count is 0.01mm]

(A)

\displaystyle 1.50

(B)

\displaystyle 1.48

(C)

\displaystyle 1.51

(D)

\displaystyle 1.54

Q36Single correctRotational Motion

The moment of inertia of a square loop made of four uniform solid cylinders, each having radius R and length L (R<L) about an axis passing through the mid points of opposite sides is (Take the mass of the entire loop as M)

(A)

\displaystyle \dfrac{3}{8}MR^2 + \dfrac{7}{12}ML^2

(B)

\displaystyle \dfrac{3}{8}MR^2 + \dfrac{1}{6}ML^2

(C)

\displaystyle \dfrac{3}{4}MR^2 + \dfrac{7}{12}ML^2

(D)

\displaystyle \dfrac{3}{4}MR^2 + \dfrac{1}{6}ML^2

SolutionAnswer: Option 2

\dfrac{3}{8}MR^2 + \dfrac{1}{6}ML^2

Approach:

Givens: a square loop assembled from four identical solid cylinders, each of radius R and length L (

R<L

), total loop mass M, so each cylinder has mass

m_1 = M/4

. The axis passes through the midpoints of two opposite sides. Two cylinders (the sides containing the midpoints) lie along the axis and spin about their own long axis; the other two cylinders lie parallel to the axis at a perpendicular distance

L/2

. Target: total moment of inertia.

Step 1:The two cylinders forming the sides through which the axis passes lie along the axis; each contributes its spin inertia about its own length.

I_{(1)} = 2\times \dfrac{m_1 R^2}{2} = m_1 R^2

Step 2:The remaining two cylinders are oriented transverse to the axis with their centres displaced by

L/2

; apply the transverse formula plus the parallel-axis shift to each.

I_{(2)} = 2\left[\dfrac{m_1 L^2}{12} + \dfrac{m_1 R^2}{4} + m_1\left(\dfrac{L}{2}\right)^2\right]

Step 3:Collect the

L^2

terms from the transverse pair:

\tfrac{1}{12} + \tfrac{1}{4} = \tfrac{1}{12} + \tfrac{3}{12} = \tfrac{4}{12} = \tfrac{1}{3}

per cylinder, doubled.

L^2\text{ terms} = 2\,m_1 L^2\left(\dfrac{1}{12} + \dfrac{1}{4}\right) = 2\,m_1 L^2\cdot\dfrac{1}{3} = \dfrac{2 m_1 L^2}{3}

Step 4:Collect all

R^2

terms:

m_1 R^2

from the axial pair plus

2\cdot\tfrac{m_1 R^2}{4} = \tfrac{m_1 R^2}{2}

from the transverse pair.

R^2\text{ terms} = m_1 R^2 + \dfrac{m_1 R^2}{2} = \dfrac{3 m_1 R^2}{2}

Step 5:Substitute

m_1 = M/4

into both grouped terms.

I = \dfrac{2}{3}\cdot\dfrac{M}{4}\,L^2 + \dfrac{3}{2}\cdot\dfrac{M}{4}\,R^2 = \dfrac{ML^2}{6} + \dfrac{3MR^2}{8}

Final answer:

\dfrac{3}{8}MR^2 + \dfrac{1}{6}ML^2

Q37Single correctWork, Energy and Power

A small bob A of mass m is attached to a massless rigid rod of length 1m pivoted at point P and kept at an angle of

60^{0}

with vertical as shown in figure. At distance of 1m below point P, an identical bob B is kept at rest on a smooth horizontal surface that extends to a circular track of radius R as shown in figure. If bob B just manages to complete the circular path of radius R upto a point Q after being hit elastically by bob A, then radius R is____ m.

A rigid rod of length 1 m is pivoted at point P at the top. The rod hangs at 60° from the vertical (a dashed vertical line drops from P). At the lower end of the rod sits bob A (a small circle labelled m) on the left. The angle 60° is marked between the rod and the dashed vertical, with the rod length '1 m' written along it. Directly below P at a vertical distance of 1 m, on a horizontal surface, sits an identical bob B (small circle labelled m). The horizontal surface extends to the right into a vertical circular loop track of radius R; point Q is marked at the top-left of the circular loop and R is marked at the bottom-right indicating the loop radius. Dashed lines indicate the centre and radius of the circular track.

(A)

\displaystyle \dfrac{1}{5}

(B)

\displaystyle \dfrac{2-\sqrt{3}}{5}

(C)

\displaystyle \dfrac{2+\sqrt{3}}{5}

(D)

\displaystyle \dfrac{3}{5}

Q38Single correctElectromagnetic Waves

Match list – I with List – II

List-I	List-II
A. $\oint \vec{E}\cdot d\vec{l} = -\dfrac{d}{dt}\int \vec{B}\cdot d\vec{a}$	I. Ampere's circuital law
B. $\oint \vec{B}\cdot d\vec{l} = \mu_0\left(I + \epsilon_0\dfrac{d\phi_E}{dt}\right)$	II. Faraday’s law of electromagnetic induction
C. $\oint \vec{E}\cdot d\vec{a} = \dfrac{1}{\epsilon_0}\int_v \rho\,dv$	III. Ampere – Maxwell law
D. $\oint \vec{B}\cdot d\vec{l} = \mu_0 I$	IV. Gauss’s law of electrostatics

(A)

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

(B)

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

(C)

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

(D)

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

Q39Single correctElectronic Devices

The following diagram shows a Zener diode as a voltage regulator. The Zener diode is rated at

V_z = 5V

and the desired current in load is 5 mA. The unregulated voltage source can supply upto 25V. Considering the Zener diode can withstand four times of the load current, the value of resistor

R_s

(Shown in circuit) should be______

\Omega

Zener voltage regulator circuit. On the left a labelled 'Unregulated Voltage' source V_in with its positive terminal at top connecting through a vertical wire to a node. From that node a series resistor R_s (drawn as a zig-zag/coil symbol) runs to the right to an output node. From the output node two parallel branches go down to the common bottom rail: a Zener diode (triangle with bent-bar cathode pointing up, connected in reverse) and a load resistor R_L (zig-zag symbol). The output node is labelled V_o 'Regulated Voltage'. Bottom rail is the common return connecting source negative, Zener anode and R_L.

(A)

\displaystyle 4000

(B)

\displaystyle 1000

(C)

\displaystyle 100

(D)

\displaystyle 10

Q40Single correctProperties of Solids and Liquids

The strain-stress plot for materials A,B,C and D is shown in the figure. Which material has the largest Young's modulus?

First-quadrant graph with vertical axis labelled 'strain' (upward arrow) and horizontal axis labelled 'stress' (rightward arrow). Four straight lines emanate from the common origin, each labelled near its upper end. From steepest to shallowest: line A is the steepest (closest to the strain axis), then B, then D, then C is the shallowest (closest to the stress axis).

(A)

\displaystyle C

(B)

\displaystyle B

(C)

\displaystyle D

(D)

\displaystyle A

Q41Single correctWork, Energy and Power

An object is projected with kinetic energy K from a point A at an angle

60^0

with the horizontal. The ratio of the difference in kinetic energies at point B and C to that at point A (See figure), in the absence of air friction is:

A projectile launched from point A at 60 degrees above the horizontal; dashed parabolic trajectory rising to apex B at the top and descending to point C on the same horizontal level as A; angle of 60 degrees marked at A between the velocity arrow and the horizontal line AC.

(A)

\displaystyle 1:2

(B)

\displaystyle 1:4

(C)

\displaystyle 3:4

(D)

\displaystyle 2:3

Q42Single correctOscillations and Waves

A simple pendulum of straight length 30cm performs 20 oscillations in 10s. The length of the string required for the pendulum to perform 40 oscillations in the same time duration is ________ cm [Assume that the mass of the pendulum remains same]

(A)

\displaystyle 15

(B)

\displaystyle 120

(C)

\displaystyle 7.5

(D)

\displaystyle 0.75

Q43Single correctOptics

Consider light travelling from a medium A to medium B separated by a plane interface, If the light undergoes total internal reflection during its travel from medium A to B and the speed of light in media A and B are

2.4\times10^8\,m/s

and

27\times10^8\,m/s

, respectively then the value of critical angle is:

(A)

\displaystyle \sin^{-1}\left(\dfrac{9}{8}\right)

(B)

\displaystyle \cot^{-1}\left(\dfrac{3}{\sqrt{13}}\right)

(C)

\displaystyle \cos^{-1}\left(\dfrac{8}{9}\right)

(D)

\displaystyle \tan^{-1}\left(\dfrac{8}{\sqrt{17}}\right)

Q44Single correctAtoms and Nuclei

In hydrogen atom spectrum (R

\rightarrow

Rydberg's constant)
A) The maximum wavelength of the radiation of Lyman series is

\dfrac{4}{3R}

B) The Balmer series lies in the visible region of the spectrum
C) The minimum wavelength of the radiation of Paschen series

\dfrac{9}{R}

D) The minimum wavelength of Lyman series is

\dfrac{5}{4R}

Choose the correct answer from the options given below:

(A)

A, B and C Only

(B)

A, B Only

(C)

B, D Only

(D)

A, B and D Only

Q45Single correctDual Nature of Matter and Radiation

The de Broglie wavelength of an oxygen molecule at

27^0 C

x\times10^{-12}\,m

. The value of x is (take Planck's constant

= 6.63\times10^{-34}\,J.s

Boltzmann constant

= 1.38\times10^{-23}\,J/K

. mass of oxygen molecule

= 5.31\times10^{-26}\,kg

)

(A)

\displaystyle 20

(B)

\displaystyle 24

(C)

\displaystyle 26

(D)

\displaystyle 30

Q46NumericalElectromagnetic Waves

The equation the electric field of an electromagnetic wave propagating through free space is given by :

E = \sqrt{377}\sin\left(6.27\times10^3 t - 2.09\times10^{-5} x\right)\,N/C

The average power of the electromagnetic wave is

\left(\dfrac{1}{\alpha}\right)

m^2

. The value of

\alpha

is___

\left(\text{Take}\ \sqrt{\dfrac{\mu_0}{\in_0}} = 377\,in\,SI\,\text{units}\right)

Q47NumericalOptics

In two separate Young's double - slit experimental set-ups and two monochromatic light sources of different wavelength are used to get fringes of equal width. The ratios of the slits separations and that of the wavelengths of light used are 2:1 and 1:2 respectively. The corresponding ratio of the distance between the slits and the respective screens

\left(D_1 / D_2\right)

is___

Q48NumericalElectrostatics

The space between the plates of a parallel plate capacitor of capacitance C (without any dielectric) is now filled with three dielectric slabs of dielectric constants

K_1 = 2, K_2 = 3

and

K_3 = 5

(as shown in figure). If new capacitance is

\dfrac{n}{3}C

then the value of n is___

Parallel plate capacitor shown as a rectangle between two horizontal plates; the top plate carries a '+' terminal and the bottom plate carries a '-' terminal. Total plate gap is labelled d on the right side. The interior is divided into an upper region of height d/2 fully occupied by dielectric K1 spanning the full plate width A (top edge labelled A, with a left vertical bracket marking d/2). The lower region of height d/2 is split vertically into two equal halves: the left half (hatched) is dielectric K2 and the right half (hatched) is dielectric K3, each spanning width A/2 (bottom bracket labels A/2).

Q49NumericalElectromagnetic Induction and Alternating Currents

Using a variable frequency a.c. voltage source the maximum current measured in the given LCR circuit is 50 mA for

V=5\sin(100t)

The values of L and R are shown in the figure. The capacitance of the capacitor (C) used is______

\mu F

Series LCR circuit drawn as a single horizontal loop. Along the top branch from left to right: an inductor (coil symbol) labelled L = 2 H, then a capacitor (two parallel plates) labelled C, then a resistor (zig-zag) labelled R = 100 Ohm. The bottom branch contains the a.c. source symbol (circle with a sine mark) labelled V = 5 sin(100 t), closing the loop with the three components in series.

Q50NumericalElectromagnetic Induction and Alternating Currents

A Simple Pendulum made of mass 10g and a metallic wire of length 10cm is suspended vertically in a uniform magnetic field of 2T. The magnetic field direction is perpendicular to the plane of oscillations of the pendulum. If the pendulum is released from an angle of

60^0

with vertical, then maximum induced EMF between the point of suspension and point of oscillations is ________mV. (Take

g=10\,m/s^2

)

Chemistry25 questions

Q51Single correctCoordination Compounds

The statements that are incorrect about the nickel (II) complex of dimethylglyoxime are:
A) It is red in colour.
B) It has a high solubility in water at pH = 9.
C) The Ni ion has two unpaired d-electrons.
D) The N

-

-

N bond angle is almost close to

90^0

E) The complex contains four five-membered metallacycles (metal containing rings) Choose the correct answer from the options given below :

(A)

A, D and B only

(B)

C and D only

(C)

C and E only

(D)

B, C and E only

SolutionAnswer: Option 4B, C and E only

Approach:

Given the bis(dimethylglyoximato)nickel(II) complex formed in ammoniacal medium, each statement is tested against its established structure, geometry, magnetic behaviour and chelate ring count to mark which statements are incorrect.

Step 1:Ni(II) is a

d^8

ion. With the strong dimethylglyoximate field the complex is square planar through

\text{dsp}^2

hybridization, which forces all eight d electrons into the four lower orbitals as pairs, leaving zero unpaired electrons (diamagnetic). Statement C asserts two unpaired electrons, contradicting this pairing.

\text{Ni}^{2+}\ (d^8),\ dsp^2,\ \text{square planar} \Rightarrow 0\ \text{unpaired}

Step 2:The complex precipitates as a bright rosy-red solid in ammoniacal solution near pH 9 and is sparingly soluble in water, which is the basis of its use as a gravimetric test for nickel. Statement A (red colour) therefore holds, while statement B (high water solubility at pH 9) contradicts the low solubility of the red precipitate.

\text{Ni(DMG)}_2 \downarrow\ (\text{red, sparingly soluble})

Step 3:Square planar coordination places the four donor nitrogen atoms at corners of a square, so each adjacent

\text{N}-\text{Ni}-\text{N}

angle is about

90^0

, confirming statement D. Each dimethylglyoximate ligand chelates through two nitrogens to close one five-membered metallacycle, giving two five-membered rings in total; the two remaining rings are six-membered, closed by intramolecular

\text{O}-\text{H}\cdots\text{O}

hydrogen bonds. Statement E claims four five-membered metallacycles, which overcounts.

\angle(\text{N-Ni-N}) \approx 90^0;\ 2\ \text{five-membered} + 2\ \text{six-membered rings}

Step 4:Collecting the incorrect statements gives B (solubility), C (unpaired electrons) and E (ring count), matching the set B, C and E only.

\text{incorrect} = \{B, C, E\}

Final answer: B, C and E only

Q52Single correctChemical Bonding and Molecular Structure

Identify the molecule (X) with maximum number of lone pairs of electrons (obtained using Lewis dot structure) among

\text{HNO}_3

\text{H}_2\text{SO}_4

\text{NF}_3

and

\text{O}_3

. Choose the correct bond angle made by the central atom of the molecule (X).

(A)

\displaystyle 120^0

(B)

\displaystyle 102^0

(C)

\displaystyle 107^0

(D)

\displaystyle 116^0

SolutionAnswer: Option 2

102^0

Approach:

Given four molecules, the total number of lone pairs in each is counted from its Lewis dot structure, the molecule X holding the maximum is identified, and the bond angle at its central atom is stated.

Step 1:In

\text{HNO}_3

the singly bonded terminal O and the O of the O

-

H carry the bulk of lone pairs; summing over the three oxygens (with one N=O, one N

-O^-

and one N

-

OH) totals 7 lone pairs on the molecule.

\text{HNO}_3:\ 7\ \text{lone pairs}

Step 2:In

\text{H}_2\text{SO}_4

the two S=O oxygens carry two lone pairs each and the two S

-O-

H oxygens carry two lone pairs each, totalling 8 lone pairs.

\text{H}_2\text{SO}_4:\ 8\ \text{lone pairs}

Step 3:In

\text{NF}_3

each fluorine carries three lone pairs (

3 \times 3 = 9

) and nitrogen carries one lone pair, totalling 10 lone pairs.

\text{NF}_3:\ 3(3) + 1 = 10\ \text{lone pairs}

Step 4:In

\text{O}_3

the central O has one lone pair, one terminal O (double bonded) has two, and the other terminal O (single bonded, bearing negative charge) has three, totalling 6 lone pairs. Comparing 7, 8, 10 and 6, the maximum belongs to

\text{NF}_3

, so X is

\text{NF}_3

\text{O}_3:\ 6;\ \max = \text{NF}_3\ (10)

Step 5:Nitrogen in

\text{NF}_3

sp^3

hybridized with one lone pair and three N

-

F bonds. The three highly electronegative fluorines pull bonding electron density away from nitrogen, shrinking the bonding-pair mutual repulsion; combined with lone-pair to bond-pair repulsion this compresses the angle well below the tetrahedral

109.5^0

to roughly

102^0

\angle(\text{F-N-F}) \approx 102^0

Final answer:

102^0

Q53Single correctSolutions

Which of the following graphs accurately represents the plot of partial pressure of

\text{CS}_2

vs its mole fraction in a mixture of acetone and

\text{CS}_2

at constant temperature?

(A)

Graph with vertical axis p/mm Hg and horizontal axis x_CS2 from 0 (pure acetone) to 1 (pure CS2). A solid curve rises from origin, bowing upward (concave down) and ending at p-degree-CS2 at right. A dashed straight line from origin to p-degree-CS2 represents Raoult's law (lower than the solid curve). A dashed straight line tangent to the curve at the origin, steeper, labelled Henry's law. Labels: Henry's law on upper steep dashed tangent, Raoult's law on lower dashed straight line.

(B)

Graph axes p/mm Hg vs x_CS2 (0 pure acetone to 1 pure CS2). Two straight solid lines forming an X: one rising from lower-left to p-degree-CS2 at upper-right, the other descending from p-degree-acetone at upper-left to lower-right. Represents ideal solution behaviour (straight Raoult lines).

(C)

Graph axes p/mm Hg vs x_CS2. Curves bowing downward (negative deviation): solid curve starting at p-degree-acetone top-left dropping below straight dashed lines, another solid curve rising to p-degree-CS2 right, both lying below their respective dashed Raoult straight lines, crossing near middle.

(D)

Graph axes p/mm Hg vs x_CS2. Curves bowing upward (positive deviation) for both components: solid curve from p-degree-acetone top-left, solid curve to p-degree-CS2 top-right, both lying above the dashed Raoult straight lines, crossing in the middle region.

SolutionAnswer: Option 1Curve showing positive deviation: partial pressure of CS2 lies above the Raoult's law line and follows the Henry's law tangent at low mole fraction.

Approach:

Given a single partial-pressure plot for one component (

\text{CS}_2

) of the acetone

-\text{CS}_2

system, the nature of deviation from ideality and the two limiting laws (Raoult's at high mole fraction, Henry's at low mole fraction) are used to identify the correct curve.

Step 1:

\text{CS}_2

is non-polar while acetone

(\text{CH}_3\text{COCH}_3)

is polar. Mixing them replaces strong acetone

-

acetone dipole interactions with weaker acetone

-\text{CS}_2

interactions, so molecules escape into the vapour more readily. This raises each partial pressure above the value predicted by Raoult's law, the signature of positive deviation.

p_{\text{CS}_2} > x_{\text{CS}_2}\,p^{\circ}_{\text{CS}_2}

Step 2:At the dilute end

x_{\text{CS}_2} \to 0

the

\text{CS}_2

molecules are surrounded almost entirely by acetone, so its partial pressure obeys Henry's law as a straight tangent of slope

K_H

. Positive deviation requires

K_H > p^{\circ}_{\text{CS}_2}

, so this tangent is steeper than the Raoult's line and lies above the actual curve near the origin.

p_{\text{CS}_2} \to K_H\,x_{\text{CS}_2}\ \text{as}\ x_{\text{CS}_2}\to 0,\ K_H > p^{\circ}_{\text{CS}_2}

Step 3:At the concentrated end

x_{\text{CS}_2} \to 1

the curve meets

p^{\circ}_{\text{CS}_2}

and becomes tangent to the Raoult's line. The required plot therefore rises from the origin, bows above its dashed Raoult straight line, carries a steeper dashed Henry tangent at low mole fraction, and terminates at

p^{\circ}_{\text{CS}_2}

. Among the options, only option 1 plots this single

\text{CS}_2

curve with both labelled tangents; option 2 is ideal (straight), option 3 is negative deviation (curve below the line), and option 4 shows both components rather than the single requested

\text{CS}_2

curve.

x_{\text{CS}_2}\to 1 \Rightarrow p_{\text{CS}_2}\to p^{\circ}_{\text{CS}_2}

Final answer: Option 1

Q54Single correctPurification and Characterisation of Organic Compounds

Given below are two statements:
Statement

-

I : Sublimation is used for the separation and purification of compounds with low melting point.
Statement

-

II : The boiling point of a liquid increases as the external pressure is reduced.
In the light of the above statements, choose the correct answer from the options given below :

(A)

Both Statement

\displaystyle -

I and Statement

\displaystyle -

II are false

(B)

Both Statement

\displaystyle -

I and Statement

\displaystyle -

II are true

(C)

Statement

\displaystyle -

I is false but Statement

\displaystyle -

II is true

(D)

Statement

\displaystyle -

I is true but Statement

\displaystyle -

II is false

Q55Single correctPurification and Characterisation of Organic Compounds

Match List – I wish List – II:

List – I (Functional group (detection))	List – II (Change observed during detection)
A. Unsaturation (Baeyer’s test)	I. Red colour appears
B. Alcoholic group (Ceric ammonium nitrite test)	II. Silver mirror appears
C. Aldehyde group (Tollen’s reagent)	III. Violet colour appears
D. Phenolic group $\text{FeCl}_3$ test	IV. Discharge of pink colour

(A)

\displaystyle -

IV; B

\displaystyle -

III; C

\displaystyle -

II; D

\displaystyle -

(B)

\displaystyle -

III; B

\displaystyle -

IV; C

\displaystyle -

I; D

\displaystyle -

(C)

\displaystyle -

III; B

\displaystyle -

IV; C

\displaystyle -

II; D

\displaystyle -

(D)

\displaystyle -

IV; B

\displaystyle -

I; C

\displaystyle -

II; D

\displaystyle -

III

SolutionAnswer: Option 4A

-

IV; B

-

I; C

-

II; D

-

III

Approach:

Given four functional-group detection tests in List I and four observations in List II, each test is matched to the colour change or visible result it produces, and the option reproducing all four pairs is chosen.

Step 1:Baeyer's test for unsaturation uses cold dilute alkaline

\text{KMnO}_4

(pink/purple). An alkene reduces

\text{MnO}_4^-

to brown

\text{MnO}_2

while forming a diol, so the pink colour is discharged. A pairs with observation IV.

3\,\text{C=C} + 2\,\text{KMnO}_4 + 4\,\text{H}_2\text{O} \to 3\,\text{diol} + 2\,\text{MnO}_2 + 2\,\text{KOH}

Step 2:The ceric ammonium nitrate test for an alcoholic

-\text{OH}

group forms a transient red alkoxy-cerium(IV) complex, turning the yellow reagent red. B pairs with observation I.

\text{ROH} + \text{Ce(IV)} \to \text{red complex}

Step 3:Tollen's reagent, the ammoniacal silver(I) complex

[\text{Ag(NH}_3)_2]^+

, is reduced by an aldehyde to metallic silver that deposits as a mirror on the glass. C pairs with observation II.

\text{RCHO} + 2[\text{Ag(NH}_3)_2]^+ + 3\,\text{OH}^- \to \text{RCOO}^- + 2\,\text{Ag}\!\downarrow + 4\,\text{NH}_3 + 2\,\text{H}_2\text{O}

Step 4:A phenol reacts with neutral

\text{FeCl}_3

to form a coloured iron(III)

-

phenolate complex that is violet (ranging to blue/green by substituent). D pairs with observation III. The full mapping A

-

IV, B

-

I, C

-

II, D

-

III matches option 4.

6\,\text{C}_6\text{H}_5\text{OH} + \text{FeCl}_3 \to [\text{Fe(OC}_6\text{H}_5)_6]^{3-}\ (\text{violet}) + 3\,\text{HCl} + 3\,\text{H}^+

Final answer: A

-

IV; B

-

I; C

-

II; D

-

III

Q56Single correctOrganic Compounds Containing Halogens

\text{(X)} \xrightarrow{?} \text{(Y)}

The correct sequence of reagents for the above conversion of X to Y is :

Reaction X to Y. X: a benzene ring (hexagon) attached to a carbon bearing a CH3 group and a Br substituent, i.e. (1-bromoethyl)benzene: Ph-CHBr-CH3. An arrow labelled with a question mark points to Y: a benzene ring attached to a CH2 group bearing a COOH group, i.e. phenylacetic acid Ph-CH2-COOH.

(A)

(i)

\displaystyle \text{B}_2\text{H}_6 / \text{H}_2\text{O}_2

(ii) NaOEt (iii) Jones reagent

(B)

(i) NaOH (aq) (ii) Jones reagent (iii)

\displaystyle \text{H}_3\text{O}

(C)

(i) NaOEt (ii)

\displaystyle \text{B}_2\text{H}_6 / \text{H}_2\text{O}_2

(iii) Jones reagent

(D)

(i) Jones reagent (ii) NaOEt (iii) Hot

\displaystyle \text{KMnO}_4 / \text{KOH}

Q57Single correctAtomic Structure

Given :
(A) n

=

m_l = -1

(B) n

=

l = 2

m_l = -1

m_s = \frac{1}{2}

The maximum number of electron(s) in an atom that can have the quantum numbers as given in (A) and (B) respectively are :

(A)

26 and 1

(B)

2 and 4

(C)

8 and 1

(D)

4 and 1

Q58Single correctOrganic Compounds Containing Nitrogen

Consider the following sequences of reactions: 4-nitrotoluene

\xrightarrow[\text{(ii) pH neutralisation}]{\text{(i) Sn, HCl}/\Delta}

(A)

\xrightarrow[\text{(ii) Br}_2/\text{AcOH}]{\text{(i) (H}_3\text{CCO)}_2\text{O}}

(B). Assuming that the reaction proceeds to completion then 137 mg of 4-nitrotoluene will produce

\ldots\ldots\ldots

mg of B. (Given molar mass in g

\text{mol}^{-1}

H : 1, C : 12, N : 14, O : 16, Br : 80 )

Reaction scheme. Starting structure: a benzene ring with a CH3 group at the top (para position) and an NO2 group at the bottom, labelled 4-nitrotoluene. An arrow with conditions (i) Sn, HCl/heat (ii) pH neutralisation leads to product (A). A second arrow with conditions (i) (H3CCO)2O acetic anhydride (ii) Br2/AcOH leads to product (B).

(A)

208

(B)

301

(C)

228

(D)

146

SolutionAnswer: Option 3228

Approach:

Given 137 mg of 4-nitrotoluene through a reduction, acetylation and mono-bromination sequence, products A and B are identified, the molecular formula and molar mass of B are computed, and the one-to-one stoichiometry converts moles of starting material to mass of B.

Step 1:Sn/HCl with heat reduces the

-\text{NO}_2

group to

-\text{NH}_2

; pH neutralisation liberates the free amine, giving 4-methylaniline (A),

\text{CH}_3\text{-C}_6\text{H}_4\text{-NH}_2

\text{4-O}_2\text{N-C}_6\text{H}_4\text{-CH}_3 \xrightarrow{\text{Sn/HCl},\ \Delta;\ \text{neutralise}} \text{4-H}_2\text{N-C}_6\text{H}_4\text{-CH}_3

Step 2:Acetic anhydride acetylates the amine to the acetamide,

\text{CH}_3\text{-C}_6\text{H}_4\text{-NHCOCH}_3

, protecting and deactivating it for controlled mono-substitution.

\text{Br}_2/\text{AcOH}

then introduces one bromine onto the ring (ortho to the strongly directing

-\text{NHCOCH}_3

), giving B as the mono-bromo acetamido toluene,

\text{CH}_3\text{-C}_6\text{H}_3(\text{Br})\text{-NHCOCH}_3

\text{A} \xrightarrow{(\text{CH}_3\text{CO})_2\text{O}} \text{ArNHCOCH}_3 \xrightarrow{\text{Br}_2/\text{AcOH}} \text{Br-substituted ArNHCOCH}_3\ (\text{B})

Step 3:Molar mass of B from

\text{C}_9\text{H}_{10}\text{BrNO}

: carbon

9 \times 12 = 108

, hydrogen

10 \times 1 = 10

, bromine 80, nitrogen 14, oxygen 16. Summing gives

108 + 10 + 80 + 14 + 16 = 228\ \text{g mol}^{-1}

M_{\text{B}} = 108 + 10 + 80 + 14 + 16 = 228\ \text{g mol}^{-1}

Step 4:Moles of 4-nitrotoluene (

M = 137\ \text{g mol}^{-1}

n = \dfrac{137 \times 10^{-3}}{137} = 1.0 \times 10^{-3}\ \text{mol}

. Each step is one-to-one, so

1.0 \times 10^{-3}

mol of B forms. Mass of B

= 1.0 \times 10^{-3} \times 228 = 0.228\ \text{g} = 228\ \text{mg}

\text{mass}_{\text{B}} = 1.0 \times 10^{-3} \times 228 = 0.228\ \text{g} = 228\ \text{mg}

Final answer: 228

Q59Single correctEquilibrium

Consider the general reaction given below at 400 K

x\text{A}(g) \rightleftharpoons y\text{B}(g)

The value of

K_p

and

K_c

are studied under the same condition of temperature but variation in x and y.
I)

K_p = 85.87

and

K_c = 2.586

appropriate units
II)

K_p = 0.862

and

K_c = 28.62

appropriate units
The values of x and y in (i) and (ii) respectively are :

(A)

(i) 3, 1 (ii) 3, 1

(B)

(i) 4, 1 (ii) 4, 1

(C)

(i) 1, 3 (ii) 2, 1

(D)

(i) 1, 2 (ii) 2, 1

Q60Single correctp-Block Elements

The correct statements from the following are:
A) Ionic radii of trivalent cations of group 13 elements decreases down the group.
B) Electro negativity of group 13 elements decreases down the group
C) Among the group 13 elements, Boron has highest first ionisation enthalpy
D) The trichloride and triiodide of group 13 elements are covalent in nature. Choose the correct answer from the options given below :

(A)

C and D only

(B)

B and D only

(C)

A and D only

(D)

A and C only

SolutionAnswer: Option 1C and D only

Approach:

Given four statements on group 13 trends, each is tested against the established periodic behaviour of ionic radius, electronegativity, first ionisation enthalpy and halide covalency, and the correct pair is chosen.

Step 1:Down group 13 the principal quantum number of the trivalent cation increases, so the ionic radius rises in the order

\text{B}^{3+} < \text{Al}^{3+} < \text{Ga}^{3+} < \text{In}^{3+} < \text{Tl}^{3+}

(with only a small contraction at Ga from poor d-shielding). Statement A claims a decrease down the group, contradicting this increase.

r(\text{B}^{3+}) < r(\text{Al}^{3+}) < r(\text{Ga}^{3+}) < r(\text{In}^{3+}) < r(\text{Tl}^{3+})

Step 2:Electronegativity in group 13 does not fall steadily; after the drop from B to Al it rises again because of poor d- and f-orbital shielding for Ga, In and Tl, giving the irregular order

\text{B} > \text{Tl} > \text{In} > \text{Ga} > \text{Al}

. Statement B claims a uniform decrease, which is false.

\text{B} > \text{Tl} > \text{In} > \text{Ga} > \text{Al}

Step 3:First ionisation enthalpy follows the irregular order

\text{B} > \text{Tl} > \text{Ga} > \text{Al} > \text{In}

. Boron, being smallest with electrons held tightest, has the highest first ionisation enthalpy, so statement C is true.

IE_1:\ \text{B} > \text{Tl} > \text{Ga} > \text{Al} > \text{In}

Step 4:The small, highly charged group 13

\text{M}^{3+}

centre strongly polarises large halide anions (Fajans' rules), so the trichlorides and triiodides are predominantly covalent (e.g. dimeric

\text{Al}_2\text{Cl}_6

, molecular

\text{BCl}_3

\text{BI}_3

). Statement D is true. Correct statements are C and D.

\text{MX}_3\ (\text{X}=\text{Cl, I}) \Rightarrow \text{covalent}

Final answer: C and D only

Q61Single correctHydrocarbons

But-2-yne and hydrogen (one mole each) are separately treated with (i)

\text{Pd}/\text{C}

and (ii)

\text{Na}/\text{liq.NH}_3

to give the products X and Y respectively. Identify the incorrect statements :
A) X and Y are stereo isomers
B) Dipole moment of X is zero
C) Boiling point of X is higher than Y
D) X and Y react with

\text{O}_3 / \text{Zn} + \text{H}_2\text{O}

to give different products Choose the correct answer from the options given below :

Reaction scheme. But-2-yne plus H2. An upper arrow labelled Pd/C points to product X. A lower arrow labelled Na/Liq.NH3 points to product Y.

(A)

A and C only

(B)

B and C only

(C)

B and D only

(D)

A and B only

SolutionAnswer: Option 3B and D only

Approach:

Given partial reduction of but-2-yne by two stereo-selective routes, products X and Y are identified by their geometry, then each statement is judged to collect the incorrect ones.

Step 1:Catalytic hydrogenation over Pd/C adds the two H atoms to the same face (syn addition), converting but-2-yne to cis-but-2-ene (X). Dissolving-metal reduction with Na in liquid

\text{NH}_3

proceeds through a trans vinyl radical/anion, giving trans-but-2-ene (Y). X and Y differ only in configuration about the C=C bond, so they are geometric (stereo) isomers.

\text{CH}_3\text{C}\!\equiv\!\text{CCH}_3 \xrightarrow{\text{Pd/C}} \text{cis-but-2-ene (X)};\ \xrightarrow{\text{Na/liq.NH}_3} \text{trans-but-2-ene (Y)}

Step 2:In cis-but-2-ene (X) the two methyl groups lie on the same side, so the two C(sp2)

-

CH3 bond dipoles do not cancel and the molecule has a small but non-zero net dipole moment. In trans-but-2-ene (Y) the symmetric arrangement makes the dipoles cancel, giving zero net moment. Statement B asserts the dipole moment of X is zero, which is false for the cis isomer.

\mu(\text{cis-X}) > 0,\quad \mu(\text{trans-Y}) = 0

Step 3:Because X (cis, polar) has stronger dipole

-

dipole intermolecular attractions than the non-polar trans isomer Y, the boiling point of cis-but-2-ene (

\approx 4\,^{\circ}\text{C}

) exceeds that of trans-but-2-ene (

\approx 1\,^{\circ}\text{C}

). Statement C (b.p. of X higher than Y) is correct.

T_b(\text{cis-X}) > T_b(\text{trans-Y})

Step 4:Ozonolysis cleaves the internal C=C of both isomers at the identical position; cis and trans but-2-ene both give two molecules of ethanal (acetaldehyde). The products are the same, so statement D (different products) is false.

\text{cis/trans-but-2-ene} \xrightarrow{\text{O}_3;\ \text{Zn},\text{H}_2\text{O}} 2\,\text{CH}_3\text{CHO}

Final answer: B and D only

Q62Single correctBiomolecules

From the given following (A to D) cyclic structures, those which will not react with Tollen's reagent are :

Four cyclic carbohydrate Haworth-style structures labelled A to D. A: pyranose six-membered ring with CH2OH at top carbon and a free OH at the anomeric carbon (hemiacetal); ring carbons bearing H and OH substituents in glucopyranose arrangement. B: same pyranose ring but the anomeric carbon bears OCH3 instead of OH (methyl glycoside / acetal), CH2OH at top. C: furanose five-membered ring with HOH2C group on one side and OCH3 at the anomeric carbon (acetal/glycoside), ring carbons bearing OH and H. D: pyranose six-membered ring with CH2OH at top and a free OH at the anomeric carbon (hemiacetal), with HO and OH substituents on ring carbons.

(A)

B and D

(B)

A and B

(C)

B and C

(D)

A and D

Q63Single correctOrganic Compounds Containing Oxygen

'x' is the product which is obtained from propanenitrile and stannous chloride in the presence of hydrochloric acid followed by hydrolysis. 'y' is the product which is obtained from the but-2-ene by the ozonolysis followed by hydrolysis. From the following, which product is not obtained when one mole of 'x' and one mole of 'y' react with each other in presence of alkali followed by heating?

(A)

Pent-2-enal

(B)

3-Methylbut-2-enal

(C)

2-Methylpent-2-enal

(D)

2-methylbut-2-enal

SolutionAnswer: Option 23-Methylbut-2-enal

Approach:

Given x from Stephen reduction of propanenitrile and y from ozonolysis of but-2-ene, both aldehydes are identified, all self- and cross-aldol condensation products under base and heat are enumerated, and the option absent from this set is selected.

Step 1:Stephen reduction treats the nitrile with

\text{SnCl}_2/\text{HCl}

to form an imine salt that hydrolyses to the aldehyde with the same carbon count. Propanenitrile

\text{CH}_3\text{CH}_2\text{CN}

(three carbons) gives propanal (x),

\text{CH}_3\text{CH}_2\text{CHO}

\text{CH}_3\text{CH}_2\text{CN} \xrightarrow{\text{SnCl}_2/\text{HCl};\ \text{H}_2\text{O}} \text{CH}_3\text{CH}_2\text{CHO}\ (x)

Step 2:Ozonolysis cleaves the symmetric internal double bond of but-2-ene

\text{CH}_3\text{CH=CHCH}_3

; reductive workup gives two molecules of ethanal (y),

\text{CH}_3\text{CHO}

\text{CH}_3\text{CH=CHCH}_3 \xrightarrow{\text{O}_3;\ \text{Zn},\text{H}_2\text{O}} 2\,\text{CH}_3\text{CHO}\ (y)

Step 3:Both propanal and ethanal carry

\alpha

-hydrogens, so under alkali each can act as either the enolate (nucleophile) or the carbonyl (electrophile). The four aldol pairings, after dehydration on heating, give: ethanal enolate + ethanal

\to

but-2-enal (

\text{CH}_3\text{CH=CHCHO}

); propanal enolate + propanal

\to

2-methylpent-2-enal; ethanal enolate + propanal

\to

pent-2-enal (

\text{CH}_3\text{CH}_2\text{CH=CHCHO}

); propanal enolate + ethanal

\to

2-methylbut-2-enal (

\text{CH}_3\text{CH=C(CH}_3)\text{CHO}

\{\text{but-2-enal},\ \text{2-methylpent-2-enal},\ \text{pent-2-enal},\ \text{2-methylbut-2-enal}\}

Step 4:3-Methylbut-2-enal,

(\text{CH}_3)_2\text{C=CHCHO}

, has a gem-dimethyl pattern at the

\beta

-carbon. Forming it would require an enolate

\alpha

-carbon bearing two methyl substituents or a ketone partner, neither of which arises from propanal/ethanal aldol pairing; the available enolates supply at most one

\alpha

-methyl. The option not obtainable is 3-methylbut-2-enal.

(\text{CH}_3)_2\text{C=CHCHO}\ \notin\ \text{aldol product set}

Final answer: 3-Methylbut-2-enal

Q64Single correctElectrochemistry

In the given electrochemical cell,

\text{Ag}(s)\,|\,\text{AgCl}(s)\,|\,\text{FeCl}_2(aq),\,\text{FeCl}_3(aq)\,|\,\text{Pt}(s)

at 298 K, the cell potential

E_{\text{cell}}

will increase when :
A) Concentration of

\text{Fe}^{2+}

is increased
B) Concentration of

\text{Fe}^{3+}

is decreased
C) Concentration of

\text{Fe}^{2+}

is decreased
D) Concentration of

\text{Fe}^{3+}

is increased
E) Concentration of

\text{Cl}^-

is increased
Choose the correct answer from the options given below :

(A)

\displaystyle \text{A and E only}

(B)

\displaystyle \text{B only}

(C)

\displaystyle \text{C, D and E only}

(D)

\displaystyle \text{A and B only}

SolutionAnswer: Option 3

\text{C, D and E only}

Approach:

Given a galvanic cell with a

\text{Ag}|\text{AgCl}|\text{Cl}^-

left electrode and a

\text{Fe}^{3+}/\text{Fe}^{2+}|\text{Pt}

right electrode; asked which concentration changes raise the measured cell potential. The cell reaction is assembled, the reaction quotient is written, and the Nernst equation determines the sign of each effect.

Step 1:At the left electrode silver is oxidised in the presence of chloride to form solid silver chloride, releasing one electron. At the right electrode ferric ion is reduced to ferrous ion accepting one electron.

\text{Ag}+\text{Cl}^-\rightarrow\text{AgCl}+e^-;\quad\text{Fe}^{3+}+e^-\rightarrow\text{Fe}^{2+}

Step 2:Adding the two half reactions gives the overall cell reaction whose reaction quotient places the product ferrous ion in the numerator and the reactant chloride and ferric ions in the denominator.

\text{Ag}+\text{Cl}^-+\text{Fe}^{3+}\rightarrow\text{AgCl}+\text{Fe}^{2+},\quad Q=\dfrac{[\text{Fe}^{2+}]}{[\text{Cl}^-][\text{Fe}^{3+}]}

Step 3:Substituting into the Nernst equation shows the cell potential rises whenever the logarithm of the quotient falls, that is whenever the quotient decreases.

E_{\text{cell}}=E^{0}_{\text{cell}}-0.059\log\dfrac{[\text{Fe}^{2+}]}{[\text{Cl}^-][\text{Fe}^{3+}]}

Step 4:Decreasing ferrous ion concentration (C), increasing ferric ion concentration (D), and increasing chloride concentration (E) each lower the quotient and therefore raise the potential. Increasing ferrous (A) or decreasing ferric (B) would raise the quotient and lower the potential.

[\text{Fe}^{2+}]\downarrow,\;[\text{Fe}^{3+}]\uparrow,\;[\text{Cl}^-]\uparrow\Rightarrow Q\downarrow\Rightarrow E_{\text{cell}}\uparrow

Final answer: C, D and E only

Q65Single correctCoordination Compounds

Given below are two statements :
Statement - I :

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

ions will absorb light of lower energy than

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

ion.
Statement - II : In

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

ion, the energy separation between the two set of d-orbitals is more than

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

ion.
In the light of the above statements, choose the correct answer from the options given below :

(A)

\displaystyle \text{Statement - I is true but Statement - II is false}

(B)

\displaystyle \text{Both Statement - I and Statement - II are true}

(C)

\displaystyle \text{Both Statement - I and Statement - II are false}

(D)

\displaystyle \text{Statement - I is false but Statement - II is true}

Q66Single correctClassification of Elements and Periodicity in Properties

The correct trend in the first ionization enthalpies of the elements in the

3^{\text{rd}}

periodic table is :

(A)

\displaystyle \text{S}<\text{Si}<\text{Al}<\text{P}<\text{Cl}

(B)

\displaystyle \text{Al}<\text{Si}<\text{S}<\text{P}<\text{Cl}

(C)

\displaystyle \text{Al}<\text{S}<\text{P}<\text{Si}<\text{Cl}

(D)

\displaystyle \text{Si}<\text{S}<\text{Al}<\text{P}>\text{Cl}

Q67Single correctChemical Thermodynamics

A cup of water at

5^{0}\text{C}

(system) is placed in a microwave oven and the oven is turned on for one minute during which the water begins to boil. Which of the following option is true?

(A)

\displaystyle q=+\text{ve},\;w=-\text{ve},\;\Delta U=+\text{ve}

(B)

\displaystyle q=+\text{ve},\;w=0,\;\Delta U=+\text{ve}

(C)

\displaystyle q=+\text{ve},\;w=+\text{ve},\;\Delta U=+\text{ve}

(D)

\displaystyle q=-\text{ve},\;w=+\text{ve},\;\Delta U=-\text{ve}

Q68Single correctSome Basic Principles of Organic Chemistry

Consider the following compounds
(a) chlorobenzene, (b) nitrobenzene, (c) anisole (methoxybenzene)
Arrange these compounds in the increasing order of reactivity with nitrating mixture.

Three benzene rings labelled a, b, c. Ring a: benzene ring with a chlorine atom (Cl) on one carbon. Ring b: benzene ring with a nitro group (NO2) on one carbon. Ring c: benzene ring with a methoxy group (OCH3) on one carbon.

(A)

\displaystyle c<b<a

(B)

\displaystyle b<c<a

(C)

\displaystyle b<a<c

(D)

\displaystyle c<a<b

Q69Single correctOrganic Compounds Containing Nitrogen

Compound 'P' undergoes the following sequence of reactions:

(A)

Cyclohexane ring with a carboxamide group: ring carbon attached to C(=O)NH2.

(B)

Cyclohexane ring attached to a carbonyl carbon C(=O) bonded to an ethyl group (CH2CH3); 1-cyclohexylpropan-1-one.

(C)

Cyclohexane ring with a carboxylic acid group: ring carbon attached to C(=O)OH.

(D)

Cyclohexane ring attached to a CH2 group bonded to an aldehyde CHO; cyclohexylacetaldehyde, drawn with the H of the CHO shown.

Q70Single correctAtomic Structure

Which of the following statements regarding the energy of the stationary state is true in the following one-electron systems?

(A)

\displaystyle -2.18\times10^{-18}\,\text{J for second orbit of He}^+ \text{ ion}

(B)

\displaystyle -1.09\times10^{-18}\,\text{J for second orbit of H atom}

(C)

\displaystyle -2.18\times10^{-18}\,\text{J for third orbit of Li}^{2+} \text{ ion}

(D)

\displaystyle -8.72\times10^{-18}\,\text{J for first orbit of He}^+ \text{ ion}

Q71NumericalSome Basic Concepts in Chemistry

x mg of pure HCl was used to make an aqueous solution 25.0 ml. of 0.1 M

\text{Ba(OH)}_2

solution is used when the HCl solution was titrated against it. The numerical value of 'x' is ......

\times10^{-1}

(nearest integer)
Given : molar mass of HCl and

\text{Ba(OH)}_2

are 36.5 and 171.0 g mo

l^{-1}

respectively).

Q72NumericalChemical Equilibrium

For the following gas phase equilibrium reaction at constant temperature.

\text{NH}_3(g)\rightleftharpoons\dfrac{1}{2}\text{N}_2(g)+\dfrac{3}{2}\text{H}_2(g)

If the total pressure is

\sqrt{3}

atm and the pressure equilibrium constant

K_p

is 9 atm, then the degree of dissociation is given as

(x\times10^{-2})^{1/2}

. The value of 'x' is ......... (nearest integer)

SolutionAnswer: 80

Approach:

For

\text{NH}_3(g)\rightleftharpoons\tfrac12\text{N}_2(g)+\tfrac32\text{H}_2(g)

at total pressure

\sqrt3

atm with

K_p=9

atm, the degree of dissociation is written as

(x\times10^{-2})^{1/2}

. Partial pressures in terms of

\alpha

are substituted into

K_p

to obtain

\alpha

, then x.

Step 1:From one mole of ammonia dissociating to degree

\alpha

, the equilibrium moles and total are obtained.

n_{\text{NH}_3}=1-\alpha,\;n_{\text{N}_2}=\tfrac{\alpha}{2},\;n_{\text{H}_2}=\tfrac{3\alpha}{2},\;n_{\text{total}}=1+\alpha

Step 2:Write each partial pressure as mole fraction times total pressure

\sqrt3

P_{\text{NH}_3}=\dfrac{1-\alpha}{1+\alpha}\sqrt3,\;P_{\text{N}_2}=\dfrac{\alpha/2}{1+\alpha}\sqrt3,\;P_{\text{H}_2}=\dfrac{3\alpha/2}{1+\alpha}\sqrt3

Step 3:Substitute into

K_p

; the numerical factors collect to a closed form.

K_p=\dfrac{3\sqrt3}{4}\cdot\dfrac{\alpha^{2}}{1-\alpha^{2}}\cdot P

Step 4:Insert

K_p=9

and

P=\sqrt3

and solve for

\alpha^2

9=\dfrac{3\sqrt3}{4}\cdot\dfrac{\alpha^{2}}{1-\alpha^{2}}\cdot\sqrt3=\dfrac{9}{4}\cdot\dfrac{\alpha^{2}}{1-\alpha^{2}}\Rightarrow\dfrac{\alpha^{2}}{1-\alpha^{2}}=4\Rightarrow\alpha^{2}=\dfrac{4}{5}

Step 5:The degree of dissociation is

(x\times10^{-2})^{1/2}

, so

\alpha^2=x\times10^{-2}

x\times10^{-2}=\dfrac{4}{5}=0.80\Rightarrow x=80

Final answer:

x=80

Q73NumericalChemical Kinetics

For the thermal decomposition of reactant AB(g), the following plot is constructed.
(A plot of

[\text{AB}]/\text{M}

on the y-axis versus time/s on the x-axis showing a straight line decreasing from 0.6 M, passing through 0.55 M at 100 s and 0.5 M at 200 s.)
The half life of the reaction is 'x' min.
x = ......... min. (nearest integer)

Line graph: vertical axis labelled [AB]/M with marked values 0.5, 0.55 and 0.6; horizontal axis labelled time/s with marked values 0, 100 and 200. A straight descending line starts at concentration 0.6 at time 0, passes through 0.55 at 100 s and 0.5 at 200 s, and continues downward beyond. Dashed guide lines connect 0.55 to 100 s and 0.5 to 200 s.

Q74NumericalCoordination Compounds

The crystal field splitting energy of

[\text{Co(oxalate)}_3]^{3-}

complex is 'n' times that of the

[\text{Cr(oxalate)}_3]^{3-}

complex. Here 'n' is ........ (Assume

\Delta_0>P

)

Q75NumericalOrganic Compounds Containing Halogens

Consider all the structural isomers with molecular formula

\text{C}_5\text{H}_{11}\text{Br}

are separately treated with KOH(aq) to give respective substitution products, without any rearrangement. The number of products which can exhibit optical isomerism from these is .........

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