NEET 2026 Official Paper (Held On: 03 May, 2026) (Previous Year Paper)

Given:

Number of turns, N = 100

Radius of coil, r = 5 cm = 0.05 m

Magnetic field at centre, B = 3.14 × 10^-3 T

μ₀ = 4π × 10^-7 T m A^-1

Formula Used:

Magnetic field at centre of circular coil:

B = μ₀NI / 2r

Magnetic moment:

M = NIA

Area of coil:

A = πr²

Calculation:

I = 2rB / μ₀N

⇒ I = 2 × 0.05 × 3.14 × 10^-3 / (4π × 10^-7 × 100) = 3.14 × 10^-4 / 4π × 10^-5 = 2.5 A

Now,

A = π × (0.05)² = π × 0.0025 ≈ 0.00785 m²

Magnetic moment:

M = NIA = 100 × 2.5 × 0.00785

⇒ M ≈ 1.96 A m² ≈ 2 A m²

So, current flowing through the coil, and the magnitude of the magnetic moment, respectively, are 2.5 A, 2 A m²

Given: We need to match List I with List II.

Concept Used:

A. E = hν

This is the energy equation of photon.

⇒ A → IV

B. Diffraction and Interference

These phenomena prove wave nature of light.

⇒ B → III

C. λ = h/p

This is de Broglie wavelength relation.

⇒ C → I

D. Compton effect

Compton effect proves particle nature of light.

⇒ D → II

Final Matching: A – IV, B – III, C – I, D – II

Calculation

Only the 4 Ω and 2 Ω branches are forward biased and conduct current.

Equivalent resistance,

(R_{eq} = \frac{4 \times 2}{4 + 2})

(\Rightarrow R_{eq} = \frac{8}{6} = \frac{4}{3},\Omega)

Given potential difference across the circuit is (10,\text{V})

Using Ohm’s law,

(I = \frac{V}{R_{eq}})

(\Rightarrow I = \frac{10}{4/3})

(\Rightarrow I = \frac{10 \times 3}{4})

(\Rightarrow I = \frac{15}{2},\text{A})

Hence, the correct answer is ( I = \frac{15}{2},\text{A}).

Given:

Speed of light in vacuum = 1 unit

Time taken by light = 6 min 40 s

Calculation:

6 min = 6 × 60= 360 s

⇒ Total time = 360 + 40 = 400 s

Distance = Speed × Time

Since speed of light = 1 unit,

⇒ Distance = 1 × 400 = 400

So, the distance between the Sun and the Earth in new unit is 400.

The correct answer is C only

Velocity-Time Graph for a Ball Thrown Vertically

• When a ball is thrown vertically upward, its velocity decreases due to the effect of gravity until it becomes zero at the highest point.
• After reaching the highest point, the ball starts falling back, and its velocity increases in the downward direction due to acceleration due to gravity.
• Correct Representation of Velocity vs Time
• In the velocity-time graph, the velocity is positive during the upward motion and negative during the downward motion.
• The graph in option C accurately shows this behavior, where the velocity decreases linearly, reaches zero at the peak, and then increases negatively (downward motion).

Given:

20 VSD = 16 MSD

and, 1 MSD = 1 mm

Formula Used:

Least Count = 1 MSD − 1 VSD

Calculation:

Given:

20 VSD = 16 MSD

⇒ 1 VSD = 16/20 MSD = 0.8 MSD

Since 1 MSD = 1 mm

⇒ 1 VSD = 0.8 mm

Now, Least Count = 1 MSD − 1 VSD

⇒ LC = 1 − 0.8 = 0.2 mm

Convert into cm: LC = 0.02 cm

Result: Least count = 0.02 cm

Given:

Resistance, R = 1 kΩ= 1000 Ω

Capacitance, C = 0.1 μF = 0.1 × 10^-6 F= 10^-7 F

Inductance, L = 1 mH = 10^-3 H

Formula Used:

Resonance frequency:

f = 1 / 2π√LC

Calculation:

 LC = 10^-3 × 10^-7 = 10^-10

⇒ √LC = 10^-5

Now,

 f = 1 / 2π × 10^-5 = 1 / 6.28 × 10^-5

⇒ f ≈ 1.59 × 10⁴ Hz ≈ 15.9 kHz

Final Answer: f = 15.9 kHz.

Given:

A long solid wire of radius a carries current I uniformly distributed across its cross-section.

Concept Used:

For a solid conductor carrying uniformly distributed current:

• Inside the conductor: B ∝ r

So, magnetic field increases linearly with distance from the axis.

• Outside the conductor: B ∝ 1/r

So, magnetic field decreases inversely with distance.

Explanation:

• At r = 0, B = 0
• From the centre to r = a, B increases linearly
• At r = a, magnetic field becomes maximum
• For r > a, B decreases as 1/r

The correct graph is

Given:

Total resistance of square wire = 4 Ω

Since square has 4 equal sides:

⇒ Resistance of each side = 4/4

Each side resistance = 1 Ω

Resistance connected between B and D = 2 Ω

Battery voltage, V = 2 V

Concept Used:

Between A and C there are two paths:

Path 1: A → B → C

⇒ Resistance = 1 + 1 = 2 Ω

Path 2: A → D → C

⇒ Resistance = 1 + 1 = 2 Ω

Also diagonal BD has 2 Ω.

Since network is symmetrical:

Potential at B = Potential at D

⇒ No current flows through BD resistance.

Therefore diagonal resistance can be ignored.

Calculation:

Now two 2 Ω branches are in parallel.

Equivalent resistance:

R = (2 × 2)/(2 + 2)= 4/4 = 1 Ω

Using Ohm’s law:

 I = V/R = 2/1 = 2 A

Result: Current I = 2 A

Given:

Nuclear density, ρ = 2.29 × 10¹⁷ kg m^-3

Mass of nucleus, M = 19.926 × 10^-27 kg

R₀ = 1.2 × 10^-15 m

4π = 12.56

Formula Used:

Density of nucleus:

ρ = Mass / Volume

Volume of nucleus:

V = (4/3)πR³

Nuclear radius:

R = R₀A^1/3

Calculation:

V = (4/3)π(R₀A^1/3)³  = (4/3)πR₀³A

Now, ρ = M / [(4/3)πR₀³A]

⇒ A = M / [ρ × (4/3)πR₀³]

Substituting values:

A = 19.926 × 10^-27 / [2.29 × 10¹⁷ × (12.56/3) × (1.2 × 10^-15)³]

⇒ A = 19.926 × 10^-27 / [2.29 × 10¹⁷ × 4.186 × 1.728 × 10^-45]

⇒ A ≈ 19.926 × 10^-27 / 16.56 × 10^-28 ≈ 12

Final Answer: Mass number A = 12

Given:

Length of loop = 8 cm = 0.08 m

Breadth of loop = 3 cm = 0.03 m

Magnetic field, B = 0.3 T

Velocity, v = 2 cm s^-1 = 0.02 m s^-1

Formula Used:

Emf induced: e = Blv

Here, l = side perpendicular to motion

Calculation:

The loop moves in a direction normal to the shorter side.

Therefore, effective length: l = 3 cm = 0.03 m

Now, e = Blv

⇒ e = 0.3 × 0.03 × 0.02 = 0.00018 V = 1.8 × 10^-4 V

Final Answer: e = 1.8 × 10^-4 Volt.

Given:

Resistance of galvanometer, G = 100 Ω

Full scale deflection current, I_g = 1 mA = 0.001 A

Required ammeter range, I = 10 A

Formula Used:

Shunt resistance: S = (I_g × G) / (I − I_g)

Calculation:

S = (0.001 × 100) / (10 − 0.001)

⇒ S = 0.1 / 9.999 ≈ 0.01 Ω

Result: Required shunt resistance = 0.01 Ω

Given:

• In Young's double slit experiment:
• Intensity at path difference λ = K units
• We need intensity when path difference = λ/3

Formula Used:

Intensity in YDSE:

⇒ I = I_max cos²(φ/2)

Phase difference:

⇒ φ = 2π(Δx/λ)

Calculation:

For path difference Δx = λ, φ = 2π(λ/λ)

⇒ φ = 2π

and, I = I_max cos²(π)

⇒ I = I_max

Given this intensity = K

⇒ I_max = K

Now for path difference Δx = λ/3, φ = 2π(1/3)

⇒ φ = 2π/3

So, I = K cos²(π/3)

⇒ I = K × (1/2)² = K/4

Result: Required intensity = K/4

Given:

Mass of body, m = 5 kg

Two mutually perpendicular forces: F₁ = 8 N and F₂ = 6 N

Formula Used:

Resultant force: F = √(F₁² + F₂²)

Acceleration: a = F / m

Direction: tan θ = Perpendicular / Base

Calculation:

F = √(8² + 6²) = √(64 + 36) = √100 = 10 N

Now,

⇒ a = 10 / 5 = 2 m s⁻²

Direction with respect to 8 N force:

tan θ = 6 / 8 = 3 / 4

⇒ θ = tan⁻¹(3/4)

Result: Acceleration = 2 m s⁻² and Direction = tan⁻¹(3/4) with 8 N force.

Given:

• C₁ = C₂ = C₃ = C₄ = 10 μF
• C₅ = 2.5 μF
• V = 50 V

Formula Used:

For series capacitors:

1/C = 1/C₁ + 1/C₂ + 1/C₃ + 1/C₄

Charge: Q = CV

Calculation:

1/C = 1/10 + 1/10 + 1/10 + 1/10

⇒ 1/C = 4/10 ⇒ C = 10/4 = 2.5 μF

This 2.5 μF combination is in parallel with C₅ = 2.5 μF

⇒ Equivalent capacitance: C_eq = 2.5 + 2.5 = 5 μF

Now, Q = CV

⇒ Q = 5 × 50 = 250 μC

Charge in the series branch:

Q = 2.5 × 50 = 125 μC

In series connection, charge on each capacitor is same.

Charge on C₁, C₂, C₃ and C₄ = 125 μC each

For C₅:

Q = 2.5 × 50 = 125 μC

Final Answer: Equivalent capacitance = 5 μF, and the charge on each capacitor is 125 μC.

Concept Used:

• In a metre bridge, balance condition depends on equality of potentials at two points.
• At balance point: No current flows through galvanometer.
• Therefore, interchange of cell and galvanometer does not affect the balance condition.
• This is based on the principle of reversibility of Wheatstone bridge.

Explanation:

• When the cell E and galvanometer G are interchanged:
• The bridge still works normally.
• On moving the jockey on different sides of balance point:
• Current direction through galvanometer changes.

Hence galvanometer shows deflection on both sides.

At exact balance point:

• Potential difference across galvanometer becomes zero.
• No deflection is observed.

Result:

 Correct option is "Both right-sided and left-sided deflection and at balance point, no deflection".

Given:

Mass, m = 1000 kg

Height, h = 20 m

Time, t = 10 s

Acceleration due to gravity, g = 9.8 m/s²

Formula Used:

Power = Work / Time

Work done against gravity: W = mgh

Calculation:

W = 1000 × 9.8 × 20 = 196000 J

⇒ Power = 196000 / 10 = 19600 W

⇒ Power = 19.6 kW

Result: Power of crane = 19.6 kW

Given:

We need to match physical quantities with their formulas.

Concept Used:

Young's Modulus:

Y = Stress / Strain

⇒ Y = (F/A) / (ΔL/L) = FL / A(ΔL)

So, A → II

Compressibility:

Compressibility is reciprocal of bulk modulus.

Compressibility = − (1/P)(ΔV/V)

So, B → III

Bulk Modulus:

B = − P(V/ΔV)

So, C → IV

Poisson’s Ratio:

Poisson’s Ratio = Lateral strain / Longitudinal strain

⇒ Δd/ΔL × L/d

So, D → I

Final Matching: A – II, B – III, C – IV, D – I

Given:

A ray of light is incident on a concave lens parallel to the principal axis.

Concept Used:

• A concave lens is a diverging lens.
• Rays parallel to the principal axis diverge after refraction.
• The refracted rays appear to come from the principal focus on the same side of the lens.

​Explanation:

When a ray parallel to the principal axis passes through a concave lens:

• The ray bends outward.
• It does not actually pass through the focus.
• Its backward extension passes through the first principal focus.

Therefore, the refracted ray appears to diverge from the first principal focus.

So, correct option is "appears to diverge from the first principal focus".

Given:

• Length of wire = L
• Linear mass density = m
• Wire is bent into a circular ring.
• Axis yy′ is tangential to the ring and lies in its plane.

Formula Used:

Total mass of ring: M = mL

Circumference of ring: L = 2πR

⇒ R = L / 2π

Moment of inertia of a ring about diameter: I_diameter = MR² / 2

Using parallel axis theorem for tangent axis: I_yy′ = I_diameter + MR²

Calculation:

 I_yy′ = MR²/2 + MR²

⇒ I_yy′ = 3MR²/2

Substituting M = mL and R = L/2π

 I_yy′ = (3/2) × (mL) × (L/2π)²

⇒ I_yy′ = (3/2) × mL × L² / 4π²

⇒ I_yy′ = 3mL³ / 8π²

Result: Moment of inertia about yy′ = 3mL³ / 8π²

Given:

Mass of cube, m = 5.580 kg

Side of cube, a = 9.0 cm ⇒ a = 0.09 m

Formula Used:

Density, ρ = Mass / Volume

Volume of cube = a³

Calculation:

Volume = (0.09)³

⇒ Volume = 0.000729 m³

⇒ ρ = 5.580 / 0.000729 = 7654.32 kg m⁻³ = 7.654 × 10³ kg m⁻³

Significant Figures:

Mass has 4 significant figures.

Side length 9.0 cm has 2 significant figures.

Final answer must have 2 significant figures.

⇒ ρ = 7.7 × 10³ kg m⁻³

Result: X = 7.7

Given:

• Wave equation: y(x,t) = 2.0 cos 2π(10 t − 0.0080 x + 0.35)
• Distance between two points: Δx = 0.5 m

Formula Used:

For a travelling wave: Phase difference, Δϕ = 2π × (coefficient of x) × Δx

Calculation:

Given coefficient of x = 0.0080

Δϕ = 2π × 0.0080 × 0.5

⇒ Δϕ = 2π × 0.004 = 0.008π rad

Therfore, phase difference = 0.008π rad

Given:

• Equilateral prism i.e. prism angle, A = 60°
• Angle of incidence, i = 50°
• Refracted ray inside prism is parallel to base BC.

Concept Used:

In an equilateral prism:

Each angle = 60°

⇒ Face AB makes 60° with base BC.

Since refracted ray QR is parallel to BC:

⇒ Angle between refracted ray and prism face AB = 60°

Angle of refraction is measured from normal.

⇒ r₁ = 90° − 60° = 30°

Similarly at second face: r₂ = 30°

Using prism relation: r₁ + r₂ = A

⇒ 30° + 30° = 60° 

Now for prism: Deviation, δ = i + e − A

Here, symmetry gives: e = i = 50°

Calculation:

δ = 50° + 50° − 60° = 40°

Result: Angle of deviation = 40°

CONCEPT:

Voltage Across an Ideal Diode

• An ideal diode behaves as an open circuit when it is reverse biased.
• An ideal diode behaves as a short circuit when it is forward biased.
• Therefore, the voltage across the diode is maximum when it is reverse biased, while it is approximately zero when it is forward biased.

EXPLANATION:

• During the positive half-cycle of the input voltage, the diode is reverse biased.
• No current flows through the circuit.
• Hence, there is no voltage drop across the resistor.
• The entire input voltage appears across the diode.
• Thus, V_D follows the positive half-cycle of the input.
• During the negative half-cycle, the diode becomes forward biased.
• The ideal diode conducts and behaves like a short circuit.
• Therefore, the voltage across the diode becomes approximately zero.
• Hence, the voltage across the diode consists of a positive half-sinusoidal waveform during the reverse-biased interval and remains zero during the forward-biased interval.

Therefore, the waveform across the diode is a positive half-wave followed by zero voltage.

Given:

Time period of simple pendulum = T

We need to identify the correct graph between K.E. and time.

Concept Used:

For a simple pendulum:

• Kinetic Energy is always positive or zero.
• Kinetic Energy becomes maximum at mean position.
• Kinetic Energy becomes zero at extreme positions.
• In one complete oscillation, K.E. becomes maximum two times.

Therefore, time period of K.E. = T/2

Explanation:

At t = 0, pendulum is at extreme position.

⇒ Velocity = 0

⇒ K.E. = 0

As pendulum moves toward mean position:

⇒ Velocity increases

⇒ K.E. increases

At mean position:

⇒ Velocity is maximum

⇒ K.E. is maximum

Again at next extreme position:

⇒ Velocity becomes zero

⇒ K.E. becomes zero

This process repeats after every T/2.

Hence, K.E. graph always remains positive and forms repeating arches.

Result: Correct graph is

Given:

Emf of battery, E = 12 V

Internal resistance, r = 2 Ω

Current, I = 0.6 A

Formula Used:

Terminal voltage: V = E − Ir

Calculation:

V = 12 − (0.6 × 2) = 12 − 1.2

⇒ V = 10.8 V

Final Answer: Terminal voltage = 10.8 V.

Given:

Mass of body = m

Radius of Earth = R

Height raised = R

Formula Used:

Gravitational potential energy: U = -GMm / r

Calculation:

Initial distance from Earth’s center:

⇒ r₁ = R

Final distance from Earth’s center:

⇒ r₂ = R + R = 2R

Work done = Increase in potential energy

W = U₂ − U₁

⇒ W = [ -GMm / (2R) ] − [ -GMm / R ]

⇒ W = GMm / R − GMm / (2R)

⇒ W = GMm / (2R)

Now, g = GM / R²

⇒ GM = gR²

Substituting, W = (gR² × m) / (2R)

⇒ W = mgR / 2

Final Answer: Work done = mgR / 2

Given:

Heat supplied per second, Q = 100 W

Work done by system per second, W = 75 J/s

Formula Used:

First law of thermodynamics:

ΔQ = ΔU + ΔW

Calculation:

ΔU = ΔQ − ΔW = 100 − 75

⇒ ΔU = 25 W

Final Answer: Rate of increase of internal energy = 25 W.

Given:

Rated power, P₁ = 400 W

Rated voltage, V₁ = 220 V

New voltage, V₂ = 200 V

Formula Used:

For a heater, resistance remains constant:

P = V² / R

Therefore, P ∝ V²

Calculation:

P₂ / P₁ = (V₂ / V₁)²

⇒ P₂ / 400 = (200 / 220)² = (10 / 11)²

⇒ P₂ / 400 = 100 / 121

⇒ P₂ = 400 × 100 / 121 ≈ 330.6 W

⇒ P₂ ≈ 331 W

Final Answer: Power consumed ≈ 331 W.

Concept

• When an object falls freely under gravity, the distance it travels is determined by its initial velocity, the acceleration due to gravity, and the time of fall.
• For a falling ruler being caught, the distance the ruler falls corresponds to the person's reaction time.
• Since the ruler is released from rest, the distance traveled is directly proportional to the square of the reaction time.

Formula Used

The equation of motion for a body starting from rest under gravity is:

(s = ut + \frac{1}{2}gt^2)

Since the ruler falls from rest:

(u = 0)

(s = \frac{1}{2}gt^2)

Since (g) is a constant ((9.8,\text{m/s}^2)), the distance (s) is directly proportional to the square of the reaction time (t):

(s \propto t^2)

Calculation

Given the reaction times for the five persons:

(t_A = 0.20,\text{s})

(t_B = 0.22,\text{s})

(t_C = 0.18,\text{s})

(t_D = 0.19,\text{s})

(t_E = 0.21,\text{s})

Comparing these reaction times in descending order (highest to lowest):

(0.22,\text{s} > 0.21,\text{s} > 0.20,\text{s} > 0.19,\text{s} > 0.18,\text{s})

This corresponds to the persons in the following order:

(B > E > A > D > C)

Since the distance (s) increases as the time (t) increases, the order of distances traveled by the ruler will be the same as the order of the reaction times.

Therefore, the correct order of distance traveled is:

(B > E > A > D > C)

Hence, the correct order is (B > E > A > D > C).

Concept

• When a charged capacitor is connected to an uncharged capacitor, charge redistribution occurs until both capacitors reach a common potential.
• In this process, energy is lost, primarily in the form of heat in the connecting wires or through electromagnetic radiation.
• The total charge remains conserved, but the total electrostatic energy decreases.

Formula Used

Energy stored in a capacitor:

(U = \frac{1}{2}CV^2)

Common potential after connection:

(V_c = \frac{C_1V_1 + C_2V_2}{C_1 + C_2})

The energy lost during the process is given by:

(\Delta U = \frac{1}{2} \frac{C_1 C_2}{C_1 + C_2} (V_1 - V_2)^2)

Calculation

Given:

Capacitance of both capacitors, (C_1 = C_2 = C = 200 \text{ pF} = 200 \times 10^{-12} \text{ F})

Potential of the first capacitor, (V_1 = 100 \text{ V})

Potential of the second capacitor (uncharged), (V_2 = 0 \text{ V})

Substituting the values into the energy loss formula:

(\Delta U = \frac{1}{2} \frac{(200 \times 10^{-12}) \times (200 \times 10^{-12})}{(200 \times 10^{-12}) + (200 \times 10^{-12})} (100 - 0)^2)

(\Rightarrow \Delta U = \frac{1}{2} \frac{(200 \times 10^{-12})^2}{400 \times 10^{-12}} \times 10000)

(\Rightarrow \Delta U = \frac{1}{2} \left( \frac{40000 \times 10^{-24}}{400 \times 10^{-12}} \right) \times 10^4)

(\Rightarrow \Delta U = \frac{1}{2} (100 \times 10^{-12}) \times 10^4)

(\Rightarrow \Delta U = 50 \times 10^{-8} \text{ J})

(\Rightarrow \Delta U = 0.5 \times 10^{-6} \text{ J})

Hence, the amount of electrostatic energy lost in the process is (0.5 \times 10^{-6} \text{ J}).

Concept

• A simple pendulum consists of a mass (bob) suspended from a fixed point that swings back and forth.
• The time period ((T)) is the time taken to complete one full oscillation.
• The time period depends on the effective length of the pendulum ((L)) and the acceleration due to gravity ((g)).

Formula Used

Time period of a simple pendulum:

(T = 2\pi \sqrt{\frac{L}{g}})

Relation between total time ((t)), number of oscillations ((n)), and time period ((T)):

(T = \frac{t}{n})

Calculation

Given:

Total time taken, (t = 60,s)

Number of oscillations, (n = 30)

Acceleration due to gravity, (g = 9.8,m/s^2)

Approximation, (\pi^2 = 9.8)

First, calculate the time period (T):

(T = \frac{60}{30} = 2,s)

Now, substitute the values into the time period formula:

(2 = 2\pi \sqrt{\frac{L}{g}})

Divide both sides by 2:

(1 = \pi \sqrt{\frac{L}{g}})

Squaring both sides:

(1^2 = \pi^2 \left( \frac{L}{g} \right))

(1 = \pi^2 \frac{L}{g})

Rearranging for (L):

(L = \frac{g}{\pi^2})

Substituting the given values (g = 9.8) and (\pi^2 = 9.8):

(L = \frac{9.8}{9.8} = 1,m)

Hence, the effective length of the simple pendulum is 1 m.

Concept

• Alternating current (AC) varies sinusoidally with time, following the equation (I = I_0 \sin(\omega t)).
• One complete cycle corresponds to a time period (T).
• Starting from zero, the current reaches its first peak value at one-fourth of the time period, i.e., (t = \frac{T}{4}).

Formula Used

Relation between time period and frequency:

(T = \frac{1}{f})

Time taken to reach peak value from zero:

(t = \frac{T}{4} = \frac{1}{4f})

Calculation

Given:

Frequency, (f = 60 \text{ Hz})

Peak current, (I_0 = 5 \text{ A})

We need to find the time (t) to reach the peak value from zero. This occurs when the phase angle is (90^\circ) or (\frac{\pi}{2}) radians.

Using the formula:

(t = \frac{1}{4f})

Substituting the value of (f):

(t = \frac{1}{4 \times 60} \text{ s})

(t = \frac{1}{240} \text{ s})

Hence, the current will take (\frac{1}{240} \text{ s}) to reach the peak value.

Concept

• Conservation of Energy: In wave optics, the principle of superposition leads to interference and diffraction. Energy is not lost; it is simply redistributed from regions of minimum intensity to regions of maximum intensity.
• Wave Nature: Interference and diffraction are fundamental characteristics of wave motion. Any entity that behaves as a wave will exhibit these phenomena.

Explanation

Statement A:

In the phenomena of interference and diffraction, the intensity of light at any point is given by the principle of superposition. At points of constructive interference, the intensity is greater than the sum of the individual intensities, and at points of destructive interference, it is less. However, the total energy across the entire pattern remains constant and equal to the sum of the energies of the individual waves. No energy is created or destroyed; it is merely redistributed. Therefore, these phenomena are perfectly consistent with the principle of conservation of energy. Hence, Statement A is true.

Statement B:

Interference and diffraction are not exclusive to light waves. They are general properties exhibited by all types of waves, including mechanical waves (such as sound waves and water waves), electromagnetic waves (such as radio waves and X-rays), and even matter waves (such as electrons). For example, sound waves undergo diffraction when they pass through a doorway. Hence, Statement B is false.

Conclusion

Based on the analysis, Statement A is correct while Statement B is incorrect.

Hence, A is true, but B is false.

Concept

• For a box to remain stationary relative to an accelerating trolley, the pseudo-force acting on the box must be balanced by the force of static friction.
• The maximum acceleration occurs when the pseudo-force equals the maximum possible static friction (limiting friction).
• Static friction depends on the coefficient of friction and the normal reaction force.

Formula Used

Maximum static friction:

(f_{s,max} = \mu_s N)

Normal force on a horizontal surface:

(N = mg)

Newton's second law (Pseudo-force):

(F_p = ma)

Condition for no slipping:

(ma \leq \mu_s mg)

Calculation

Given:

• Mass of the box, (m = 15,kg)
• Coefficient of static friction, (\mu_s = 0.12)
• Acceleration due to gravity, (g = 10,m/s^2)

To keep the box stationary over the trolley, the horizontal acceleration (a) must satisfy:

(ma \leq f_{s,max})

(ma \leq \mu_s mg)

The mass (m) cancels out from both sides:

(a_{max} = \mu_s g)

Substituting the given values:

(a_{max} = 0.12 \times 10)

(a_{max} = 1.2,m/s^2)

Hence, the maximum acceleration with which the trolley can be moved horizontally is (1.2,m/s^2).

Concept

• The total mechanical energy of a simple pendulum, which is the sum of its kinetic energy (KE) and potential energy (PE), remains constant during its motion (neglecting air resistance).
• At the equilibrium position (the lowest point of the swing), the potential energy is at its minimum. If we take this point as the reference level, the potential energy is zero.
• Consequently, at the equilibrium position, all the total energy of the pendulum is converted into kinetic energy.

Formula Used

Total mechanical energy ((E)):

(E = KE + PE)

Kinetic energy at the equilibrium position:

(KE_{max} = \frac{1}{2}mv^2)

Where:

• (m) is the mass of the bob.
• (v) is the speed of the bob at the equilibrium position.

Calculation

Given:

Total energy, (E = 0.02,J)

Mass of the bob, (m = 20,g = 0.02,kg)

By the law of conservation of energy, the total energy at the equilibrium position is purely kinetic:

(E = \frac{1}{2}mv^2)

Substituting the given values into the formula:

(0.02 = \frac{1}{2} \times 0.02 \times v^2)

Multiplying both sides by 2:

(0.04 = 0.02 \times v^2)

Solving for (v^2):

(v^2 = \frac{0.04}{0.02})

(v^2 = 2)

Taking the square root:

(v = \sqrt{2} \approx 1.414,m/s)

Rounding to two decimal places, we get approximately (1.41,m/s).

Hence, the speed of the simple pendulum bob at the equilibrium position is approximately 1.41 m/s.

Concept

• The radius of a nucleus (R) is related to its mass number (A).
• The volume of a nucleus, assuming it is spherical, depends on the cube of its radius.
• The mass defect is a measure of the binding energy of the nucleus, representing the difference between the total mass of individual nucleons and the actual mass of the nucleus.

Formula Used

Nuclear radius:

(R = R_0 A^{1/3})

Nuclear volume:

(V = \frac{4}{3} \pi R^3)

Mass defect ((\Delta m)):

(\Delta m = [Z m_p + (A - Z) m_n] - M_{nucleus})

Explanation

For Statements A and B:

The radius of a nucleus is given by (R = R_0 A^{1/3}). The volume (V) of the nucleus is:

(V = \frac{4}{3} \pi (R_0 A^{1/3})^3)

(V = \frac{4}{3} \pi R_0^3 A)

This shows that (V \propto A). Therefore, Statement A is false and Statement B is true.

For Statements C and D:

The mass defect is defined as the difference between the sum of the masses of the constituents (protons and neutrons) and the actual mass of the nucleus. The difference between the mass of an atom and its nucleus is approximately the mass of the electrons, which is not the mass defect.

Therefore, Statement C is false and Statement D is true.

Conclusion:

Statements B and D are true, while A and C are false.

Hence, B and D are true, but A and C are false.

Concept

• The motion of a flywheel is an example of rotational motion with constant angular acceleration.
• Angular displacement can be calculated using the average angular velocity when acceleration is uniform.
• The number of revolutions is the total angular displacement divided by (2\pi) radians.

Formula Used

Angular speed in rad/s:

(\omega = \frac{2\pi N}{60}) (where (N) is in rpm)

Average angular velocity:

(\omega_{avg} = \frac{\omega_1 + \omega_2}{2})

Angular displacement:

(\theta = \omega_{avg} \times t)

Number of revolutions ((n)):

(n = \frac{\theta}{2\pi})

Calculation

Given:

Initial speed (N_1 = 600,rpm)

Final speed (N_2 = 1200,rpm)

Time (t = 10,s)

First, calculate initial and final angular velocities in rad/s:

(\omega_1 = \frac{2\pi \times 600}{60} = 20\pi,rad/s)

(\omega_2 = \frac{2\pi \times 1200}{60} = 40\pi,rad/s)

Now, find the average angular velocity:

(\omega_{avg} = \frac{20\pi + 40\pi}{2} = \frac{60\pi}{2} = 30\pi,rad/s)

Calculate total angular displacement ((\theta)):

(\theta = \omega_{avg} \times t)

(\theta = 30\pi \times 10 = 300\pi,rad)

Finally, find the number of revolutions ((n)):

(n = \frac{\theta}{2\pi})

(n = \frac{300\pi}{2\pi} = 150)

Hence, the number of revolutions completed by the flywheel is 150.

Concept

• The absolute pressure at a depth in a fluid is the sum of the atmospheric pressure at the surface and the hydrostatic pressure exerted by the fluid column.
• Atmospheric pressure ((P_{atm})) is typically 1 atm unless otherwise specified.
• Hydrostatic pressure is given by the product of depth, density, and gravitational acceleration.

Formula Used

Absolute pressure formula:

(P_{abs} = P_{atm} + \rho gh)

Where:

• (P_{abs}) = Absolute pressure
• (P_{atm}) = Atmospheric pressure
• (\rho) = Density of the fluid
• (g) = Acceleration due to gravity
• (h) = Depth below the surface

Calculation

Given:

(P_{abs} = 100 \text{ atm} = 100 \times 10^5 \text{ Pa})

(P_{atm} = 1 \text{ atm} = 1 \times 10^5 \text{ Pa})

(\rho = 1000 \text{ kg/m}^3)

(g = 10 \text{ m/s}^2)

Using the absolute pressure relation:

(P_{abs} = P_{atm} + \rho gh)

Substitute the values:

(100 \times 10^5 = 1 \times 10^5 + (1000 \times 10 \times h))

Subtracting atmospheric pressure from both sides to find gauge pressure:

(99 \times 10^5 = 10^4 \times h)

Solving for (h):

(h = \frac{99 \times 10^5}{10^4})

(h = 99 \times 10)

(h = 990 \text{ m})

Hence, the submarine can go to a depth of 990 m below the water surface.

Concept

• Electromagnetic waves are categorized by their frequency and wavelength, and each type is produced through different physical processes.
• Microwaves are generated by oscillating currents in special vacuum tubes.
• Visible light is produced by the excitation of electrons in atoms.
• Gamma rays are of nuclear origin, produced during transitions within the atomic nucleus.
• Infrared rays are often referred to as heat waves and are produced by the thermal motion of atoms and molecules.

Explanation

• A. Microwaves: These are produced by special vacuum tubes called klystrons, magnetrons, or Gunn diodes. Thus, A matches with IV.
• B. Visible light: Visible light is emitted when electrons in atoms move from a higher energy state to a lower energy state (electronic transitions). Thus, B matches with I.
• C. Gamma rays: These high-frequency waves are produced during the radioactive decay of a nucleus or in nuclear reactions. Thus, C matches with II.
• D. Infra-red rays: These are produced by the vibration and rotation of atoms and molecules in a substance. Thus, D matches with III.

Conclusion

By matching the electromagnetic waves with their respective production methods, we get:

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

Hence, the correct match is A - IV, B - I, C - II, D - III.

Concept

• Electrostatics of Conductors: Conductors contain mobile charge carriers (electrons) that redistribute themselves until they reach an equilibrium state.
• Internal Electric Field: In a static situation, the electric field inside a conductor is zero. If it weren't, the free charges would move.
• Charge Distribution: According to Gauss’s Law, since the electric field inside is zero, the net flux through any internal surface is zero, meaning no net charge resides inside the conductor.
• Surface Field: The electric field at the surface must be perpendicular to the surface; otherwise, a tangential component would cause surface charges to move.
• Electrostatic Potential: Since the electric field inside is zero ($E = -dV/dr = 0$), the potential remains constant throughout the volume and on the surface of the conductor.

Explanation

Let us evaluate each statement:

• Statement A: Inside a conductor, the electrostatic field is zero. This is a fundamental property of conductors in electrostatic equilibrium. (Correct)
• Statement B: The electric field at the surface of a charged conductor is given by (E = \frac{\sigma}{\epsilon_0}), where (\sigma) is the surface charge density. Thus, it depends on the surface charge density. (Incorrect)
• Statement C: By Gauss’s law, if the electric field is zero everywhere inside, any Gaussian surface within the conductor encloses zero net charge. All excess charge resides on the surface. (Correct)
• Statement D: If the electric field were not normal to the surface, there would be a tangential component that would exert force on surface charges and cause them to move. In a static situation, the field must be normal. (Correct)
• Statement E: The electrostatic potential is constant (equipotential) throughout the volume and surface of the conductor because the field is zero. It is not necessarily zero unless the conductor is grounded. (Incorrect)

Based on the analysis, statements A, C, and D are the correct ones.

Hence, the correct answer is A, C and D only.

Concept

• The photoelectric effect occurs only if the energy of the incident photon ((E)) is greater than or equal to the work function ((\phi)) of the metal.
• Energy of a photon is inversely proportional to its wavelength ((\lambda)).
• The maximum wavelength that can cause photoelectric emission is called the threshold wavelength ((\lambda_0)). If the incident wavelength is greater than the threshold wavelength ((\lambda > \lambda_0)), no photoelectric effect occurs.

Formula Used

Energy of incident photon:

(E = \frac{hc}{\lambda})

Threshold wavelength:

(\lambda_0 = \frac{hc}{\phi})

Conversion of energy from eV to Joules:

(1\text{ eV} = 1.6 \times 10^{-19}\text{ J})

Constants:

(h = 6.6 \times 10^{-34}\text{ J s})

(c = 3 \times 10^8\text{ m/s})

Calculation

Given:

Work function, (\phi = 6.6\text{ eV} = 6.6 \times 1.6 \times 10^{-19}\text{ J})

Calculating the threshold wavelength ((\lambda_0)):

(\lambda_0 = \frac{hc}{\phi})

(\lambda_0 = \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{6.6 \times 1.6 \times 10^{-19}})

Canceling (6.6) from numerator and denominator:

(\lambda_0 = \frac{3 \times 10^{-26}}{1.6 \times 10^{-19}})

(\lambda_0 = 1.875 \times 10^{-7}\text{ m})

Converting to nanometers ((1\text{ nm} = 10^{-9}\text{ m})):

(\lambda_0 = 187.5\text{ nm})

For the photoelectric effect to not occur, the incident wavelength (\lambda) must be greater than (\lambda_0):

(\lambda > 187.5\text{ nm})

Comparing with the given options:

• 200 nm: (200\text{ nm} > 187.5\text{ nm}) (Emission will not occur)
• 150 nm: (150\text{ nm} < 187.5\text{ nm}) (Emission will occur)
• 100 nm: (100\text{ nm} < 187.5\text{ nm}) (Emission will occur)
• 50 nm: (50\text{ nm} < 187.5\text{ nm}) (Emission will occur)

Hence, the wavelength that does not give rise to the photoelectric effect is 200 nm.

Concept

• According to the Bohr model of the hydrogen atom, the total energy of an electron is equal to half of its potential energy and negative of its kinetic energy.
• The total energy of an electron in an orbit of radius (r) is related to the electrostatic potential energy between the electron and the nucleus.
• For a hydrogen atom (Z = 1), the total energy is given by (E = -\frac{ke^2}{2r}), where (k = \frac{1}{4\pi\epsilon_0}).

Formula Used

The total energy of an electron in a hydrogen atom is:

(E = -\frac{1}{4\pi\epsilon_0}\frac{e^2}{2r})

Rearranging for the radial distance:

(r = \frac{1}{4\pi\epsilon_0}\frac{e^2}{2|E|})

Calculation

Given:

Energy in the first excited state, (E = -3.4,\text{eV})

Converting energy into joules:

(|E| = 3.4 \times 1.6 \times 10^{-19},\text{J})

Charge of electron:

(e = 1.6 \times 10^{-19},\text{C})

Coulomb constant:

(\frac{1}{4\pi\epsilon_0} = 9 \times 10^9,\text{N m}^2/\text{C}^2)

Substituting the values:

(r = \frac{9 \times 10^9 \times (1.6 \times 10^{-19})^2}{2 \times (3.4 \times 1.6 \times 10^{-19})})

Cancelling common terms:

(r = \frac{9 \times 10^9 \times 1.6 \times 10^{-19}}{2 \times 3.4})

(r = \frac{14.4 \times 10^{-10}}{6.8})

(r \approx 2.117 \times 10^{-10},\text{m})

Therefore,

(r \approx 2.1 \times 10^{-10},\text{m})

Hence, the radial distance of the electron from the hydrogen nucleus is approximately (2.1 \times 10^{-10},\text{m}).

Concept

• In forward bias, the positive terminal of the battery is connected to the p-side and the negative terminal to the n-side of the p-n junction diode.
• Threshold voltage (also known as cut-in or knee voltage) is the minimum voltage required for the diode to conduct current significantly in forward bias.
• Reverse saturation current is the current that flows due to minority charge carriers when the diode is in reverse bias.

Formula Used

The current through a p-n junction diode is given by the diode equation:

(I = I_0 (e^{\frac{eV}{\eta kT}} - 1))

Where:

• (I) = Diode current
• (I_0) = Reverse saturation current
• (V) = Applied voltage
• (e) = Charge of an electron
• (k) = Boltzmann constant
• (T) = Temperature

Explanation

Analysis of Statement A:

When a p-n junction diode is forward biased, the applied voltage opposes the internal barrier potential. As the forward voltage increases and exceeds the threshold voltage (approximately (0.7,V) for Silicon and (0.3,V) for Germanium), the depletion layer width decreases significantly, allowing majority charge carriers to cross the junction. This leads to an exponential increase in the diode current. Thus, Statement A is true.

Analysis of Statement B:

The significant current that flows during forward bias is called the forward current. Reverse saturation current is the very small current (in the order of microamperes or nanoamperes) that flows when the diode is reverse biased, caused by the drift of minority charge carriers across the junction. Therefore, Statement B is false.

Hence, Statement A is true, but Statement B is false.

Given:

Mass ratio of Argon : Chlorine = 2 : 1

Molar mass of Ar = 40 u

Molar mass of Cl₂ = 70 u

Temperature = 27°C

Formula Used:

For a gas,

(v_{rms}=\sqrt{\frac{3RT}{M}})

⇒ v_rms ∝ 1/√M

Calculation:

Required ratio:

⇒ v_rms^Ar / v_rms^Cl₂ = √(M_Cl₂ / M_Ar) = √(70 / 40) = √(7 / 4) = √7 / 2

Final Answer: v_rms^Ar / v_rms^Cl₂ = √7 / 2

CONCEPT:

Conversion of Organic Compounds Using Reagents

• Different reagents are used for specific organic transformations such as oxidation, reduction, hydration, and aromatic substitution.
• Important reactions used in the question are:
• Oxidation of alkyl benzene using O₂ followed by acidic hydrolysis produces phenol.
• Carboxylic acids can be reduced to alcohols using catalytic hydrogenation in acidic medium.
• Alcohols undergo dehydration in presence of concentrated H₂SO₄ and on subsequent hydration give alcohols according to Markovnikov’s rule.
• Benzene can be converted into phenol through sulphonation using oleum followed by fusion with NaOH and acidification.

EXPLANATION:

Reaction A:

Isopropyl benzene (cumene) on oxidation with O₂ followed by acidic hydrolysis forms phenol.

Hence, A matches with:

Reaction B:

Acetic acid is converted into ethanol by reduction using CH₃OH, H⁺ followed by catalytic hydrogenation.

Hence, B matches with:

Reaction C:

1-Propanol on dehydration using concentrated H₂SO₄ and heat forms propene, which on hydration gives 2-propanol.

Hence, C matches with:

​

Reaction D:

Benzene undergoes sulphonation with oleum, followed by fusion with NaOH and acidification to produce phenol.

Hence, D matches with:

• Therefore, the correct matching is:

Hence, the correct matching of List-I with List-II is A-II, B-III, C-IV, D-I.

CONCEPT:

• Reaction with AgNO₂: Alkyl halides react with silver nitrite (AgNO₂), which is a covalent compound, primarily to form nitroalkanes because the lone pair on the nitrogen atom acts as the nucleophilic site.
• Reduction of Nitro Groups: The nitro group (-NO₂) can be reduced to a primary amino group (-NH₂) using a metal and a concentrated acid, such as tin and hydrochloric acid (Sn/HCl).
• Diazotization and Decomposition: Primary aliphatic amines react with nitrous acid (HNO₂, generated from NaNO₂ and HCl) at low temperatures (0-5°C) to form aliphatic diazonium salts. These salts are highly unstable and decompose immediately in aqueous solutions to yield alcohols and nitrogen gas.

EXPLANATION:

• Step 1: Conversion of C₂H₅Br to X

Ethyl bromide reacts with AgNO₂ through a nucleophilic substitution reaction to form nitroethane.

C₂H₅Br + AgNO₂ → C₂H₅NO₂ + AgBr

Product X is Nitroethane (C₂H₅NO₂).

• Step 2: Conversion of X to Y

Nitroethane is reduced by the Sn/HCl mixture to form the corresponding primary amine.

C₂H₅NO₂ + 6[H] → C₂H₅NH₂ + 2H₂O

Product Y is Ethylamine (C₂H₅NH₂).

• Step 3: Conversion of Y to Z

Ethylamine reacts with nitrous acid (HNO₂) at 0-5°C. An unstable diazonium salt is formed as an intermediate, which decomposes in the presence of water to form ethanol.

C₂H₅NH₂ + HNO₂ → [C₂H₅N₂⁺Cl^-] → C₂H₅OH + N₂↑ + HCl

Product Z is Ethanol (C₂H₅OH).

Therefore, the major product Z formed in the sequence is C₂H₅OH.

CONCEPT:

Solubility Product (K) and pH Calculation

• For a sparingly soluble base or salt in equilibrium with its ions, the equilibrium constant (K) is determined by the concentrations of the dissolved ions.
• For the reaction: BiO(OH)(s) ⇌ BiO⁺(aq) + OH^-(aq), the expression is:

K = [BiO⁺][OH^-]

• The pOH is calculated as: pOH = -log[OH^-].
• The pH is calculated from pOH at 298 K using the relation: pH + pOH = 14.

EXPLANATION:

• The given equilibrium reaction is:

BiO(OH)(s) ⇌ BiO⁺(aq) + OH^-(aq)

• Let the molar solubility of BiO(OH) be 's'. According to the stoichiometry:
• [BiO⁺] = s
• [OH^-] = s
• Given the equilibrium constant K = 4 × 10^-10:
• K = [BiO⁺][OH^-]
• 4 × 10^-10 = s × s
• s² = 4 × 10^-10
• s = √(4 × 10^-10)
• s = 2 × 10^-5 M
• The concentration of hydroxide ions is [OH^-] = 2 × 10^-5 M.
• Calculating pOH:
• pOH = -log[OH^-]
• pOH = -log(2 × 10^-5)
• pOH = -(log 2 + log 10^-5)
• pOH = -(0.3010 - 5)
• pOH = 4.699
• Calculating pH:
• pH = 14 - pOH
• pH = 14 - 4.699
• pH = 9.301

Therefore, the pH of the solution when equilibrium exists is 9.301.

CONCEPT:

Boudouard Reaction and Stoichiometry

• When carbon dioxide (CO₂) gas is passed over red-hot coke (solid carbon), a chemical reaction occurs to form carbon monoxide (CO) gas.
• The balanced chemical equation for this reaction is:

CO₂(g) + C(s) → 2CO(g)

• According to the stoichiometry of the reaction, 1 mole (or 1 volume at constant T and P) of CO₂ gas reacts with solid coke to produce 2 moles (or 2 volumes) of CO gas.
• The total volume of the resulting mixture depends on the extent of the reaction and the amount of unreacted CO₂ remaining.

EXPLANATION:

• Let the initial volume of CO₂ gas be 1 dm³.
• Let 'x' dm³ be the volume of CO₂ that reacts with the coke.
• Using the balanced equation: CO₂(g) + C(s) → 2CO(g)
• Volume of CO₂ reacted = x dm³
• Volume of CO produced = 2x dm³
• After the reaction is complete, the gaseous mixture consists of:
• Volume of remaining (unreacted) CO₂ = (1 - x) dm³
• Volume of CO produced = 2x dm³
• The total volume of the gaseous mixture is given as 1.4 dm³. Therefore:
• Total Volume = (Volume of unreacted CO₂) + (Volume of CO produced)
• 1.4 = (1 - x) + 2x
• 1.4 = 1 + x
• x = 1.4 - 1 = 0.4 dm³
• Now, determine the composition of the mixture by substituting the value of x:
• Volume of CO = 2x = 2 × 0.4 = 0.8 dm³
• Volume of CO₂ = 1 - x = 1 - 0.4 = 0.6 dm³

Therefore, the composition of the gaseous mixture at STP is 0.8 dm³ of CO and 0.6 dm³ of CO₂.

CONCEPT:

Quantum Numbers and Orbitals

• Principal Quantum Number (n): It designates the main energy level or shell in which the electron resides. It can have positive integer values (1, 2, 3, ...).
• Azimuthal Quantum Number (l): Also known as the orbital angular momentum quantum number, it defines the three-dimensional shape of the orbital. For a given value of n, l can have values from 0 to (n-1).
• The values of l are associated with specific subshell notations:
• l = 0 corresponds to an s orbital.
• l = 1 corresponds to a p orbital.
• l = 2 corresponds to a d orbital.
• l = 3 corresponds to an f orbital.
• An orbital is identified by writing the principal quantum number followed by the letter symbol of the subshell (e.g., for n=2 and l=1, the orbital is 2p).

EXPLANATION:

• A. n = 2, l = 1:
• The principal shell is 2.
• Since l = 1, it represents a p-subshell.
• The orbital is 2p. (Match: II)
• B. n = 4, l = 0:
• The principal shell is 4.
• Since l = 0, it represents an s-subshell.
• The orbital is 4s. (Match: III)
• C. n = 5, l = 3:
• The principal shell is 5.
• Since l = 3, it represents an f-subshell.
• The orbital is 5f. (Match: IV)
• D. n = 3, l = 2:
• The principal shell is 3.
• Since l = 2, it represents a d-subshell.
• The orbital is 3d. (Match: I)

Therefore, the correct matching is A - II, B - III, C - IV, D - I.

CONCEPT:

Chlorination of Benzene

• Benzene undergoes different reactions with chlorine depending upon the reaction conditions.
• In the presence of anhydrous AlCl₃ under dark and cold conditions, electrophilic substitution occurs forming benzene hexachloride derivative with six chlorine atoms attached.
• In the presence of UV light and high temperature, addition reaction occurs and benzene forms benzene hexachloride (BHC), which also contains six chlorine atoms.

EXPLANATION:

• Formation of X:
• The product X contains six chlorine atoms.
• Formation of Y:
• The product Y is benzene hexachloride (BHC), which also contains six chlorine atoms.
• Therefore, the number of chlorine atoms present in X and Y are 6 and 6 respectively.

Hence, the number of chlorine atoms present in products X and Y are 6 and 6 respectively.

CONCEPT:

Reaction of Alcohol with PCl₅ and Dehydrohalogenation

• Alcohols react with PCl₅ to form alkyl chlorides.
• During this reaction, phosphorus oxychloride (POCl₃) and HCl are formed as by-products.
• Alkyl chlorides on heating with alcoholic KOH undergo β-elimination (dehydrohalogenation) to form alkenes.
• Alkenes react with HBr in the presence of peroxide ((C₆H₅CO)₂O₂) through anti-Markovnikov addition, giving bromoalkanes.

EXPLANATION:

• Hence, compound X is:

POCl₃

• Propyl chloride on heating with alcoholic KOH undergoes elimination reaction to form propene:

CH₃CH₂CH₂Cl  (\ \rightarrow[Δ]{alc.\ KOH} )CH₃CH=CH₂

• The alkene formed reacts with HBr in presence of peroxide.
• Due to peroxide effect, HBr adds according to anti-Markovnikov rule:

CH₃CH=CH₂ (\rightarrow) ([(C_{6}H_{5}CO){2}O{2}]{HBr} CH3CH2CH2Br)

• Therefore, compound Z is: CH₃CH₂CH₂Br

Therefore, X = POCl₃ and Z = CH₃CH₂CH₂Br.

CONCEPT:

Catalytic Properties of Transition Metals and their Compounds

• Transition metals and their compounds are known for their catalytic activity. This is attributed to their ability to adopt multiple oxidation states and their ability to form complexes.
• They provide a surface for reactants to be adsorbed, which brings the reactant molecules closer together and lowers the activation energy of the reaction.
• Specific catalysts are used for industrial processes like the Haber process, Contact process, and Wacker process.

EXPLANATION:

• A. V₂O₅ (Vanadium pentoxide):
• It is used as a catalyst in the Contact Process for the oxidation of sulphur dioxide (SO₂) to sulphur trioxide (SO₃), which is a key step in the preparation of sulphuric acid (H₂SO₄).
• Match: A - III
• B. Fe (Iron):
• Finely divided iron is used as a catalyst in the Haber Process for the industrial synthesis of ammonia from a nitrogen (N₂) and hydrogen (H₂) mixture.
• Match: B - I
• C. PdCl₂ (Palladium chloride):
• Palladium salts like PdCl₂ are used in the Wacker Process. While typically used for ethene, it is also involved in the catalytic oxidation of alkynes/alkenes to carbonyl compounds like ethanal.
• Match: C - IV
• D. Ni complex (Nickel complexes):
• Nickel complexes (such as those used in Reppe chemistry) are effective catalysts for the polymerisation of alkynes to form cyclic or linear polymers.
• Match: D - II

Therefore, the correct matching is A-III, B-I, C-IV, D-II.

CONCEPT:

Valence Shell Electron Pair Repulsion (VSEPR) Theory

• The geometry of a molecule depends on the total number of valence shell electron pairs (bond pairs and lone pairs) surrounding the central atom.
• The Steric Number (SN) is calculated as:

SN = (Number of bond pairs) + (Number of lone pairs)

• For a steric number of 5, the electron geometry is trigonal bipyramidal. If lone pairs are present, they occupy equatorial positions to minimize electronic repulsion, leading to different molecular shapes.

EXPLANATION:

• In the molecule ClF3ClF3ClF3

  :

• The central atom is Chlorine (ClClCl

  ), which belongs to Group 17 and has 7 valence electrons.

• There are 3 Fluorine (FFF

  ) atoms bonded to the central Chlorine atom, forming 3 bond pairs.

• The number of remaining valence electrons on Chlorine = 7 - 3 = 4 electrons.

• These 4 electrons form 42=242=242=2

  lone pairs.

• Calculation of Steric Number:

• Steric Number = 3 (bond pairs) + 2 (lone pairs) = 5.

• A steric number of 5 implies sp3dsp3dsp3d

  hybridization and a trigonal bipyramidal electron geometry.

• Molecular Geometry:

• With 2 lone pairs and 3 bond pairs, the two lone pairs occupy the equatorial positions of the trigonal bipyramid to minimize repulsion (lp-lp and lp-bp).

• This arrangement results in a molecular shape known as T-shaped geometry.

Therefore, ClF3ClF3ClF3

has a T-shaped geometry with two lone pairs on the ClClCl

atom.

CONCEPT:

Nernst Equation for a Half-Cell

• The electrode potential of a half-cell depends on the standard electrode potential, temperature, concentration of ions, and pressure of gases involved.
• For the hydrogen electrode represented as an oxidation electrode:

H₂(g) → 2H⁺(aq) + 2e^-

• The electrode potential is calculated using the Nernst equation:

E = E^o - (\dfrac{0.059}{n} log \dfrac{[H^+]^2}{P_{H_2}})

• Here,
• n = number of electrons transferred
• [H⁺] = concentration of hydrogen ions
• P_H2 = pressure of hydrogen gas

EXPLANATION:

• The given half-cell is:

Pt(s) | H₂(g, 2 atm) | HCl(aq, 0.02 M)

• The oxidation reaction taking place is:

H₂(g) → 2H⁺(aq) + 2e^-

• Given values:
• E^o = 0 V
• P_H2 = 2 atm
• [H⁺] = 0.02 M = 2 × 10^-2 M
• n = 2
• Using the Nernst equation:

E = 0 -( \dfrac{0.059}{2} log \dfrac{(2 \times 10^{-2})^2}{2})

• Simplifying:

(2 × 10^-2)² = 4 × 10^-4

(\dfrac{4 \times 10^{-4}}{2} = 2 \times 10^{-4})

E = -0.0295 ( \log(2 \times 10^{-4}))

• Using logarithmic properties:

(\log(2 \times 10^{-4}) = \log 2 + \log 10^{-4})

= 0.3010 - 4

= -3.699

• Substituting:

E = -0.0295 x (-3.699)

E = 0.109 V

Therefore, the electrode potential of the given half-cell is 0.109 V.

CONCEPT:

Units of Rate Constant (k)

• The rate law for a reaction of order n is expressed as: Rate = k[Concentration]ⁿ.
• To find the unit of the rate constant k, the formula used is:

Unit of k = (mol L^-1)^1-n s^-1

• Alternatively, the units can be written as mol^1-n L^n-1 s^-1.

EXPLANATION:

• A. Zero order (n = 0):
• Applying the formula: Unit = (mol L^-1)^1-0 s^-1
• Unit = mol L^-1 s^-1
• This matches with IV.
• B. First order (n = 1):
• Applying the formula: Unit = (mol L^-1)^1-1 s^-1
• Unit = (mol L^-1)⁰ s^-1 = s^-1
• This matches with III.
• C. Second order (n = 2):
• Applying the formula: Unit = (mol L^-1)^1-2 s^-1
• Unit = (mol L^-1)^-1 s^-1 = mol^-1 L s^-1
• This matches with I.
• D. Third order (n = 3):
• Applying the formula: Unit = (mol L^-1)^1-3 s^-1
• Unit = (mol L^-1)^-2 s^-1 = mol^-2 L² s^-1
• This matches with II.

Therefore, the correct matching is A-IV, B-III, C-I, D-II.

CONCEPT:

Spin-only Magnetic Moment (μ)

• The magnetic moment of transition metal ions is primarily determined by the number of unpaired electrons present in their d-orbitals.
• It can be calculated using the spin-only formula:

μ = √[n(n + 2)] BM

• Where:
• n is the number of unpaired electrons.
• BM stands for Bohr Magneton, which is the unit of magnetic moment.

EXPLANATION:

• The given ion is Ti²⁺.
• The atomic number of Titanium (Ti) is 22. Its ground state electronic configuration is:

Ti: [Ar] 3d² 4s²

• To form the Ti²⁺ ion, two electrons are removed from the outermost 4s orbital:

Ti²⁺: [Ar] 3d² 4s⁰ (or simply 3d²)

• In the 3d² configuration, there are 2 unpaired electrons (n = 2) in the d-subshell according to Hund's rule.
• Applying the spin-only magnetic moment formula:
• μ = √[n(n + 2)]
• μ = √[2(2 + 2)]
• μ = √[2 × 4]
• μ = √8
• μ ≈ 2.828 BM
• Rounding to two decimal places, we get approximately 2.84 BM.

Therefore, the calculated 'spin-only' magnetic moment of Ti²⁺ is 2.84 BM.

CONCEPT:

Friedel-Crafts Alkylation and Nitration of Toluene

• Benzene reacts with CH₃Cl in the presence of anhydrous AlCl₃ through Friedel-Crafts alkylation to form toluene.
• Toluene undergoes nitration with dilute HNO₃ and dilute H₂SO₄ to form a mixture of ortho-nitrotoluene and para-nitrotoluene.
• Ortho and para products are liquid isomers having different boiling points.
• Such liquid mixtures are separated by fractional distillation.

EXPLANATION:

• In the first step, benzene reacts with methyl chloride in the presence of anhydrous AlCl₃:

C₆H₆ + CH₃Cl &xrightarrow{anhyd.\ AlCl_{3}} C₆H₅CH₃

The product formed is toluene.

• T oluene undergoes nitration with dilute nitric acid and dilute sulphuric acid:

C₆H₅CH₃ &xrightarrow[dil.\ H_{2}SO_{4}]{dil.\ HNO_{3}} o\text{-nitrotoluene} + p\text{-nitrotoluene}

• The methyl group is an electron-donating group and directs substitution mainly at the ortho and para positions.
• The products obtained are:
• Ortho-nitrotoluene (X)
• Para-nitrotoluene (Y)
• Since both compounds are liquids with different boiling points, they can be separated by:

Fractional distillation

Therefore, products X and Y can be separated using fractional distillation.

CONCEPT:

• Power and Energy: Power (P) is the rate at which energy is emitted or consumed per unit time. 1 Watt = 1 Joule per second.
• Efficiency (η): It is the ratio of useful energy output to the total energy input. In this case, the useful energy is the light energy.
• Photon Count: The total energy emitted in the form of light is the product of the number of photons (n) and the energy of a single photon (E_p).

Total Light Energy = n × E_p

EXPLANATION:

• Given data:
• Power of the bulb (P_total) = 150 W = 150 J/s
• Efficiency (η) = 8% = 0.08
• Energy of one photon (E_p) = 4.42 × 10^-19 J
• Calculate the energy converted into light per second:
• Energy emitted as light per second (E_light) = Total Power × Efficiency
• E_light = 150 J/s × 0.08
• E_light = 12 J/s
• Calculate the number of photons emitted per second (n):
• The number of photons emitted per second is the total light energy per second divided by the energy of one photon.
• n = E_light / E_p
• n = 12 / (4.42 × 10^-19)
• n = (12 / 4.42) × 10¹⁹
• n ≈ 2.7149 × 10¹⁹

Therefore, the number of photons emitted by the bulb per second is 2.71 × 10¹⁹.

CONCEPT:

Identification of Anions (Acidic Radicals)

• In qualitative inorganic analysis, dilute sulphuric acid (H₂SO₄) is used to detect certain anions based on the characteristics of the gases they evolve.
• Each anion reacts with the acid to produce a specific gas or vapor with unique properties such as color, odor, and its effect on indicators like litmus paper.

EXPLANATION:

• When dilute H₂SO₄ is added to a salt containing the acetate ion (CH₃COO^-), acetic acid (CH₃COOH) is formed.

2CH₃COO^- + H₂SO₄ → 2CH₃COOH + SO₄^2-

• Vinegar Smell: The resulting acetic acid vapors have a very characteristic sharp smell, similar to vinegar.
• Litmus Test: Since acetic acid is an acid, its vapors turn moist blue litmus paper red.
• Analysis of other options:
• Carbonate (CO₃^2-): Reacts with dilute H₂SO₄ to produce CO₂ gas, which is colorless and odorless, and turns lime water milky.
• Sulphide (S^2-): Reacts with dilute H₂SO₄ to produce H₂S gas, which has a distinct smell of rotten eggs.
• Sulphate (SO₄^2-): Does not react with dilute H₂SO₄ because it is the anion of the acid itself.

Therefore, the anion that produces colorless vapors with the smell of vinegar and turns blue litmus red is Acetate (CH₃COO^-).

CONCEPT:

Reduction of Nitriles to Primary Amines

• The reduction of nitriles (R-C≡N) involves the addition of hydrogen across the carbon-nitrogen triple bond to form primary amines (R-CH₂NH₂).
• This transformation can be achieved using strong reducing agents, catalytic hydrogenation, or specific chemical reduction methods like the Mendius reaction.

EXPLANATION:

• A. (i) LiAlH₄; (ii) H₂O: Lithium aluminium hydride is a strong nucleophilic reducing agent. It reduces nitriles to primary amines. The second step (hydrolysis) is necessary to release the final amine from its complex.

R-C≡N + [H] → R-CH₂NH₂

• B. Sn + HCl: This reagent combination is primarily used for the reduction of nitro groups (-NO₂) into primary amines. It is not typically used to reduce nitriles to primary amines.
• C. H₂/Ni: Catalytic hydrogenation using hydrogen gas in the presence of a nickel catalyst (such as Raney Nickel) is a standard method to reduce nitriles to primary amines.
• D. Na(Hg)/C₂H₅OH: The reduction of nitriles using sodium amalgam in ethanol is known as the Mendius reduction. It effectively produces primary amines from nitriles.
• E. Br₂/aq. NaOH: These are the reagents for the Hoffmann Bromamide Degradation reaction. This reaction converts an amide into a primary amine with one fewer carbon atom and is not used for nitriles.

Comparing the given reagents, A, C, and D are the correct reagents for the conversion of nitriles to primary amines.

Therefore, the correct answer is A, C and D only.

CONCEPT:

Properties of p-Block Elements

• The chemical and physical properties of p-block elements are influenced by their atomic size, electronegativity, and electronic configuration.
• Catenation: This is the property of atoms of the same element to link together to form long chains or rings. It depends on the strength of the element-element bond.
• Oxidation State: This represents the number of electrons an atom loses, gains, or appears to use when joining with other atoms in compounds.
• Multiple Bonding: Smaller atoms of the second period (like Carbon) have the ability to form stable ( p\pi-p\pi ) multiple bonds.

EXPLANATION:

• Carbon's Multiple Bonding: Carbon has a small atomic size and high electronegativity, allowing it to form strong ( p\pi-p\pi ) multiple bonds with itself (as seen in alkenes and alkynes) and with other atoms like Nitrogen or Oxygen. This statement is correct.
• Trihalides of Boron and Aluminum: ( \text{BCl}_3 ) exists as a monomer because the small Boron atom cannot easily accommodate four large Chlorine atoms in a dimerized structure. ( \text{AlCl}_3 ), however, exists as a dimer (( \text{Al}_2\text{Cl}_6 )) in vapor and non-polar solvents to complete the octet of Aluminum. This statement is correct.
• Oxidation States of Oxygen: Oxygen typically exhibits a -2 oxidation state in most of its compounds (oxides). However, it exhibits other oxidation states as well:
• In peroxides (e.g., ( \text{H}_2\text{O}_2 )), the oxidation state is -1.
• In superoxides (e.g., ( \text{KO}_2 )), the oxidation state is -1/2.
• In ( \text{OF}_2 ), the oxidation state is +2.
• In ( \text{O}_2\text{F}_2 ), the oxidation state is +1.

Since Oxygen exhibits several oxidation states besides -2, the statement that it exhibits only -2 is incorrect.

• Catenation in Group 14: Catenation tendency decreases down the group as the atomic size increases and the element-element bond enthalpy decreases. The order is ( \text{C} \gg \text{Si} > \text{Ge} \approx \text{Sn} ). This statement is correct.

Therefore, the incorrect statement is: Oxygen exhibits only -2 oxidation state.

CONCEPT:

Oxidation States of Lanthanoids

• Lanthanoids predominantly exhibit a +3 oxidation state. However, some elements show +2 or +4 oxidation states because of the extra stability associated with attaining an empty (f⁰), half-filled (f⁷), or fully filled (f¹⁴) f-subshell.
• Cerium (Ce) is the first element of the lanthanoid series with an atomic number of 58.

EXPLANATION:

• The electronic configuration of Cerium (Z = 58) in its ground state is:

Ce: [Xe] 4f¹ 5d¹ 6s²

• In the +3 oxidation state (Ce³⁺), the configuration is:

Ce³⁺: [Xe] 4f¹

• In the +4 oxidation state (Ce⁴⁺), the atom loses one more electron from the 4f orbital:

Ce⁴⁺: [Xe] 4f⁰

• The ( 4f^0 ) electronic configuration represents an empty f-subshell, which is highly stable as it achieves the noble gas configuration of Xenon ([Xe]).
• While Ce⁴⁺ is a strong oxidizing agent (tending to revert to the more stable +3 state in aqueous solution), the initial formation of the +4 state is facilitated by this noble gas configuration.

Therefore, Cerium shows a +4 oxidation state because after losing one more electron (from the +3 state), it acquires a ( 4f^0 ) electronic configuration.

CONCEPT:

Lassaigne's Test (Sodium Fusion Test)

• Lassaigne's test is a qualitative analysis used to detect the presence of nitrogen, sulphur, and halogens in an organic compound.
• Organic compounds are generally covalent in nature, which means the elements are bonded covalently and do not ionize easily in water.
• To identify these elements using inorganic reagents, they must first be converted into water-soluble ionic compounds.

EXPLANATION:

• In Lassaigne's test, the organic compound is fused with a small piece of sodium metal in a fusion tube.
• Sodium metal is highly reactive and converts the elements present in the organic compound into their corresponding ionic sodium salts:
• Sodium + Carbon + Nitrogen → Sodium cyanide (NaCN)
• 2 × Sodium + Sulphur → Sodium sulphide (Na₂S)
• Sodium + Halogen (X) → Sodium halide (NaX)
• In the organic compound, nitrogen, sulphur, and halogens are present in the covalent form.
• After fusion with sodium metal, these elements are converted into the ionic form (Na⁺, CN^-, S^2-, X^-).
• The resulting ionic salts are then dissolved in distilled water to prepare the Lassaigne's extract (sodium fusion extract) for further chemical tests.

Therefore, during Lassaigne's test, the elements present in an organic compound are converted from covalent form to ionic form.

CONCEPT:

Mole Concept and Stoichiometry

• The number of moles (n) of a substance is calculated using the formula:

n = Mass / Molar mass

• One mole of any substance contains Avogadro's number (N_A) of entities (molecules or atoms), which is approximately 6.022 × 10²³.
• To find the total number of atoms of a specific element in a sample, multiply the number of moles of the compound by Avogadro's number and then by the number of atoms of that element present in one molecule of the compound.

EXPLANATION:

• The chemical formula for urea is NH₂CONH₂ (or CH₄N₂O).
• One molecule of urea contains 4 hydrogen (H) atoms.
• Given data:
• Mass of urea = 5.4 g
• Molar mass of urea = 60 g mol^-1
• N_A = 6.022 × 10²³ mol^-1
• Step 1: Calculate the number of moles of urea:
• Moles of urea = Mass / Molar mass
• Moles of urea = 5.4 / 60 = 0.09 mol
• Step 2: Calculate the number of molecules of urea:
• Number of molecules = Moles × N_A
• Number of molecules = 0.09 × 6.022 × 10²³
• Step 3: Calculate the total number of hydrogen atoms:
• Since 1 molecule of urea has 4 H atoms,
• Total H atoms = 4 × Number of molecules
• Total H atoms = 4 × 0.09 × 6.022 × 10²³
• Total H atoms = 0.36 × 6.022 × 10²³
• Total H atoms = 2.16792 × 10²³

Therefore, the number of hydrogen atoms present in 5.4 g of urea is 2.168 × 10²³.

CONCEPT:

Metamerism

• Metamerism is a type of structural isomerism shown by compounds having the same molecular formula and the same functional group but different alkyl groups on either side of a polyvalent atom such as oxygen, sulphur, or nitrogen.
• Ethers commonly show metamerism because different alkyl groups can be attached on both sides of the oxygen atom.
• For metamers:
• Molecular formula must be the same.
• Functional group must be the same.
• Distribution of carbon atoms around the functional group must be different.

EXPLANATION:

• The compounds: CH₃OCH₂CH₂CH₃  and  CH₃CH₂OCH₂CH₃
  

are ethers having the same molecular formula: C₄H₁₀O

• In the first ether, the alkyl groups attached to oxygen are:
• Methyl group (CH₃-)
• Propyl group (CH₃CH₂CH₂-)
• In the second ether, the alkyl groups attached to oxygen are:
• Ethyl group (CH₃CH₂-)
• Ethyl group (CH₃CH₂-)
• Thus, both compounds have the same functional group but different distribution of carbon chains around the oxygen atom.
• Hence, they are metamers.

Therefore, CH₃OCH₂CH₂CH₃ and CH₃CH₂OCH₂CH₃ are metamers.

CONCEPT:

Electronic Configuration and Bonding in Group 15 Elements

• Nitrogen (N): Being a second-period element, nitrogen has the valence electronic configuration 2s² 2p³. It lacks vacant d-orbitals in its valence shell.
• Heavier Elements: Phosphorus (P), Arsenic (As), and Antimony (Sb) have vacant d-orbitals in their valence shells (n ≥ 3), which allows them to expand their octet and participate in d-orbital bonding.
• Pi (π) Bonding: The type of π-bonding (pπ-pπ, pπ-dπ, or dπ-dπ) depends on the availability and size of the orbitals on the bonding atoms.

EXPLANATION:

• Bonding with Transition Metals: Triethylphosphine (P(C₂H₅)₃) and Triphenylarsine (As(C₆H₅)₃) can act as ligands that form dπ-dπ bonds. This occurs when filled d-orbitals of a transition metal overlap with the vacant d-orbitals of Phosphorus or Arsenic (back-bonding).
• Inability of Nitrogen to form dπ Bonds: Because nitrogen does not possess d-orbitals, it is physically impossible for it to form dπ-pπ bonds. In oxides of nitrogen (such as NO₂ or NO₃^-), any π-bonding that occurs is of the pπ-pπ type.
• Multiple Bonds in Nitrogen: Nitrogen is small enough to allow for effective lateral overlap of its p-orbitals, enabling it to form strong pπ-pπ multiple bonds with itself, as seen in the diatomic N₂ molecule.
• Catenation Property: Catenation is the ability of atoms of the same element to form chains or rings. Phosphorus (P₄), arsenic, and antimony show this property significantly, whereas nitrogen shows it only to a very limited extent.

Therefore, the statement that Nitrogen can form dπ-pπ bonds with oxygen is incorrect.

CONCEPT:

Acid-Base Indicators (Phenolphthalein)

• Indicators are chemical substances that change color depending on the pH of the medium.
• Phenolphthalein is a synthetic indicator that is widely used in acid-base titrations, especially those involving strong bases.
• It has a pH range of approximately 8.2 to 10.0:
• In acidic and neutral solutions (pH < 8.2), phenolphthalein is colorless.
• In alkaline/basic solutions (pH > 8.2), phenolphthalein turns pink or pinkish-red.

EXPLANATION:

• In the given titration:
• Sodium hydroxide (NaOH) is a strong base.
• Oxalic acid (H₂C₂O₄) is a weak organic acid.
• The phrase 'titration of sodium hydroxide against a standard solution of oxalic acid' indicates that the sodium hydroxide solution is being analyzed (placed in the conical flask) and the oxalic acid is the titrant (placed in the burette).
• Since the sodium hydroxide solution is in the conical flask, the initial medium is alkaline.
• When phenolphthalein is added to the sodium hydroxide solution, it turns pink.
• As the standard oxalic acid is added from the burette, the hydroxide ions (OH^-) are neutralized by the hydrogen ions (H⁺) from the acid.
• At the equivalence point, the base is completely neutralized. As the pH drops below the indicator's transition range (approximately 8.2), the solution becomes colorless.
• Therefore, the color change observed at the endpoint is from pink to colorless.

The correct observation for the color change during this titration is pink to colourless.

CONCEPT:

• Sigma (σ) Bond: A covalent bond formed by the head-on overlap of atomic orbitals. All single bonds are σ bonds.
• Pi (π) Bond: A covalent bond formed by the lateral overlap of atomic orbitals. A double bond contains one σ and one π bond, while a triple bond contains one σ and two π bonds.
• Lone Pair: A pair of valence electrons that are not shared with another atom in a covalent bond.

EXPLANATION:

• A. C₂H₄ (Ethene):
• The Lewis structure is H₂C=CH₂.
• There are 4 C-H single bonds (4 σ bonds) and 1 C=C double bond (1 σ and 1 π bond).
• Total: 5 σ bonds and 1 π bond. This matches with IV.
  

• B. C₂H₂ (Ethyne):
• The Lewis structure is HC≡CH.
• There are 2 C-H single bonds (2 σ bonds) and 1 C≡C triple bond (1 σ and 2 π bonds).
• Total: 3 σ bonds and 2 π bonds. This matches with I.
  

• C. CH₄ (Methane):
• The Lewis structure consists of a central carbon atom bonded to four hydrogen atoms by single bonds.
• Total: 4 σ bonds. This matches with III.
  

• D. NH₃ (Ammonia):
• The nitrogen atom forms three single bonds with three hydrogen atoms (3 σ bonds).
• Nitrogen has 5 valence electrons, 3 are used for bonding, leaving 2 electrons as 1 lone pair.
• Total: 3 σ bonds and 1 lone pair. This matches with II.
  

Matching the pairs: A-IV, B-I, C-III, D-II.

CONCEPT:

First Law of Thermodynamics

• The First Law of Thermodynamics states that the change in the internal energy (ΔU) of a system is equal to the heat (q) added to the system plus the work (w) done on the system.
• The mathematical expression is:

ΔU = q + w

• Sign Conventions:
• Heat (q): Positive (+) if heat is absorbed by the system, negative (−) if heat is released by the system.
• Work (w): Positive (+) if work is done on the system (compression), negative (−) if work is done by the system (expansion).

EXPLANATION:

• According to the problem:
• Heat absorbed by the system (q) = +500 J
• Work done by the system (w) = -200 J (Since work is done by the system, the sign is negative according to IUPAC convention).
• Using the First Law of Thermodynamics formula:
• ΔU = q + w
• ΔU = 500 J + (-200 J)
• ΔU = 500 J - 200 J
• ΔU = 300 J

Therefore, the change in internal energy of the system is 300 J.

CONCEPT:

Steam Reforming of Methane

• Steam reforming is the process by which hydrocarbons like methane react with steam at high temperatures in the presence of a catalyst to produce dihydrogen.
• This reaction is used industrially for the large-scale production of hydrogen gas and synthesis gas (syngas).

EXPLANATION:

• The balanced chemical equation for the reaction of methane with steam is:

CH₄(g) + H₂O(g) → CO(g) + 3H₂(g)

• Reactants: Methane (CH₄) and Steam (H₂O).
• Conditions: Temperature of 1273 K and a Nickel (Ni) catalyst.
• Products: Carbon monoxide (CO) and Hydrogen gas (H₂).
• Methane reacts with steam at 1273 K using Nickel as a catalyst to yield a mixture of carbon monoxide and hydrogen gas.
• The resulting mixture of CO and H₂ is often called synthesis gas or water gas.

Therefore, methane reacts with steam at 1273 K in the presence of nickel catalyst to form CO and H₂.

CONCEPT:

Identification of Aldehyde and Ketone Compounds

• 2,4-DNP reagent gives a red-orange precipitate with aldehydes and ketones due to the presence of the carbonyl group (>C=O).
• Fehling’s reagent is reduced by aliphatic aldehydes, but aromatic ketones do not reduce Fehling’s solution.
• Strong oxidation of alkyl side chains attached to a benzene ring converts them into carboxylic acid groups.
• Carboxylic acids react with aqueous NaHCO₃ to produce brisk effervescence due to the evolution of CO₂ gas.

EXPLANATION:

Therefore, compound P is acetophenone and compound Q is benzoic acid.

CONCEPT:

Faraday's First Law of Electrolysis

• The mass (m) of a substance deposited or liberated at any electrode is directly proportional to the quantity of electricity (Q) passed through the electrolyte.
• It is mathematically expressed as:

m = Z × Q = Z × I × t

• The formula to calculate the mass of a substance deposited is:

m = (M × I × t) / (n × F)

• m = mass of the substance deposited (in grams)
• M = molar mass of the substance (in g mol^-1)
• I = current (in amperes)
• t = time (in seconds)
• n = number of electrons involved in the redox reaction (valency)
• F = Faraday constant (approximately 96487 C mol^-1)

EXPLANATION:

• Step 1: Identify the given values
• Current (I) = 1.5 A
• Time (t) = 10 minutes = 10 × 60 seconds = 600 s
• Molar mass of Copper (M) = 63 g mol^-1
• Faraday constant (F) = 96487 C mol^-1
• Step 2: Determine the number of electrons (n)
• The electrolysis of copper sulphate (CuSO₄) involves the reduction of Cu²⁺ ions at the cathode:
• Cu²⁺(aq) + 2e^- → Cu(s)
• Therefore, the number of electrons involved, n = 2.
• Step 3: Calculate the mass deposited
• Using the formula: m = (M × I × t) / (n × F)
• m = (63 × 1.5 × 600) / (2 × 96487)
• m = 56700 / 192974
• m ≈ 0.2938 g

Therefore, the mass of copper deposited at the cathode is 0.2938 g.

CONCEPT:

Phthalein Dye Test

• The phthalein dye test is a qualitative chemical analysis used to identify the presence of a phenolic functional group in an organic compound.
• It involves a condensation reaction between a phenol and phthalic anhydride in the presence of a dehydrating agent.

EXPLANATION:

• When a phenolic compound is heated with phthalic anhydride and a few drops of concentrated sulfuric acid (H₂SO₄), it undergoes condensation to form a phthalein derivative.
• For example, when phenol itself reacts with phthalic anhydride, it forms phenolphthalein.
• The reaction can be represented as:

Phenol + Phthalic Anhydride → Phenolphthalein + H₂O

• Upon adding dilute sodium hydroxide (NaOH) to the reaction product, a characteristic color is produced due to the formation of a sodium salt with a highly conjugated system.
• Phenol produces a pink/red color.
• Resorcinol produces a green fluorescence (due to the formation of fluorescein).
• Alcohols, aldehydes, and carboxylic acids do not produce these specific colored dyes when subjected to these reaction conditions.

Therefore, the functional group that can be identified through the phthalein dye test is phenolic.

CONCEPT:

Secondary Structure of Nucleic Acids

• DNA (Deoxyribonucleic Acid): The secondary structure of DNA is a double-stranded helix. Two polynucleotide chains are coiled around each other and held together by hydrogen bonds between specific base pairs. DNA contains four nitrogenous bases: Adenine (A), Guanine (G), Cytosine (C), and Thymine (T).
• RNA (Ribonucleic Acid): The secondary structure of RNA typically consists of a single polynucleotide strand (single-stranded helix). RNA contains four nitrogenous bases: Adenine (A), Guanine (G), Cytosine (C), and Uracil (U).

EXPLANATION:

• Analyzing the structural differences:
• DNA is double-stranded and contains the base thymine.
• RNA is single-stranded and contains the base uracil instead of thymine.
• Reviewing the characteristics:
• The statement that DNA possesses a double strand helix structure and contains thymine as one of the four bases is correct.
• The description of DNA as a single-stranded helix is incorrect.
• The description of RNA as having thymine is incorrect because RNA contains uracil.
• The description of RNA as a double strand helix is generally incorrect for its standard secondary structure.

Therefore, the correct statement is that DNA possesses a double strand helix structure and contains thymine as one of the four bases.

CONCEPT:

• Molality (m): It is defined as the number of moles of solute per kilogram (kg) of the solvent.

Molality (m) = (Mass of solute / Molar mass of solute) / Mass of solvent in kg

• Molarity (M): It is defined as the number of moles of solute per litre of the solution.

Molarity (M) = (Mass of solute / Molar mass of solute) / Volume of solution in L

• Henry Law and Gas Solubility:
• The solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid. Therefore, solubility increases with an increase in pressure.
• The solubility of gases in liquids typically decreases as the temperature increases. This makes cold water richer in dissolved oxygen compared to warm water.
• Mole Fraction (χ): For a binary mixture of A and B, the mole fraction of component B is the ratio of moles of B to the total moles in the solution.

χ_B = n_B / (n_A + n_B)

EXPLANATION:

• Statement A:
• Mass of ethanoic acid = 2.5 g, Molar mass = 60 g mol^-1
• Moles of ethanoic acid = 2.5 / 60 = 0.04167 mol
• Mass of benzene (solvent) = 75 g = 0.075 kg
• Molality = 0.04167 mol / 0.075 kg = 0.556 m
• Therefore, statement A is correct.
• Statement B:
• Mass of NaOH = 5 g, Molar mass = 40 g mol^-1
• Moles of NaOH = 5 / 40 = 0.125 mol
• Volume of solution = 450 mL = 0.450 L
• Molarity = 0.125 mol / 0.450 L = 0.278 M
• Therefore, statement B is correct.
• Statement C: Solubility of oxygen in water increases as the temperature decreases. Because cold water holds more dissolved oxygen, aquatic species are more comfortable in it. Therefore, statement C is correct.
• Statement D: According to Henry Law, the solubility of a gas increases with an increase in pressure. Statement D is incorrect.
• Statement E: The mole fraction of B is χ_B = n_B / (n_A + n_B). The given formula χ_B = n_A / (n_A + n_B) actually defines the mole fraction of component A. Therefore, statement E is incorrect.

Based on the above analysis, the correct statements are A, B and C only.

CONCEPT:

Negative Deviation from Raoult's Law

• A non-ideal solution shows a negative deviation from Raoult's law when the vapor pressure of the mixture is lower than the value calculated using Raoult's law.
• This behavior occurs when the intermolecular attractive forces between the different components (A-B interactions) are stronger than the forces between the like molecules (A-A and B-B interactions).
• Stronger A-B interactions lead to a decrease in the escaping tendency of the molecules into the vapor phase.

EXPLANATION:

• Chloroform (CHCl₃) and acetone (CH₃COCH₃) molecules exhibit specific interactions when mixed.
• The hydrogen atom in chloroform is bonded to a carbon atom that is attached to three highly electronegative chlorine atoms. This makes the hydrogen atom electron-deficient (acidic).
• The acetone molecule contains a carbonyl oxygen (C=O) with lone pairs of electrons.
• When these two substances are mixed, a hydrogen bond is formed between the hydrogen atom of chloroform and the oxygen atom of acetone.
• The formation of this hydrogen bond represents a stronger intermolecular attraction (A-B) compared to the dipole-dipole interactions found in pure chloroform (A-A) or pure acetone (B-B).
• Because the molecules are held more strongly together in the mixture, fewer molecules escape to the vapor phase, resulting in a lower vapor pressure and thus a negative deviation from Raoult's law.

Therefore, the mixture of chloroform and acetone shows negative deviation from Raoult's law due to the formation of hydrogen bonding between acetone and chloroform.

CONCEPT:

Henderson-Hasselbalch Equation

• For an acidic buffer solution, the pH is calculated using the Henderson-Hasselbalch equation:

pH = pK_a + log([Conjugate Base] / [Acid])

• The relationship between the acid dissociation constant (K_a) and the base dissociation constant (K_b) for a conjugate acid-base pair is:

K_a × K_b = K_w

where K_w is the ionic product of water, which is 10^-14 at 298 K.

EXPLANATION:

• Given data:
• K_b for X^- = 10^-10
• Concentration of salt [X^-] = Concentration of acid [HX]
• Temperature = 298 K
• Step 1: Find the value of K_a for HX:
• K_a = K_w / K_b
• K_a = 10^-14 / 10^-10
• K_a = 10^-4
• Step 2: Calculate the pK_a value:
• pK_a = -log(K_a)
• pK_a = -log(10^-4)
• pK_a = 4
• Step 3: Calculate the pH of the buffer using the Henderson-Hasselbalch equation:
• pH = pK_a + log([X^-] / [HX])
• Since [X^-] = [HX], the ratio [X^-] / [HX] = 1.
• pH = 4 + log(1)
• Since log(1) = 0:
• pH = 4 + 0
• pH = 4

Therefore, the pH of the buffer solution is 4.

CONCEPT:

• IUPAC Nomenclature of Elements: Elements with atomic numbers greater than 100 are named systematically using roots for each digit: 1 (un), 0 (nil), and 7 (sept), ending with the suffix 'ium'.
• Atomic and Ionic Radii Trends:
• In a period, the atomic radius generally decreases from left to right due to the increasing effective nuclear charge.
• For isoelectronic species (atoms or ions with the same number of electrons), the radius decreases as the atomic number increases because the nucleus exerts a stronger pull on the same number of electrons.
• Cations are always smaller than their parent neutral atoms due to the loss of an electron shell or increased effective nuclear charge.
• Diagonal Relationship: This refers to the similarity in chemical properties between certain elements of the second period and the elements of the third period placed diagonally to them, such as Lithium (Li) and Magnesium (Mg).
• Oxidation State and Covalency: The oxidation state is the charge assigned to an atom in a compound, while the covalency in coordination complexes typically refers to the coordination number (the number of ligand donor atoms bonded to the central metal).

EXPLANATION:

• Naming of Element 107: Using IUPAC roots, 1 is 'un', 0 is 'nil', and 7 is 'sept'. Combining these gives Un + nil + sept + ium = Unnilseptium. This statement is correct.
• Comparison of Sizes:
• Between ( Mg ) (Z=12) and ( Al ) (Z=13), both are in the third period. Since atomic size decreases across the period, ( Mg ) is larger than ( Al ).
• Between ( Mg^{2+} ) and ( Al^{3+} ), both are isoelectronic with 10 electrons. ( Al^{3+} ) has a higher nuclear charge (+13) compared to ( Mg^{2+} ) (+12), making ( Al^{3+} ) the smaller ion.
• Neutral atoms are larger than their corresponding cations. Therefore, the overall size order is ( Mg > Al > Mg^{2+} > Al^{3+} ).
• The largest species is ( Mg ) and the smallest is ( Al^{3+} ). The statement identifying ( Al ) as the largest and ( Mg^{2+} ) as the smallest is incorrect.
• Diagonal Relationship: Li and Mg show similar properties such as the formation of nitrides and the thermal instability of their carbonates because they have similar ionic sizes and polarizing power. This statement is correct.
• Aluminum Complex: In ( [AlCl(H_2O)_5]^{2+} ):
• The oxidation state of Al (x) is calculated as: ( x + (-1) + 5(0) = +2 \Rightarrow x = +3 ).
• The coordination number (covalency) is the total number of ligands, which is ( 1 (Cl^-) + 5 (H_2O) = 6 ). This statement is correct.

Therefore, the incorrect statement is that the largest and smallest species among the given group are Al and Mg²⁺, respectively.

CONCEPT:

• Metallic Character: It refers to the ease with which an atom can lose its valence electrons to form positive ions (cations).
• Periodic Trends:
• Across a Period (Left to Right): Metallic character decreases because the effective nuclear charge increases, pulling the electrons closer to the nucleus and making them harder to remove.
• Down a Group (Top to Bottom): Metallic character increases because the atomic size increases and the valence electrons are farther from the nucleus, making them easier to remove.

EXPLANATION:

• Identify the positions of the given elements in the periodic table:
• Na (Sodium): Group 1, Period 3
• Mg (Magnesium): Group 2, Period 3
• Be (Beryllium): Group 2, Period 2
• Si (Silicon): Group 14, Period 3
• P (Phosphorus): Group 15, Period 3
• Comparing Period 3 elements (Na, Mg, Si, P):
• Metallic character decreases from Group 1 to Group 15.
• Order: Na > Mg > Si > P.
• Comparing Group 2 elements (Be, Mg):
• Metallic character increases down the group.
• Order: Mg > Be.
• Relative Metallic Nature:
• P is a non-metal (least metallic).
• Si is a metalloid (more metallic than P).
• Be is a group 2 metal. Although it is in Period 2, it is more metallic than the metalloid Si.
• Mg is below Be in Group 2, so it is more metallic than Be.
• Na is to the left of Mg in Period 3, so it is the most metallic.
• Combining the trends, the increasing order of metallic character is:

P < Si < Be < Mg < Na

Therefore, the correct order is P < Si < Be < Mg < Na.

CONCEPT:

IUPAC Rules for Naming Alkanes:

• Longest Chain Rule: The parent structure is the longest continuous chain of carbon atoms.
• Lowest Locant Rule: The principal chain is numbered from the end that gives the lowest possible locants (position numbers) to the substituents.
• Alphabetical Order Rule: If two different substituents are present at equivalent positions from either end of the chain, the numbering is done such that the substituent that comes first alphabetically is assigned the lower number.

EXPLANATION:

• Identification of the longest chain:

In the given compound, if we select the horizontal chain, it contains 6 carbon atoms (hexane). However, by including the carbons of the ethyl groups, a longer chain of 7 carbon atoms (heptane) can be identified.

• Identifying substituents:

When the 7-carbon chain is selected as the parent chain, the groups remaining outside the chain are a methyl group (-CH₃) and an ethyl group (-CH₂CH₃).

• 
• Assembling the name:

Combining the locants, substituent names, and the parent chain name (heptane) gives: 3-ethyl-5-methylheptane.

Therefore, the correct IUPAC name of the compound is 3-ethyl-5-methylheptane.

CONCEPT:

Coordination Polyhedron and Geometry

• The spatial arrangement of the ligand atoms which are directly attached to the central atom/ion defines a coordination polyhedron about the central atom.
• The most common coordination polyhedra are octahedral, square planar, tetrahedral, and trigonal bipyramidal.
• The geometry is determined by the coordination number (CN) and the hybridization of the central metal ion:
• CN = 4: Tetrahedral (sp³) or Square planar (dsp²).
• CN = 5: Trigonal bipyramidal (dsp³).
• CN = 6: Octahedral (d²sp³ or sp³d²).

EXPLANATION:

• A. [Pt(Cl₂)(NH₃)₂]:
• Central metal: Pt²⁺ (d⁸ configuration).
• Coordination number: 4.
• For Platinum(II) complexes, the crystal field stabilization energy is very high, making 4-coordinate Pt(II) complexes almost exclusively square planar (III).
• B. [Co(NH₃)₆]Cl₃:
• Central metal: Co³⁺ (d⁶ configuration).
• Coordination number: 6.
• A coordination number of 6 corresponds to an octahedral geometry (I).
• C. [NiCl₄]^2-:
• Central metal: Ni²⁺ (d⁸ configuration).
• Coordination number: 4.
• Cl^- is a weak field ligand, resulting in sp³ hybridization and a tetrahedral geometry (IV).
• D. [Fe(CO)₅]:
• Central metal: Fe⁰ (d⁸ configuration after electron pairing).
• Coordination number: 5.
• A coordination number of 5 for d⁸ systems with strong field ligands like CO leads to a trigonal bipyramidal geometry (II).

Based on the above analysis, the correct matching is: A - III, B - I, C - IV, D - II.

CONCEPT:

Integrated Rate Laws

• The order of a reaction is determined by the relationship between the concentration of reactants and time.
• For a zero-order reaction, the rate of reaction is independent of the concentration of reactants. The integrated rate equation is:

[R]=−kt+[R]0[R]=−kt+[R]0[R]=−kt+[R]0

where [R][R][R]

is the concentration at time ttt

, kkk

is the rate constant, and [R]0[R]0[R]0

is the initial concentration.

• For a first-order reaction, the integrated rate equation is:

ln[R]=−kt+ln[R]0ln[R]=−kt+ln[R]0ln⁡[R]=−kt+ln⁡[R]0

EXPLANATION:

• The given plot is concentration [R][R][R]

  versus time ttt

  .

• From the zero-order integrated rate equation:

[R]=−kt+[R]0[R]=−kt+[R]0[R]=−kt+[R]0

• This equation is in the form of a straight line, y=mx+cy=mx+cy=mx+c

  , where:

• y=[R]y=[R]y=[R]

• x=tx=tx=t

• m=−km=−km=−k

  (Slope)

• c=[R]0c=[R]0c=[R]0

  (Intercept)

• Since the graph of [R][R][R]

  vs ttt

  is a straight line with a negative slope (equal to −k−k−k

  ), the reaction must be of zero order.

• In contrast, for a first-order reaction, a plot of ln[R]ln[R]ln⁡[R]

  vs ttt

  would be linear, and for a second-order reaction, a plot of 1/[R]1/[R]1/[R]

  vs ttt

  would be linear.

Therefore, the order of the reaction is 0.

CONCEPT:

Ambidentate Ligands

• Ligands that possess more than one donor atom but coordinate to the central metal atom or ion through only one of those donor atoms at a time are called ambidentate ligands.
• Common examples include the thiocyanate ion (SCN^-), the nitrite ion (NO₂^-), and the cyanide ion (CN^-).

EXPLANATION:

• Thiocyanate (SCN^-): This ligand has two potential donor atoms: Sulfur (S) and Nitrogen (N). It can coordinate to the metal ion through the sulfur atom (M-SCN, known as thiocyanato) or through the nitrogen atom (M-NCS, known as isothiocyanato). Because it has two different ways to bind, it is an ambidentate ligand.
• Ethylenediaminetetraacetate ion (EDTA): This is a hexadentate ligand. It coordinates to a metal ion through six donor atoms (two nitrogen atoms and four oxygen atoms) simultaneously, forming a stable chelate complex.
• Oxalate (C₂O₄^2-): This is a bidentate ligand. it coordinates to the central metal ion through two oxygen atoms at the same time.
• Ethane-1,2-diamine (en): This is also a bidentate ligand, coordinating to the central metal ion through two nitrogen atoms simultaneously.

Therefore, Thiocyanate is an ambidentate ligand.

CONCEPT:

• The relationship between the standard enthalpy change (ΔH°) and the standard internal energy change (ΔU°) for a gaseous reaction is given by:

ΔH° = ΔU° + Δn_gRT

• The standard Gibbs free energy change (ΔG°) is related to ΔH° and ΔS° by the Gibbs-Helmholtz equation:

ΔG° = ΔH° - TΔS°

• Spontaneity Criteria:
• If ΔG° < 0, the reaction is spontaneous.
• If ΔG° > 0, the reaction is non-spontaneous.
• If ΔG° = 0, the reaction is at equilibrium.

EXPLANATION:

• For the given reaction:

2A(g) + B(g) → 2D(g)

• Δn_g = (moles of gaseous products) - (moles of gaseous reactants)
• Δn_g = 2 - (2 + 1) = 2 - 3 = -1
• Given values:
• ΔU° = -10 kJ mol^-1 = -10,000 J mol^-1
• ΔS° = -44 J K^-1 mol^-1
• T = 298 K
• R = 8.31 J mol^-1 K^-1
• Step 1: Calculate the standard enthalpy change (ΔH°):

ΔH° = ΔU° + Δn_gRT

ΔH° = -10,000 J mol^-1 + [(-1) × 8.31 J mol^-1 K^-1 × 298 K]

ΔH° = -10,000 - 2476.38

ΔH° = -12,476.38 J mol^-1

• Step 2: Calculate the standard Gibbs free energy change (ΔG°):

ΔG° = ΔH° - TΔS°

ΔG° = -12,476.38 J mol^-1 - [298 K × (-44 J K^-1 mol^-1)]

ΔG° = -12,476.38 + 13,112

ΔG° = +635.62 J mol^-1

ΔG° = +0.63562 kJ mol^-1 ≈ +0.63568 kJ mol^-1

• Since ΔG° is positive (+0.63568 kJ mol^-1), the reaction is non-spontaneous at 298 K.

Therefore, the standard Gibbs free energy change (ΔG°) for the reaction is +0.63568 kJ mol^-1 and the reaction is non-spontaneous.

CONCEPT:

Formal Charge

• Formal charge is the theoretical charge assigned to an atom in a molecule, assuming that electrons in all chemical bonds are shared equally between atoms, regardless of relative electronegativity.
• The formal charge (FC) of an atom in a Lewis structure can be calculated using the following formula:

FC = [V] - [L] - [B/2]

Where:

• V = Number of valence electrons of the atom in its ground state.
• L = Number of non-bonding (lone pair) electrons.
• B = Number of bonding (shared) electrons.

EXPLANATION:

• In the Lewis structure of the ozone (O₃) molecule, there are three oxygen atoms. For any oxygen atom, the number of valence electrons (V) is 6.
• Based on the numbering in the structure:
• For Oxygen atom 2 (Double-bonded oxygen):
• Number of lone pair electrons (L) = 4 (2 lone pairs)
• Number of bonding electrons (B) = 4 (from 1 double bond)
• FC = 6 - 4 - (4/2) = 6 - 4 - 2 = 0
• For Oxygen atom 1 (Central oxygen):
• Number of lone pair electrons (L) = 2 (1 lone pair)
• Number of bonding electrons (B) = 6 (from 1 double bond and 1 single bond)
• FC = 6 - 2 - (6/2) = 6 - 2 - 3 = +1
• For Oxygen atom 3 (Single-bonded oxygen):
• Number of lone pair electrons (L) = 6 (3 lone pairs)
• Number of bonding electrons (B) = 2 (from 1 single bond)
• FC = 6 - 6 - (2/2) = 6 - 6 - 1 = -1
• The formal charges on atoms 2, 1, and 3 are 0, +1, and -1 respectively.

Therefore, the correct formal charges on oxygen atoms numbered 2, 1 and 3 respectively are 0, +1, -1.

CONCEPT:

Relationship between K_p and K_c

• The equilibrium constant in terms of partial pressure (K_p) and the equilibrium constant in terms of molar concentration (K_c) are related by the following equation:

K_p = K_c(RT)^Δn_g

• Where:
• R is the universal gas constant.
• T is the absolute temperature (in Kelvin).
• Δn_g is the change in the number of moles of gaseous species, calculated as:

Δn_g = (Sum of stoichiometric coefficients of gaseous products) - (Sum of stoichiometric coefficients of gaseous reactants)

• From the equation, we can conclude that:
• If Δn_g = 0, then K_p = K_c.
• If Δn_g ≠ 0, then K_p ≠ K_c.

EXPLANATION:

• To identify the reaction where K_p ≠ K_c, we must calculate Δn_g for each reaction:
• For H₂(g) + I₂(g) ⇌ 2HI(g):

Δn_g = 2 - (1 + 1) = 2 - 2 = 0. Thus, K_p = K_c.

• For N₂(g) + O₂(g) ⇌ 2NO(g):

Δn_g = 2 - (1 + 1) = 2 - 2 = 0. Thus, K_p = K_c.

• For N₂(g) + 3H₂(g) ⇌ 2NH₃(g):

Δn_g = 2 - (1 + 3) = 2 - 4 = -2. Since Δn_g ≠ 0, K_p ≠ K_c.

• For H₂O(g) + CO(g) ⇌ H₂(g) + CO₂(g):

Δn_g = (1 + 1) - (1 + 1) = 2 - 2 = 0. Thus, K_p = K_c.

Therefore, the reaction for which K_p ≠ K_c is N₂(g) + 3H₂(g) ⇌ 2NH₃(g).

CONCEPT:

Arrhenius Equation

• The temperature dependence of the rate constant (k) for a reaction is given by the Arrhenius equation:

k = A e^−E_a/RT

• Taking the natural logarithm (ln) on both sides of the equation, we get the linear form:

ln k = ln A − E_a / (RT)

• Where:
• k is the rate constant.
• A is the pre-exponential factor or frequency factor.
• E_a is the activation energy of the reaction.
• R is the universal gas constant (1.987 cal mol⁻¹ K⁻¹).
• T is the absolute temperature in Kelvin.

EXPLANATION:

• Given the expression for the first-order reaction:

ln k = 14.34 − (1.25 × 10⁴) / T

• Comparing this expression with the standard logarithmic form of the Arrhenius equation:

ln k = ln A − E_a / (RT)

• By comparing the terms that include temperature (T), we can establish the following relationship:

E_a / R = 1.25 × 10⁴

• Rearranging to solve for the activation energy (E_a):

E_a = 1.25 × 10⁴ × R

• Substituting the given value of R = 1.987 cal mol⁻¹ K⁻¹:

E_a = 1.25 × 10⁴ × 1.987

E_a = 12500 × 1.987

E_a = 24837.5 cal mol⁻¹

• To find the activation energy in kcal mol⁻¹, we divide the value by 1000:

E_a = 24837.5 / 1000 kcal mol⁻¹

E_a = 24.8375 kcal mol⁻¹

• Rounding the result to two decimal places, we get approximately 24.84 kcal mol⁻¹.

Therefore, the energy of activation for the reaction is 24.84 kcal mol⁻¹.

CONCEPT:

Formation of Cyanides and Isocyanides

• Alkyl halides react with KCN to form alkyl cyanides (R−CN).
• Alkyl halides react with AgCN to form alkyl isocyanides (R−NC).
• Isocyanides have a very unpleasant foul smell.
• Primary amines on heating with chloroform and alcoholic KOH give isocyanides through the carbylamine reaction.

EXPLANATION:

• In the first reaction:

C₂H₅Cl + AgCN → C₂H₅NC

• Silver cyanide is mainly covalent, so the nucleophilic attack occurs through nitrogen, producing an isocyanide.
• The product formed is:

C₂H₅NC

Ethyl isocyanide

• Therefore, reagent X is:

AgCN

• In the second reaction, propanamide undergoes Hofmann bromamide degradation:

C₂H₅CONH₂ &xrightarrow{Br_{2}/NaOH} C₂H₅NH₂

• The primary amine formed reacts with CHCl₃ and alcoholic KOH on heating:

C₂H₅NH₂ &xrightarrow[\Delta]{CHCl_{3}/alc.\ KOH} C₂H₅NC

• This is the carbylamine reaction which produces foul smelling isocyanides.
• Thus, the foul smelling product Z is:

C₂H₅NC

Therefore, X = AgCN and Z = C₂H₅NC.

CONCEPT:

Isomerism in Coordination Compounds:

• Geometrical Isomerism: This type of isomerism arises in heteroleptic complexes due to different possible geometric arrangements of the ligands around the central metal atom. Common in square planar and octahedral complexes.
• Optical Isomerism: This occurs when the complex molecules are non-superimposable mirror images of each other. These are called enantiomers and are typical in octahedral complexes containing didentate ligands.
• Linkage Isomerism: This arises in coordination compounds containing ambidentate ligands (ligands that can bind through more than one donor atom, such as NO₂^- or SCN^-).
• Solvate (Hydrate) Isomerism: This is a form of structural isomerism where the solvent molecule (typically water) can act as either a ligand within the coordination sphere or as a molecule of crystallization outside the sphere.

EXPLANATION:

• A. [Pt(NH₃)₂Cl₂]:
• This is a square planar complex of the type MA₂B₂.
• It exhibits Geometrical isomerism as the two chlorine ligands can be adjacent to each other (cis-isomer) or opposite to each other (trans-isomer).
• Match: A - III
  

• B. [Co(en)₃]³⁺:
• This octahedral complex contains three bidentate ethylenediamine (en) ligands.
• Due to the arrangement of the chelate rings, it lacks a plane of symmetry and forms non-superimposable mirror images (dextro and laevo forms). Thus, it exhibits Optical isomerism.
• Match: B - I
  

• C. [Co(NH₃)₅NO₂]Cl₂:
• The NO₂^- ligand is an ambidentate ligand.
• It can bind to the cobalt atom via the nitrogen atom (-NO₂, nitro) or the oxygen atom (-ONO, nitrito), leading to Linkage isomerism.
• Match: C - IV
• D. [Cr(H₂O)₆]Cl₃:
• This complex involves water as the ligand.
• It can exhibit Solvate isomerism where water molecules can be replaced by chloride ions within the coordination sphere (e.g., [Cr(H₂O)₅Cl]Cl₂·H₂O).
• Match: D - II

By matching the lists, we get: A - III, B - I, C - IV, D - II.

The correct answer is - Habitat loss and fragmentation; over-exploitation; Alien species invasions; Co-extinctions

Key Points

• Habitat loss and fragmentation
• Occurs when large natural habitats are divided into smaller fragments by human activities like urbanization and deforestation.
• Leads to the isolation of species populations, reducing their ability to survive and reproduce.
• Over-exploitation
• Involves excessive hunting, fishing, or harvesting of species, which disrupts ecological balance.
• Example: Overfishing leading to the collapse of marine ecosystems.
• Alien species invasions
• Occurs when non-native species are introduced to an area, often outcompeting native species for resources.
• Example: Invasive plants like Lantana and water hyacinth disrupt native ecosystems.
• Co-extinctions
• Happens when the extinction of one species leads to the extinction of another dependent species.
• Example: The extinction of a host plant can lead to the extinction of its pollinator species.

Additional Information

• The "Evil Quartet"
• The term was introduced by biologists to describe the four major causes of biodiversity loss.
• These factors often interact, amplifying their negative effects on ecosystems.
• Importance of biodiversity
• Biodiversity ensures ecosystem stability by supporting processes like pollination, nutrient cycling, and climate regulation.
• Loss of biodiversity can lead to ecosystem collapse and impact human survival.
• Conservation efforts
• Include habitat restoration, controlling invasive species, and enforcing laws to prevent over-exploitation.
• International agreements like the Convention on Biological Diversity (CBD) aim to protect global biodiversity.

The correct answer is - Nucleolus

Key Points

• Nucleolus
• The nucleolus is a specialized structure within the cell nucleus where ribosomal RNA (rRNA) is synthesized.
• It is the primary site for rRNA transcription, processing, and the initial stages of ribosome assembly.
• Contains specific regions, such as the fibrillar center, where rRNA genes are located, and the dense fibrillar component, where rRNA processing occurs.
• This structure ensures the production of rRNA, which is essential for forming functional ribosomes required for protein synthesis.

Additional Information

• Other cellular components
• Chromatin: Composed of DNA and proteins, it is involved in gene regulation but does not directly synthesize rRNA.
• Centrosome: An organelle involved in organizing microtubules during cell division, not linked to rRNA synthesis.
• Kinetochore: A protein structure on chromosomes involved in chromosome segregation during mitosis, unrelated to rRNA production.
• Significance of rRNA
• rRNA forms the structural and functional core of ribosomes, facilitating the translation of mRNA into proteins.
• rRNA contributes to the catalytic activity of ribosomes, specifically in peptide bond formation.

The correct answer is - A - II, B - III, C - IV, D - I

Key Points

• G₁ phase
• During the G₁ phase (Gap 1 phase), the cell is metabolically active and grows continuously.
• However, DNA replication does not occur in this phase.
• S phase
• This is the phase where DNA synthesis occurs, leading to the doubling of DNA content in the cell.
• Chromosome replication takes place, preparing the cell for mitosis.
• G₂ phase
• In the G₂ phase (Gap 2 phase), the cell continues to grow and synthesizes proteins required for mitosis.
• This phase ensures that the cell is fully prepared for division.
• M phase
• The M phase (Mitotic phase) is where the cell undergoes actual division to form two daughter cells.
• It consists of mitosis (nuclear division) and cytokinesis (cytoplasm division).

Additional Information

• Interphase
• The combination of G₁, S, and G₂ phases is referred to as interphase, during which the cell prepares for division.
• It accounts for about 90% of the cell cycle.
• Checkpoints in the cell cycle
• Specific checkpoints in the G₁, G₂, and M phases ensure that the cell is ready to proceed to the next stage.
• These checkpoints verify DNA integrity, proper replication, and readiness for division.
• Cell cycle regulation
• The cell cycle is tightly regulated by proteins such as cyclins and cyclin-dependent kinases (CDKs).
• These proteins ensure proper progression and prevent errors during the cycle.

The correct answer is - A - III, B - I, C - IV, D - II

Key Points

• Productivity
• Refers to the rate of biomass production in an ecosystem.
• It is categorized as primary productivity (production by autotrophs) and secondary productivity (production by heterotrophs).
• Net Primary Productivity (NPP)
• Defined as Gross Primary Productivity (GPP) minus respiration losses.
• Represents the available organic matter for consumption by herbivores and decomposers.
• Gross Primary Productivity (GPP)
• Refers to the rate of production of organic matter during photosynthesis.
• It is the total energy captured by autotrophs before losses due to respiration.
• Secondary Productivity
• Describes the rate of formation of new organic matter by consumers in an ecosystem.
• It is directly influenced by the efficiency of energy transfer from primary producers to consumers.

Additional Information

• Primary Productivity Types
• Gross Primary Productivity (GPP): The total rate at which solar energy is converted into chemical energy by autotrophs.
• Net Primary Productivity (NPP): The energy remaining after autotrophs have used a portion for respiration.
• Secondary Productivity
• Involves the production of biomass by heterotrophic organisms such as herbivores, carnivores, and decomposers.
• Depends on the energy and nutrient availability in the ecosystem.
• Importance of Productivity
• Critical for understanding energy flow and nutrient cycling in ecosystems.
• Provides insights into ecosystem efficiency and sustainability.

The correct answer is - A, B and D only

Key Points

• Habitat loss
• The Amazon rainforest being cut and cleared for soybean cultivation is a clear example of habitat loss.
• Deforestation disrupts ecosystems and displaces many species dependent on the forest for survival.
• Over-exploitation of species
• Steller's sea cow and passenger pigeon became extinct due to excessive hunting and exploitation by humans.
• Such practices reduce population sizes drastically, leading to extinction.
• Invasive species
• Water hyacinth is a well-known invasive aquatic species.
• It spreads rapidly, choking water bodies and disrupting aquatic ecosystems.

Additional Information

• Incorrect statements
• The Nile perch introduced into Lake Victoria actually led to a decline in cichlid fish populations due to predation, not population growth.
• When a species becomes extinct, the associated plant and animal species are often affected, contrary to statement E.
• Impact of invasive species
• Invasive species like the water hyacinth can outcompete native species for resources.
• They often cause ecological and economic harm by altering ecosystems and reducing biodiversity.
• Human activities and habitat destruction
• Human-induced activities such as deforestation, agriculture, and urbanization are major causes of habitat loss.
• Efforts like afforestation and conservation programs are crucial to mitigate habitat destruction.

The correct answer is - B, C and E only

Key Points

• Proteins are polypeptides
• Proteins are composed of long chains of amino acids, linked by peptide bonds.
• These amino acid chains fold into specific three-dimensional structures, giving proteins their functionality.
• Polysaccharides are long chains of sugars
• Polysaccharides are made of repeated units of monosaccharides, connected by glycosidic bonds.
• Examples include starch, cellulose, and glycogen, which serve as energy storage or structural components.
• Almost all enzymes are proteins
• Enzymes are biological catalysts primarily composed of proteins, although some RNA molecules (ribozymes) also exhibit catalytic activity.
• They accelerate chemical reactions by lowering the activation energy required.

Additional Information

• Incorrect Statements
• Statement A - Lipids are generally not water-soluble. They are hydrophobic molecules and dissolve in nonpolar solvents like chloroform or ether.
• Statement D - Adenine and guanine are purines, not pyrimidines. They are nitrogenous bases found in nucleic acids.
• Enzymes and Biomolecules
• While most enzymes are proteins, some ribozymes (RNA molecules) also act as catalysts in specific biological processes like RNA splicing.
• Proteins, polysaccharides, and lipids play distinct roles in cellular structure and metabolism.

The correct answer is - 18 ATP and 12 NADPH

Key Points

• Calvin Cycle
• The Calvin cycle, also known as the C3 pathway, is the process through which plants fix carbon dioxide into glucose.
• Energy Requirement
• To synthesize one molecule of glucose, the Calvin cycle requires 18 ATP and 12 NADPH molecules.
• This energy is used in the reduction phase and regeneration phase of the cycle.
• Six CO₂ Molecules
• The Calvin cycle operates six times to fix six CO₂ molecules, which are needed to produce one glucose molecule (C₆H₁₂O₆).

Additional Information

• Phases of the Calvin Cycle
• Carbon Fixation
• CO₂ combines with ribulose-1,5-bisphosphate (RuBP) to form 3-phosphoglycerate (3-PGA).
• Reduction
• ATP and NADPH are used to convert 3-PGA into glyceraldehyde-3-phosphate (G3P).
• Regeneration
• ATP is utilized to regenerate RuBP from G3P, allowing the cycle to continue.
• Source of ATP and NADPH
• ATP and NADPH are produced during the light-dependent reactions of photosynthesis in the thylakoid membranes.
• Importance of the Calvin Cycle
• The Calvin cycle is essential for converting inorganic carbon into organic compounds, which are used as energy sources by the plant and other organisms.

The correct answer is - C and D only

Key Points

• Restriction endonucleases
• These enzymes are often referred to as molecular scissors because they can cut DNA molecules at specific sites.
• They play a key role in restricting the growth of foreign DNA, such as bacteriophage DNA, in E. coli and other organisms by recognizing and cleaving specific sequences.
• They recognize palindromic sequences in the DNA but do not necessarily cut only at the center of these sequences. Some enzymes cut at specific points within or near the palindromic site.
• They are endonucleases, meaning they cleave DNA at internal positions, not just at the ends of DNA fragments like exonucleases.
• The statements C ("They cut the DNA only at the center of the palindromic sites") and D ("They remove nucleotides only from the ends of DNA fragments") are incorrect because:
• Restriction endonucleases cut at specific recognition sites, but not necessarily at the center of palindromic sites.
• They do not remove nucleotides from the ends of DNA fragments; this is the role of exonucleases.