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With one of the highest numbers of candidates registered for the NEET 2024 exam, the NEET exam previous year question paper 2024 serves as an essential reference for the medical entrance exams’ current competitive nature.
A logic circuit provides the output Y as per the following truth table:
| A | B | Y |
|---|---|---|
| 0 | 0 | 1 |
| 0 | 1 | 0 |
| 1 | 0 | 1 |
| 1 | 1 | 0 |
Answer: c) \bar{B}
Explanation:
Observe the truth table: Y = 1 when B = 0, and Y = 0 when B = 1
The output is independent of the value of A
Therefore, Y = \bar{B} (NOT B)
A wheel is rolling on a level road. If its linear speed is v, which point moves faster – P (highest point) or Q (lowest point)?
Answer: b) Point P moves faster than point Q
Explanation:
In pure rolling motion:
The topmost point (P) has velocity = 2v (v + v)
The lowest point (Q) has velocity = 0 (v – v)
Therefore, point P moves faster than point Q
Equivalent capacitance between terminal A and B:
Answer: 2μF
Explanation:
The given circuit is balanced (Wheatstone bridge configuration)
Redraw the circuit to simplify
The equivalent capacitance is calculated as:
Two parallel branches each with two capacitors in series
C_AB = (1/2 + 1/2)⁻¹ + (1/2 + 1/2)⁻¹ = 2μF
Instantaneous power when displacement x = 2t – 1 and force = 5N:
Answer: a) 10 W
Explanation:
x = 2t – 1
Velocity, v = dx/dt = 2 m/s
Power, P = F × v = 5 × 2 = 10 W
Assertion and Reason about electric dipole potential:
Answer: c) A is true but R is false
Explanation:
Potential at axial point of dipole: V = KP cosθ/r²
At θ = 0°: V = KP/r² = 9×10⁹ × 4×10⁻⁶ / 4 = 9×10³ V
At θ = 180°: V = -9×10³ V
Assertion A is true: V = ±9×10³ V
Reason R is false: The formula should be V = KP/r², not 2P/(4πε₀r²)
Ratio V₁:V₂ for completely inelastic collision:
Answer: b) 2:1
Explanation:
Before collision: Body A (mass m) moving with velocity V₁, Body B at rest
After completely inelastic collision: Both move together with velocity V₂
Conservation of linear momentum:
m × V₁ + 0 = (m + m) × V₂
V₁ = 2V₂
V₁:V₂ = 2:1
Magnetic moment of needle from oscillation:
Answer: d) 1280π²
Explanation:
Time period, T = 2π√(I/MB)
Given: T = 5/20 = 0.25 s, I = 9.8×10⁻⁶ kgm², B = 0.049 T
M = (2π/T)² × I/B
M = (4π² × 16 × 9.8×10⁻⁶) / 0.049
M = 1280π² × 10⁻⁵ Am²
Unpolarised light at Brewster’s angle:
Answer: d) The reflected light will be completely polarised but the refracted light will be partially polarised
Explanation:
At Brewster’s angle, reflected ray is completely polarised
The refracted ray is partially polarised
This is a fundamental property of Brewster’s law
Statements about solar cell and reverse biased pn junction:
Answer: a) A is correct but B is incorrect
Explanation:
Statement A is correct: Solar cell I-V characteristics lie in the IV quadrant
Statement B is incorrect: In reverse biased pn junction, current is due to minority charge carriers, not majority
Graph of (1/λ²) vs kinetic energy E:
Answer: d)
Explanation:
De Broglie wavelength: λ = h/p = h/√(2mE)
1/λ² = (2m/h²)E
This is a linear relationship: 1/λ² ∝ E
The graph is a straight line passing through origin
Terminal voltage of battery with emf 10V and internal resistance 1Ω connected to external 4Ω:
Answer: c) 8V
Explanation:
Current, I = E/(R+r) = 10/(4+1) = 2A
Terminal voltage, V = E – Ir = 10 – 2(1) = 8V
Quantities with same dimensions as solid angle:
Answer: a) strain and angle
Explanation:
Solid angle is a dimensionless quantity
Strain (change in length/original length) is dimensionless
Angle (arc/radius) is dimensionless
Stress has dimensions of pressure (ML⁻¹T⁻²)
This appears to be missing/incomplete in the PDF
This appears to be missing/incomplete in the PDF
SHM: x = 5 sin(πt + π/3) m
Answer: b) 5m, 2s
Explanation:
Amplitude = 5m
ω = π = 2π/T
T = 2π/π = 2s
Acceleration due to gravity on planet with mass 1/10 Earth and radius 1/2 Earth:
Answer: d) 3.92 m/s²
Explanation:
g’ = GM’/R’² = G(M/10)/(R/2)²
= G(M/10)/(R²/4) = 4GM/(10R²)
= 0.4 × g_Earth = 0.4 × 9.8 = 3.92 m/s²
Force exerted by block A on block B (masses 2kg and 3kg, force 10N):
Answer: c) 6N
Explanation:
a = F/(m₁+m₂) = 10/(2+3) = 2 m/s²
Force on B = m₂a = 3 × 2 = 6N
Work done along path bc in thermodynamic cycle:
Answer: Zero
Explanation:
Path bc is an isochoric process (constant volume)
Work done in isochoric process = 0 (W = PΔV, ΔV = 0)
Statements about atoms:
Answer: c) Statement I is correct but Statement II is incorrect
Explanation:
Statement I is correct: Atoms are electrically neutral (equal protons and electrons)
Statement II is incorrect: Atoms are NOT all stable; many are radioactive and emit characteristic spectra only when excited
Nuclear emission sequence:
Answer: 286, 81
Explanation:
²⁹⁰₈₂X → α → ²⁸⁶₈₀Y (α decay: A-4, Z-2)
²⁸⁶₈₀Y → e⁺ → ²⁸⁶₇₉Z (β⁺ decay: Z-1)
²⁸⁶₇₉Z → β⁻ → ²⁸⁶₈₀P (β⁻ decay: Z+1)
²⁸⁶₈₀P → e⁻ → ²⁸⁶₈₁Q (electron capture: Z+1)
Particle moving with uniform speed in circular path:
Answer: d) Varying velocity and varying acceleration
Explanation:
Speed is constant but direction changes continuously
Velocity changes due to change in direction
Acceleration changes direction continuously (centripetal acceleration)
Refractive index of prism:
Answer: b) √5/2
Explanation:
Apply Snell’s law at incidence surface:
1 × sin30° = μ × sin(90° – c)
1/2 = μ × cos c
sin c = 1/μ, cos c = √(μ²-1)/μ
1/2 = μ × √(μ²-1)/μ = √(μ²-1)
Squaring: 1/4 = μ² – 1
μ² = 5/4
μ = √5/2
Vernier constant:
Answer: b) 1/{100(N+1)}
Explanation:
Given: (N+1) VSD = N MSD
VSD = N/(N+1) MSD
V.C = MSD – VSD = MSD/(N+1)
1 MSD = 0.1 mm = 0.01 cm
V.C = 0.01/(N+1) = 1/{100(N+1)} cm
Young’s double slit with white light:
Answer: Coloured fringes will be observed
Explanation:
White light produces coloured interference fringes
Central fringe is white, surrounded by coloured fringes
Different wavelengths have different fringe widths
Pattern does not disappear
Statements about photon:
Answer: b) A, B, C and D only
Explanation:
A: Energy of photon = hν ✓
B: Velocity of photon = c ✓
C: Momentum p = hν/c ✓
D: Both energy and momentum conserved in photon-electron collision ✓
E: Photon does NOT possess positive charge ✗ (photon is neutral)
Output of logic gate:
Answer: d) AND gate
Explanation:
The given combination of gates is equivalent to an AND gate
This can be verified using De Morgan’s laws
Match spectral lines of hydrogen:
| Transition | Wavelength (nm) |
|---|---|
| n₂=3 to n₁=2 | 656.3 |
| n₂=4 to n₁=2 | 486.1 |
| n₂=5 to n₁=2 | 434.1 |
| n₂=6 to n₁=2 | 410.2 |
Answer: b) A-III, B-IV, C-II, D-I
Excess force to lift disc from water:
Answer: a) 19.8 mN
Explanation:
Excess force = T × 2πR
= 0.07 × 2 × 3.14 × 0.045
= 197.82 × 10⁻⁴ N
= 19.8 × 10⁻³ N
= 19.8 mN
Ideal transformer turns ratio:
Answer: b) 2:1
Explanation:
Np/Ns = 1/2
Vs/Vp = Ns/Np = 2:1
All Question Paper PDF NEET 2024 PYQ
| Section | NEET UG 2024 Questions Paper Solve |
|---|---|
| Physics | 2024-Physics Question Paper solve |
| Chemistry | 2024-Physics Question Paper solve |
Maximum elongation of steel wire:
Answer: a) 4mm
Explanation:
δ_max = σ_elastic × L / Y
= (8×10⁸ × 1) / (2×10¹¹)
= 4×10⁻³ m
= 4 mm
Potential difference between points C and P of charged spherical shell:
Answer: d) Zero
Explanation:
For a uniformly charged spherical shell:
Potential is constant inside and on the surface
V = kq/R for r ≤ R
Therefore, Vc = Vp
Vc – Vp = 0
Resistance of combination:
Answer: b) 52 ohm
Explanation:
Wire of 100Ω divided into 10 equal parts: each part = 10Ω
First 5 parts in series: R₁ = 5 × 10 = 50Ω
Next 5 parts in parallel: R₂ = 10/5 = 2Ω
Series combination: R = 50 + 2 = 52Ω
Magnetic field at centre of coil:
Answer: c) 4.4 mT
Explanation:
B = μ₀IN/(2R)
= (4π×10⁻⁷ × 7 × 100) / (2 × 0.1)
= 4.4 × 10⁻³ T
= 4.4 mT
Match materials with susceptibility:
| Material | Susceptibility (χ) |
|---|---|
| Diamagnetic | 0 > χ ≥ -1 |
| Ferromagnetic | χ >> 1 |
| Paramagnetic | 0 < χ < ε |
| Non-magnetic | χ = 0 |
Answer: a) A-II, B-III, C-IV, D-I
Tension in string when speed doubles:
Answer: b) 4T
Explanation:
Tension provides centripetal force: T = mω²r
If ω becomes 2ω, T’ = m(2ω)²r = 4mω²r = 4T
Current in capacitor circuit:
Answer: b) Displacement current of magnitude equal to I flows in the same direction as I
Explanation:
In the gap between capacitor plates, there is no conduction current
But displacement current exists: I_d = ε₀(dφ_E/dt)
I_d equals the conduction current I in magnitude
Flows in the same direction as I
Peak current in capacitor circuit:
Answer: b) 0.93 A
Explanation:
C = 10μF = 10×10⁻⁶F
V_rms = 210V, f = 50Hz
Xc = 1/(2πfC) = 1/(2×3.14×50×10×10⁻⁶)
I_rms = V_rms/Xc
I_peak = I_rms × √2 ≈ 0.93A
New time period of pendulum:
Answer: b) √2
Explanation:
T = 2π√(L/g)
Mass does not affect time period
New length = L/2
T’ = 2π√(L/2g) = (1/√2) × 2π√(L/g) = T/√2
Given T’ = x/2 × T
x/2 = 1/√2
x = √2
T-V curves for ideal gas:
Answer: d) P₁ > P₂ > P₃
Explanation:
At same temperature, higher volume corresponds to lower pressure
From graph: V₃ > V₂ > V₁
Therefore: P₁ > P₂ > P₃
Minimum energy to launch satellite at altitude 2R:
Answer: a) 5GmM/6R
Explanation:
Energy required = E_final – E_initial
E_initial = -GMm/R (on surface)
E_final = -GMm/(3R) (at radius 3R)
ΔE = -GMm/3R – (-GMm/R) = GMm/R – GMm/3R = 2GMm/3R
But for circular orbit, total energy = -GMm/(2r) where r = 3R
E_orbit = -GMm/(2×3R) = -GMm/6R
Energy required = (-GMm/6R) – (-GMm/R) = 5GMm/6R
Property NOT of electromagnetic wave:
Answer: d) They originate from charges moving with uniform speed
Explanation:
EM waves originate from accelerating charges, not uniform motion
They are transverse ✓
Energy densities in E and B fields are equal ✓
Speed = 1/√(μ₀ε₀) ✓
Circuit for bridge balance:
Answer: a)
Explanation:
Bridge balance condition: R₁/R₂ = R₃/R₄
In option a): 10/15 = 10/(5+RD)
This condition can be satisfied with appropriate RD
Dimensionless factor:
Answer: b) αt/β
Explanation:
F = αt² + βt
By principle of homogeneity: [αt²] = [F] and [βt] = [F]
[α] = [F]/[t²] and [β] = [F]/[t]
[αt]/[β] = ([F]/[t²] × [t])/([F]/[t]) = [F]/[t] × [t]/[F] = dimensionless
Ratio of power outputs in series and parallel:
Answer: b) 2:9
Explanation:
Heater A: 1kW, Heater B: 2kW
R_A = V²/P_A = V²/1, R_B = V²/P_B = V²/2
In series: P_series = V²/(R_A+R_B) = V²/(V² + V²/2) = 2/3 × V²/P_A? Let’s calculate properly:
P_series = V²/(R_A+R_B) = V²/(V² + V²/2) = 2/3 × V²/V² = 2/3 W
P_parallel = V²/R_A + V²/R_B = 1 + 2 = 3 W
P_series:P_parallel = 2:9
Magnetic moment of bent iron bar:
Answer: b) M/2
Explanation:
Original magnetic moment: M = m × L (pole strength × length)
After bending at 60°, each half has length L/2
Magnetic moment = m × (L/2 + L/2 × cos60°) = m × L/2 × (1+1/2)? Actually:
Distance between poles = 2 × (L/2) × cos(30°) = L × cos30°
M_new = m × L × cos30° = M × cos30° = M × √3/2
Wait, let’s reconsider: For a bar bent at middle at 60°:
The magnetic moment = m × (L/2 + L/2 × cos60°) = m × L/2 × (1+1/2) = 3mL/4 = 3M/4
But the correct answer is M/2
Compressive force developed in heated metallic bar:
Answer: b) 50 × 10³ N
Explanation:
Thermal strain = αΔT = 10⁻⁵ × 100 = 10⁻³
Thermal stress = Y × strain = 0.5×10¹¹ × 10⁻³ = 5×10⁷ N/m²
Force = stress × area = 5×10⁷ × 10⁻³ = 5×10⁴ = 50×10³ N
v-t graph of motion:
Answer: (This appears incomplete in the PDF)
Parallel plate capacitor plates moved closer:
Answer: b) A, C and E only
Explanation:
When plates are moved closer (d decreases):
A is correct: Charge increases (Q = CV, V constant)
B is incorrect: Energy increases (U = ½CV², C increases)
C is correct: Capacitance increases (C = ε₀A/d)
D is incorrect: Q/V = C increases (not constant)
E is correct: Q×V increases (Q increases, V constant)
Magnifying power of telescope:
Answer: b) 28
Explanation:
For distant object: m = f₀/fₑ
= 140/5 = 28
Force needed for sheet near magnetic pole:
Answer: b) A and C only
Explanation:
A is correct: Magnetic sheet experiences force from magnet
B is incorrect: Non-magnetic sheet experiences no force
C is correct: Conducting sheet moving through magnetic field experiences eddy currents (opposes motion)
D is incorrect: Non-conducting, non-polar sheet experiences no force
Arrange elements in increasing electronegativity: N, O, F, C, Si
Answer: (1) Si < C < N < O < F
Explanation:
Electronegativity increases across a period (left to right)
Electronegativity decreases down a group
Order: Si (Group 14, Period 3) < C (Group 14, Period 2) < N < O < F
Identify reagents for transformation:
Answer: (2) (i) BH₃ (ii) H₂O₂/OH⁻ (iii) PCC
Explanation:
BH₃ + H₂O₂/OH⁻ = Hydroboration-oxidation (anti-Markovnikov addition)
PCC = Mild oxidizing agent (converts primary alcohol to aldehyde)
Compound with fastest Sₙ1 reaction:
Answer: (4)
Explanation:
Sₙ1 reaction rate depends on carbocation stability
More stable carbocation → faster Sₙ1 reaction
Stability order: 3° > 2° > 1° > methyl
Reaction quotient analysis:
Answer: (3) Reaction has tendency to go backward
Explanation:
Qc = [B][C]/[A]² = (2×10⁻³)²/(2×10⁻³)² = 1
Kc = 4×10⁻³
Qc > Kc → Reaction proceeds backward
Highest number of helium atoms:
Answer: (1) 4 mol of helium
Explanation:
4 mol = 4NA atoms
4u = 1 atom
4g = 1 mol = NA atoms
2.271 L at STP = 0.1 mol = 0.1NA atoms
Match thermodynamic processes:
Answer: (4) A-II, B-III, C-IV, D-I
| Process | Condition |
|---|---|
| Isothermal | Constant temperature |
| Isochoric | Constant volume |
| Isobaric | Constant pressure |
| Adiabatic | No heat exchange |
Energy of electron in Be³⁺:
Answer: (1) -x
Explanation:
For He⁺: E₁ = -4R_H = -x → R_H = x/4
For Be³⁺ (n=2): E₂ = -R_H(4²/2²) = -4R_H = -x
Match molecules with carbon-carbon bonds:
Answer: (3) A-III, B-IV, C-II, D-I
| Molecule | Bonds |
|---|---|
| Ethane | One σ-bond |
| Ethene | One σ, one π |
| C₂ | Two π-bonds |
| Ethyne | One σ, two π |
Match compounds with shapes:
Answer: (1) A-I, B-IV, C-II, D-III
| Compound | Shape |
|---|---|
| NH₃ | Trigonal Pyramidal |
| BrF₅ | Square Pyramidal |
| XeF₄ | Square Planar |
| SF₆ | Octahedral |
E° value for Mn³⁺/Mn²⁺:
Answer: (3) d⁴ to d⁵ configuration
Explanation:
Mn³⁺: [Ar]3d⁴, Mn²⁺: [Ar]3d⁵
d⁵ is half-filled (extra stable)
Change from d⁴ to d⁵ provides extra stability
Section | Number of Questions | Marks for Correct Answer | Negative Marking | Total Marks |
Physics | 45 | +4 | −1 | 180 |
Chemistry | 45 | +4 | −1 | 180 |
Biology (Botany + Zoology) | 90 | +4 | −1 | 360 |
Total | 180 | 720 |
Kp and Kc not equal:
Answer: (1) PCl₅(g) ⇌ PCl₃(g) + Cl₂(g)
Explanation:
Kp = Kc(RT)^Δn_g
For PCl₅: Δn_g = 2-1 = 1 ≠ 0
For others: Δn_g = 0
Group 16 element not showing -2 oxidation state:
Answer: (4) Po
Explanation:
Polonium shows only +2 and +4 oxidation states
Other elements show -2, +2, +4, +6
Same spin-only magnetic moment:
Answer: (1) B and D only
Explanation:
Spin-only moment = √(n(n+2)) BM
Cr²⁺: n=4, Fe²⁺: n=4 → Same moment
Ti³⁺: n=1, Mn²⁺: n=5, Sc³⁺: n=0
Compound with two tertiary carbons:
Answer: (3) 2,3-dimethylbutane
Explanation:
Structure: CH₃-CH(CH₃)-CH(CH₃)-CH₃
Both middle carbons are tertiary
Other options have 0 or 1 tertiary carbons
Boiling point of isomeric pentanes:
Answer: (1) Both statements correct
Explanation:
n-pentane > isopentane > neopentane
Branching → spherical shape → less surface area → weaker IMF → lower BP
These questions are present in the Biology section of the PDF
Most stable carbocation:
Answer: (4)
Explanation:
Stability depends on hyperconjugation extent
More hyperconjugative structures → more stability
Order: (4) > (2) > (1) > (3)
Purification technique for sublimation:
Answer: (2) Sublimation
Explanation:
Sublimation: Solid → Vapour (without liquid state)
Used for substances like iodine, camphor, NH₄Cl
Match complexes with isomerism:
Answer: (1) A-II, B-III, C-IV, D-I
| Complex | Isomerism |
|---|---|
| [Co(NH₃)₅(NO₂)]Cl₂ | Linkage |
| [Co(NH₃)₅(SO₄)]Br | Ionization |
| [Co(NH₃)₆][Cr(CN)₆] | Coordination |
| [Co(H₂O)₆]Cl₃ | Solvate |
Glucose does not react with:
Answer: (3) B and E
Explanation:
Glucose does NOT give Schiff’s test (B)
Glucose does NOT form bisulphite addition product (E)
Despite having aldehyde group, cyclic hemiacetal structure prevents these reactions
Match reactions:
Answer: (3) A-IV, B-I, C-II, D-III
Activation energy calculation:
Answer: (4) rate constant at two different temperatures
Explanation:
Arrhenius equation: log(k₂/k₁) = (Ea/2.303R)(1/T₁ – 1/T₂)
Match conversions with Faraday:
Answer: (1) A-II, B-IV, C-I, D-II
| Conversion | Faraday |
|---|---|
| H₂O → O₂ | 2F |
| MnO₄⁻ → Mn²⁺ | 5F |
| 1.5 mol Ca | 3F |
| FeO → Fe₂O₃ | 1F |
Intramolecular hydrogen bonding:
Answer: (1)
Explanation:
Intramolecular H-bonding occurs in o-nitrophenol
Requires proper geometry (5- or 6-membered ring)
Statements about aniline:
Answer: (1) Both statements true
Explanation:
Aniline does NOT undergo Friedel-Crafts (forms salt with AlCl₃)
Aniline cannot be prepared by Gabriel synthesis (aryl halides don’t undergo SN)
Alcohol reacting instantaneously with Lucas reagent:
Answer: (4) Tertiary alcohol
Explanation:
Tertiary alcohols → immediate turbidity with Lucas reagent
Secondary → after 5-10 minutes
Primary → slowly or not at all
Increasing order of first ionization enthalpy:
Answer: (2) Li < B < Be < C < N
Explanation:
General trend increases across period
Exception: Be (2s²) > B (2p¹) due to stability of full orbital
Which reaction is NOT redox:
Answer: (4) BaCl₂ + Na₂SO₄ → BaSO₄ + 2NaCl
Explanation:
No change in oxidation states
Double displacement/precipitation reaction
Match quantum numbers:
Answer: (2) A-III, B-IV, C-I, D-II
| Quantum Number | Information |
|---|---|
| m | Orientation of orbital |
| mₛ | Orientation of spin |
| l | Shape of orbital |
| n | Size of orbital |
Processes where entropy increases:
Answer: (3) A, C and D
Explanation:
A: Liquid → Vapour (ΔS > 0)
B: Cooling from 130K to 0K (ΔS < 0)
C: More gas moles formed (ΔS > 0)
D: 1 mol → 2 mol (ΔS > 0)
Mass of NaOH left unreacted:
Answer: (2) 250 mg
Explanation:
Moles HCl = 0.75 × 25/1000 = 0.01875 mol
Mass HCl = 0.01875 × 36.5 = 0.684 g
36.5g HCl reacts with 40g NaOH
0.684g HCl reacts with (40/36.5)×0.684 = 0.750g NaOH
NaOH left = 1 – 0.750 = 0.250g = 250mg
Products A and B:
Answer: (4) H₃PO₃ and POCl₃
Explanation:
3ROH + PCl₃ → 3RCl + H₃PO₃ (A)
ROH + PCl₅ → RCl + HCl + POCl₃ (B)
This appears incomplete in the PDF
Major product C:
Answer: (1) propylamine
Explanation:
Hofmann bromamide degradation: RCONH₂ → RNH₂
Converts amide to amine with one less carbon
Acid added to Mohr’s salt:
Answer: (4) dilute sulphuric acid
Explanation:
Dilute H₂SO₄ prevents hydrolysis of Fe²⁺
HCl and HNO₃ can oxidize Fe²⁺
Degree of dissociation of NO:
Answer: (4) 0.717
Explanation:
Kc = 1.607 from equilibrium concentrations
For 2NO(g) ⇌ N₂(g) + O₂(g)
Using ICE table: Kc = (0.05α)²/[0.1(1-α)]²
α/(1-α) = 2.54
α = 0.717
Work done in reversible isothermal expansion:
Answer: (2) -413.14 calories
Explanation:
W = -2.303 nRT log(P₁/P₂)
= -2.303 × 1 × 2 × 298 × log(20/10)
= -2.303 × 1 × 2 × 298 × log 2
= -413.14 calories
Mass of copper deposited:
Answer: (2) 0.315 g
Explanation:
Cu²⁺ + 2e⁻ → Cu
w = MIt/(nF)
= 63 × 9.6487 × 100/(2 × 96487)
= 0.315 g
Activation energy:
Answer: (1) 38.04 kJ/mol
Explanation:
log(k₂/k₁) = (Ea/2.303R)(1/T₁ – 1/T₂)
log 4 = (Ea/2.303×8.314)(1/300 – 1/330)
Ea = 3.804×10⁴ J/mol = 38.04 kJ/mol
Temperature from osmotic pressure:
Answer: (1) 37°C
Explanation:
π = CRT
Slope = RT
T = 25.73/0.083 = 310 K = 37°C
Increasing group number of cations:
Answer: (1) B, A, D, C, E
Explanation:
Cu²⁺ (Group II) < Al³⁺ (Group III) < Co²⁺ (Group IV) < Ba²⁺ (Group V) < Mg²⁺ (Group VI)
Homoleptic/heteroleptic complexes:
Answer: (1) Both statements true
Explanation:
[Co(NH₃)₆]³⁺: Only NH₃ ligands → Homoleptic
[Co(NH₃)₄Cl₂]⁺: Two types of ligands → Heteroleptic
Diamagnetic lanthanoid ions:
Answer: (1) Ce⁴⁺ and Yb²⁺
Explanation:
Ce⁴⁺: f⁰ (0 unpaired) → Diamagnetic
Yb²⁺: f¹⁴ (0 unpaired) → Diamagnetic
All others have unpaired electrons → Paramagnetic
This appears incomplete in the PDF
Correct statement:
Answer: (4) Three canonical forms can be drawn for CO₃²⁻ ion
Explanation:
Ozone: 2 resonance structures (not 3)
BF₃: Zero dipole moment (symmetrical)
NH₃ dipole > NF