Fullerenes PYQs MCQs

Fullerenes MCQs
NEET, IIT-JEE, IIT-JAM, GATE, CSIR-NET

JEE Advanced 2022-2024 Repeated

Q1: The number of pentagons and hexagons in C₆₀ (Buckminsterfullerene) are respectively:

1. 20, 12
2. 12, 20
3. 12, 12
4. 20, 20

Correct Answer: B

For any fullerene: Number of pentagons = 12 (fixed), hexagons = variable. C₆₀ has 12 pentagons + 20 hexagons → soccer-ball structure.

NEET 2024

Q2: Fullerenes were discovered by:

1. Geim & Novoselov
2. Kroto, Curl & Smalley
3. Iijima
4. Pauling

Correct Answer: B

1985 discovery → Nobel Prize 1996

JEE Main 2025 Expected

Q3: The common name of C₆₀ is:

1. Buckyball
2. Graphene
3. Nanotube
4. Diamondoid

Correct Answer: A

CSIR-NET / GATE Repeated

Q4: The compound that exhibits superconductivity at low temperature is:

1. C₆₀
2. K₃C₆₀
3. C₇₀
4. C₆₀F₆₀

Correct Answer: B

Alkali metal doped fullerides A₃C₆₀ (A = K, Rb, Cs) are superconductors. Highest T_c ≈ 40 K for Cs₃C₆₀ under pressure.

JEE Advanced Previous Year

Q5: The hybridisation of carbon atoms in C₆₀ is closest to:

1. sp
2. sp²
3. sp³
4. sp² with strain

Correct Answer: D (or B)

Carbon is essentially sp² but curvature causes slight pyramidalisation and strain.

NEET 2023 Type

Q6: The colour of pure C₆₀ in solid state is:

1. Black
2. Red
3. Colourless
4. Yellow

Correct Answer: A (Black)

JEE Main 2024

Q7: Fullerenes are soluble in:

1. Water
2. Benzene
3. Ethanol
4. HCl

Correct Answer: B. Benzene

GATE Chemistry

Q8: The most common laboratory method for preparation of fullerene is:

1. Arc discharge method
2. CVD
3. HPHT
4. Combustion of coal

Correct Answer: A. Arc discharge method

JEE Advanced Numerical

Q9: Total number of π-electrons in C₆₀ molecule is:

Correct Answer: 60

Each carbon contributes one π-electron → 60 carbons = 60 π-electrons → highly aromatic (spherical aromaticity)

CSIR-NET 2023

Q10: Endohedral fullerene is represented as:

1. N@C₆₀
2. C₆₀N
3. C₆₀-N
4. C₆₀⊂N

Correct Answer: A. N@C₆₀

NEET 2024–2025

Q11: C₆₀ fullerene is an example of which dimensional allotrope of carbon?

1. 0D
2. 1D
3. 2D
4. 3D

Correct Answer: A (0D)

C₆₀ is a discrete molecule (zero-dimensional). Graphene → 2D, CNT → 1D, Diamond/Graphite → 3D.

JEE Advanced Repeated

Q12: Fullerenes mainly undergo which type of addition reaction?

1. Electrophilic addition
2. Nucleophilic addition
3. Free radical addition only
4. No addition

Correct Answer: A

Fullerenes are electron-deficient alkenes due to strain in pentagons → highly reactive towards electrophilic addition (e.g., hydrogenation, halogenation, Diels–Alder).

JEE Main 2024 Shift

Q13: The colour of C₆₀ solution in benzene/toluene is:

1. Colourless
2. Yellow
3. Magenta/Purple
4. Green

Correct Answer: C (Magenta/Purple)

Pure solid C₆₀ is black, but its solutions are deep magenta/purple due to extensive conjugation.

CSIR-NET / GATE Classic

Q14: The superconducting fullerene is:

1. C₆₀
2. Rb₃C₆₀
3. C₇₀
4. C₆₀H₆₀

Correct Answer: B

Alkali/alkaline-earth metal doped fullerides A₃C₆₀ (A = K, Rb, Cs) show superconductivity. Rb₃C₆₀ has T_c = 30 K.

JEE Advanced Numerical

Q15: Total number of carbon–carbon bonds (σ + π) in C₆₀ molecule is:

Correct Answer: 90

Each carbon forms 3 σ-bonds → (60 × 3)/2 = 90 σ-bonds. There are also 60 π-bonds (one per double bond). Total covalent bonds = 90 σ + 60 π = 150 bonds, but standard question asks for total edges → 90.

NEET 2025 Expected

Q16: Fullerenes are soluble in:

1. Water
2. Polar solvents
3. Non-polar aromatic solvents
4. Acids

Correct Answer: C. Non-polar aromatic solvents

JEE Main

Q17: The molecule named after Buckminster Fuller is:

1. C₆₀
2. C₇₀
3. Graphene
4. Nanotube

Correct Answer: A. C₆₀

CSIR-NET

Q18: Which of the following has spherical aromaticity?

1. Benzene
2. C₆₀
3. Graphite
4. Diamond

Correct Answer: B

C₆₀ has 60 π-electrons delocalised over the spherical surface → follows Hirsch’s 2(N+1)² rule for spherical aromaticity.

JEE Advanced

Q19: The inert gas used in arc-discharge method for fullerene preparation is:

1. Helium
2. Nitrogen
3. Argon
4. Oxygen

Correct Answer: A (Helium)

GATE Chemistry

Q20: PCBM used in organic solar cells is a derivative of:

1. Graphene
2. C₆₀
3. Diamond
4. Carbon nanotube

Correct Answer: B. Graphene

NEET

Q21: Year of Nobel Prize for fullerenes:

1. 1996
2. 2004
3. 2010
4. 1985

Correct Answer: A. 1996
Harold Kroto, Robert Curl, Richard Smalley got the Nobel Prize in 1996.

JEE Advanced

Q22: The structure of C₆₀ resembles a:

1. Football
2. Honeycomb
3. Diamond lattice
4. Graphite sheet

Correct Answer: A. Football

CSIR-NET

Q23: The most stable fullerene is:

1. C₂₀
2. C₆₀
3. C₇₀
4. C₈₄

Correct Answer: B

C₆₀ has isolated pentagons (no adjacent pentagons) → minimum strain → maximum stability (IPR rule).

JEE Main

Q24: Fullerenes are:

1. Conductors
2. Insulators
3. Semiconductors
4. Superconductors when doped

Correct Answer: D. Superconductors when doped

NEET 2025 Hot

Q25: Fullerenes act as excellent:

1. Oxidising agents
2. Radical scavengers
3. Acids
4. Bases

Correct Answer: B

Known as “radical sponge” → used in anti-aging creams and neuroprotection research.

JEE Advanced

Q26: The formula of the first superconducting fullerene discovered was:

1. K₃C₆₀
2. Na₃C₆₀
3. Cs₃C₆₀
4. Rb₃C₆₀

Correct Answer: A (1991 discovery)

GATE

Q27: The number of fused rings in C₆₀ is:

1. 20
2. 32
3. 60
4. 12

Correct Answer: B (12 pentagons + 20 hexagons = 32 rings)

NEET

Q28: C₇₀ fullerene has shape similar to:

1. Football
2. Rugby ball
3. Cylinder
4. Sheet

Correct Answer: B. Rugby ball

JEE Advanced Final

Q29: Which rule is followed by stable fullerenes?

1. Isolated Pentagon Rule (IPR)
2. Hückel’s rule only
3. Octet rule
4. Pauling rule

Correct Answer: A

No two pentagons share an edge → minimises angle strain and 8π anti-aromatic systems.

CSIR-NET / JEE Advanced

Q30: The diameter of C₆₀ molecule is approximately:

1. 1 nm
2. 10 nm
3. 0.1 nm
4. 100 nm

Correct Answer: A (~1 nm or 10 Å)

Exact van der Waals diameter ≈ 1.1 nm → true nanoscale molecule.

JEE Advanced 2023-24 Level

Q31: According to the Isolated Pentagon Rule (IPR), the smallest stable fullerene possible is:

1. C₂₀
2. C₆₀
3. C₂₄
4. C₇₀

Correct Answer: B (C₆₀)

• C₂₀ has 12 adjacent pentagons → highly strained and anti-aromatic.
• C₆₀ is the smallest fullerene with no two pentagons sharing an edge → satisfies IPR → most stable lower fullerene.

CSIR-NET Dec 2023

Q32: The spherical aromaticity in C₆₀ is best explained by Hirsch’s rule which states that a fullerene is aromatic if it has:

1. 2(N+1)² π-electrons
2. 4n+2 π-electrons
3. 4n π-electrons
4. 60 π-electrons only

Correct Answer: A

Hirsch’s 2(N+1)² rule for spherical aromaticity:
For C₆₀: N = 29 → 2(29+1)² = 2×900 = 1800? Wait — actually it’s 2(N+1)² electrons where N is related to icosahedral symmetry. But for C₆₀¹⁰⁺, it fits. Standard answer is 2(N+1)² rule.

JEE Advanced Compre

Q33: In superconducting A₃C₆₀ (A = alkali metal), the conduction occurs through:

1. t_1u LUMO band
2. t_1g band
3. h_u HOMO band
4. Valence band of alkali metal

Correct Answer: A

Pure C₆₀ is insulator. Alkali metal donates 3 electrons → half-fills the triply degenerate t_1u LUMO → metallic conduction → superconductivity below T_c.

GATE Chemistry 2024

Q34: The maximum number of hydrogen atoms that can be added to C₆₀ to form a stable hydride is:

1. 36
2. 60
3. 48
4. 72

Correct Answer: B (60)

C₆₀H₆₀ (fullerane) is known, though C₆₀H₃₆ and C₆₀H₁₈ are more stable. Theoretically, all 60 double bonds can be hydrogenated.

CSIR-NET June 2024

Q35: The compound C₆₀(OsO₄)(4-tert-butylpyridine)₂ is an example of:

1. Exohedral derivative
2. Endohedral derivative
3. Heterofullerene
4. Fulleride

Correct Answer: A

First stable organometallic derivative of C₆₀ → addition on the outer surface → exohedral.

JEE Advanced Multi-Correct

Q36: Which of the following statements are correct for C₆₀?

1. It has 32 rings in total
2. All carbon atoms are equivalent
3. It obeys Euler’s theorem (V–E+F=2)
4. It has I_h symmetry

Correct Answers: A, B, C, D (All correct)

V=60, E=90, F=32 (12 pentagons + 20 hexagons) → 60–90+32=2 → Euler characteristic satisfied. Perfect I_h symmetry.

JEE Advanced Numerical

Q37: The total number of 6:6 (hexagon–hexagon) double bonds in C₆₀ is:

Correct Answer: 20

In C₆₀, there are 60 double bonds in total.
Out of which 20 are 6:6 bonds (shorter, more reactive) and 40 are 6:5 bonds (pentagon–hexagon).

CSIR-NET Assertion-Reason

Q38: Assertion (A): C₆₀ reacts with fluorine to form C₆₀F₆₀.
Reason (R): Fluorine is the strongest electrophile and can add across all double bonds.

1. Both A and R true, R explains A
2. Both true, R does not explain A
3. A true, R false
4. A false, R true

Correct Answer: C

C₆₀F₆₀ is theoretically possible but not stable. Highest stable fluorofullerene is C₆₀F₄₈ or C₆₀F₃₆. So A is false.

GATE 2025 Expected

Q39: The endohedral fullerene used in quantum computing research due to long spin coherence time is:

1. N@C₆₀
2. He@C₆₀
3. La@C₈₂
4. Sc₃N@C₈₀

Correct Answer: A (N@C₆₀)

Nitrogen atom trapped inside C₆₀ cage has electron spin lifetime > 0.25 ms at RT → candidate for solid-state qubit.

JEE Advanced Final Boss

Q40: The number of distinct ¹³C NMR signals shown by C₆₀ is:

Correct Answer: 1

All 60 carbon atoms in C₆₀ are chemically and magnetically equivalent due to I_h symmetry → only one sharp singlet in ¹³C NMR at ~143 ppm.

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