JEE Main 2021 · 16 March · evening · Single correct · +4 / −1
The INCORRECT statement regarding the structure of C₆₀ is :
- (a) The five-membered rings are fused only to six-membered rings.
- (b) It contains 12 six-membered rings and 24 five-membered rings.
- (c) Each carbon atom forms three sigma bonds.
- (d) The six-membered rings are fused to both six and five-membered rings.
Answer and solution
Answer (b) It contains 12 six-membered rings and 24 five-membered rings.
Solution
Chemistry ma’am · 3 steps
Think of a football: 12 black pentagons, 20 white hexagons, and no two pentagons ever touch.
- Bonding at each carbon. Every carbon in C₆₀ is sp² hybridised and bonded to three other carbons. That gives three σ bonds per carbon, with the leftover electron in a delocalised π system.
- Count the rings. There are 60 atoms with 3 bonds each, so 60 × 3 / 2 = 90 C–C bonds. Euler's rule for a closed cage gives 60 - 90 + F = 2, so F = 32 rings. With p pentagons and h hexagons, p + h = 32 and 5p + 6h = 2 × 90 = 180.
5p + 6(32 - p) = 180 ⇒ p = 12, h = 20
Statement Check Verdict (a) Pentagons fused only to hexagons Each pentagon is surrounded by 5 hexagons, and no two pentagons share an edge correct (b) 12 hexagons and 24 pentagons Actually 20 hexagons and 12 pentagons incorrect (c) Three σ bonds per carbon Each carbon is sp² with 3 neighbours correct (d) Hexagons fused to hexagons and pentagons Each hexagon touches 3 pentagons and 3 hexagons correct - So. Only (b) gets the ring count wrong. C₆₀ has 20 six-membered and 12 five-membered rings, so (b) is the incorrect statement.
The idea to keep. For C₆₀, remember the football: 20 hexagons, 12 pentagons, every carbon sp² with three σ bonds, and pentagons always isolated from each other.
Answer from the NTA final answer key. Solution by PrepWiser.