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Question: Answered & Verified by Expert

Match list-I with List-II :

List - I List - II
A. Coke I. Carbon atoms are sp3 hybridised
B. Diamond II. Used as dry lubricants
C. Fullerene III. Used as a reducing agent
D. Graphite IV. Cage like molecules

Choose the correct answer from the options given below:

Chemistryp Block Elements (Group 15, 16, 17 & 18)NEETNEET 2023 (All India)
Options:
  • A A-III, B-IV, C-I, D-II
  • B A-II, B-IV, C-I, D-III
  • C A-IV, B-I, C-II, D-III
  • D A-III, B-I, C-IV, D-II
Solution:
1571 Upvotes Verified Answer
The correct answer is: A-III, B-I, C-IV, D-II

Coke when reacted with moderately reactive metal oxides at high temperature it forms CO and free metal during the metallurgical process. In this way coke play the role of removal of oxygen from metal oxide and hence act as reducing agent. 

In diamond each carbon atom undergoes sp3 hybridisation and linked to four other carbon atoms by using hybridised orbitals in tetrahedral fashion.

Fullerenes are the only pure form of carbon because they have smooth structure without having ‘dangling’ bonds. Fullerenes are cage like molecules. 

Graphite cleaves easily between the layers and, therefore, it is very soft and slippery. For this reason graphite is used as a dry lubricant in machines running at high temperature, where oil cannot be used as a lubricant. 

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