Sankalp NEET Full Test-1 Question-72 Solution

Question: 72. Match List I with List II.

List-I
(Molecule)
List-II
(Number and type of bond/s between two carbon atoms)
A. ethaneI. one \sigma-bond and two \pi-bonds
B. etheneII. two \pi-bonds
C. carbon molecule, \mathrm{C}_{2}III. one \sigma-bond
D. ethyneIV. one \sigma-bond and one \pi-bond

Choose the correct answer from the options given below:

(1) A-I, B-IV, C-II, D-III

(2) A-IV, B-III, C-II, D-I

(3) A-III, B-IV, C-II, D-I

(4) A-III, B-IV, C-I, D-II

Answer: Option (3)

Explanation:

In ethane, the two carbon atoms are joined by a single covalent bond,

which consists of one \sigma-bond only.

Therefore, ethane corresponds to III.

In ethene, the two carbon atoms are connected by a double bond,

which consists of one \sigma-bond and one \pi-bond.

Hence, ethene corresponds to IV.

The carbon molecule \mathrm{C}_{2} has two carbon atoms bonded by two \pi-bonds without any \sigma-bond.

Thus, \mathrm{C}_{2} corresponds to II.

In ethyne, the two carbon atoms are joined by a triple bond,

which consists of one \sigma-bond and two \pi-bonds.

Therefore, ethyne corresponds to I.

The correct matching is A-III, B-IV, C-II, D-I, which is option (3).

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