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Tamil Nadu Board of Secondary EducationHSC Science Class 11

Explain the bond formation in ethylene. - Chemistry

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Question

Explain the bond formation in ethylene.

Answer in Brief

Solution

The bonding in ethylene can be explained using the hybridization concept. The molecular formula of ethylene is C2H4. The valency of carbon is 4. The electronic configuration of valence shell of carbon in ground state is [He] 2s2 2px1 2py1 2pz0. To satisfy the valency of carbon promote an electron from 2s orbital to 2pz orbital in the excited state.

In ethylene both the carbon atoms undergo sp2 hybridization involving 2s, 2px, and 2py, orbitals, resulting in three equivalent sp2 hybridised orbitals lying in the XY plane at an angle of 120° to each other. The unhybridized 2pz orbital lies perpendicular to the XY plane.

Formation of sigma bond:

One of the sp2 hybridised orbitals of each carbon lying on the molecular axis (x-axis) linearly overlaps with each other resulting in the formation of a C-C sigma bond. The other two sp2 hybridised orbitals of both carbons linearly overlap with the four is orbitals of four hydrogen atoms leading to the formation of two C-H sigma bonds on each carbon.

Formation of Pi (π ) bond:

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Bond Parameters
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Chapter 10: Chemical bonding - Evaluation [Page 107]

APPEARS IN

Samacheer Kalvi Chemistry - Volume 1 and 2 [English] Class 11 TN Board
Chapter 10 Chemical bonding
Evaluation | Q II. 49. a. | Page 107
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