
The angle between\[s{p^2}\]orbitals in ethylene is
A.\[90^\circ \]
B.\[120^\circ \]
C.\[180^\circ \]
D.\[109.5^\circ \]
Answer
164.4k+ views
Hint: Ethylene is a hydrocarbon that has the formula\[{C_2}{H_4}\] which is the simplest alkene.
The bond angle can be found by using VSEPR (valence shell electron pair repulsion theory).
Complete step by step solution:Ethylene also known as ethene is an alkene that is a hydrocarbon possessing a carbon-carbon double bond.
Hybridization is the combination of atomic orbitals to construct a unique set of equivalent orbitals called hybrid orbitals.
The shape formed by the hybridized orbitals is predicted by VSEPR theory.
VSEPR theory provides a basic idea about the shape of a molecule relying upon the electron pair present in the valence shell.
A molecule that undergoes\[s{p^2}\]hybridization, has one s orbital and one p orbital combined to form an sp2 hybridized orbital which has the same orbital energy as the s and p orbitals.
Here ethylene molecules undergo sp2 hybridization.
Carbon is the central atom of the molecule with electronic configuration \[1{s^2}2{s^2}2{p_x}^12{p_y}^1\].
Carbon extends its valency by moving one electron from 2s to 2pz orbital.
Now the final configuration of carbon becomes \[1{s^2}2{s^1}2{p_x}^12{p_y}^12{p_z}^1\] with valency 4.
One 2s orbital and two 2p orbitals of carbon undergo hybridization to form an\[s{p^2}\] hybridized orbital.
These hybridized orbitals overlap with the 1s orbital of four hydrogen atoms to form ethylene.
The remaining \[2{p_z}\] lies unhybridized and perpendicular to the plane of the \[s{p^2}\]hybridized orbital.
Hydrogen atoms are placed at\[120^\circ \] with the central metal atom to form a shape known as trigonal planar.

Image: Ethylene
So, the angle between \[s{p^2}\]orbitals in ethylene is \[120^\circ \].
So, option B is correct.
Note: Ethylene is a hormone that influences the ripening and flowering of many plants.
It is widely utilized to regulate freshness in horticulture and fruits.
The hydroformylation of ethylene gives propionaldehyde which is used for preparing propionic acid and n-propyl alcohol.
The bond angle can be found by using VSEPR (valence shell electron pair repulsion theory).
Complete step by step solution:Ethylene also known as ethene is an alkene that is a hydrocarbon possessing a carbon-carbon double bond.
Hybridization is the combination of atomic orbitals to construct a unique set of equivalent orbitals called hybrid orbitals.
The shape formed by the hybridized orbitals is predicted by VSEPR theory.
VSEPR theory provides a basic idea about the shape of a molecule relying upon the electron pair present in the valence shell.
A molecule that undergoes\[s{p^2}\]hybridization, has one s orbital and one p orbital combined to form an sp2 hybridized orbital which has the same orbital energy as the s and p orbitals.
Here ethylene molecules undergo sp2 hybridization.
Carbon is the central atom of the molecule with electronic configuration \[1{s^2}2{s^2}2{p_x}^12{p_y}^1\].
Carbon extends its valency by moving one electron from 2s to 2pz orbital.
Now the final configuration of carbon becomes \[1{s^2}2{s^1}2{p_x}^12{p_y}^12{p_z}^1\] with valency 4.
One 2s orbital and two 2p orbitals of carbon undergo hybridization to form an\[s{p^2}\] hybridized orbital.
These hybridized orbitals overlap with the 1s orbital of four hydrogen atoms to form ethylene.
The remaining \[2{p_z}\] lies unhybridized and perpendicular to the plane of the \[s{p^2}\]hybridized orbital.
Hydrogen atoms are placed at\[120^\circ \] with the central metal atom to form a shape known as trigonal planar.

Image: Ethylene
So, the angle between \[s{p^2}\]orbitals in ethylene is \[120^\circ \].
So, option B is correct.
Note: Ethylene is a hormone that influences the ripening and flowering of many plants.
It is widely utilized to regulate freshness in horticulture and fruits.
The hydroformylation of ethylene gives propionaldehyde which is used for preparing propionic acid and n-propyl alcohol.
Recently Updated Pages
JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Types of Solutions

Degree of Dissociation and Its Formula With Solved Example for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

NCERT Solutions for Class 11 Chemistry In Hindi Chapter 1 Some Basic Concepts of Chemistry

NCERT Solutions for Class 11 Chemistry Chapter 7 Redox Reaction

Instantaneous Velocity - Formula based Examples for JEE

JEE Advanced 2025 Notes
