What is the shape of ethene molecule?
A) Planar
B) Tetrahedral
C) Bipyramidal
D) None of these
Answer
540k+ views
Hint: It is easy to identify the shape of a molecule if we know the hybridization of that molecule. Alkanes, Alkenes and Alkynes are the hydrocarbons that contain bonds between carbon and hydrogen atoms.
Complete step by step answer:
We have to know that the ethene has a molecular formula \[{C_2}{H_4}\]. It has a double bond and therefore belongs to alkene.
We can write the general formula of Alkene as: \[{C_n}{H_{2n}}\]
We need to know that the electronic configuration of C in ground state: \[1{s^2}2{s^2}2{p^2}\]
Excitation of electron from s orbital to vacant p orbital so,
We must know that the electronic configuration of C in excited state: \[1{s^2}2{s^1}2{p^3}\]
Therefore, Hybridization of Ethene= \[s{p^2}\] one electron will be unpaired that is an unhybridized orbital and 3 \[s{p^2}\]orbitals.
We need to remember that the ethene has geometry is Planar
We can draw the structure of ethane molecule as,
Option A) this is a correct option as the molecule has a planar shape as it has \[s{p^2}\] hybridization.
Option B) This is an incorrect option as tetrahedral geometry is for \[s{p^3}\] hybridization.
Option C) this is an incorrect option as bipyramidal geometry has a hybridization of \[s{p^3}d\].
Option D) this is an incorrect option as we got a correct option as ethene has planar shape.
Hence, the correct answer is, ‘Option A’.
Note: We need to remember that the ethane has molecular formula \[{C_2}{H_6}\]; Ethene has a molecular formula \[{C_2}{H_4}\] and ethylene has molecular formula \[{C_2}{H_2}\] having slightly sweet smell. We need to know that the ethylene is a nonpolar molecule and soluble in non-polar solvents.
Complete step by step answer:
We have to know that the ethene has a molecular formula \[{C_2}{H_4}\]. It has a double bond and therefore belongs to alkene.
We can write the general formula of Alkene as: \[{C_n}{H_{2n}}\]
We need to know that the electronic configuration of C in ground state: \[1{s^2}2{s^2}2{p^2}\]
Excitation of electron from s orbital to vacant p orbital so,
We must know that the electronic configuration of C in excited state: \[1{s^2}2{s^1}2{p^3}\]
Therefore, Hybridization of Ethene= \[s{p^2}\] one electron will be unpaired that is an unhybridized orbital and 3 \[s{p^2}\]orbitals.
We need to remember that the ethene has geometry is Planar
We can draw the structure of ethane molecule as,
Option A) this is a correct option as the molecule has a planar shape as it has \[s{p^2}\] hybridization.
Option B) This is an incorrect option as tetrahedral geometry is for \[s{p^3}\] hybridization.
Option C) this is an incorrect option as bipyramidal geometry has a hybridization of \[s{p^3}d\].
Option D) this is an incorrect option as we got a correct option as ethene has planar shape.
Hence, the correct answer is, ‘Option A’.
Note: We need to remember that the ethane has molecular formula \[{C_2}{H_6}\]; Ethene has a molecular formula \[{C_2}{H_4}\] and ethylene has molecular formula \[{C_2}{H_2}\] having slightly sweet smell. We need to know that the ethylene is a nonpolar molecule and soluble in non-polar solvents.
Recently Updated Pages
Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Chemistry: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

State and prove Bernoullis theorem class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

Discuss the various forms of bacteria class 11 biology CBSE

