(C) In $CH_4$,before bond formation,the $2s$ orbital and three $2p$ orbitals of carbon undergo $sp^3$ hybridization. This results in the formation of

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(C) In $CH_4$,before bond formation,the $2s$ orbital and three $2p$ orbitals of carbon undergo $sp^3$ hybridization. This results in the formation of four equivalent $sp^3$ hybrid orbitals directed towards the corners of a regular tetrahedron. Therefore,the bonds are not formed using the original $p$-orbitals at $90^{\circ}$,but rather through these $sp^3$ hybrid orbitals,resulting in a bond angle of $109.5^{\circ}$.

Class 11 Chemistry Chemical Bonding and Molecular Structure Hybridisation

Easy

Why is the bond angle in $CH_4$ not $90^{\circ}$,even though the outermost $p$-orbitals of carbon are at $90^{\circ}$ to each other?

A
Due to $sp$ hybridization
B
Due to $sp^2$ hybridization
C
Due to $sp^3$ hybridization
D
Due to $dsp^2$ hybridization

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Class 11 Questions

Class 11

Chemistry Questions

Chapter

Chemical Bonding and Molecular Structure

Subtopic

Hybridisation

Assertion : Among the carbon allotropes,diamond is an insulator,whereas,graphite is a good conductor of electricity.
Reason : Hybridization of carbon in diamond and graphite are $sp^3$ and $sp^2$,respectively.

DifficultJEE MAIN 2016

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If $p_x$ and $p_y$ orbitals participate in $sp^2$ hybridisation of the sulphur atom in $SO_3$,then the $d$ orbitals which are used in $p\pi-d\pi$ bonds are

Medium

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Assuming pure $2s$ and $2p$ orbitals of carbon are used in forming $CH_4$ molecule,which of the following statements is false?

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Which of the following shows a linear structure?

Medium

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Define the following terms:
$(i)$ Hybridisation
$(ii)$ $sp$ Hybridisation
$(iii)$ $sp^2$ Hybridisation
$(iv)$ $sp^3$ Hybridisation
$(v)$ Hydrogen bonding

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Why is the bond angle in $CH_4$ not $90^{\circ}$,even though the outermost $p$-orbitals of carbon are at $90^{\circ}$ to each other?

Similar Questions

Assertion : Among the carbon allotropes,diamond is an insulator,whereas,graphite is a good conductor of electricity.Reason : Hybridization of carbon in diamond and graphite are $sp^3$ and $sp^2$,respectively.

If $p_x$ and $p_y$ orbitals participate in $sp^2$ hybridisation of the sulphur atom in $SO_3$,then the $d$ orbitals which are used in $p\pi-d\pi$ bonds are

Assuming pure $2s$ and $2p$ orbitals of carbon are used in forming $CH_4$ molecule,which of the following statements is false?

Which of the following shows a linear structure?

Define the following terms:$(i)$ Hybridisation$(ii)$ $sp$ Hybridisation$(iii)$ $sp^2$ Hybridisation$(iv)$ $sp^3$ Hybridisation$(v)$ Hydrogen bonding

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Assertion : Among the carbon allotropes,diamond is an insulator,whereas,graphite is a good conductor of electricity.
Reason : Hybridization of carbon in diamond and graphite are $sp^3$ and $sp^2$,respectively.

Define the following terms:
$(i)$ Hybridisation
$(ii)$ $sp$ Hybridisation
$(iii)$ $sp^2$ Hybridisation
$(iv)$ $sp^3$ Hybridisation
$(v)$ Hydrogen bonding