Draw the lewis structure for the molecule ch2chch3. how many sigma and pi bonds does it 30) contain
1 answer:
Answer:
1-Propene has eight sigma bonds and one pi bond.
Explanation:
The lewis structure of 1-Propene is shown below. All sigma bonds are highlighted by blue color and the pi bond is highlighted by red color.
Carbon having four unpaired electrons in its excited state can form four covalent (sigma bonds) bonds. In given structure the orbitals containing unpaired electron in sp³ hybridized C undergo head to head overlap with three hydrogen atoms having single unpaired electron and orbital of other sp² hybridized carbon in same fashion. So, four sigma bonds are formed by sp³ hybridized carbon atom.
The doubly bonded carbon atoms also form sigma bond but two unpaired electrons of these two carbons undergo the formation of pi bond. This pi bond forms when the orbitals overlap along the plane perpendicular to the existing sigma bond resulting in the formation of single pi bond.
1-Propene has eight sigma bonds and one pi bond.
Explanation:
The lewis structure of 1-Propene is shown below. All sigma bonds are highlighted by blue color and the pi bond is highlighted by red color.
Carbon having four unpaired electrons in its excited state can form four covalent (sigma bonds) bonds. In given structure the orbitals containing unpaired electron in sp³ hybridized C undergo head to head overlap with three hydrogen atoms having single unpaired electron and orbital of other sp² hybridized carbon in same fashion. So, four sigma bonds are formed by sp³ hybridized carbon atom.
The doubly bonded carbon atoms also form sigma bond but two unpaired electrons of these two carbons undergo the formation of pi bond. This pi bond forms when the orbitals overlap along the plane perpendicular to the existing sigma bond resulting in the formation of single pi bond.
You might be interested in
Answer : The ratio of the protonated to the deprotonated form of the acid is, 100
Explanation : Given,
pH = 6.0
To calculate the ratio of the protonated to the deprotonated form of the acid we are using Henderson Hesselbach equation :
Now put all the given values in this expression, we get:
As per question, the ratio of the protonated to the deprotonated form of the acid will be:
Therefore, the ratio of the protonated to the deprotonated form of the acid is, 100