Question #aad44
1 Answer
Sep 17, 2016
Here's what I got.
Explanation:
The thing to remember here is that
- a single bond
#-># contains one sigma bond- a double bond
#-># contains one sigma bond and one pi bond- a triple bond
#-># contains one sigma bond and two pi bonds
In order to find the number of sigma and pi bonds present in a molecule of cyanogen,
As you can see, the cyanogen molecule contains one
This means that you will have
- a total of three sigma bonds
Here you have one sigma bond from the
#"C"-"C"# single bond and one sigma bond from each of the two#"C"-="N" # triple bonds
- a total of four pi bonds
Here you get two pi bonds from each of the two
#"C" -= "N"# triple bonds