Question

Question #aad44

Answer 1

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, `"C"_2"N"_2`, you must examine its Lewis structure.

As you can see, the cyanogen molecule contains one `"C"-"C"` single bond and two `"C"-="N"` triple bonds.

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