Sep 17, 2016

Here's what I got.

#### Explanation:

The thing to remember here is that

• a single bond $\to$ contains one sigma bond
• a double bond $\to$ contains one sigma bond and one pi bond
• a triple bond $\to$ 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, ${\text{C"_2"N}}_{2}$, you must examine its Lewis structure.

As you can see, the cyanogen molecule contains one $\text{C"-"C}$ single bond and two $\text{C"-="N}$ triple bonds.

This means that you will have

• a total of three sigma bonds

Here you have one sigma bond from the $\text{C"-"C}$ single bond and one sigma bond from each of the two $\text{C"-="N}$triple bonds

• a total of four pi bonds

Here you get two pi bonds from each of the two $\text{C" -= "N}$ triple bonds