Question #aad44

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Question #aad44

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Answer 1 · 2016-09-17T15:47:46

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, $C_{2} N_{2}$ $"C"_2"N"_2$ , you must examine its Lewis structure.

![http://chemistry.about.com/od/factsstructures/ig/Chemical-Structures---C/Cyanogen.htm]()

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

This means that you will have

a total of three sigma bonds

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

a total of four pi bonds

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

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Question #aad44

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