Question

answer the questions below about the highlighted atom in this lewis structure: h h | | h - c - c - c ≡ n : h h in how many sigma bonds does the highlighted atom participate? in how many pi bonds does the highlighted atom participate? what is the orbital hybridization of the highlighted atom?

Explanation:

Step1: Identify sigma bonds

A single - bond is a sigma bond, and a double or triple bond has one sigma bond. The highlighted carbon atom has 4 single - bonds. So the number of sigma bonds is 4.

Step2: Identify pi bonds

A double bond has 1 pi bond and a triple bond has 2 pi bonds. Since the highlighted carbon is only in single - bonds, the number of pi bonds is 0.

Step3: Determine hybridization

The number of electron - groups around the highlighted carbon atom is 4 (4 single - bonds). When an atom has 4 electron - groups, its hybridization is $sp^{3}$.

Answer:

In how many sigma bonds does the highlighted atom participate? 4
In how many pi bonds does the highlighted atom participate? 0
What is the orbital hybridization of the highlighted atom? $sp^{3}$