Question

answer the questions below about the highlighted atom in this lewis structure: h | h - c - c ≡ c - 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. The highlighted carbon atom has 4 single - bonds (3 C - H and 1 C - C). So, it participates in 4 sigma bonds.

Step2: Identify pi bonds

Pi bonds are present in double and triple bonds. The highlighted carbon atom is not part of any double or triple bonds, so it participates in 0 pi bonds.

Step3: Determine orbital hybridization

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