Question

answer the questions below about the highlighted atom in this lewis structure: h h h - c - c - c≡n: 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 and pi bonds

A single - bond is a sigma bond, a double - bond has 1 sigma and 1 pi bond, and a triple - bond has 1 sigma and 2 pi bonds. The highlighted carbon atom has 4 single bonds.

Step2: Count sigma bonds

Since each single bond is a sigma bond, the number of sigma bonds the highlighted atom participates in is 4.

Step3: Count pi bonds

There are no double or triple bonds attached to the highlighted atom, so the number of pi bonds is 0.

Step4: Determine hybridization

The number of electron - group around the highlighted carbon atom is 4 (4 single bonds). According to the VSEPR theory and hybridization rules, 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? $sp^{3}$