Question

answer the questions below about the highlighted atom in this lewis structure:
h-o-c=o
h-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: Count sigma bonds

A single bond is a sigma bond. The highlighted carbon atom forms 4 single bonds (three with H and one with C). So, the number of sigma bonds is 4.

Step2: Count pi bonds

A double bond has one pi bond and a single bond has no pi bond. The highlighted carbon atom has no double or triple bonds. So, the number of pi bonds is 0.

Step3: Determine hybridization

For an atom with 4 sigma bonds (and no lone pairs), the hybridization is \(sp^{3}\). The formula for hybridization is \(H=\frac{V + M - C + A}{2}\), where \(V\) is the number of valence electrons (for C, \(V = 4\)), \(M\) is the number of monovalent atoms (here \(M=4\) H atoms), \(C\) is the charge (0), and \(A\) is the anion charge (0). \(H=\frac{4 + 4}{2}=4\). When \(H = 4\), the 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? \(sp^{3}\)