Question

7. The Lewis Structure of an organic compound is shown below. Answer the following questions H a. which is the most polar bond? (circle or just write the two atoms involved) Н. b. How many sp hybridized atoms are on the compound? x c. How many sigma bonds are there? Н d. How many pi bonds are there? e. Using valence bond theory, describe how a Catom is bonded to a Hatom 11

Answer 1

a. S-F is the most polar bond. This could be understood on the basis of electronegativity (EN).

For S-F (EN of S ~ 2.58, F ~ 3.98, ▲EN ~ 1.4)

For S-O (EN of S ~ 2.58, O ~ 3.44, ▲EN ~ 0.86)

For S-C (EN of S ~ 2.58, C ~ 2.55, ▲EN ~ 0.03)

As the electronegativity difference is more in case of S-F (▲EN ~ 1.4) amongst all, therefore, S-F is the most polar bond.

b. There are 6 sp² hybridized atoms in the compound i.e. all the 6 C atoms of the benzene ring are sp².

c. There are 14 sigma bonds in the compound. Every single bond is considered to be a sigma bond (6 C-C bonds, 5 C-H bonds, 1 C-S bond, 1 S-O bond, 1 S-F bond).

d. There are 3 pi bonds in the compound. The double bonds in the benzene ring (C=C) comprise of 1 sigma and 1 pi bond each. As there are 3 double bonds, therefore, there are 3 pi bonds.

e. According to the valence bond theory, each C atom has 3 sp² hybridized orbitals. 1 sp² orbital of each C atom bonds with the s orbital of H atom to give C-H (sp²-s) sigma bonds; with adjoining sp² orbital of C atom to give C-C (sp²-sp²) sigma bonds.

The remaining one sp² orbital of each C atom bonds sideways with sp² orbital of another C to give C=C (sp²-sp²) pi bonds.