Question

The bond angles in a regular polygon with nsides are equal to
${180}^o-\frac{{360}^o}{n}$

a. What are the bond angles in a regular octagon?

b. What are the bond angles in a regular nonagon?

Short answer

a. For octagon, n = 8 and therefore the bond angle is ${135}^{\circ }$.

b. For nonagon, n = 9 and therefore the bond angle is ${140}^{\circ }$.

Step-by-step solution

Explanation of formula

This formula is known as the Baeyer formula which basically calculates the angle of a particular molecule. All the carbons in cyclic rings like propane-butane are formed with sp³ carbon which has a known angle of 109.5°, but those cyclic rings don’t have this angle in their structure and with increasing the no of the carbon atom, the angle strain decreases but after 5 carbon atoms, the angle again increases and as well as angle strain.

n in this formula is the number of carbon atoms in a compound.

Subpart (a) The bond angle in a regular octagon.

For octagon, n = 8

Therefore, the bond angle is

$$\begin{gathered} ={180}^o-\frac{{360}^o}{n} \\ ={180}^o-\frac{{360}^o}{8} \\ ={180}^o-{45}^o \\ ={135}^o \end{gathered}$$

Subpart (b) The bond angle in a regular nonagon.

For nonagon, n = 9

Therefore, the bond angle is

$$\begin{gathered} ={180}^o-\frac{{360}^o}{n} \\ ={180}^o-\frac{{360}^o}{9} \\ ={180}^o-{40}^{\circ } \\ ={140}^{\circ } \end{gathered}$$