Q10.49P - Chemistry: Molecular Nature Of Matter And Change

Question

State an ideal value for each of the bond angles in each molecule, and note where you expect deviations:

(a)  

(b)  

(c)  

Step-by-Step Solution

Answer

The shapes and bond angles of molecules can be determined from the VSEPR theory. The deviations in observed bond angles arise mainly due to lone pair repulsion around the central atoms.

Step 1: The molecule is Nitric Acid

The nitrogen atom is ${\mathbf{sp}}^2$   hybridised with a bond angle of  $\mathbf{120}\mathbf{°}$ .  Since the nitrogen has no lone pair on the atom, the bond angle may be as expected. The oxygen of the OH group is sp³ hybridised with an expected bond angle of $\mathbf{109}\mathbf{.}\mathbf{5}\mathbf{°}$ , but the lone pairs on oxygen decrease the bond angle from the expected value. 

Step 2: The molecule is Acetaldehyde

The methyl carbon atom is  ${\mathbf{sp}}^3$ hybridized with a bond angle of  data-custom-editor="chemistry" $\mathbf{109}\mathbf{.}\mathbf{5}\mathbf{°}$  . The carbonyl carbon atom is data-custom-editor="chemistry" ${\mathbf{sp}}^2$   hybridized with a bond angle of data-custom-editor="chemistry" $\mathbf{120}\mathbf{°}$ . Since carbon atoms have no lone pairs on them the bond angles are as expected.

Step 3: The molecule is Formic Acid

The carbonyl carbon atom is ${\mathbf{sp}}^2$   hybridized with bond angle of  $\mathbf{120}\mathbf{°}$  . The oxygen of the OH group is ${\mathbf{sp}}^3$  hybridized with an expected bond angle of  $\mathbf{109}\mathbf{.}\mathbf{5}\mathbf{°}$ , but the lone pairs on oxygen decrease the bond angle from  $\mathbf{109}\mathbf{.}\mathbf{5}\mathbf{°}$.