Q10.48P - 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: H 2 CNOH

The carbon atom is ${\mathbf{sp}}^2$   hybridized with an expected bond angle of 120⁰. Similarly, nitrogen is   hybridized with an expected bond angle of $\mathbf{120}\mathbf{°}$  ^.  Deviations occur due to the presence of two and one lone pair on oxygen and nitrogen respectively. 

 

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

Step 2: H 3 COCH 3

The carbon atoms are ${\mathbf{sp}}^3$  hybridised with a bond angle of  $\mathbf{109}\mathbf{.}{\mathbf{5}}^o$  . Each oxygen atom is are ${\mathbf{sp}}^3$  hybridized with an expected bond angle of  $\mathbf{109}\mathbf{.}{\mathbf{5}}^o$ , but the lone pairs of oxygen decrease the bond angle from  $\mathbf{109}\mathbf{.}{\mathbf{5}}^o$ .

Step 3: H 3 BO 3

The geometry around boron is ${\mathbf{sp}}^2$   hybridized and has a bond angle of  ${\mathbf{120}}^o$ . The oxygen atoms are ${\mathbf{sp}}^3$  hybridized with an expected bond angle of  $\mathbf{109}\mathbf{.}{\mathbf{5}}^o$ , but the lone pairs on oxygen decrease the bond angle from  $\mathbf{109}\mathbf{.}{\mathbf{5}}^o$ .