Question

Correct order of bond angles in the given compounds is/are:

• A. $\gamma < \gamma'$
• B. $\gamma < \beta = \alpha$
• C. $\alpha > \alpha'$
• D. $\beta > \beta'$

Answer 1

Answer: (A)

a) $\gamma < \gamma'$

Answer 2

In molecules like $CF_2O$ and $CCl_2O$, the central carbon atom is $sp^2$ hybridized, leading to a trigonal planar electron geometry. The $C=O$ double bond exerts greater electron repulsion than the $C-X$ single bonds (where X is F or Cl). This compression effect reduces the angles involving the $C=O$ bond, making them less than $120^\circ$, while the angle between the two $C-X$ bonds is expanded, making it greater than $120^\circ$. Thus, in both molecules, the angle between the two single bonds ($\gamma$ and $\gamma'$) is larger than the angle between the $C=O$ bond and a single bond ($\beta$ and $\beta'$).

Comparing $CF_2O$ and $CCl_2O$: Fluorine is more electronegative and smaller than chlorine. The greater electronegativity of fluorine leads to a more polarized $C-F$ bond and potentially stronger electron withdrawal. This can result in a smaller bond angle between the $C=O$ bond and the $C-F$ bonds ($\beta < \beta'$), and also a smaller bond angle between the two $C-F$ bonds compared to the two $C-Cl$ bonds ($\gamma < \gamma'$). Therefore, the correct order is $\gamma < \gamma'$.