The correct order of decreasing H-C-H bond an | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) To determine the $H-C-H$ bond angle, we look at the hybridization and geometry of the central carbon atom in each molecule:$I$ (Cyclopropylidene):

Vedclass · Accepted Answer

$II > I > III$

Vedclass · Answer

(B) To determine the $H-C-H$ bond angle, we look at the hybridization and geometry of the central carbon atom in each molecule:$I$ (Cyclopropylidene): The central carbon atom is $sp^2$ hybridized. Due to the ring strain in the three-membered ring, the $H-C-H$ bond angle is approximately $115^{\circ}$.$II$ (Ethene): The carbon atom is $sp^2$ hybridized. In a standard alkene, the $H-C-H$ bond angle is approximately $120^{\circ}$.$III$ (Methane): The carbon atom is $sp^3$ hybridized with a tetrahedral geometry. The $H-C-H$ bond angle is $109^{\circ} 28^{\prime}$ (or $\approx 109.5^{\circ}$).Comparing these values: $120^{\circ} (II) > 115^{\circ} (I) > 109.5^{\circ} (III)$.Therefore, the correct decreasing order is $II > I > III$.

The correct order of decreasing $H-C-H$ bond angle in the following molecules is:

Similar Questions

As the $s$-character of the hybrid orbital decreases,the bond angle ...........

What is the hybridization of $C$ atoms in graphite?

Which of the following overlapping is not present in $XeO_3$ molecule?

The linear shape of $CO_2$ is due to

In which of the following pairs do both the species have $sp^{3}$ hybridization?

Vedclass Test Series

Exam Paper Generator

Online Exam Module