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

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(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):

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$II > I > III$

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(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:

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