The correct order of bond angles (smallest fi | vedclass

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(C) To determine the bond angles,we analyze the hybridization and molecular geometry of each molecule:$1$. $H_2S$: The central atom $S$ has two bond p

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$H_2S < NH_3 < SiH_4 < BF_3$

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(C) To determine the bond angles,we analyze the hybridization and molecular geometry of each molecule:$1$. $H_2S$: The central atom $S$ has two bond pairs and two lone pairs. Due to the presence of lone pairs,the bond angle is significantly reduced from the tetrahedral angle,approximately $92.6^o$.$2$. $NH_3$: The central atom $N$ has three bond pairs and one lone pair. The lone pair-bond pair repulsion reduces the bond angle to approximately $107^o$.$3$. $SiH_4$: The central atom $Si$ is $sp^3$ hybridized with four bond pairs and no lone pairs,resulting in a perfect tetrahedral geometry with a bond angle of $109^o 28'$.$4$. $BF_3$: The central atom $B$ is $sp^2$ hybridized with three bond pairs and no lone pairs,resulting in a trigonal planar geometry with a bond angle of $120^o$.Thus,the correct order of bond angles (smallest first) is $H_2S < NH_3 < SiH_4 < BF_3$ $(92.6^o < 107^o < 109^o 28' < 120^o)$.

The correct order of bond angles (smallest first) in $H_2S, NH_3, BF_3$ and $SiH_4$ is

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