The compound that has the largest H-M-H bond | vedclass

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(B) To determine the largest $H-M-H$ bond angle,we analyze the hybridization and lone pairs of the central atom in each molecule:$1$. In $H_{2}O$ ($O$

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$CH_{4}$

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(B) To determine the largest $H-M-H$ bond angle,we analyze the hybridization and lone pairs of the central atom in each molecule:$1$. In $H_{2}O$ ($O$ is central atom),the hybridization is $sp^{3}$ with $2$ lone pairs,resulting in a bond angle of $104^{\circ}5'$.$2$. In $CH_{4}$ ($C$ is central atom),the hybridization is $sp^{3}$ with $0$ lone pairs,resulting in a perfect tetrahedral geometry with a bond angle of $109^{\circ}28'$.$3$. In $NH_{3}$ ($N$ is central atom),the hybridization is $sp^{3}$ with $1$ lone pair,resulting in a bond angle of $107^{\circ}$.$4$. In $H_{2}S$ ($S$ is central atom),the central atom is in the $3rd$ period,and the bond angle is close to $90^{\circ}$ due to the lack of hybridization (Drago's rule).Comparing these values,$CH_{4}$ has the largest bond angle of $109^{\circ}28'$.Thus,the correct option is $B$.

List-$I$ (molecule/ion)	List-$II$ (Number of lone pairs of electrons on the central atom)
$(A)$ $XeF_2$	$(I)$ $2$
$(B)$ $XeO_3$	$(II)$ $0$
$(C)$ $XeF_4$	$(III)$ $3$
$(D)$ $PF_6^-$	$(IV)$ $1$

The compound that has the largest $H-M-H$ bond angle $(M = N, O, S, C)$ is :

Similar Questions

Discuss the shape of the following molecules using the $VSEPR$ model: $BeCl_{2}, BCl_{3}, SiCl_{4}, AsF_{5}, H_{2}S, PH_{3}$

Select from each set the molecule or ion having the smallest bond angle:
$(i)$ $NH_3, PH_3$ or $AsH_3$
$(ii)$ $O_3^+, O_3$
$(iii)$ $NO_2^-$ or $O_3$
$(iv)$ $X-S-X$ angle in $SOCl_2$ and $SOF_2$

Identify the molecular geometry of $XeF_4$.

In which of the following sets,all the given species are isostructural?

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