The bond angle H-X-H is the greatest in the c | vedclass

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(B) The bond angle depends on the hybridization of the central atom and the repulsion between lone pairs and bond pairs. In $CH_4$,the central atom $C

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$CH_4$

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(B) The bond angle depends on the hybridization of the central atom and the repulsion between lone pairs and bond pairs. In $CH_4$,the central atom $C$ is $sp^3$ hybridized with no lone pair,resulting in a tetrahedral geometry with a bond angle of $109.5^{\circ}$. In $NH_3$,$H_2O$,and $PH_3$,the central atoms have lone pairs which cause bond pair-lone pair repulsion,leading to a decrease in bond angle from the ideal tetrahedral angle. Specifically,$PH_3$ has a bond angle of approximately $93.5^{\circ}$ due to the lack of hybridization and the large size of the $P$ atom. Therefore,$CH_4$ has the greatest bond angle among the given options.

List-$I$ (Compound)	List-$II$ (Structure)
$A$. $ClF_3$	$1$. Square planar
$B$. $PCl_5$	$2$. Tetrahedral
$C$. $IF_5$	$3$. Trigonal bipyramidal
$D$. $CCl_4$	$4$. Square pyramidal
$E$. $XeF_4$	$5$. $T$-shaped

The bond angle $H-X-H$ is the greatest in the compound

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