The H-C-H bond angle in CH_4 is 109.5^{circ}. | vedclass

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(C) According to the Valence Shell Electron Pair Repulsion $(VSEPR)$ theory,the geometry of a molecule is determined by the repulsion between electron

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(C) According to the Valence Shell Electron Pair Repulsion $(VSEPR)$ theory,the geometry of a molecule is determined by the repulsion between electron pairs in the valence shell of the central atom.$1.$ In $CH_4$,there are $4$ bonding pairs and $0$ lone pairs,resulting in a tetrahedral geometry with a bond angle of $109.5^{\circ}$.$2.$ In $H_2O$,the oxygen atom has $2$ bonding pairs and $2$ lone pairs.$3.$ According to $VSEPR$ theory,the order of repulsion is: $	ext{lone pair-lone pair} > 	ext{lone pair-bonding pair} > 	ext{bonding pair-bonding pair}$.$4.$ The strong repulsion between the two lone pairs on the oxygen atom pushes the bonding pairs closer together,causing the $H-O-H$ bond angle to decrease from the ideal tetrahedral angle of $109.5^{\circ}$ to approximately $104.5^{\circ}$.

List-$I$ (Molecule)	List-$II$ (Shape)
$A. SF_4$	$I. T\text{-shaped}$
$B. ClF_3$	$II. \text{Square planar}$
$C. BrF_5$	$III. \text{See-saw}$
$D. XeF_4$	$IV. \text{Square pyramidal}$

The $H-C-H$ bond angle in $CH_4$ is $109.5^{\circ}$. Due to lone pair repulsion,the $H-O-H$ angle in $H_2O$ will:

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