Which of the following compounds has the smal | vedclass

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(C) $1$. For $OSX_2$ series: As the electronegativity of the halogen $X$ decreases $(F > Cl > Br)$,the bond pair electrons move closer to the central

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$OSF_2, SbI_3$ and $PI_3$

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(C) $1$. For $OSX_2$ series: As the electronegativity of the halogen $X$ decreases $(F > Cl > Br)$,the bond pair electrons move closer to the central atom $S$. This increases the repulsion between bond pairs,but the bond angle is primarily determined by the electronegativity of the substituent. Lower electronegativity of $X$ leads to a smaller bond angle. Thus,$OSF_2$ has the smallest bond angle.$2$. For $SbX_3$ series: As the size of the halogen increases $(Cl $3$. For $AI_3$ series: As the electronegativity of the central atom $A$ decreases $(P > As > Sb)$,the bond pair electrons move further away from the central atom,reducing the repulsion between them. Thus,$SbI_3$ has the smallest bond angle.Correct option is $C$.

Which of the following compounds has the smallest bond angle $(X-A-X)$ in each series respectively?

$A$. $OSF_2, OSCl_2, OSBr_2$

$B$. $SbCl_3, SbBr_3, SbI_3$

$C$. $PI_3, AsI_3, SbI_3$

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Which of the following compounds has the smallest bond angle $(X-A-X)$ in each series respectively? $A$. $OSF_2, OSCl_2, OSBr_2$ $B$. $SbCl_3, SbBr_3, SbI_3$ $C$. $PI_3, AsI_3, SbI_3$