The decreasing order of bond angle in the fol | vedclass

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(D) In $CH_4$,the central atom $C$ has $4$ bond pairs and $0$ lone pairs,resulting in a tetrahedral geometry with a bond angle of $109^{\circ} 28'$.In

Vedclass · Accepted Answer

$CH_4 > NH_3 > H_2O$

Vedclass · Answer

(D) In $CH_4$,the central atom $C$ has $4$ bond pairs and $0$ lone pairs,resulting in a tetrahedral geometry with a bond angle of $109^{\circ} 28'$.In $NH_3$,the central atom $N$ has $3$ bond pairs and $1$ lone pair,which causes repulsion and reduces the bond angle to $107^{\circ}$.In $H_2O$,the central atom $O$ has $2$ bond pairs and $2$ lone pairs,leading to greater repulsion and a further reduction in the bond angle to $104.5^{\circ}$.Therefore,the decreasing order of bond angle is $CH_4 > NH_3 > H_2O$.

$A$. $SnCl_2$	$I$. Trigonal planar
$B$. $NH_3$	$II$. Linear
$C$. $I_3^{-}$	$III$. Angular
$D$. $SO_3$	$IV$. Trigonal pyramidal

The decreasing order of bond angle in the following molecules is

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Match the following:
$A$. $SnCl_2$ $I$. Trigonal planar
$B$. $NH_3$ $II$. Linear
$C$. $I_3^{-}$ $III$. Angular
$D$. $SO_3$ $IV$. Trigonal pyramidal

The correct answer is:

The number of bond pairs,lone pairs present on the central atom of $SF_4$ molecule and its shape respectively are

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