Which of the following is NOT tetrahedral in | vedclass

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(C) To determine the shape,we calculate the hybridization and number of lone pairs for each species:$1$. $NH_4^+$: Central atom $N$ has $5$ valence el

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$SF_4$

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(C) To determine the shape,we calculate the hybridization and number of lone pairs for each species:$1$. $NH_4^+$: Central atom $N$ has $5$ valence electrons. $H = (5+4-1)/2 = 4$. Hybridization is $sp^3$ with $0$ lone pairs,so it is tetrahedral.$2$. $SiCl_4$: Central atom $Si$ has $4$ valence electrons. $H = (4+4)/2 = 4$. Hybridization is $sp^3$ with $0$ lone pairs,so it is tetrahedral.$3$. $SF_4$: Central atom $S$ has $6$ valence electrons. $H = (6+4)/2 = 5$. Hybridization is $sp^3d$ with $1$ lone pair. Due to the presence of a lone pair,its shape is see-saw,not tetrahedral.$4$. $SO_4^{2-}$: Central atom $S$ has $6$ valence electrons. $H = (6+0+2)/2 = 4$. Hybridization is $sp^3$ with $0$ lone pairs,so it is tetrahedral.Thus,$SF_4$ is the molecule that is not tetrahedral.

List-$I$ (Compound / Species)	List-$II$ (Shape / Geometry)
$A$. $SF_4$	$I$. Tetrahedral
$B$. $BrF_3$	$II$. Pyramidal
$C$. $BrO_3^{-}$	$III$. See-saw
$D$. $NH_4^{+}$	$IV$. Bent $T$-shape

Which of the following is $NOT$ tetrahedral in shape?

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The correct order of bond angles of the following is: $H_2O$ $(I)$,$NH_3$ $(II)$,$CH_4$ $(III)$,$SO_2$ $(IV)$

Match List-$I$ with List-$II$ :

List-$I$ (Compound / Species) List-$II$ (Shape / Geometry)

$A$. $SF_4$ $I$. Tetrahedral

$B$. $BrF_3$ $II$. Pyramidal

$C$. $BrO_3^{-}$ $III$. See-saw

$D$. $NH_4^{+}$ $IV$. Bent $T$-shape

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