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

Vedclass · Accepted Answer

$SF_4$

Vedclass · Answer

(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-$I$ Molecule/ion	List-$II$ Bond pair : lone pair (on the central atom)
$A. ICl_2^-$	$I. 4 : 2$
$B. H_2O$	$II. 4 : 1$
$C. SO_2$	$III. 2 : 3$
$D. XeF_4$	$IV. 2 : 2$

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

Similar Questions

Match the List-$I$ with List-$II$.

List-$I$ Molecule/ion List-$II$ Bond pair : lone pair (on the central atom)

$A. ICl_2^-$ $I. 4 : 2$

$B. H_2O$ $II. 4 : 1$

$C. SO_2$ $III. 2 : 3$

$D. XeF_4$ $IV. 2 : 2$

Choose the correct answer from the options given below:

Which of the following molecules has a regular geometry as expected?

Which species has the same shape as $NH_3$?

Between the dichloronitronium ion $[ONCl_2]^+$ and thionyl chloride $(OSCl_2)$,which has a higher $Cl-\hat{X}-Cl$ bond angle? (where $X$ is the central atom $N$ or $S$)

Which of the following compounds have the same number of lone pairs on their central atom?
$I. XeF_5^- \quad II. BrF_3 \quad III. XeF_2 \quad IV. H_3S^{+} \quad V. CH_2$ (Methylene)

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