The number of bond pairs,lone pairs present o | vedclass

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(D) In $SF_4$,the central atom is sulfur $(S)$.Sulfur has $6$ valence electrons.It forms $4$ single bonds with $4$ fluorine atoms,using $4$ valence el

Vedclass · Accepted Answer

$4$,$1$,see-saw

Vedclass · Answer

(D) In $SF_4$,the central atom is sulfur $(S)$.Sulfur has $6$ valence electrons.It forms $4$ single bonds with $4$ fluorine atoms,using $4$ valence electrons.This leaves $6 - 4 = 2$ electrons,which form $1$ lone pair on the sulfur atom.Thus,there are $4$ bond pairs and $1$ lone pair.The steric number is $4 + 1 = 5$,which corresponds to $sp^3d$ hybridization.Due to the presence of $1$ lone pair in the equatorial position of a trigonal bipyramidal geometry,the shape of the $SF_4$ molecule is see-saw.

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

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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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

Choose the correct answer from the options given below :

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