Which of the following possess two lone pairs | vedclass

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(C) To determine the geometry and lone pairs,we use the formula for steric number: $SN = \frac{1}{2}(V + M - C + A)$,where $V$ is valence electrons,$M

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$III, IV$

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(C) To determine the geometry and lone pairs,we use the formula for steric number: $SN = \frac{1}{2}(V + M - C + A)$,where $V$ is valence electrons,$M$ is monovalent atoms,$C$ is cationic charge,and $A$ is anionic charge.$I. SF_4$: $SN = \frac{1}{2}(6 + 4) = 5$ ($sp^3d$ hybridization). It has $1$ lone pair and a see-saw shape.$II. XeO_4$: $SN = \frac{1}{2}(8 + 0) = 4$ ($sp^3$ hybridization). It has $0$ lone pairs and a tetrahedral shape.$III. XeF_4$: $SN = \frac{1}{2}(8 + 4) = 6$ ($sp^3d^2$ hybridization). It has $2$ lone pairs and a square planar shape.$IV. ICl_4^-$: $SN = \frac{1}{2}(7 + 4 + 1) = 6$ ($sp^3d^2$ hybridization). It has $2$ lone pairs and a square planar shape.Thus,$III$ and $IV$ possess two lone pairs and are square planar.

$A$. Trigonal planar	$I$. $PCl_5$
$B$. Tetrahedral	$II$. $SF_6$
$C$. Trigonal bipyramidal	$III$. $BF_3$
$D$. Octahedral	$IV$. $CCl_4$

Which of the following possess two lone pairs of electrons on the central atom and are square planar in shape?
$I. SF_4, II. XeO_4, III. XeF_4, IV. ICl_4^-$

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Which of the following possess two lone pairs of electrons on the central atom and are square planar in shape?$I. SF_4, II. XeO_4, III. XeF_4, IV. ICl_4^-$