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(A) $I$: $SiH_4$ has $Si$ in $sp^3$ hybridization with $0$ lone pairs,resulting in a tetrahedral geometry. This is correct.$II$: $BeCl_2$ has $Be$ in

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$I$

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(A) $I$: $SiH_4$ has $Si$ in $sp^3$ hybridization with $0$ lone pairs,resulting in a tetrahedral geometry. This is correct.$II$: $BeCl_2$ has $Be$ in $sp$ hybridization with $0$ lone pairs,resulting in a linear geometry. The table incorrectly states $sp^2$ and $1$ lone pair.$III$: $SF_4$ has $S$ in $sp^3d$ hybridization with $1$ lone pair,resulting in a see-saw geometry. The table incorrectly states $dsp^3$ and square planar geometry.$IV$: $SnCl_2$ has $Sn$ in $sp^2$ hybridization with $1$ lone pair,resulting in an angular (bent) geometry. The table incorrectly states $sp$ and linear geometry.Therefore,only set $I$ is correct.

Which of the following sets is correct based on the table provided below?
Molecule Hybridization Geometry Number of lone pairs of electrons on central atom
$I$ $SiH_4$ $sp^3$ tetrahedral $0$
$II$ $BeCl_2$ $sp^2$ linear $1$
$III$ $SF_4$ $dsp^3$ square planar $1$
$IV$ $SnCl_2$ $sp$ linear $0$

Similar Questions

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	Molecule	Hybridization	Geometry	Number of lone pairs of electrons on central atom
$I$	$SiH_4$	$sp^3$	tetrahedral	$0$
$II$	$BeCl_2$	$sp^2$	linear	$1$
$III$	$SF_4$	$dsp^3$	square planar	$1$
$IV$	$SnCl_2$	$sp$	linear	$0$

Which of the following sets is correct based on the table provided below?MoleculeHybridizationGeometryNumber of lone pairs of electrons on central atom$I$$SiH_4$$sp^3$tetrahedral$0$$II$$BeCl_2$$sp^2$linear$1$$III$$SF_4$$dsp^3$square planar$1$$IV$$SnCl_2$$sp$linear$0$