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(D) $1$. For $BrF_5$: The central atom $Br$ has $7$ valence electrons. It forms $5$ bonds with $F$ and has $1$ lone pair. Steric number = $5 + 1 = 6$,

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$I$ & $II$

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(D) $1$. For $BrF_5$: The central atom $Br$ has $7$ valence electrons. It forms $5$ bonds with $F$ and has $1$ lone pair. Steric number = $5 + 1 = 6$,which corresponds to $sp^3d^2$ hybridization. The geometry is square pyramidal. This is correctly matched.$2$. For $XeF_6$: The central atom $Xe$ has $8$ valence electrons. It forms $6$ bonds with $F$ and has $1$ lone pair. Steric number = $6 + 1 = 7$,which corresponds to $sp^3d^3$ hybridization. The geometry is distorted octahedral. This is correctly matched.$3$. For $SF_4$: The central atom $S$ has $6$ valence electrons. It forms $4$ bonds with $F$ and has $1$ lone pair. Steric number = $4 + 1 = 5$,which corresponds to $sp^3d$ hybridization. The geometry is see-saw. This is incorrectly matched.$4$. For $PbCl_2$: $Pb$ is in group $14$. It forms $2$ bonds with $Cl$ and has $1$ lone pair. Steric number = $2 + 1 = 3$,which corresponds to $sp^2$ hybridization. The geometry is bent. This is incorrectly matched.Therefore,sets $I$ and $II$ are correctly matched.

Molecule	Hybridization	Geometry
$I$. $BrF_5$	$sp^3d^2$	Square pyramidal
$II$. $XeF_6$	$sp^3d^3$	Distorted octahedral
$III$. $SF_4$	$dsp^2$	Square planar
$IV$. $PbCl_2$	$sp$	Linear

Which of the following sets are correctly matched?
Molecule Hybridization Geometry
$I$. $BrF_5$ $sp^3d^2$ Square pyramidal
$II$. $XeF_6$ $sp^3d^3$ Distorted octahedral
$III$. $SF_4$ $dsp^2$ Square planar
$IV$. $PbCl_2$ $sp$ Linear

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Which of the following sets are correctly matched? MoleculeHybridizationGeometry$I$. $BrF_5$$sp^3d^2$Square pyramidal$II$. $XeF_6$$sp^3d^3$Distorted octahedral$III$. $SF_4$$dsp^2$Square planar$IV$. $PbCl_2$$sp$Linear