Match the species in List-I with their geomet | vedclass

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(B) To determine the geometry, we calculate the hybridization and electron pair arrangement:$1$. $PCl_5$: Phosphorus has $5$ valence electrons, formin

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$A-III, B-IV, C-I, D-II$

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(B) To determine the geometry, we calculate the hybridization and electron pair arrangement:$1$. $PCl_5$: Phosphorus has $5$ valence electrons, forming $5$ bonds with $Cl$. Hybridization is $sp^3d$, resulting in a Trigonal bipyramidal geometry $(A-III)$.$2$. $BrF_5$: Bromine has $7$ valence electrons. It forms $5$ bonds with $F$ and has $1$ lone pair. Steric number is $6$ $(sp^3d^2)$, leading to a Square pyramidal geometry $(B-IV)$.$3$. $BF_4^-$: Boron has $3$ valence electrons, plus $1$ from the negative charge, forming $4$ bonds. Hybridization is $sp^3$, resulting in a Tetrahedral geometry $(C-I)$.$4$. $[Ni(CN)_4]^{2-}$: Nickel is in $+2$ oxidation state $(d^8)$. $CN^-$ is a strong field ligand, causing pairing. Hybridization is $dsp^2$, resulting in a Square Planar geometry $(D-II)$.Therefore, the correct matching is $A-III, B-IV, C-I, D-II$.

List-$I$	List-$II$
$A. PCl_5$	$I. \text{Tetrahedral}$
$B. BrF_5$	$II. \text{Square Planar}$
$C. BF_4^-$	$III. \text{Trigonal bipyramidal}$
$D. [Ni(CN)_4]^{2-}$	$IV. \text{Square pyramidal}$

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Match the species in List-$I$ with their geometry in List-$II$:List-$I$List-$II$$A. PCl_5$$I. \text{Tetrahedral}$$B. BrF_5$$II. \text{Square Planar}$$C. BF_4^-$$III. \text{Trigonal bipyramidal}$$D. [Ni(CN)_4]^{2-}$$IV. \text{Square pyramidal}$Choose the correct answer from the options given below: