Match the List-I with List-IIList-I (Molecule | vedclass

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(A) $PF_5$: The central atom $P$ has $5$ bonding pairs and $0$ lone pairs,so steric number is $5$,which corresponds to $sp^3d$ hybridisation.$SF_6$: T

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

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(A) $PF_5$: The central atom $P$ has $5$ bonding pairs and $0$ lone pairs,so steric number is $5$,which corresponds to $sp^3d$ hybridisation.$SF_6$: The central atom $S$ has $6$ bonding pairs and $0$ lone pairs,so steric number is $6$,which corresponds to $sp^3d^2$ hybridisation.$Ni(CO)_4$: $Ni$ is in $0$ oxidation state. The $CO$ ligand is a strong field ligand,causing pairing of electrons in $3d$ orbitals. The hybridisation is $sp^3$.$[PtCl_4]^{2-}$: $Pt$ is in $+2$ oxidation state. The $Cl^-$ ligand is a weak field ligand,but due to the high crystal field splitting energy of $5d$ series elements,it undergoes $dsp^2$ hybridisation.

List-$I$ (Molecules/ion)	List-$II$ (Hybridisation of central atom)
$A. PF_5$	$I. dsp^2$
$B. SF_6$	$II. sp^3d$
$C. Ni(CO)_4$	$III. sp^3d^2$
$D. [PtCl_4]^{2-}$	$IV. sp^3$

Match the List-$I$ with List-$II$
List-$I$ (Molecules/ion) List-$II$ (Hybridisation of central atom)
$A. PF_5$ $I. dsp^2$
$B. SF_6$ $II. sp^3d$
$C. Ni(CO)_4$ $III. sp^3d^2$
$D. [PtCl_4]^{2-}$ $IV. sp^3$
Choose the correct answer from the options given below $:$

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Match the List-$I$ with List-$II$List-$I$ (Molecules/ion)List-$II$ (Hybridisation of central atom)$A. PF_5$$I. dsp^2$$B. SF_6$$II. sp^3d$$C. Ni(CO)_4$$III. sp^3d^2$$D. [PtCl_4]^{2-}$$IV. sp^3$Choose the correct answer from the options given below $:$