Match List-I with List-II and select the corr | vedclass

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(A) $1. [Ni(NH_3)_6]^{2+}$: $Ni^{2+}$ is $3d^8$. $NH_3$ is a weak field ligand,so it forms an outer-orbital complex with $sp^3d^2$ hybridization.$2. [

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(A) $1. [Ni(NH_3)_6]^{2+}$: $Ni^{2+}$ is $3d^8$. $NH_3$ is a weak field ligand,so it forms an outer-orbital complex with $sp^3d^2$ hybridization.$2. [PtCl_4]^{2-}$: $Pt^{2+}$ is a $5d^8$ metal ion. $4d$ and $5d$ series metals always form square planar complexes with $dsp^2$ hybridization regardless of the ligand strength.$3. [Ni(CO)_4]$: $Ni$ is $3d^8 4s^2$. $CO$ is a strong field ligand,causing pairing,resulting in $sp^3$ hybridization.$4. [Co(ox)_3]^{3-}$: $Co^{3+}$ is $3d^6$. $ox^{2-}$ is a strong field ligand,causing pairing,resulting in $d^2sp^3$ hybridization.Therefore,the correct matching is $A-2, B-3, C-1, D-4$.

List-$I$ (Compound)	List-$II$ (Hybridization of Central atom)
$A. [Ni(NH_3)_6]^{2+}$	$1. sp^3$
$B. [PtCl_4]^{2-}$	$2. sp^3d^2$
$C. [Ni(CO)_4]$	$3. dsp^2$
$D. [Co(ox)_3]^{3-}$	$4. d^2sp^3$

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Match List-$I$ with List-$II$ and select the correct answer using the codes given below the lists. List-$I$ (Compound) List-$II$ (Hybridization of Central atom) $A. [Ni(NH_3)_6]^{2+}$ $1. sp^3$ $B. [PtCl_4]^{2-}$ $2. sp^3d^2$ $C. [Ni(CO)_4]$ $3. dsp^2$ $D. [Co(ox)_3]^{3-}$ $4. d^2sp^3$ $A-B-C-D$