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)
$A. [Ni(NH_3)_6]^{2+}$	$1. sp^3$
$B. [PtCl_4]^{2-}$	$2. sp^3d^2$
$C. [Ni(CO)_4]$	$3. dsp^2$
$D. [Co(ox)_3]^{3-}$	$4. d^2sp^3$

List $-I$	List $-II$
$A$. $[PtCl_4]^{2-}$	$I$. $sp^3d$
$B$. $BrF_5$	$II$. $d^2sp^3$
$C$. $PCl_5$	$III$. $dsp^2$
$D$. $[Co(NH_3)_6]^{3+}$	$IV$. $sp^3d^2$

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Hybridizations of the central atom in $[Fe(H_2O)_6]^{3+}$ and $[Co(NH_3)_6]^{3+}$,respectively,are

Match List $-I$ with List $-II$:

List $-I$ List $-II$

$A$. $[PtCl_4]^{2-}$ $I$. $sp^3d$

$B$. $BrF_5$ $II$. $d^2sp^3$

$C$. $PCl_5$ $III$. $dsp^2$

$D$. $[Co(NH_3)_6]^{3+}$ $IV$. $sp^3d^2$

Which of the following statements is correct regarding the complexes $[CoF_6]^{3-}$ and $[Co(NH_3)_6]^{3+}$?

Which of the following has a square planar geometry? $(Co = 27, Ni = 28, Fe = 26, Pt = 78)$

According to the valence bond theory,the hybridization of the central metal atom is $dsp^{2}$ for which one of the following compounds?

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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) $A. [Ni(NH_3)_6]^{2+}$ $1. sp^3$ $B. [PtCl_4]^{2-}$ $2. sp^3d^2$ $C. [Ni(CO)_4]$ $3. dsp^2$ $D. [Co(ox)_3]^{3-}$ $4. d^2sp^3$ $A-B-C-D$