Match List-I with List-II: List-I Lis | vedclass

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(D) To determine the hybridization,we analyze the electronic configuration of the central metal ion and the nature of the ligand:$A. K_2[Ni(CN)_4]$: $

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

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(D) To determine the hybridization,we analyze the electronic configuration of the central metal ion and the nature of the ligand:$A. K_2[Ni(CN)_4]$: $Ni^{2+}$ is $[Ar] 3d^8$. $CN^-$ is a strong field ligand $(SFL)$,causing pairing of electrons in $3d$ orbitals,resulting in $dsp^2$ hybridization.$B. [Ni(CO)_4]$: $Ni$ is $[Ar] 3d^8 4s^2$. $CO$ is a strong field ligand,causing pairing,resulting in $sp^3$ hybridization.$C. [Co(NH_3)_6]Cl_3$: $Co^{3+}$ is $[Ar] 3d^6$. $NH_3$ is a strong field ligand for $Co^{3+}$,causing pairing,resulting in $d^2sp^3$ hybridization.$D. Na_3[CoF_6]$: $Co^{3+}$ is $[Ar] 3d^6$. $F^-$ is a weak field ligand $(WFL)$,so no pairing occurs,resulting in $sp^3d^2$ hybridization.Matching the results: $A-III, B-I, C-IV, D-II$.

List-$I$	List-$II$
$A. K_2[Ni(CN)_4]$	$I. sp^3$
$B. [Ni(CO)_4]$	$II. sp^3d^2$
$C. [Co(NH_3)_6]Cl_3$	$III. dsp^2$
$D. Na_3[CoF_6]$	$IV. d^2sp^3$

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

List-$I$ List-$II$

$A. K_2[Ni(CN)_4]$ $I. sp^3$

$B. [Ni(CO)_4]$ $II. sp^3d^2$

$C. [Co(NH_3)_6]Cl_3$ $III. dsp^2$

$D. Na_3[CoF_6]$ $IV. d^2sp^3$

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Match List-$I$ with List-$II$: List-$I$ List-$II$ $A. K_2[Ni(CN)_4]$ $I. sp^3$ $B. [Ni(CO)_4]$ $II. sp^3d^2$ $C. [Co(NH_3)_6]Cl_3$ $III. dsp^2$ $D. Na_3[CoF_6]$ $IV. d^2sp^3$ Choose the correct answer from the options given below: