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(D) For $[Co(NH_3)_6]^{3+}$: $Co$ is in $+3$ oxidation state. The electronic configuration of $Co^{3+}$ is $[Ar] 3d^6$. Since $NH_3$ is a strong field

Vedclass · Accepted Answer

$2$

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(D) For $[Co(NH_3)_6]^{3+}$: $Co$ is in $+3$ oxidation state. The electronic configuration of $Co^{3+}$ is $[Ar] 3d^6$. Since $NH_3$ is a strong field ligand,it causes pairing of electrons in $d$-orbitals,resulting in $t_{2g}^6 e_g^0$. Thus,the number of unpaired electrons is $0$.For $[NiCl_4]^{2-}$: $Ni$ is in $+2$ oxidation state. The electronic configuration of $Ni^{2+}$ is $[Ar] 3d^8$. Since $Cl^-$ is a weak field ligand,the electrons remain unpaired in the $d$-orbitals. The configuration is $e^4 t_2^4$,which results in $2$ unpaired electrons.Total number of unpaired electrons = $0 + 2 = 2$.

List-$I$ (Complex)	List-$II$ (Type of isomerism)
$A$. $[Co(NH_3)_5(NO_2)]Cl_2$	$I$. Solvate isomerism
$B$. $[Co(NH_3)_5(SO_4)]Br$	$II$. Linkage isomerism
$C$. $[Co(NH_3)_6][Cr(CN)_6]$	$III$. Ionization isomerism
$D$. $[Co(H_2O)_6]Cl_3$	$IV$. Coordination isomerism

Total number of unpaired electrons in the complex ions $[Co(NH_3)_6]^{3+}$ and $[NiCl_4]^{2-}$ is...............

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