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(D) The correct answer is $[Ni(NH_3)_6]^{2+}$.$1$. $[Fe(CN)_6]^{4-}$: $Fe^{2+}$ is $3d^6$. $CN^-$ is a strong field ligand,so all $6$ electrons are pa

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$[Ni(NH_3)_6]^{2+}$

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(D) The correct answer is $[Ni(NH_3)_6]^{2+}$.$1$. $[Fe(CN)_6]^{4-}$: $Fe^{2+}$ is $3d^6$. $CN^-$ is a strong field ligand,so all $6$ electrons are paired in $t_{2g}$ orbitals $(t_{2g}^6 e_g^0)$. It is diamagnetic.$2$. $[Fe(CN)_6]^{3-}$: $Fe^{3+}$ is $3d^5$. $CN^-$ is a strong field ligand,so $t_{2g}^5 e_g^0$. It has $1$ unpaired electron in $t_{2g}$ orbitals.$3$. $[Zn(NH_3)_6]^{2+}$: $Zn^{2+}$ is $3d^{10}$. All orbitals are fully filled. It is diamagnetic.$4$. $[Ni(NH_3)_6]^{2+}$: $Ni^{2+}$ is $3d^8$. $NH_3$ is a weak field ligand for $Ni^{2+}$. The crystal field splitting results in $t_{2g}^6 e_g^2$ configuration. Here,all $6$ electrons in $t_{2g}$ orbitals are paired,and there are $2$ unpaired electrons in $e_g$ orbitals,making it paramagnetic.

List-$I$	List-$II$
$A$. $[MnBr_4]^{2-}$	$I$. $d^2sp^3$ and diamagnetic
$B$. $[FeF_6]^{3-}$	$II$. $sp^3d^2$ and paramagnetic
$C$. $[Co(C_2O_4)_3]^{3-}$	$III$. $sp^3$ and diamagnetic
$D$. $[Ni(CO)_4]$	$IV$. $sp^3$ and paramagnetic

The paramagnetic complex ion which has no unpaired electrons in $t_{2g}$ orbitals is

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