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(D) The magnetic moment $\mu = \sqrt{n(n+2)} \ BM$. For $\mu = \sqrt{3} \ BM$,the number of unpaired electrons $n = 1$.$1$. $[Mn(CN)_6]^{4-}$: $Mn^{2+

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$[Cu(H_2O)_6]^{2+}$

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(D) The magnetic moment $\mu = \sqrt{n(n+2)} \ BM$. For $\mu = \sqrt{3} \ BM$,the number of unpaired electrons $n = 1$.$1$. $[Mn(CN)_6]^{4-}$: $Mn^{2+}$ is $3d^5$. $CN^-$ is a strong field ligand,so $n=1$. Hybridization is $d^2sp^3$ (inner $d$-orbitals).$2$. $[Fe(NH_3)_6]^{3+}$: $Fe^{3+}$ is $3d^5$. $NH_3$ is a strong field ligand,so $n=1$. Hybridization is $d^2sp^3$ (inner $d$-orbitals).$3$. $[Co(CO)_4]$: $Co$ is $3d^7 4s^2$. $CO$ is a strong field ligand,$n=1$. Hybridization is $sp^3$.$4$. $[Cu(H_2O)_6]^{2+}$: $Cu^{2+}$ is $3d^9$. $H_2O$ is a weak field ligand,$n=1$. Hybridization is $sp^3d^2$ (outer $d$-orbitals).Thus,$[Cu(H_2O)_6]^{2+}$ satisfies both conditions: $\mu = \sqrt{3} \ BM$ and uses outer $d$-orbitals $(sp^3d^2)$.

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

In which of the following complex ions is the value of the magnetic moment (spin only) $\sqrt{3} \ BM$ and outer $d$-orbitals are used in hybridization?

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