Identify the complex having the highest numbe | vedclass

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(B) $1$. For $[Co(NH_3)_6]^{3+}$: $Co^{3+}$ is $3d^6$. $NH_3$ is a strong field ligand, causing pairing. Unpaired electrons $n = 0$.$2$. For $[CoF_6]^

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$[CoF_6]^{3-}$

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(B) $1$. For $[Co(NH_3)_6]^{3+}$: $Co^{3+}$ is $3d^6$. $NH_3$ is a strong field ligand, causing pairing. Unpaired electrons $n = 0$.$2$. For $[CoF_6]^{3-}$: $Co^{3+}$ is $3d^6$. $F^-$ is a weak field ligand, no pairing. $n = 4$.$3$. For $[NiCl_4]^{2-}$: $Ni^{2+}$ is $3d^8$. $Cl^-$ is a weak field ligand, no pairing. $n = 2$.$4$. For $[Ni(CN)_4]^{2-}$: $Ni^{2+}$ is $3d^8$. $CN^-$ is a strong field ligand, causing pairing. $n = 0$.Comparing the values, $[CoF_6]^{3-}$ has the highest number of unpaired electrons $(n = 4)$.

List-$I$ (Complex/Species)	List-$II$ (Shape and magnetic moment)
$A$. $[Ni(CO)_4]$	$I$. Tetrahedral,$2.8 \ BM$
$B$. $[Ni(CN)_4]^{2-}$	$II$. Square planar,$0 \ BM$
$C$. $[NiCl_4]^{2-}$	$III$. Tetrahedral,$0 \ BM$
$D$. $[MnBr_4]^{2-}$	$IV$. Tetrahedral,$5.9 \ BM$

Identify the complex having the highest number of unpaired electrons from the following:

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