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(D) $1$. In $[CoF_6]^{3-}$,$Co$ is in $+3$ oxidation state $(3d^6)$. Since $F^-$ is a weak field ligand,electrons do not pair up,resulting in $4$ unpa

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$[CoF_6]^{3-}, [Fe(H_2O)_6]^{2+}$

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(D) $1$. In $[CoF_6]^{3-}$,$Co$ is in $+3$ oxidation state $(3d^6)$. Since $F^-$ is a weak field ligand,electrons do not pair up,resulting in $4$ unpaired electrons.$2$. In $[FeF_6]^{3-}$,$Fe$ is in $+3$ oxidation state $(3d^5)$. Since $F^-$ is a weak field ligand,there are $5$ unpaired electrons.$3$. In $[Fe(CN)_6]^{3-}$,$Fe$ is in $+3$ oxidation state $(3d^5)$. Since $CN^-$ is a strong field ligand,electrons pair up,resulting in $1$ unpaired electron.$4$. In $[Fe(CN)_6]^{4-}$,$Fe$ is in $+2$ oxidation state $(3d^6)$. Since $CN^-$ is a strong field ligand,all electrons pair up,resulting in $0$ unpaired electrons.$5$. In $[Ni(CN)_4]^{2-}$,$Ni$ is in $+2$ oxidation state $(3d^8)$. Since $CN^-$ is a strong field ligand,electrons pair up,resulting in $0$ unpaired electrons.$6$. In $[Fe(H_2O)_6]^{2+}$,$Fe$ is in $+2$ oxidation state $(3d^6)$. Since $H_2O$ is a weak field ligand,there are $4$ unpaired electrons.Comparing the pairs:- Pair $A$: $4$ and $5$ (Not equal)- Pair $B$: $1$ and $0$ (Not equal)- Pair $C$: $1$ and $0$ (Not equal)- Pair $D$: $4$ and $4$ (Equal)Therefore,the correct option is $D$.

Complex	Magnetic Moment $(BM)$
$K_4[Mn(CN)_6]$	$2.2$
$[Fe(H_2O)_6]^{2+}$	$5.3$
$K_2[MnCl_4]$	$5.9$

$A$. Tetrahedral $NiCl_2Br_2^{2-}$	$B$. Square planar $Pt(NH_3)_2Cl_2$
$C$. Octahedral $Co(NH_3)_3Cl_3$	$D$. Square planar $Pd(NH_3)_3Br^{+}$
$E$. Octahedral $Co(en)_3^{3+}$

Which of the following pairs of compounds has the same number of unpaired electrons?

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What can be inferred from the magnetic moment values of the following complex species?

Complex Magnetic Moment $(BM)$

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$[Fe(H_2O)_6]^{2+}$ $5.3$

$K_2[MnCl_4]$ $5.9$

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