The correct order of [FeF_6]^{3-}, [CoF_6]^{3 | vedclass

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Question

(A) To determine the number of unpaired electrons,we analyze the electronic configuration of the metal ion in each complex:$1$. $[FeF_6]^{3-}$: $Fe^{3

Vedclass · Accepted Answer

$[FeF_6]^{3-} > [CoF_6]^{3-} > [Ni(CN)_4]^{2-} = [Ni(CO)_4]$

Vedclass · Answer

(A) To determine the number of unpaired electrons,we analyze the electronic configuration of the metal ion in each complex:$1$. $[FeF_6]^{3-}$: $Fe^{3+}$ is $3d^5$. Since $F^-$ is a weak field ligand,electrons remain unpaired. Number of unpaired electrons = $5$.$2$. $[CoF_6]^{3-}$: $Co^{3+}$ is $3d^6$. Since $F^-$ is a weak field ligand,it forms a high-spin complex. Number of unpaired electrons = $4$.$3$. $[Ni(CN)_4]^{2-}$: $Ni^{2+}$ is $3d^8$. $CN^-$ is a strong field ligand,causing pairing of electrons. Number of unpaired electrons = $0$.$4$. $[Ni(CO)_4]$: $Ni$ is $3d^8 4s^2$. $CO$ is a strong field ligand,causing pairing of all electrons in $3d$ orbital. Number of unpaired electrons = $0$.Thus,the order is $[FeF_6]^{3-} (5) > [CoF_6]^{3-} (4) > [Ni(CN)_4]^{2-} (0) = [Ni(CO)_4] (0)$.

Column $I$	Column $II$
$(A)$ $sp^3$	$(i)$ $[Co(NH_3)_6]^{3+}$
$(B)$ $dsp^2$	$(ii)$ $[Ni(CO)_4]$
$(C)$ $sp^3d^2$	$(iii)$ $[Pt(NH_3)_2Cl_2]$
$(D)$ $d^2sp^3$	$(iv)$ $[CoF_6]^{3-}$
	$(v)$ $[Fe(CO)_5]$

The correct order of $[FeF_6]^{3-}, [CoF_6]^{3-}, [Ni(CO)_4]$ and $[Ni(CN)_4]^{2-}$ complex species based on the number of unpaired electrons present is $:$

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