Among the following,the pair of paramagnetic | vedclass

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(A) The correct answer is $A$.$1$. $K_{3}[Fe(CN)_{6}]$: The central metal ion is $Fe^{3+}$,which has a $[Ar] 3d^{5}$ configuration as a free ion.In th

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$K_{3}[Fe(CN)_{6}]$ and $K_{3}[CoF_{6}]$

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(A) The correct answer is $A$.$1$. $K_{3}[Fe(CN)_{6}]$: The central metal ion is $Fe^{3+}$,which has a $[Ar] 3d^{5}$ configuration as a free ion.In the strong ligand field of $CN^{-}$,the $d$-orbitals split to become $t_{2g}^{5} e_{g}^{0}$.There is one unpaired electron,therefore it is paramagnetic.$2$. $K_{3}[CoF_{6}]$: The central metal ion is $Co^{3+}$.The free metal ion configuration is $[Ar] 3d^{6}$.In the weak ligand field of $F^{-}$,the $d$-orbitals split to become $t_{2g}^{4} e_{g}^{2}$.There are four unpaired electrons,therefore it is paramagnetic.$3$. In $K_{4}[Fe(CN)_{6}]$,the central ion is $Fe^{2+}$ $(d^{6})$,and in $[Co(NH_{3})_{6}]Cl_{3}$,the central ion is $Co^{3+}$ $(d^{6})$.Both $CN^{-}$ and $NH_{3}$ are strong field ligands,so $\Delta_{0} > P$.Thus,the $d^{6}$ configuration splits to $t_{2g}^{6} e_{g}^{0}$.There are no unpaired electrons,therefore both are diamagnetic.

Among the following,the pair of paramagnetic complexes is

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