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Chemistry Question on d -and f -Block Elements

[Fe(CN)6]4−,[V(CN)6]4−,[Cr(NH3)6]2+ {[Fe(CN)6]^{4-}, [V(CN)6]^{4-}, [Cr(NH3)6]^{2+}} and [Ru(CN)6]3− {[Ru(CN)6]^{3-}} Find the decreasing order of spin only magnetic moment

A

\ce[V(CN)6]4−>[Cr(NH3)6]2+>[Ru(CN)6]3−>[Fe(CN)6]4−\ce{[V(CN)6]^{4-} > [Cr(NH3)6]^{2+} > [Ru(CN)6]^{3-} > [Fe(CN)6]^{4-}}

B

\ce[Fe(CN)6]4−>[V(CN)6]4−>[Cr(NH3)6]2+>[Ru(CN)6]3−\ce{ [Fe(CN)6]^{4-} > [V(CN)6]^{4-} > [Cr(NH3)6]^{2+} > [Ru(CN)6]^{3-} }

C

\ce[Ru(CN)6]3−>[Fe(CN)6]4−>[V(CN)6]4−>[Cr(NH3)6]2+\ce{[Ru(CN)6]^{3-} > [Fe(CN)6]^{4-} > [V(CN)6]^{4-} > [Cr(NH3)6]^{2+}}

D

\ce[Cr(NH3)6]2+>[Ru(CN)6]3−>[Fe(CN)6]4−>[V(CN)6]4−\ce{ [Cr(NH3)6]^{2+} > [Ru(CN)6]^{3-} > [Fe(CN)6]^{4-} > [V(CN)6]^{4-}}

Answer

\ce[V(CN)6]4−>[Cr(NH3)6]2+>[Ru(CN)6]3−>[Fe(CN)6]4−\ce{[V(CN)6]^{4-} > [Cr(NH3)6]^{2+} > [Ru(CN)6]^{3-} > [Fe(CN)6]^{4-}}

Explanation

Solution

CN−  &  NH3 {CN^{-} \; \& \; NH_3} are strong field ligands. V2+−>[Ar]3d3−>3  unpaired  electrons { V^{2+} -> [Ar] 3d^3 -> 3 \; unpaired \; electrons} Cr2+−>[Ar]3d4−>2  unpaired  electrons { Cr^{2+} -> [Ar] 3d^4 -> 2 \; unpaired\; electrons } Ru3+−>[Kr]4d5−>1  unpaired  electron { Ru^{3+} -> [Kr] 4d^5 -> 1 \; unpaired \;electron } Fe2+−>[Ar]3d6−>0  unpaired  electrons { Fe^{2+} -> [Ar] 3d^6 -> 0 \; unpaired \; electrons}