Question

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

• A. $\ce{[V(CN)6]^{4-} > [Cr(NH3)6]^{2+} > [Ru(CN)6]^{3-} > [Fe(CN)6]^{4-}}$
• B. $\ce{ [Fe(CN)6]^{4-} > [V(CN)6]^{4-} > [Cr(NH3)6]^{2+} > [Ru(CN)6]^{3-} }$
• C. $\ce{[Ru(CN)6]^{3-} > [Fe(CN)6]^{4-} > [V(CN)6]^{4-} > [Cr(NH3)6]^{2+}}$
• D. $\ce{ [Cr(NH3)6]^{2+} > [Ru(CN)6]^{3-} > [Fe(CN)6]^{4-} > [V(CN)6]^{4-}}$

Answer 1

Answer: (A)

$\ce{CN^{-} \; \& \; NH_3}$ are strong field ligands.
$\ce{ V^{2+} -> [Ar] 3d^3 -> 3 \; unpaired \; electrons}$
$\ce{ Cr^{2+} -> [Ar] 3d^4 -> 2 \; unpaired\; electrons }$
$\ce{ Ru^{3+} -> [Kr] 4d^5 -> 1 \; unpaired \;electron }$
$\ce{ Fe^{2+} -> [Ar] 3d^6 -> 0 \; unpaired \; electrons}$