Question

Compounds with spin-only magnetic moment equivalent to five unpaired electrons are

• A. $K_{4}\left[Mn\left(CN\right)_{6}\right]$
• B. $\left[Fe\left(H_{2}O\right)_{6}\right]Cl_{3}$
• C. $K_{3}\left[FeF_{6}\right]$
• D. $K_{4}\left[MnF_{6}\right]$

Answer 1

Answer: (D)

$K_{4}\left[MnF_{6}\right]$

Answer 2

(A) In $K_{4}$ $\left[Mn\left(CN\right)_{6}\right],$Mn has +2 O.S. and $Mn^{+2} =\left[Ar\right]3d^{5}4S^{0},$ and $CN^{-}$is strong field ligand. Pairing takes place and the complex has only one unpaired electron.
(B) In $\left[Fe\left(H_{2}O\right)_{6}\right]Cl_{3},$Fe has +3 O.S. and $Fe^{+3}=\left[Ar\right]3d^{5}4S^{0},$ the number of unpaired $e^{-}$ = 5
($H_{2}O$ is a weak field ligand thus no pairing)
(C) In $K_{3}\left[FeF_{6}\right],$Fe has +3 O.S. and $Fe^{+3}=\left[Ar\right]3d^{5}4S^{0},$thus number of unpaired $e^{-}$ =5
(F- is a weak field ligand thus no pairing)
(D) In $K_{4}\left[MnF_{6}\right],$ Mn has O.S. of +2 thus $Mn^{+2}=\left[Ar\right]3d^{5}4S^{0}$is number of unpaired $e^{-}$=5
(F- is a weak field ligand thus no pairing)