Question

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

• A. $K_4\left[Mn{\left(CN\right)}_6\right]$
• B. $\left[Fe{\left(H_{2}O\right)}_6\right]C{l}_3$
• C. $K_3\left[Fe{F}_6\right]$
• D. $K_4\left[Mn{F}_6\right]$

Answer 1

Answer: (B), (C), (D)

(A) In $K_4$$\left[Mn{\left(CN\right)}_6\right],$Mn has +2 O.S. and $M{n}^{+2}=\left[Ar\right]3d^{5}4S^0,$ and $C{N}^{-}$is strong field ligand. Pairing takes place and the complex has only one unpaired electron.
(B) In $\left[Fe{\left(H_{2}O\right)}_6\right]C{l}_3,$Fe has +3 O.S. and $F{e}^{+3}=\left[Ar\right]3d^{5}4S^0,$ the number of unpaired $e^{-}$ = 5
($H_{2}O$ is a weak field ligand thus no pairing)
(C) In $K_3\left[Fe{F}_6\right],$Fe has +3 O.S. and $F{e}^{+3}=\left[Ar\right]3d^{5}4S^0,$thus number of unpaired $e^{-}$ =5
(F- is a weak field ligand thus no pairing)
(D) In $K_4\left[Mn{F}_6\right],$ Mn has O.S. of +2 thus $M{n}^{+2}=\left[Ar\right]3d^{5}4S^0$is number of unpaired $e^{-}$=5
(F- is a weak field ligand thus no pairing)