Question

The correct order of the spin-only magnetic moments of the following complexes is : $(I)\, [Cr(H_2O)_6]Br_2$ $(II)\, Na_4[Fe(CN)_6]$ $(III)\, Na_3[Fe(C_2O_4)_3] (\Delta_0 > P)$ $(IV)\, (Et_4N)_2[CoCl_4]$

• A. (III) > (I) > (II) > (IV)
• B. (III) > (I) > (IV) > (II)
• C. (I) > (IV) > (III) > (II)
• D. (II) $\approx$ (I) > (IV) > (III)

Answer 1

The Correct Option is C

$I \left[Cr\left(H_{2}O\right)_{6}\right]^{2+}$
$Cr^{+2} \Rightarrow\left[Ar\right]3d^{4}$
$H_{2}O\to$ Weak field ligand
Unpaired $e^{-}=4$
Magnetic moment=$\sqrt{24} BM$
$=4.89 BM$
II $\left[Fe\left(CN\right)_{6}\right]^{4-}$
$Fe^{+2} \Rightarrow \left[Ar\right]3d^{6}$
$CN^{-} \to $ Strong field ligand
Unpaired $e^{-}=0$
Magnetic moment = 0 BM=0 BM
III $\left[Fe\left(C_{2}O_{4}\right)_{3}\right]^{3-}$
$Fe^{+3} \Rightarrow \left[Ar\right]3d^{5}$
$As\, \Delta_{0} >\,P$
$Unpaired \, e^{-}=1$
$Magnetic\,moment=\sqrt{3} BM =1.73 BM$
$IV \left(Et_{4}N\right)^{+} \left[CoCl_{4}\right]^{2-}$
$Co^{+2} \Rightarrow\left[Ar\right]3d^{7}$
$\left[CoCl_{4}\right]^{-2}=$
$Unpaired \,electrons = 3$
$Magnetic \,moment =\sqrt{15} BM =3.87 BM$
$Hence\, order \,of \,magnetic\, moment\,is\,$
$I > IV > III > II$