Question

The correct order of the spin-only magnetic moments of the following complexes is :
$(I)[Cr(H_{2}O{)}_6]B{r}_2$
$(II)N{a}_4[Fe(CN{)}_6]$
$(III)N{a}_3[Fe(C_2{O}_4{)}_3]({\Delta }_0>P)$
$(IV)(E{t}_{4}N{)}_2[CoC{l}_4]$

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

Answer 1

Answer: (C)

$I{\left[Cr{\left(H_{2}O\right)}_6\right]}^{2+}$
$C{r}^{+2}\Rightarrow \left[Ar\right]3d^4$
$H_{2}O\to$ Weak field ligand
Unpaired $e^{-}=4$
Magnetic moment=$\sqrt{24}BM$
$=4.89BM$
II ${\left[Fe{\left(CN\right)}_6\right]}^{4-}$
$F{e}^{+2}\Rightarrow \left[Ar\right]3d^6$
$C{N}^{-}\to$ Strong field ligand
Unpaired $e^{-}=0$
Magnetic moment = 0 BM=0 BM
III ${\left[Fe{\left(C_2{O}_4\right)}_3\right]}^{3-}$
$F{e}^{+3}\Rightarrow \left[Ar\right]3d^5$
$As{\Delta }_0>P$
$Unpaired{e}^{-}=1$
$Magneticmoment=\sqrt{3}BM=1.73BM$
$IV{\left(E{t}_{4}N\right)}^{+}{\left[CoC{l}_4\right]}^{2-}$
$C{o}^{+2}\Rightarrow \left[Ar\right]3d^7$
${\left[CoC{l}_4\right]}^{-2}=$
$Unpairedelectrons=3$
$Magneticmoment=\sqrt{15}BM=3.87BM$
$Henceorderofmagneticmomentis$
$I>IV>III>II$