Question

Arrange the following species in the increasing order of their magnetic moment.
(I) $VC{l}_3$ (II) $VOS{O}_4$ (III) $N{a}_{3}V{O}_4$ (IV)

$\left[V{\left(H_{2}O\right)}_6\right]S{O}_4.H_{2}O$

• A. III < II < I < IV
• B. III < IV < III < I
• C. II < III < I < IV
• D. IV < I < II < III

Answer 1

The correct option is A III < II < I < IV

The compound which has the maximum number of unpaired electrons will have higher magnetic moment.
The complexes and the number of unpaired electrons present in them are given below:

$VC{l}_3;V^{3+};\left[Ar\right]3d^2$ = 2 unpaired electrons

$VO{}^{2+};V^{4+};\left[Ar\right]3d^1$ = 1 unpaired electron
${\left[V{O}_4\right]}^{3-}$; $V^{5+};\left[Ar\right]3d^0$ = no unpaired electron
${\left[V{\left(H_{2}O\right)}_6\right]}^{2+}$; $V^{2+};\left[Ar\right]3d^2$ = 3 unpaired electrons
Hence, the correct order is III < II < I < IV