Question

Arrange the following species in the increasing order of their magnetic moment.
$\left(I\right)VC{l}_3\left(II\right)VOS{O}_4\left(III\right)N{a}_{3}V{O}_4\left(IV\right)\left[V\right(H_{2}O{)}_6]S{O}_4.H_{2}O$

• A. $III<II<I<IV$
• B. $III<IV<II<I$
• 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:
$VC{l}_3;V^{3+};[Ar{]}^{18}3d^2=2$ unpaired electrons
$V{O}^{2+};V^{4+};[Ar{]}^{18}3d^1=1$ unpaired electron
$[V{O}_4{]}^{3-};V^{5+};[Ar{]}^{18}3d^0=$ no unpaired electron
$\left[V\right(H_{2}O{)}_6{]}^{2+};V^{2+};[Ar{]}^{18}3d^3=3$ unpaired electrons
Hence, the correct order is $III<II<I<IV$.