Question

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

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

Answer 1

The correct option is C $\left(I\right)>\left(IV\right)>\left(III\right)>\left(II\right)$

Complex (I) has the central metal ion as $C{r}^{2+}$with weak field ligands. Configuration of $C{r}^{2+}=\left[Ar\right]3d^4$
As weak field ligands are present, pairing does not take place. There will be 4 unpaired electrons and hence the magnetic moment $\sqrt{24}\text{B.M}$

Complex (II) has the central metal ion as $F{e}^{2+}$ with strong field ligands. Configuration of $F{e}^{2+}=\left[Ar\right]3d^6$
Strong field ligands will pair up all the electrons and hence the magnetic moment will be zero.

Complex (III) has the central metal ion as $F{e}^{3+}$ with strong field ligands.
Configuration of $F{e}^{3+}=\left[Ar\right]3d^5$
As it is given that ${\Delta }_0>P$ hence the electrons will pair up but as we have a $\left[Ar\right]3d^5$ configuration, one electron will remain unpaired and hence the magnetic moment will be $\sqrt{3}\text{B.M}$.

Complex (IV) has the central metal ion as $C{o}^{2+}$ with weak field ligands.
Configuration of $C{o}^{2+}=\left[Ar\right]3d^7$
As weak field ligands are present no pairing can occur. There will be 3 unpaired electrons and hence the magnetic moment $\sqrt{15}\text{B.M}$.