The correct order of the spin-only magnetic moments of the following complexes is :
(I) $\left[\mathrm{Cr}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]{\mathrm{Br}}_2$
(II) ${\mathrm{Na}}_4\left[\mathrm{Fe}(\mathrm{CN}{)}_6\right]$
(III) ${\mathrm{Na}}_3\left[\mathrm{Fe}{\left({\mathrm{C}}_2{\mathrm{O}}_4\right)}_3\right]\left({\mathrm{\Delta }}_0>\mathrm{P}\right)$

$$\begin{gathered} \left(I\right) {\left[Cr{\left(H_{2}O\right)}_6\right]}^{2+} C{r}^{2+}= \left[Ar\right]3d^4 \\ {H}_{2}O\hspace{0.17em}is a weak field ligand \end{gathered}$$

4 unpaired electrons 

Magnetic moment $\mu =\sqrt{24 }= 4.89BM$

$$\begin{gathered} \left(II\right)\hspace{0.17em}{\left[Fe{\left(CN\right)}_6\right]}^{4-} F{e}^{2+}=\left[Ar\right]3d^6 \\ C{N}^{-} is a strong field ligand \end{gathered}$$

Unpaired electrons = 0 

magnetic moment = 0 

$$\begin{gathered} \left(III\right)\hspace{0.17em}{\left[Fe{\left(C_2{O}_4\right)}_3\right]}^{3-} F{e}^{3+}= \left[Ar\right]3d^5 \\ As {\Delta }_0 > P \end{gathered}$$

Nummber of unpaired electron = 1 

Magnetic moment $\mu =\sqrt{1(1+2) }=1.732BM$

$$\begin{gathered} \left(IV\right)\hspace{0.17em}{\left(EtN\right)}^{+}\left(CoC{l}_4\right) \\ C{o}^{2+}=\left[Ar\right]3d^7 \end{gathered}$$

Number of unpaired electrons=3 

Magnetic moment $\mu =\sqrt{3(3+2) }=3.87BM$