This compound can be identified and named as follows
$\text{Cr}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\text{C}{\text{l}}_{\mathrm{n}}$
As ${\left({\mathrm{\mu }}_{\text{complex}}\right)}_{\mathrm{s}\mathrm{p}\mathrm{i}\mathrm{n}}=3.8\mathrm{ }\mathrm{B}.\mathrm{M}.\mathrm{ }=\sqrt{\mathrm{n}(\mathrm{n}+2)}$
On solving $\mathrm{n}=3$
Oxidation No. of $\text{Cr}=+3$
As compound also shows Geometrical isomerism so it will be $[\text{Cr}\left({\mathrm{H}}_2\mathrm{O}{)}_4\text{C}{\text{l}}_2\right]\text{Cl}.\text{2}{\text{H}}_2\mathrm{O}$
It is named as $\to$ Tetraaquadichloridochromium(III)chloride dihydrate