This compound can be identified and named as follows
$\mathrm{C}\mathrm{r}({\mathrm{H}}_2\mathrm{O}{)}_6\mathrm{B}{\mathrm{r}}_{\mathrm{n}}$
As $({\mathrm{\mu }}_{\mathrm{c}\mathrm{o}\mathrm{m}\mathrm{p}\mathrm{l}\mathrm{e}\mathrm{x}}{)}_{\mathrm{s}\mathrm{p}\mathrm{i}\mathrm{n}}=$ 3.8 B.M. $=\sqrt{\mathrm{n}(\mathrm{n}+2)}$
On solving $\mathrm{n}=3$
Hence the complex is $\left[\mathrm{C}\mathrm{r}\right({\mathrm{H}}_2\mathrm{O}{)}_6]\mathrm{B}{\mathrm{r}}_3$
oxidation number of Cr $=+3.$
As compound also shows geometrical isomerism so it will be $\left[\mathrm{C}\mathrm{r}\right({\mathrm{H}}_2\mathrm{O}{)}_4\mathrm{B}{\mathrm{r}}_2]\mathrm{B}\mathrm{r}.2{\mathrm{H}}_2\mathrm{O}$
It is named as $\to$ Tetraaquadibromido chromium (III) bromide dihydrate