Complexes $\left[\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_5{\mathrm{SO}}_4\right]\mathrm{Br}$  and $\left[\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_5{\mathrm{Br}}^3{\mathrm{SO}}_4\right.$ can be distinguished by

The complexes  $\left[\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_5{\mathrm{SO}}_4\right]\mathrm{Br}$ and $\left[\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_5{\mathrm{Br}}^3{\mathrm{SO}}_4\right.$ can be used to differentiate include  ${\mathrm{AgNO}}_3$ 

Sulfate ions are produced by  $\left[\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_5\mathrm{Br}\right]{\mathrm{SO}}_4$ in the solution, which react with  ${\mathrm{BaCI}}_2$ to create white  ppt of ${\mathrm{BaSO}}_4$ 

$\left[\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_5{\mathrm{Br}}^{-{\mathrm{SO}}_4}+{\mathrm{BaCl}}_2\to \left[\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_5{\mathrm{Br}}^3\right]{\mathrm{Cl}}_2+\underset{\text{ white }}{{\mathrm{BaSO}}_4}\downarrow \right.$   

In the solution $\left[\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_5{\mathrm{SO}}_4\right]\mathrm{Br}$ produces bromide ions that react with ${\mathrm{AgNO}}_3$ to produce yellow ppt  of $\mathrm{AgBr}.$

$\left[\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_5{\mathrm{SO}}_4\right]\mathrm{Br}+{\mathrm{AgNO}}_3\to \left[\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_5{\mathrm{SO}}_4\right]{\mathrm{NO}}_3+\underset{\text{ yellow }}{\mathrm{AgBr}}\downarrow$

Thus, the option D is correct.