Idea of the test reagents

• $BaCl_2$ gives a white precipitate if free $SO_4^{2-}$ ions are present in solution, due to formation of: $Ba^{2+} + SO_4^{2-} \rightarrow BaSO_4(s)$
• $AgNO_3$ gives a precipitate if free halide ions like $Cl^-$ or $Br^-$ are present: $Ag^+ + Cl^- \rightarrow AgCl(s)$ $Ag^+ + Br^- \rightarrow AgBr(s)$

Therefore, the correct compound must have:

• $SO_4^{2-}$ as counter ion (outside coordination sphere), and
• no free $Cl^-$ or $Br^-$ ions outside the coordination sphere.

Check each option

Option A: $[Co(NH_3)_5Br]SO_4$

• Coordination sphere: $[Co(NH_3)_5Br]^{2+}$
• Counter ion: $SO_4^{2-}$
• So in solution, it gives free $SO_4^{2-}$.
• The $Br^-$ is coordinated inside the complex, so it is not free.

Hence:

• With $BaCl_2$: forms white $BaSO_4$ precipitate
• With $AgNO_3$: no precipitate due to no free halide ion

So, A satisfies the condition.

Option B: $[Co(NH_3)_5SO_4]Br$

• Here $SO_4^{2-}$ is coordinated inside the complex.
• $Br^-$ is the counter ion, so free $Br^-$ is present in solution.

Hence:

• No free $SO_4^{2-}$ for $BaCl_2$
• $AgNO_3$ will give $AgBr$ precipitate

So, B is incorrect.

Option C: $[Pt(NH_3)_4Cl_2]Br_2$

• $Cl^-$ ions are coordinated inside the complex.
• $Br^-$ ions are outside as counter ions.

Hence:

• No free $SO_4^{2-}$, so no $BaSO_4$
• Free $Br^-$ gives $AgBr$ with $AgNO_3$

So, C is incorrect.

Option D: $[Pt(NH_3)_4Br_2]Cl_2$

• $Br^-$ ions are coordinated inside the complex.
• $Cl^-$ ions are outside as counter ions.

Hence:

• No free $SO_4^{2-}$, so no white precipitate with $BaCl_2$
• Free $Cl^-$ gives $AgCl$ with $AgNO_3$

So, D is incorrect.

Final conclusion

The only compound that gives a white precipitate with $BaCl_2$ but not with $AgNO_3$ is: $[Co(NH_3)_5Br]SO_4$

Comparison with stored answer

Stored correct answer: A

Our derived answer: A

They agree.