Two isomeric compounds, \(A: [Co(NH_3)_5Cl]SO_4\) and \(B: [Co(NH_3)_5SO_4]Cl\) can be distinguished using which of the following?

aq. \(BaCl_2\) solution

The correct answer is option 1. aq. \(BaCl_2\) solution.

Let us explore the ionization isomers \(A: [Co(NH_3)_5Cl]SO_4\) and \(B: [Co(NH_3)_5SO_4]Cl\) in detail and how they can be distinguished using different reagents.

Compound \(A: [Co(NH_3)_5Cl]SO_4\)

In this compound, the sulfate ion (\(SO_4^{2-}\)) is the counterion outside the coordination sphere, while the chloride ion (\(Cl^-\)) is inside the coordination sphere.

Upon dissolving in water, it dissociates as follows:

\([Co(NH_3)_5Cl]SO_4 \rightarrow [Co(NH_3)_5Cl]^+ + SO_4^{2-}\)

The solution contains free sulfate ions (\(SO_4^{2-}\)).

Compound \(B: [Co(NH_3)_5SO_4]Cl\)

Here, the chloride ion (\(Cl^-\)) is the counterion outside the coordination sphere, while the sulfate ion (\(SO_4^{2-}\)) is inside the coordination sphere.

Upon dissolving in water, it dissociates as follows:

\([Co(NH_3)_5SO_4]Cl \rightarrow [Co(NH_3)_5SO_4]^+ + Cl^-\)

The solution contains free chloride ions (\(Cl^-\)).

Distinguishing Using Reagents

1. Aqueous \(BaCl_2\) Solution

\(BaCl_2\) is commonly used to test for the presence of sulfate ions (\(SO_4^{2-}\)) because it reacts with sulfate to form a white precipitate of barium sulfate (\(BaSO_4\)).

For Compound A:

\(SO_4^{2-}\) is present in the solution.

When \(BaCl_2\) is added, a white precipitate of \(BaSO_4\) will form:

\(SO_4^{2-} + BaCl_2 \rightarrow BaSO_4 \ (\text{white precipitate}) + 2Cl^- \)

For Compound B:

There are no free \(SO_4^{2-}\) ions in the solution.

When \(BaCl_2\) is added, no reaction occurs, so no precipitate forms.

\(BaCl_2\) solution will produce a white precipitate with compound A but not with compound B, making it a reliable way to distinguish between the two.

2. Nessler's Reagent

Nessler's reagent is used to detect ammonia (\(NH_3\)) or ammonium ions (\(NH_4^+\)), and it forms a brown precipitate or a yellow coloration in the presence of these ions. Since both compounds involve \(NH_3\) as ligands but do not release free \(NH_3\) or \(NH_4^+\) in solution, Nessler's reagent is irrelevant for distinguishing these isomers.

3. Uric Ammonium Nitrate Reagent

This reagent is more commonly associated with biochemical assays or tests for uric acid, and is not applicable to these coordination compounds.

4. Passing \(H_2S\) Gas Through Their Aqueous Solution

\(H_2S\) gas is typically used to precipitate metal ions as sulfides (like \(PbS\), \(ZnS\), etc.) in qualitative analysis. Neither \(A\) nor \(B\) contains metal ions that would precipitate with \(H_2S\), so this method is not useful for distinguishing these compounds.

Conclusion

The most effective way to distinguish between these ionization isomers is by using aqueous \(BaCl_2\) solution:

Compound A will react with \(BaCl_2\) to form a white precipitate of \(BaSO_4\).

Compound B will not react with \(BaCl_2\), and no precipitate will form.

Thus, the correct answer is Option 1: aq. \(BaCl_2\) solution.