Problem 85 - Chemistry and Chemical Reactivity

Question

Two different coordination compounds containing one cobalt(III) ion, five ammonia molecules, one bromide ion, and one sulfate ion exist. The dark violet form (A) gives a precipitate upon addition of aqueous \(\mathrm{BaCl}_{2}\). No reaction is seen upon addition of aqueous \(\mathrm{BaCl}_{2}\) to the violet- red form \((\mathrm{B})\) Suggest structures for these two compounds, and write a chemical equation for the reaction of (A) with aqueous \(\mathrm{BaCl}_{2}\).

Step-by-Step Solution

Answer

A: [Co(NH₃)₅Br]SO₄; B: [Co(NH₃)₅(SO₄)]Br. A reacts with BaCl₂ to form BaSO₄.

Step 1: Identify Compound Components

Each compound contains one cobalt(III) ion, five ammonia molecules, one bromide ion, and one sulfate ion. These components can be rearranged to form different coordination compounds.

Step 2: Analyze the Dark Violet Form (A)

The dark violet form (A) gives a precipitate with \(\mathrm{BaCl}_{2}\). This suggests that the sulfate ion is not bound to the cobalt ion and is free to react with \(\mathrm{Ba}^{2+}\) to form \(\mathrm{BaSO}_{4}\). Therefore, the chemical formula for A is likely \([\mathrm{Co(NH}_3)_5\mathrm{Br}]\mathrm{SO}_4\).

Step 3: Analyze the Violet-Red Form (B)

No reaction occurs with \(\mathrm{BaCl}_{2}\) for the violet-red form (B), indicating that the sulfate ion is coordinated to the cobalt. Therefore, a probable structure is \([\mathrm{Co(NH}_3)_5(\mathrm{SO}_4)]\mathrm{Br}\), where the sulfate is within the coordination sphere.

Step 4: Write the Reaction of Compound (A)

For compound A \(\mathrm{BaCl}_{2}\) reacting with \(\mathrm{Ba}^{2+}\) and \(\mathrm{SO}_4^{2-}\) will give the precipitate \(\mathrm{BaSO}_{4}\): \[\mathrm{[Co(NH}_3)_5Br]SO_4_{(aq)} + \mathrm{BaCl}_2_{(aq)} \rightarrow [Co(NH}_3)_5Br]Cl_{(aq)} + \mathrm{BaSO}_4_{(s)}\]

Step 5: Summary

Compound A likely has the formula \([\mathrm{Co(NH}_3)_5\mathrm{Br}]\mathrm{SO}_4\), in which sulfate ions are free and can react with \(\mathrm{BaSO}_4\). Compound B is likely \([\mathrm{Co(NH}_3)_5(\mathrm{SO}_4)]\mathrm{Br}\), where sulfate remains in the coordination sphere.

Key Concepts

Cobalt(III) Complexes

Cobalt(III) complexes are coordination compounds that consist of a cobalt ion in the +3 oxidation state. This ion typically coordinates with ligands, which are molecules or ions that donate electron pairs to the central metal ion. In our scenario, the cobalt(III) ion is surrounded by ammonia (\(\mathrm{NH}_3\)) ligands, a bromide ion (\(\mathrm{Br}^-\)), and a sulfate ion (\(\mathrm{SO}_4^{2-}\)).

Coordination compounds form when these ligands attach to a central metal ion. This attachment is due to coordinate covalent bonds, where the ligand provides both electrons.

The structure and properties of these complexes can vary significantly, resulting in different stability, solubility, and reactivity properties. The two cobalt(III) complexes in this exercise both have five ammonia molecules attached, showcasing how different arrangements of the same components can lead to distinct chemical behaviors.

Precipitation Reaction

A precipitation reaction occurs when solutions of ionic substances are mixed together, and an insoluble solid, known as a precipitate, forms. In the case of complex (A) from this exercise, such a reaction occurs when aqueous \(\mathrm{BaCl}_2\) is added.

Precipitation reactions are crucial in chemistry for separating and identifying compounds. They rely on the fact that certain combinations of ions form insoluble products.

In solution, ions move randomly, but when certain ion pairs meet, they form neutral molecules that aggregate into a solid. This solid can then be separated from the solution, often as a means of purifying a compound or removing unwanted ions. This characteristic is utilized in complex (A) to identify the sulfate ions not bound within the coordination sphere.

BaCl2 Reaction

The reaction with \(\mathrm{BaCl}_2\) is a classic test for the presence of free sulfate ions. When compound (A) is tested with \(\mathrm{BaCl}_2\), a precipitate forms. This signifies a chemical interaction where the sulfate ion, unbound in the coordination sphere, reacts with \(\mathrm{Ba}^{2+}\) ions to create \(\mathrm{BaSO}_4\).

\(\mathrm{BaSO}_4\) is insoluble in water, hence it forms a precipitate. The chemical equation for this process is:

• \([\mathrm{Co(NH}_3)_5\mathrm{Br}]\mathrm{SO}_4 + \mathrm{BaCl}_2 \rightarrow [\mathrm{Co(NH}_3)_5\mathrm{Br}]{Cl} + \mathrm{BaSO}_4 (\text{s})\)

This reaction helps in confirming the arrangement of ions in compound (A) and shows that sulfate ions are outside the primary sphere of coordination. Compound (B), conversely, shows no precipitate as the sulfate ion is bound within the coordination sphere, preventing formation of \(\mathrm{BaSO}_4\).

Chemical Structures

Understanding the chemical structures of complexes is essential to predicting their chemical behavior and reactions. In coordination chemistry, particularly with cobalt(III), structural arrangement can dictate the reactivity towards other compounds.

These complexes have a core structure where ligands surround a central metal in a specific geometric arrangement, often octahedral for cobalt(III). The way ligands like ammonia, bromide, and sulfate are arranged in the coordination sphere affects the compound's properties.

The structural differences in compounds (A) and (B) are primarily due to whether the sulfate ion is inside or outside this coordination sphere. These small changes in structure can significantly alter their chemical reactions, as seen in their different responses to \(\mathrm{BaCl}_2\).

Sulfate Coordination

In coordination chemistry, the position of the sulfate ion plays a crucial role in determining the properties of the complex. In the exercise, we see how sulfate can either be a part of the coordination sphere or remain as an external ion.

When sulfate is coordinated to cobalt as observed in compound (B), it occupies one of the coordination sites typically taken by a ligand like ammonia. This internal binding prevents it from reacting with \(\mathrm{Ba}^{2+}\) ions, hence no precipitate forms.

In compound (A), sulfate is free in solution. This freedom allows it to interact with barium ions from \(\mathrm{BaCl}_2\), resulting in the formation of a precipitate. The coordination sphere's configuration strongly influences the compound's overall reactivity and can be a beneficial property for specific applications or reactions.