Question

Aluminum forms the complex ions \(\mathrm{AlCl}_{4}^{-}\) and \(\mathrm{AlF}_{6}^{3-}\). Describe the shapes of these ions. \(\mathrm{AlCl}_{6}^{3-}\) does not form. Why? (Hint: Consider the relative sizes of \(\mathrm{Al}^{3+}, \mathrm{F}^{-},\) and \(\mathrm{Cl}^{-}\) ions.)

Short answer

AlCl4^- is tetrahedral, AlF6^3- is octahedral; AlCl6^3- doesn't form due to steric hindrance.

Step-by-step solution

Understanding the Complex Ion Shapes

Aluminum can form different complex ions depending on how many ligands are coordinated to it. For \( \mathrm{AlCl}_{4}^{-} \), four chloride ions are coordinated to the aluminum ion. This results in a tetrahedral shape due to the arrangement of four ligands around the metal center. For \( \mathrm{AlF}_{6}^{3-} \), six fluoride ions are coordinated, resulting in an octahedral shape. This is because the coordination number is six, which typically forms an octahedral geometry around the central metal atom.

Examining Ion Size and Coordination

The size of the central metal ion and the ligands affects the formation of complexes. The \( \mathrm{Al}^{3+} \) ion is relatively small, and fluoride ions are smaller than chloride ions. This allows \( \mathrm{F}^{-} \) to fit around the \( \mathrm{Al}^{3+} \) in a way that \( \mathrm{Cl}^{-} \) cannot, due to steric hindrance, if attempting to coordinate six \( \mathrm{Cl}^{-} \) ions.

Reasoning for Non-formation of \( \mathrm{AlCl}_{6}^{3-} \)

\( \mathrm{AlCl}_{6}^{3-} \) does not form because the chloride ion \( \mathrm{Cl}^{-} \) is larger than the fluoride ion \( \mathrm{F}^{-} \), leading to steric hindrance around the relatively small \( \mathrm{Al}^{3+} \) ion that prevents six chloride ions from fitting in an octahedral arrangement. Thus, only the \( \mathrm{AlCl}_{4}^{-} \) tetrahedral complex can form.

Key concepts

Tetrahedral Geometry

Tetrahedral geometry occurs when a central atom is surrounded by four ligands or substituents arranged in a space that forms the corners of a tetrahedron, similar to a tripod stand. Each ligand pushes itself away from the others equally, and this balance results in a compact symmetrical shape.
This geometry is essential in coordination chemistry and can be observed in the complex ion \( \mathrm{AlCl}_{4}^{-} \). Here, the aluminum atom is at the center, and the four chloride ions surround it, forming a tetrahedron.

• The bond angles in a tetrahedral geometry are approximately \( 109.5^{\circ} \).
• It is a saturated structure, meaning no additional ligands can fit without enlarging the central metal ion or changing the ligands.
• This shape provides a stable arrangement for the coordination of four ligands.

The tetrahedral geometry for \( \mathrm{AlCl}_{4}^{-} \) is favored because it allows maximum space utilization, minimizing repulsion between the larger \( \mathrm{Cl}^{-} \) ions, leading to a more stable complex.

Octahedral Geometry

Octahedral geometry is a configuration where a central atom is coordinated by six ligands arranged at the vertices of what resembles an octahedron. In the case of \( \mathrm{AlF}_{6}^{3-} \), the aluminum ion achieves a coordination number of six by interacting with six fluoride ions.
The small size of \( \mathrm{F}^{-} \) ions compared to \( \mathrm{Cl}^{-} \) ions allows this arrangement to be possible without significant steric interference.

• In an octahedral geometry, all bond angles are \( 90^{\circ} \), offering symmetry and uniformity.
• This structure is most common for metal ions with six ligands due to its efficient space utilization.
• The shape provides good stability because it can accommodate many small ligands around a small central ion like \( \mathrm{Al}^{3+} \).

The size compatibility between \( \mathrm{Al}^{3+} \) and smaller ligands such as \( \mathrm{F}^{-} \) enables the formation of the octahedral complex \( \mathrm{AlF}_{6}^{3-} \), despite the small size of the central ion.

Steric Hindrance

Steric hindrance is a concept in chemistry where the size of atoms or groups in a molecule prevents chemical reactions that would occur if the molecules were less crowded.
This effect greatly influences the geometry of complex ions, particularly in the formation of \( \mathrm{AlCl}_{4}^{-} \) and the absence of \( \mathrm{AlCl}_{6}^{3-} \).When ligands are large or close together around a central atom, they may repel each other due to their electron clouds, leading to steric hindrance.
When \( \mathrm{Al}^{3+} \) attempts to form \( \mathrm{AlCl}_{6}^{3-} \), the large chloride ions encounter steric hindrance as there's simply not enough space around the small \( \mathrm{Al}^{3+} \) ion to accommodate six of them.

• Steric hindrance can restrict the number of ligands around a central atom.
• It affects the stability of complexes, making some configurations impossible, like the \( \mathrm{AlCl}_{6}^{3-} \).
• Understanding steric effects is crucial for predicting the structure and reactivity of molecules.

In summary, steric hindrance prevents \( \mathrm{AlCl}_{6}^{3-} \) from forming because the chloride ions' size and repulsion forces prohibit an octahedral arrangement around the aluminum ion.