Question

Describe the change in hybridization (if any) of the \(\mathrm{Al}\) atom in the following reaction: $$ \mathrm{AlCl}_{3}+\mathrm{Cl}^{-} \longrightarrow \mathrm{AlCl}_{4}^{-} $$

Short answer

Hybridization changes from \(\mathrm{sp}^2\) to \(\mathrm{sp}^3\).

Step-by-step solution

Understand Initial Hybridization

In the compound \(\mathrm{AlCl}_3\), the aluminum (\(\mathrm{Al}\)) atom forms three covalent bonds with chlorine (\(\mathrm{Cl}\)) atoms. Initially, aluminum exhibits \(\mathrm{sp}^2\) hybridization to form a trigonal planar geometry with bond angles of approximately 120 degrees.

Analyze the Reaction

In the reaction, \(\mathrm{Cl}^-\) (chloride ion) donates an electron pair to form an additional bond with the aluminum atom, resulting in the compound \(\mathrm{AlCl}_4^-\). The aluminum atom is now surrounded by four chlorine atoms.

Determine New Hybridization

With the addition of the chloride ion and the formation of a fourth bond, the hybridization of aluminum changes from \(\mathrm{sp}^2\) to \(\mathrm{sp}^3\). This \(\mathrm{sp}^3\) hybridization results in a tetrahedral geometry with bond angles of approximately 109.5 degrees.

Key concepts

sp2 hybridization

In chemistry, hybridization is a concept used to explain the formation of chemical bonds in atoms. One common type, known as \( sp^2 \) hybridization, occurs in atoms that form three sigma bonds. This happens when one \( s \) orbital mixes with two \( p \) orbitals from the same atom to form three equivalent \( sp^2 \) hybrid orbitals. These orbitals are directed towards the corners of an equilateral triangle, resulting in a trigonal planar geometry.
In \( sp^2 \) hybridization, the bond angles are approximately 120 degrees. This type of hybridization is common in molecules where the central atom needs to form three bonds with surrounding atoms or groups, such as in ethylene (\(C_2H_4\)) or, as seen in the exercise, in \( \mathrm{AlCl}_3 \) where aluminum forms bonds with three chlorine atoms.
Knowing \( sp^2 \) hybridization is useful in predicting the shape and bond angles of molecules. It helps in visualizing the molecular geometry as flat with all atoms in the same plane.

sp3 hybridization

An important transformation in molecular geometry is the shift to \( sp^3 \) hybridization. This occurs when one \( s \) orbital mixes with three \( p \) orbitals, creating four equivalent \( sp^3 \) hybrid orbitals. These orbitals are equally distributed in space, forming a three-dimensional shape called a tetrahedral geometry.
Unlike \( sp^2 \) hybridization, \( sp^3 \) results in bond angles of about 109.5 degrees. This arrangement occurs in many different compounds including methane (\( CH_4 \)) and, as demonstrated in the original exercise, the \( \mathrm{AlCl}_4^- \) ion. Here, adding an extra bond, due to the chloride ion attachment, causes the aluminum atom to switch from \( sp^2 \) to \( sp^3 \) hybridization.
This hybridization expands the arrangement into a more stable tetrahedral configuration, providing an efficient distribution for minimizing electron pair repulsion.

tetrahedral geometry

Tetrahedral geometry is a three-dimensional molecular shape with four bonds directed towards the corners of a tetrahedron. This shape is influenced by \( sp^3 \) hybridization, where one \( s \) and three \( p \) orbitals hybridize to form four bonds.
In a tetrahedral arrangement, the bond angles are about 109.5 degrees. This geometry is crucial for molecules requiring optimal bonding spatial arrangements, reducing repulsion among electron pairs. Common examples include methane (\( CH_4 \)) and, in the exercise's context, the \( \mathrm{AlCl}_4^- \) ion where the aluminum atom adopts this geometry after reaction with chloride ion.
Understanding tetrahedral geometry is vital in predicting molecule shapes and behaviors, which is essential for analyzing chemical reactions and creating new compounds efficiently.