Problem 13 - Chemistry Principles and Practice

Question

Identify the hybrid orbitals used by boron in \(\mathrm{BCl}_{3}\) and in \(\mathrm{BCl}_{4}^{-},\) the ion formed from the reaction of \(\mathrm{BCl}_{3}\) and \(\mathrm{Cl}^{-} .\) Explain your choices.

Step-by-Step Solution

Answer

Boron in \\(\text{BCl}_3\\) uses sp² hybrid orbitals; in \\(\text{BCl}_4^-\\), it uses sp³ hybrid orbitals.

Step 1: Understanding Boron's Electron Configuration

The atomic number of boron is 5, making its ground state electron configuration 1s² 2s² 2p¹. This tells us boron has three valence electrons available for bonding, found in the 2s and 2p orbitals.

Step 2: Hybridization in BCl₃

In \( ext{BCl}_3\), boron forms three sigma bonds with three chlorine atoms. This requires the mixing of the 2s and two 2p orbitals to form three sp² hybrid orbitals. Thus, boron uses sp² hybridized orbitals in \( ext{BCl}_3\).

Step 3: Hybridization in BCl₄⁻

The ion \(\text{BCl}_4^-\) is formed when \( ext{BCl}_3\) accepts a \(\text{Cl}^-\) ion. Boron forms four sigma bonds with four chlorine atoms. Boron uses the 2s, three 2p orbitals, resulting in four sp³ hybrid orbitals. Thus, boron is sp³ hybridized in \(\text{BCl}_4^-\).

Key Concepts

Boron Electron Configuration

Boron is an element with atomic number 5. This means it has 5 electrons. Electrons fill orbitals according to specific rules. In the case of boron, the electron configuration is 1s² 2s² 2p¹.

• The 1s orbital is filled first with two electrons.
• The 2s orbital also gets filled with two electrons.
• Finally, the remaining one electron occupies a spot in the 2p orbital.

These three electrons in the second energy level are boron's valence electrons, which are important for forming bonds. Two are in the s orbital and one is in the p orbital. This configuration plays a critical role in how boron forms bonds, including its hybridization behavior.

Sigma Bonds

Sigma bonds are a type of covalent bond. They are formed by the head-on overlap of atomic orbitals. Boron, in compounds like \( \text{BCl}_3 \) and \( \text{BCl}_4^- \), forms sigma bonds with chlorine atoms.

• A sigma bond allows for free rotation around the bond axis.
• It is the strongest type of covalent bond, providing significant stability.
• In boron's compounds, sigma bonds help achieve more stable electron configurations.

In \( \text{BCl}_3 \), three sigma bonds are formed using sp² hybrid orbitals. In contrast, \( \text{BCl}_4^- \) involves four sigma bonds made by sp³ hybrid orbitals.

sp² Hybrid Orbitals

When boron forms \( \text{BCl}_3 \), it uses sp² hybridization. This involves mixing one 2s and two 2p orbitals. The result is three equivalent sp² hybrid orbitals.
Each of the orbitals will participate in the formation of a sigma bond with a chlorine atom in \( \text{BCl}_3 \).

• sp² hybrid orbitals lie in a plane, 120° apart.
• This arrangement allows for a trigonal planar shape.
• Such geometry is crucial for minimizing electron repulsion.

Therefore, sp² hybridization is essential for creating the correct bond angles and optimizing spatial arrangement. In \( \text{BCl}_3 \), this results in a stable, planar triangle.

sp³ Hybrid Orbitals

In \( \text{BCl}_4^- \), boron undergoes sp³ hybridization. This occurs after \( \text{BCl}_3 \) accepts a \( \text{Cl}^- \) ion, creating a new compound.
Here, boron utilizes its 2s and all three of its 2p orbitals for hybridization.

• sp³ hybrid orbitals form a tetrahedral geometry.
• They are oriented 109.5° apart to minimize electron-pair repulsion.
• This geometry allows boron to form four equivalent sigma bonds.

The resulting structure of \( \text{BCl}_4^- \) is a symmetry-driven tetrahedron. This structure enhances stability by distributing the atoms evenly in three-dimensional space. Understanding sp³ hybridization is key for predicting molecular shapes and bonding patterns in tetrahedral compounds.