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Chapter 10: Problem 13

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.

Short Answer

Expert verified
Boron in \\(\text{BCl}_3\\) uses sp虏 hybrid orbitals; in \\(\text{BCl}_4^-\\), it uses sp鲁 hybrid orbitals.

Step by step solution

01

Understanding Boron's Electron Configuration

The atomic number of boron is 5, making its ground state electron configuration 1s虏 2s虏 2p鹿. This tells us boron has three valence electrons available for bonding, found in the 2s and 2p orbitals.
02

Hybridization in BCl鈧�

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 sp虏 hybrid orbitals. Thus, boron uses sp虏 hybridized orbitals in \( ext{BCl}_3\).
03

Hybridization in BCl鈧勨伝

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 sp鲁 hybrid orbitals. Thus, boron is sp鲁 hybridized in \(\text{BCl}_4^-\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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 1s虏 2s虏 2p鹿.
  • 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 sp虏 hybrid orbitals. In contrast, \( \text{BCl}_4^- \) involves four sigma bonds made by sp鲁 hybrid orbitals.
sp虏 Hybrid Orbitals
When boron forms \( \text{BCl}_3 \), it uses sp虏 hybridization. This involves mixing one 2s and two 2p orbitals. The result is three equivalent sp虏 hybrid orbitals.
Each of the orbitals will participate in the formation of a sigma bond with a chlorine atom in \( \text{BCl}_3 \).
  • sp虏 hybrid orbitals lie in a plane, 120掳 apart.
  • This arrangement allows for a trigonal planar shape.
  • Such geometry is crucial for minimizing electron repulsion.
Therefore, sp虏 hybridization is essential for creating the correct bond angles and optimizing spatial arrangement. In \( \text{BCl}_3 \), this results in a stable, planar triangle.
sp鲁 Hybrid Orbitals
In \( \text{BCl}_4^- \), boron undergoes sp鲁 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.
  • sp鲁 hybrid orbitals form a tetrahedral geometry.
  • They are oriented 109.5掳 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 sp鲁 hybridization is key for predicting molecular shapes and bonding patterns in tetrahedral compounds.

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Most popular questions from this chapter

Write a Lewis struicture for each of the following molecules. Indicate all of the bond angles as predicted by the VSEPR model. Deduce the skeleton structure from the way each formula is written. (a) \(\mathrm{H}_{3} \mathrm{CCCH}\) (b) \(\mathrm{Br}_{2} \mathrm{CCH}_{2}\) (c) \(\mathrm{H}_{3} \mathrm{CNH}_{2}\)

Identify two homonuclear diatomic molecules or ions with each of the following molecular orbital electron configurations. Are these species stable? (a) \(\left(\sigma_{2 s}\right)^{2}\left(\sigma_{2 s}^{*}\right)^{2}\left(\pi_{2 p}\right)^{4}\left(\sigma_{2 p}\right)^{2}\left(\pi_{2 p}^{*}\right)^{3}\) (b) \(\left(\sigma_{2 s}\right)^{2}\left(\sigma_{2 s}^{*}\right)^{2}\left(\pi_{2 p}\right)^{4}\left(\sigma_{2 p}\right)^{2}\) (c) \(\left(\sigma_{2 s}\right)^{2}\left(\sigma_{2 s}^{*}\right)^{2}\)

Ionization energies can be determined for molecules and atoms. Draw the molecular orbital diagrams for \(\mathrm{NO}\) and \(\mathrm{CO}\), and predict which compound has the lower ionization energy.

Give the bonded-atom lone-pair arrangement expected for a central atom that has (a) three bonded atoms and no lone pairs. (b) two bonded atoms and two lone pairs. (c) four bonded atoms and no lone pairs. (d) four bonded atoms and one lone pair.

Draw the three possible arrangements of the fluorine atoms about the iodine atom in \(\mathrm{IF}_{3}\). Choose the shape predicted by VSEPR theory, and explain wby this arrangement is favored.
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