Question

When \(\mathrm{BCl}_{3}\) gas is passed through an electric discharge, small amounts of the reactive molecule \(\mathrm{B}_{2} \mathrm{Cl}_{4}\) are produced. (The molecule has a \(\mathrm{B}-\mathrm{B}\) covalent bond.) (a) Draw a Lewis electron dot structure for \(\mathrm{B}_{2} \mathrm{Cl}_{4}\) (b) Describe the hybridization of the B atoms in the molecule and the geometry around each B atom.

Short answer

(a) Lewis structure: B atoms linked, each bonded to 2 Cl atoms. (b) B atoms: sp² hybridized with trigonal planar geometry.

Step-by-step solution

Drawing the Lewis Structure

First, draw the skeleton structure of \( \mathrm{B}_{2} \mathrm{Cl}_{4} \). Place the two boron (B) atoms in the center, connected by a B-B covalent bond. Surround each B atom with two chlorine (Cl) atoms. Each Cl atom forms a single bond with a B atom. Boron forms three single bonds, and chlorine forms one single covalent bond. This leads to each B atom completing its octet with the help of the surrounding Cl atoms.

Account for Valence Electrons

Each boron atom has 3 valence electrons, contributing a total of 6 (3 per B atom) for boron. Each chlorine atom has 7 valence electrons, contributing a total of 28 (7 per Cl atom) for chlorine. This accumulates to 34 valence electrons. We use these electrons to ensure all atoms achieve a full outer shell.

Assign Electrons as Lone Pairs

After forming the single bonds, allocate the remaining valence electrons as lone pairs on the chlorine atoms to complete their octet. Each chlorine already has one bond pair, so assign three lone pairs (6 electrons) to each Cl to fulfill the octet rule. Boron, being electron-deficient, does not require fulfilling an octet.

Hybridization of Boron Atoms

Each B atom in \( \mathrm{B}_{2} \mathrm{Cl}_{4} \) is bonded to two Cl atoms and one B atom with single bonds. This requires forming 3 equivalent orbitals due to the presence of three regions of electron density around each B atom. Consequently, the hybridization state of each B atom is \(\text{sp}^2\).

Determine Geometry Around B Atoms

With \(\text{sp}^2\) hybridization, the molecular geometry around each B atom is trigonal planar. This results from having three bonding sites (one B-B bond and two B-Cl bonds) around each B atom, spatially oriented to minimize repulsion according to VSEPR theory.

Key concepts

Lewis Electron Dot Structure

The Lewis electron dot structure is a way to visualize the distribution of electrons in molecules. For the molecule \( \mathrm{B}_2\mathrm{Cl}_4 \), you start by arranging the atoms so that each boron (B) is at the center, connected by a single \( \mathrm{B}-\mathrm{B} \) bond. Surrounding each B atom are two chlorine (Cl) atoms that form single covalent bonds with boron.

Each line connecting B and Cl in the structure represents a pair of shared electrons, symbolizing the covalent bonds. Boron, with its unique electron-deficient nature, typically forms three single bonds without needing a full octet for stability.

Meanwhile, chlorine satisfies the octet rule, needing six more electrons beyond the one shared in the bond. Therefore, each chlorine atom receives three lone pairs (six electrons) that are not involved in bonding.

• Place each of the boron atoms in the center.
• Connect the boron atoms with a \( \mathrm{B}-\mathrm{B} \) bond.
• Attach two chlorine atoms to each boron using single bonds.
• Assign three lone pairs to each Cl to complete their octet.

Hybridization

Hybridization is a concept that helps explain the shapes and bonding properties of molecules by mixing atomic orbitals into new hybrid orbitals. In \( \mathrm{B}_2\mathrm{Cl}_4 \), each boron atom exhibits \( \text{sp}^2 \) hybridization. Here's how it works:

Boron has three regions of electron density—one from the single bond with the other boron atom and one from each chlorine atom bonded to it. These are areas where electrons are shared. To accommodate this, boron blends one \( s \) orbital and two \( p \) orbitals to form three equivalent \( \text{sp}^2 \) hybrid orbitals.

• The \( s \) and \( p \) orbitals mix to create three equal-energy hybrid orbitals.
• The \( \text{sp}^2 \) hybridization accounts for the trigonal planar geometry.
• This allows the formation of stable bonds with the adjacent Cl and B atoms.

Electron Geometry

Electron geometry considers the spatial arrangement of all electron pairs (bonding and non-bonding) around a central atom. For \( \mathrm{B}_2\mathrm{Cl}_4 \), each boron atom forms a trigonal planar shape due to its \( \text{sp}^2 \) hybridization.

In a trigonal planar arrangement, all the atoms and electron pairs lie in the same plane, creating 120-degree angles between them. This configuration is optimal to minimize repulsion between the pairs of electrons, adhering to the Valence Shell Electron Pair Repulsion (VSEPR) theory.

• Results from \( \text{sp}^2 \) hybridization and involves three bonding regions.
• Angles of 120 degrees between each bond make the structure symmetrical.
• The planar shape helps minimize electron repulsion.

Valence Electrons

Valence electrons are the electrons available for bonding in the outermost shell of an atom. In the molecule \( \mathrm{B}_2\mathrm{Cl}_4 \), understanding the distribution of valence electrons is key to drawing its Lewis structure.

Boron is unique because it has only three valence electrons, making it electron-deficient. As a result, it can bond only via three single covalent bonds. Chlorine, on the other hand, has seven valence electrons and forms one single bond per atom, using lone pairs to complete its octet.

In total, boron contributes 6 valence electrons (3 from each boron atom), while chlorine contributes 28 valence electrons (7 from each chlorine atom), totaling 34 valence electrons for the molecule.

• Boron: 3 valence electrons per atom, focusing on efficiency, no octet.
• Chlorine: 7 valence electrons, completes its octet with lone pairs.
• Total: 34 valence electrons are accounted for in bonds and lone pairs.