Problem 66 - Chemistry and Chemical Reactivity

Question

When 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 \(\mathbf{B}_{2} \mathrm{Cl}_{4}\) (b) Describe the hybridization of the B atoms in the molecule and the geometry around each \(\mathrm{B}\) atom.

Step-by-Step Solution

Answer

In \(\mathbf{B}_2 \mathrm{Cl}_4\), each B is \(\text{sp}^2\) hybridized with trigonal planar geometry.

Step 1: Draw the Skeletal Structure

To draw the Lewis structure for \(\mathbf{B}_2 \mathrm{Cl}_4\), start by arranging the two boron atoms bonded directly to each other, as the compound has a \(\mathrm{B}-\mathrm{B}\) covalent bond. Place the chlorine atoms around the boron atoms, with each boron atom having two chlorine atoms bonded to it.

Step 2: Count the Valence Electrons

Each boron (B) atom has 3 valence electrons, and each chlorine (Cl) atom has 7 valence electrons. This gives a total of \(2 \times 3\) for boron and \(4 \times 7\) for chlorine, making a total of 34 valence electrons to work with for the molecule.

Step 3: Draw the Bonds and Distribute the Electrons

Each \(\mathrm{B-B}\) bond and \(\mathrm{B-Cl}\) bond will use 2 electrons. With 5 bonds formed (2 \(\mathrm{B-Cl}\) bonds per boron and 1 \(\mathrm{B-B}\) bond), 10 electrons are used. Distribute the remaining electrons to complete the octets of the more electronegative chlorine atoms, and remember boron can be stable with less than an octet.

Step 4: Check the Formal Charges

Ensure the formal charges are minimized in the structure. Each Cl should have a complete octet with 3 lone pairs, and each B should not have more than 6 electrons, considering the sharing process involves 1 lone electron left per boron after forming bonds.

Step 5: Determine the Hybridization

Each boron atom in \(\mathrm{B}_2 \mathrm{Cl}_4\) is bonded to three groups (2 Cl atoms and 1 B atom). The hybridization of boron is \(\mathrm{sp}^2\), which corresponds to the formation of 3 sigma bonds.

Step 6: Determine the Geometry

The \(\mathrm{sp}^2\) hybridization of B leads to a trigonal planar geometry around each boron. This means each boron atom forms a planar triangle with the 2 Cl atoms and the other B atom.

Key Concepts

Boron Chemistry

Boron is a fascinating element, playing a unique role in chemistry. Unlike many other elements, boron often does not follow the octet rule, making its compounds interesting to study. Boron typically forms three covalent bonds, as it has three valence electrons. This allows it to form clusters and complexes that are rich in electrons and often have unusual bonding arrangements.

• Boron's three valence electrons lead it to form compounds that can involve two-centre two-electron bonds or even multi-centre bonding.
• In the compound \((\mathrm{B}_2 \mathrm{Cl}_4)\), boron forms a \(\mathrm{B}-\mathrm{B}\) covalent bond and connects to chlorine atoms.

In boron chemistry, compounds tend to be reactive and can form different structures depending on the surrounding atoms and conditions. This makes borons distinctive in forming stable yet sometimes underfilled electron shells.

Hybridization

Hybridization is a concept that explains the mixing of atomic orbitals in an atom to form new hybrid orbitals. These are used to form covalent bonds in molecules. In \(\mathrm{B}_2 \mathrm{Cl}_4\), the central boron atoms are crucial to understanding the molecular structure.
For each boron in this compound:

• The boron atom forms three bonds; thus it undergoes \(\mathrm{sp}^2\) hybridization.
• The \(\mathrm{sp}^2\) hybridization involves one \(s\) orbital and two \(p\) orbitals, forming three equivalent \(\mathrm{sp}^2\) hybrid orbitals.

These hybrid orbitals arrange themselves as far apart as possible, leading to maximum spacing and thus stability in the molecule. This characteristic arrangement plays a significant role in defining the geometry of boron compounds.

Molecular Geometry

The molecular geometry of a compound is determined by the arrangement of electron pairs around the central atom, which in turn affects the shape of the molecule. In \(\mathrm{B}_2 \mathrm{Cl}_4\), each boron atom is central to this arrangement.

• Due to the \(\mathrm{sp}^2\) hybridization, each boron atom exhibits a trigonal planar geometry.
• This means that all atoms bonded to boron are in a plane, and the bond angles between them are approximately \(120^\circ\).

Trigonal planar geometry affects how these molecules interact with others, as the spatial arrangement influences chemical reactivity and interactions with other molecules.

Covalent Bonds

Covalent bonding is a type of chemical bond where pairs of electrons are shared between atoms. It is a foundational concept necessary to understand molecular structures. In \(\mathrm{B}_2 \mathrm{Cl}_4\):

• The \(\mathrm{B-Cl}\) bonds involve the sharing of electrons between boron and chlorine atoms, allowing each to attain a more stable electronic state.
• The formation of the \(\mathrm{B-B}\) bond signifies a direct linkage between the boron atoms, showcasing how covalent bonds can form complex molecules beyond simple binary units.

Covalent bonds provide the stability necessary for the creation and persistence of diverse molecular forms, while also influencing the physical properties, such as melting and boiling points, of these molecules.