Question

Borazine, \(\mathrm{B}_{3} \mathrm{~N}_{3} \mathrm{H}_{6}\), a compound that has been called "inorganic benzene" because of its similar hexagonal structure (but with alternating \(B\) and \(N\) atoms in place of \(C\) atoms), is the basis of a large class of boron-nitrogen compounds. Write its Lewis structure and predict the composition of the hybrid orbitals used by each \(B\) and \(N\) atom.

Short answer

In borazine, boron atoms use sp2 hybridization for bonding with nitrogen and hydrogen atoms. Nitrogen atoms also use sp2 hybridization for bonds with boron, a hydrogen atom, and to accommodate a lone pair. The Lewis structure consists of a hexagonal ring with alternating B and N atoms, each B bonded to two Hs, and each N bonded to one H and possessing one lone pair.

Step-by-step solution

Drawing the Skeleton Structure

Draw a hexagonal ring, alternating between boron (B) and nitrogen (N) atoms. Connect each B to an N with a single bond, since they will share a pair of electrons to form a sigma bond each. There will be three B atoms and three N atoms, hence the chemical formula \(B_3N_3H_6\).

Placing the Hydrogen Atoms

Boron has three valence electrons and nitrogen has five valence electrons. For borazine, each boron atom will form three bonds: one with an adjacent nitrogen in the ring and two with hydrogen atoms. Each nitrogen atom will form three bonds as well: one with an adjacent boron in the ring, one with a hydrogen atom, and a lone pair of electrons.

Complete the Octets for Nitrogen

Make sure each nitrogen atom in the structure has an octet. Each nitrogen will share one pair of electrons with boron, one with hydrogen, and will have one lone pair, therefore having 3 bonds and one lone pair in total.

Account for the Valence Electrons

Count the total number of valence electrons that should be present in the molecule. In borazine, each boron atom contributes three valence electrons and each nitrogen atom contributes five, plus there is one additional electron for each hydrogen atom. This gives a total of \(3(3B) + 5(3N) + 6(1H) = 36\) valence electrons to be placed in the Lewis structure.

Draw the Lewis Structure

Utilize the electrons to make sure every B and N atom gets an octet of electrons (or for B, as close to an octet as possible since it can be stable with six electrons) by distributing the electrons accordingly. Since B tends to be electron-deficient, there is no need for B to have a complete octet, thus each B will be associated with 6 electrons and each N will have 8 electrons surrounding them in the structure.

Predict the Composition of the Hybrid Orbitals

Boron atoms in borazine will use sp2 hybrid orbitals for sigma bonding with the adjacent nitrogen atoms and hydrogen atoms. Nitrogen atoms will also use sp2 hybridization for the sigma bonds with boron atoms, one hydrogen atom, and to accommodate the lone pair of electrons. Both B and N will have an unhybridized p orbital which can overlap to form pi-bonds, although in reality, the pi system is delocalized over the whole ring.

Key concepts

Chemical Bonding

Chemical bonding is a fundamental concept in chemistry that explains how atoms connect to form molecules. It involves the attraction between the positive nuclei of atoms and the negative electrons that they share or transfer to each other. There are three main types of chemical bonds: ionic, covalent, and metallic.

In the context of borazine, the focus is on covalent bonds, where atoms share pairs of electrons to obtain stable electronic configurations. Borazine, with the formula \( \mathrm{B}_3 \mathrm{N}_3 \mathrm{H}_6 \), has a unique structure that necessitates the sharing of electrons between boron and nitrogen atoms. Each boron atom shares electrons with adjacent nitrogen atoms and hydrogen atoms to satisfy its valence electron requirements, even though it does not achieve a full octet. Nitrogen, on the other hand, shares electrons not only to form bonds but also retains a lone pair, successfully reaching an octet.

In borazine, every two adjacent atoms in the hexagonal ring form a strong sigma bond, with the possibility of pi-bonding between the boron and nitrogen atoms. These bonds make borazine a stable compound, which is often compared to benzene due to its hexagonal structure and delocalized bonding, even though the two molecules have different elements and bonding patterns.

Hybrid Orbitals

Hybrid orbitals arise from the concept of hybridization, which describes the merging of atomic orbitals on an atom to form new, equivalent orbitals suitable for chemical bonding. The type of hybridization depends on the number of electron pairs surrounding an atom, which in turn influences the bond angles and molecular geometry.

For borazine's boron atoms, they undergo sp2 hybridization. This hybridization involves mixing one s orbital and two p orbitals to create three equivalent sp2 hybrid orbitals. Each sp2 orbital in boron is used to form a sigma bond: one with an adjacent nitrogen atom and two with hydrogen atoms. Nitrogen atoms also use sp2 hybridization. One sp2 hybrid orbital forms a sigma bond with boron, another with a hydrogen atom, and the third accommodates the lone pair of electrons. Due to the presence of the lone pair, nitrogen's sp2 hybridization is somewhat different from boron's.

Furthermore, both boron and nitrogen retain an unhybridized p orbital. In a perfect planar hexagonal structure, these p orbitals can overlap side-by-side to form pi bonds. This overlapping contributes to the molecular stability and the delocalized electron network across the borazine ring, creating a bond structure analogous to that seen in benzene.

Molecular Geometry

Molecular geometry refers to the three-dimensional arrangement of atoms in a molecule. It plays a critical role in determining a compound's properties, including reactivity, polarity, color, and phase of matter. The geometry is influenced by the number of bonds and lone pairs of electrons around the central atom.

In borazine, each boron and nitrogen atom is in a planar triangular (trigonal planar) environment due to their sp2 hybridization. Since the molecule is composed of a ring structure with alternating boron and nitrogen atoms, the overall geometry of borazine is a flat hexagonal ring, similar to the structure of benzene. This hexagonal planar geometry is not a coincidence; it is a direct result of the hybridization state of the atoms involved.

The trigonal planar geometry around each atom also means that all bond angles are approximately 120 degrees, which is characteristic of sp2 hybridized atoms. This geometry allows for the delocalization of electrons and the stabilization of the molecular structure through overlapping p orbitals in the ring, contributing to the remarkable stability and aromaticity of borazine—a key comparison point with benzene, despite the differences in their atomic composition.