Question

(a) Identify the Lewis acid and the Lewis base when ammonia \(\left(\mathrm{NH}_{3}\right)\) reacts with borane \(\left(\mathrm{BH}_{3}\right)\) to form the hydrogen storage material ammonia borane \(\left(\mathrm{NH}_{3} \mathrm{BH}_{3}\right) .\) (b) Draw the electron- dot structure for ammonia borane. Give the hybridization of \(\mathrm{N}\) and \(\mathrm{B}\) and the bond angles predicted by VSEPR theory.

Short answer

(a) NH₃ is the Lewis base, and BH₃ is the Lewis acid. (b) NH₃BH₃ retains \(\mathrm{sp}^3\) nitrogen and \(\mathrm{sp}^2\) boron hybridization. Bond angles: 109.5° (N) and 120° (B).

Step-by-step solution

Identify Lewis Acid and Base

In the reaction between ammonia \((\mathrm{NH}_3)\) and borane \((\mathrm{BH}_3)\), a Lewis acid is the compound that accepts an electron pair, while the Lewis base is the compound that donates an electron pair. Ammonia \((\mathrm{NH}_3)\) has a lone pair of electrons on the nitrogen atom, which allows it to donate electron density, making it the Lewis base. Borane \((\mathrm{BH}_3)\), on the other hand, has an electron-deficient boron atom, making it able to accept electron pairs, thus it is the Lewis acid.

Draw Electron-Dot Structure

To draw the electron-dot structure for ammonia borane \((\mathrm{NH}_3 \mathrm{BH}_3)\), start by pairing the lone pair of electrons from the nitrogen atom in \(\mathrm{NH}_3\) with the empty orbital on boron in \(\mathrm{BH}_3\). Represent the bonds by pairing single electrons from each hydrogen atom with electrons from nitrogen and boron, finalizing the structure: \[\mathrm{H}_3\mathrm{N} -> \mathrm{BH}_3\]. Here the arrow indicates the donation of the electron pair from nitrogen to boron.

Determine Hybridization

For nitrogen in ammonia \((\mathrm{NH}_3)\), the hybridization is \(\mathrm{sp}^3\) due to four regions of electron density (three bonds and one lone pair). For boron in borane \((\mathrm{BH}_3)\), the hybridization is \(\mathrm{sp}^2\) as it forms three regions of electron density (three bonds with hydrogen). In the formed compound \((\mathrm{NH}_3\mathrm{BH}_3)\), the hybridization states do not change from the individual molecules as the structures maintain similar bonding patterns.

Predict Bond Angles Using VSEPR

Using VSEPR theory, predict the bond angles. For nitrogen in ammonia \((\mathrm{NH}_3)\), the ideal bond angle is approximately 109.5 degrees due to the \(\mathrm{sp}^3\) hybridization (tetrahedral geometry). For boron in borane \((\mathrm{BH}_3)\), the bond angle is 120 degrees due to the \(\mathrm{sp}^2\) hybridization (trigonal planar geometry). In \(\mathrm{NH}_3\mathrm{BH}_3\), the coordination around nitrogen remains closer to a tetrahedral, and around boron, it remains closer to a planar geometry.

Key concepts

Electron-Dot Structures

Electron-dot structures, also known as Lewis structures, are a visual representation of the valence electrons in molecules. These diagrams help us understand how atoms bond in a molecule. To create an electron-dot structure, you need to know the valence electrons of each atom involved.

In the electron-dot structure of ammonia borane \(\mathrm{NH}_3\mathrm{BH}_3\), nitrogen has a lone pair of electrons. This pair is important in showing how nitrogen acts as a Lewis base. It donates this pair to the electron-deficient boron atom in borane \(\mathrm{BH}_3\). Boron can accept these electrons because it's a Lewis acid.

To visualize this, draw nitrogen with its lone pair. Then, connect this pair to boron with an arrow. This arrow shows electron donation, which forms a coordinate covalent bond. Each hydrogen atom in ammonia borane shares its single electron with boron or nitrogen, completing the octets or filling the bonding requirements.

VSEPR Theory

Valence Shell Electron Pair Repulsion (VSEPR) theory is a model used to predict the geometric arrangement of electron pairs around a central atom. It is based on the idea that electron pairs, whether in bonds or as lone pairs, repel each other and will arrange themselves as far apart as possible.

In ammonia \(\mathrm{NH}_3\), nitrogen has \(\mathrm{sp}^3\) hybridization, indicating four areas of electron density—three bonds with hydrogen and one lone pair. According to VSEPR theory, these regions will roughly form a tetrahedral geometry, resulting in bond angles close to 109.5 degrees.

Borane \(\mathrm{BH}_3\), on the other hand, has \(\mathrm{sp}^2\) hybridization with three bonds to hydrogen. This results in a trigonal planar shape with 120-degree bond angles. In the coordinate molecule \(\mathrm{NH}_3\mathrm{BH}_3\), these geometries remain largely intact, with the nitrogen's tetrahedral arrangement interacting with boron's planar shape.

Hybridization

Hybridization is a concept in chemistry that describes the mixing of atomic orbitals to form new hybrid orbitals. These hybrid orbitals are what atoms use to form bonds in molecules. The type of hybridization influences the molecule’s shape and bond angles.

Nitrogen in ammonia \(\mathrm{NH}_3\) undergoes \(\mathrm{sp}^3\) hybridization, blending one s orbital with three p orbitals. This results in four equivalent orbitals optimized for tetrahedral bonding. This arrangement allows for one lone pair and three bonds, illustrating nitrogen’s ability to form stable structures with three hydrogen atoms.

In borane \(\mathrm{BH}_3\), boron undergoes \(\mathrm{sp}^2\) hybridization. Here, one s orbital and two p orbitals mix to create three planarly arranged hybrid orbitals. These orbitals facilitate equal bonding with hydrogen atoms and highlight boron's electron deficiency—a key factor in its role as a Lewis acid when reacting with \(\mathrm{NH}_3\) to form ammonia borane. This understanding of hybridization is crucial for predicting the bonding patterns and geometries of molecules.