Problem 51 - Chemistry and Chemical Reactivity

Question

Suppose you carry out the following reaction of ammonia and boron trifluoride in the laboratory. (a) What is the geometry of the boron atom in \(\mathrm{BF}_{3} ? \mathrm{In} \mathrm{H}_{3} \mathrm{~N} \rightarrow \mathrm{BF}_{3} ?\) (b) What is the hybridization of the boron atom in the two compounds? (c) Considering the structures and bonding of \(\mathrm{NH}_{3}\) and \(\mathrm{BF}_{3}\), why do you expect the nitrogen on \(\mathrm{NH}_{3}\) to donate an electron pair to the \(\mathrm{B}\) atom of \(\mathrm{BF}_{3} ?\) (d) \(\mathrm{BF}_{3}\) also reacts readily with water. Based on the ammonia reaction above, speculate on how water can interact with \(\mathrm{BF}_{3}\).

Step-by-Step Solution

Answer

In \( \mathrm{BF}_{3} \), boron is trigonal planar and \( \mathrm{sp}^2 \) hybridized; in \( \mathrm{H}_{3} \mathrm{N} \rightarrow \mathrm{BF}_{3} \), boron becomes tetrahedral and \( \mathrm{sp}^3 \) hybridized. \( \mathrm{NH}_{3} \), as a Lewis base, donates an electron pair to electron-deficient \( \mathrm{BF}_{3} \). Water likely interacts similarly with \( \mathrm{BF}_{3} \) via lone pair donation.

Step 1: Determine the Geometry of Boron in BF3

Boron trifluoride, \( \mathrm{BF}_{3} \), has a central boron atom surrounded by three fluorine atoms. With three bonding pairs and no lone pairs around boron, the structure adopts a trigonal planar geometry.

Step 2: Determine the Geometry in NH3 → BF3

When ammonia \( \mathrm{NH}_{3} \) reacts with \( \mathrm{BF}_{3} \), a coordinate covalent bond is formed between the nitrogen of \( \mathrm{NH}_{3} \) and the boron of \( \mathrm{BF}_{3} \), resulting in \( \mathrm{H}_{3} \mathrm{N} \rightarrow \mathrm{BF}_{3} \). This interaction gives boron a new geometry and transforms it into a tetrahedral structure.

Step 3: Determine the Hybridization of Boron in BF3

In \( \mathrm{BF}_{3} \), boron's electron configuration allows it to undergo \( \mathrm{sp}^2 \) hybridization, aligning with its trigonal planar geometry. This involves the mixing of one \( s \) and two \( p \) orbitals.

Step 4: Determine the Hybridization of Boron in NH3 → BF3

When \( \mathrm{NH}_{3} \) donates an electron pair to \( \mathrm{BF}_{3} \), the resulting \( \mathrm{H}_{3} \mathrm{N} \rightarrow \mathrm{BF}_{3} \) complex causes boron to adopt \( \mathrm{sp}^3 \) hybridization, corresponding to a tetrahedral geometry.

Step 5: Explain Electron Donation from NH3 to BF3

\( \mathrm{NH}_{3} \) has a lone electron pair on the nitrogen atom, making it a good Lewis base. \( \mathrm{BF}_{3} \) is electron-deficient because boron has an incomplete octet, making it a good Lewis acid. Thus, nitrogen donates an electron pair to boron.

Step 6: Speculate on BF3 Interaction with Water

Like ammonia, water \( \mathrm{H}_{2} \mathrm{O} \) contains lone pairs of electrons that can be donated to \( \mathrm{BF}_{3} \). This interaction likely forms a coordinate covalent bond between oxygen and boron, assisting in stabilizing \( \mathrm{BF}_{3} \)'s electronic structure.

Key Concepts

Boron Trifluoride

Boron trifluoride, represented as \( \mathrm{BF}_{3} \), is a compound where a boron atom is bonded to three fluorine atoms. This compound is well-known in chemistry as an example of a Lewis acid. One of its key features is its geometry. The boron atom, being surrounded by three fluorine atoms with only bonding pairs and no lone pairs, adopts a trigonal planar structure. This shape minimizes electron repulsion between the bonds, resulting in bond angles of 120 degrees.

Understanding the trigonal planar shape is crucial as it plays a vital role in the reactivity and properties of \( \mathrm{BF}_{3} \). The boron atom in \( \mathrm{BF}_{3} \) is electron-deficient, having only six valence electrons, which explains its strong tendency to accept electrons through bond formation with other molecules.

• Boron trifluoride acts as a Lewis acid, seeking electron pairs because of its incomplete octet.
• The molecular geometry allows efficient overlap of orbitals during reactions, facilitating interactions like electron pair donations.

Hybridization

Hybridization is a concept in chemistry used to describe the mixing of atomic orbitals to form new hybrid orbitals, which provide explanations for molecular shapes and bonding properties. In \( \mathrm{BF}_{3} \), boron undergoes \( \mathrm{sp}^2 \) hybridization. This process involves combining one \( s \) orbital and two \( p \) orbitals to create three equivalent hybrid orbitals, corresponding to the trigonal planar geometry. These hybrid orbitals align in the plane, each at 120-degree angles from the other, perfectly aligning with \( \mathrm{BF}_{3} \)'s structure.

When \( \mathrm{BF}_{3} \) reacts with ammonia \( \mathrm{NH}_{3} \), forming the \( \mathrm{H}_{3} \mathrm{N} \rightarrow \mathrm{BF}_{3} \) complex, the hybridization of boron shifts to \( \mathrm{sp}^3 \). This shift happens because the electron pair donation from nitrogen changes boron's coordination number, leading to a tetrahedral geometry.

• Boron's \( \mathrm{sp}^3 \) hybridization involves mixing one \( s \) orbital with three \( p \) orbitals.
• This results in four hybrid orbitals, allowing the formation of four bonds, accommodating the extra electron pair from \( \mathrm{NH}_{3} \).

Electron Pair Donation

Electron pair donation is a fundamental concept, especially in the realm of Lewis acid-base chemistry. In this context, \( \mathrm{NH}_{3} \) acts as a Lewis base. Ammonia has a lone pair of electrons on its nitrogen atom, which makes it eager to donate to an electron-deficient acceptor, such as \( \mathrm{BF}_{3} \). The donation of this electron pair forms a coordinate covalent bond with the boron atom.

The driving force behind this electron donation is the incomplete valence shell of boron in \( \mathrm{BF}_{3} \). Having only six electrons, boron readily accepts the electron pair to stabilize its electronic configuration. This interaction highlights the complementary nature of Lewis acids and bases, where one species provides electron density, and the other requires it.

• Boron trifluoride's role as an electron acceptor stems from its electron-deficient state.
• Ammonia's donation ability is due to its lone pair that it can spare for bond formation.

This concept also explains the behavior of \( \mathrm{BF}_{3} \) with other molecules like water, where oxygen can donate an electron pair to boron, further stabilizing \( \mathrm{BF}_{3} \).