Borane Reactions: Videos & Practice Problems

In chemistry, boron is unique due to its electron deficiency, as it does not fulfill the octet rule with only three valence electrons. This characteristic makes boron compounds, such as diborane (B₂H₆), highly reactive, especially in reactions with water and in Lewis acid-base interactions. Diborane reacts with water to form boric acid (H₃BO₃) and hydrogen gas, a notable reaction that highlights its reactivity. Furthermore, boranes are considered Lewis acids because they can accept an electron pair. For example, in a reaction with ammonia (a Lewis base), boron can form an adduct by accepting a lone pair from the nitrogen atom in ammonia, resulting in a new bond and formal charges on the nitrogen (+1) and boron (-1) atoms. This process exemplifies a Lewis acid-base reaction, where boron's ability to act as an electron pair acceptor is due to its high electron deficiency.

Borane Reactions

Borane, represented as B(H₃), consists of a boron atom bonded to three hydrogen atoms. In the context of Lewis dot structures, elements typically strive to fulfill the octet rule, which states that atoms tend to form bonds until they are surrounded by eight valence electrons. However, boron has only three valence electrons, limiting it to forming three bonds. This electron deficiency makes boron highly reactive, as it seeks to achieve a more stable electronic configuration.

When discussing boron reactions, it is essential to note that they are primarily driven by the high electron deficiency of boranes and the reactivity of diboranes, which are formed by bridging hydrogens connecting two boron atoms. These diboranes exhibit unusual structures and are characterized by their high reactivity and instability.

In this context, two significant types of reactions involving boron will be explored: reactions with water and Lewis acid-base reactions. The reactivity of boron compounds in these reactions is a direct consequence of their electron deficiency. Understanding these interactions is crucial for grasping the behavior of boron in various chemical environments.

Reaction with H2O

Diborane, a highly reactive compound, readily interacts with liquid water, leading to a significant chemical reaction. When gaseous diborane (B₂H₆) comes into contact with water, it produces boric acid (H₃BO₃) and hydrogen gas (H₂). This reaction can be represented by the balanced equation:

2 B₂H₆ + 6 H₂O → 4 H₃BO₃ + 6 H₂

In this equation, the coefficients indicate the stoichiometric relationships between the reactants and products, highlighting that two moles of diborane react with six moles of water to yield four moles of boric acid and six moles of hydrogen gas. Understanding this reaction is crucial, as it illustrates the reactivity of diborane and its ability to produce useful compounds like boric acid while releasing hydrogen gas, which can be harnessed in various applications.

Borane Reactions Example

In this thermochemical scenario, we are tasked with determining the energy released when 150 grams of diborane (B₂H₆) reacts with excess water. The balanced chemical equation for this reaction indicates that 1 mole of diborane reacts with 6 moles of water to produce 2 moles of boric acid and 6 moles of hydrogen gas. The enthalpy change (ΔH) for this reaction is given as -466 kilojoules per mole, indicating that this amount of energy is released during the reaction.

To find the total energy released, we first need to convert the mass of diborane into moles. The molar mass of diborane can be calculated by adding the atomic masses of its constituent elements: 2 boron atoms (B) and 6 hydrogen atoms (H). The molar mass is calculated as follows:

\[\text{Molar mass of B}_2\text{H}_6 = (2 \times 10.81 \, \text{g/mol}) + (6 \times 1.008 \, \text{g/mol}) = 27.668 \, \text{g/mol}\]

Next, we convert 150 grams of diborane to moles using the formula:

\[\text{Moles of B}_2\text{H}_6 = \frac{150 \, \text{g}}{27.668 \, \text{g/mol}} \approx 5.42 \, \text{moles}\]

Now, we can calculate the total energy released by multiplying the number of moles of diborane by the energy released per mole:

\[\text{Energy released} = \text{Moles of B}_2\text{H}_6 \times \Delta H = 5.42 \, \text{moles} \times (-466 \, \text{kJ/mol}) \approx -2530.12 \, \text{kJ}\]

Considering significant figures, we round the final answer to three significant figures, resulting in an energy release of approximately -2530 kilojoules. This value represents the heat released when 150 grams of diborane reacts with excess water.

Lewis Acid-Base Reaction

A Lewis acid is defined as an electron pair acceptor, while a Lewis base is an electron pair donor. In this context, boranes are classified as Lewis acids due to their high electron deficiency. For example, consider the ammonia molecule, where nitrogen possesses a lone pair of electrons that it can readily share, making it a Lewis base.

In the case of borane, boron forms three bonds because it has three valence electrons, but it does not fulfill the octet rule, leaving it with the capacity to accept an electron pair. When nitrogen donates its lone pair to boron, a new bond is formed, resulting in the creation of an adduct. An adduct is the product formed from the reaction between a Lewis acid and a Lewis base.

To analyze the formal charges in this interaction, nitrogen, which shares its lone pair, has a formal charge of +1, while boron, which gains electrons, has a formal charge of -1. The overall charge of the molecule remains neutral, as the positive and negative charges balance each other out. This illustrates how boranes, due to their electron deficiency, can effectively act as Lewis acids by accepting lone pairs from Lewis bases, such as ammonia.

Borane Reactions Example

In the reaction between borane (BH₃) and a hydroxide ion (OH^-), an adduct product is formed through the interaction of these two species. Borane, which consists of a boron atom bonded to three hydrogen atoms, reacts with the hydroxide ion, where the oxygen atom has three lone pairs of electrons.

During this reaction, one of the lone pairs from the oxygen atom is utilized to form a new bond with the boron atom. As a result, the boron atom, which initially has a formal charge of 0, now shares electrons with the oxygen, leading to a formal charge of -1 on the boron. The oxygen, having formed two bonds (one with boron and one with hydrogen), maintains a formal charge of 0, as it has two remaining lone pairs.

This newly created adduct product illustrates the interaction between borane and hydroxide, showcasing how boron can accept electrons to achieve a more stable electronic configuration. The overall structure reflects the ideal bonding scenario for both elements, with boron gaining additional electrons and oxygen fulfilling its bonding requirements.

Here’s what students ask on this topic:

What is the structure of diborane (B₂H₆) and why is it highly reactive?

Diborane (B₂H₆) has a unique structure featuring two boron atoms and six hydrogen atoms. Four of the hydrogen atoms are terminal, each bonded to a boron atom, while the remaining two hydrogen atoms are bridging, forming bonds between the two boron atoms. This structure is unusual because the bridging hydrogen atoms create three-center two-electron bonds, which are not common in most molecules. Diborane is highly reactive due to its electron deficiency; the boron atoms do not fulfill the octet rule, making the molecule eager to react with other substances to achieve a more stable electronic configuration.

How does diborane react with water, and what are the products of this reaction?

Diborane (B₂H₆) reacts readily with water due to its high reactivity. When gaseous diborane reacts with liquid water, it produces boric acid (H₃BO₃) and hydrogen gas (H₂). The balanced chemical equation for this reaction is:

$1 B_2{H}_6 + 6 H{2}_{\mathrm{O}} \to 2 H_3{\mathrm{BO}}_3 + 6 H{2}_2$

This reaction highlights the high reactivity of diborane and its tendency to form more stable products.

What is a Lewis acid and how do boranes act as Lewis acids?

A Lewis acid is defined as an electron pair acceptor. Boranes, such as BH₃, act as Lewis acids due to their electron deficiency. Boron in boranes has only three valence electrons and forms three bonds, leaving it with an incomplete octet. This deficiency allows boranes to accept electron pairs from Lewis bases (electron pair donors). For example, in a reaction with ammonia (NH₃), the nitrogen atom in ammonia donates a lone pair of electrons to the boron atom in borane, forming an adduct. This interaction exemplifies a Lewis acid-base reaction, where borane acts as the Lewis acid by accepting the electron pair from ammonia.

What are the formal charges on nitrogen and boron in the adduct formed between ammonia and borane?

In the adduct formed between ammonia (NH₃) and borane (BH₃), the formal charges on the nitrogen and boron atoms can be determined as follows:

1. Nitrogen in ammonia has a lone pair of electrons that it donates to boron. By sharing this lone pair, nitrogen effectively loses one electron, resulting in a formal charge of +1.

2. Boron in borane accepts the lone pair from nitrogen, gaining two additional electrons. This results in a formal charge of -1.

Overall, the adduct is neutral, but the formal charges on nitrogen and boron are +1 and -1, respectively, reflecting the electron transfer in the Lewis acid-base reaction.

Why are boron compounds considered electron deficient?

Boron compounds are considered electron deficient because boron has only three valence electrons, which allows it to form only three covalent bonds. This results in boron having only six electrons in its valence shell, falling short of the octet rule, which states that atoms tend to have eight electrons in their valence shell to achieve stability. This electron deficiency makes boron compounds highly reactive, as they seek to accept additional electrons to fulfill the octet rule. This characteristic is particularly evident in boranes, which readily participate in reactions to achieve a more stable electronic configuration.