Problem 8 - A Complete Resource Book in Chemistry for JEE Main

Question

Which of the following reacts with \(\mathrm{BCl}_{3}\) to form diborane? (a) \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) (b) \(\mathrm{NH}_{3}\) (c) \(\mathrm{NaHg}\) (d) \(\mathrm{LiAlH}_{4}\)

Step-by-Step Solution

Answer

(d) \(\mathrm{LiAlH}_{4}\) reacts with \(\mathrm{BCl}_{3}\) to form diborane.

Step 1: Understanding the Reaction

We need to find which compound can reduce boron trichloride (\(\mathrm{BCl}_{3}\)) to produce diborane (\(\mathrm{B}_{2}\mathrm{H}_{6}\)). Diborane formation typically requires a reducing agent that can provide hydride ions (\(\mathrm{H}^-\)).

Step 2: Analyzing the Options

Let's analyze each option: (a) \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) - This is a strong oxidizing agent and will not supply hydride ions.(b) \(\mathrm{NH}_3\) - Ammonia is a weak base and is not known to provide hydride ions.(c) \(\mathrm{NaHg}\) - This is a sodium amalgam with mercury, used sometimes in reductions but not known for providing hydride ions.(d) \(\mathrm{LiAlH}_{4}\) - Lithium aluminum hydride is a well-known reducing agent that provides hydride ions.

Step 3: Identifying the Correct Reagent

Only \(\mathrm{LiAlH}_{4}\) is known to effectively provide the hydride ions necessary for reducing \(\mathrm{BCl}_{3}\) to \(\mathrm{B}_{2}\mathrm{H}_{6}\). This makes option (d) the correct answer.

Key Concepts

Boron trichloride reduction

Boron trichloride, or \( \mathrm{BCl}_{3} \), is a colorless gas that plays a crucial role in many chemical reactions. To convert boron trichloride into diborane, \( \mathrm{B}_{2} \mathrm{H}_{6} \), a reducing agent is needed. The process involves the transfer of electrons to \( \mathrm{BCl}_{3} \) to facilitate the breaking and reforming of bonds, eventually forming diborane. This transformation is not simply a matter of mixing chemicals; it requires a specific agent capable of donating hydride ions. Hydride ions, \( \mathrm{H}^- \), are negative hydrogen ions that play a pivotal part in such reductions. This is where the choice of the right reagent becomes critical. Without the appropriate donating agent, \( \mathrm{BCl}_{3} \) cannot be effectively converted into \( \mathrm{B}_{2} \mathrm{H}_{6} \).

Reducing agents in chemistry

Reducing agents are substances that donate electrons in a chemical reaction. This is a key function in reduction reactions where electrons are transferred from one species to another. Essential reducing agents in chemistry include compounds or elements like hydrogen, lithium aluminum hydride (\( \mathrm{LiAlH}_{4} \)), and sodium, among others. Each of these agents has unique properties that make them suitable for specific types of reactions.

• Hydrogen gas \( (\mathrm{H}_{2}) \): Often used in industrial processes due to its ability to add hydrogen.
• \( \mathrm{LiAlH}_{4} \): A strong reducing agent often used in organic chemistry for the precise donation of hydride ions.

They each serve different roles, but their main job is to reduce other compounds. That means they help other elements or compounds gain electrons during a reaction, often leading to the breaking of bonds or creation of new ones.

Role of hydride ions

Hydride ions, noted as \( \mathrm{H}^- \), are negatively charged hydrogen ions. These ions are crucial in reduction reactions where they donate electrons. The hydride ion is essentially a hydrogen atom that has gained an extra electron, giving it a negative charge.
In the case of diborane formation from boron trichloride, the hydride ions act as electron donors. They help in the breaking of \( \mathrm{B}-\mathrm{Cl} \) bonds and contribute to the formation of \( \mathrm{B}-\mathrm{H} \) bonds, a necessary step for diborane synthesis.

• Electron Donors: Hydride ions provide a pair of electrons, enabling the transformation in organometallic reactions.
• Binds to Boron: Once the \( \mathrm{B}-\mathrm{Cl} \) bonds are broken, hydride ions assist in establishing \( \mathrm{B}-\mathrm{H} \) bonds.

Without these ions, the reduction of \( \mathrm{BCl}_{3} \) to \( \mathrm{B}_{2} \mathrm{H}_{6} \) would not be possible. Thus, they play a fundamental role in this chemical process.