Problem 150 - A Complete Resource Book in Chemistry for JEE Main

Question

Compound \(\mathrm{X}\) on reduction with \(\mathrm{LiAlH}_{4}\) gives a hydride Y containing \(21.72 \%\) hydrogen along with other products. The compound \(\mathrm{Y}\) reacts with air explosively resulting in boron trioxide. Identify \(\mathrm{Y}\). (a) \(\mathrm{B}_{4} \mathrm{H}_{10}\) (b) \(\mathrm{B}_{2} \mathrm{H}_{6}\) (c) \(\mathrm{Na}_{2} \mathrm{~B}_{4} \mathrm{O}_{7}\) (d) both (a) and (b)

Step-by-Step Solution

Answer

The compound \( Y \) is \( \mathrm{B}_{2} \mathrm{H}_{6} \).

Step 1: Determine the Options

The problem presents four chemical compounds as possible identities for \( Y \). These compounds are \( \mathrm{B}_{4} \mathrm{H}_{10} \) (tetraborane), \( \mathrm{B}_{2} \mathrm{H}_{6} \) (diborane), \( \mathrm{Na}_{2} \mathrm{B}_{4} \mathrm{O}_{7} \) (sodium tetraborate), and both (a) and (b) together.

Step 2: Calculate Hydrogen Percentage

We need to calculate the percentage of hydrogen in \( Y \). Since \( Y \) contains \( 21.72 \% \) hydrogen, we will find if \( \mathrm{B}_{4} \mathrm{H}_{10} \) or \( \mathrm{B}_{2} \mathrm{H}_{6} \) matches this percentage.

Step 3: Analyze \( \mathrm{B}_{2} \mathrm{H}_{6} \)

Calculate the percentage of hydrogen in \( \mathrm{B}_{2} \mathrm{H}_{6} \). Its molar mass is \[(2 \times 10.81 + 6 \times 1.008) = 27.66 \, \mathrm{g/mol} \]. The mass of hydrogen is \( 6 \times 1.008 = 6.048 \, \mathrm{g/mol} \) giving a percentage hydrogen content of \( \frac{6.048}{27.66} \times 100 = 21.86 \% \).

Step 4: Analyze \( \mathrm{B}_{4} \mathrm{H}_{10} \)

Calculate the percentage of hydrogen in \( \mathrm{B}_{4} \mathrm{H}_{10} \). The molar mass is \[(4 \times 10.81 + 10 \times 1.008) = 53.32 \, \mathrm{g/mol} \]. The hydrogen mass is \( 10 \times 1.008 = 10.08 \, \mathrm{g/mol} \) giving a percentage hydrogen content of \( \frac{10.08}{53.32} \times 100 = 18.91 \% \).

Step 5: Identify \( Y \)

The hydrogen percentage in \( Y \) was calculated to match more closely to \( \mathrm{B}_{2} \mathrm{H}_{6} \) since \( 21.86 \% \) is quite close to \( 21.72 \% \). \( \mathrm{B}_{2} \mathrm{H}_{6} \) also combusts in air to form boron trioxide \( \mathrm{B}_{2} \mathrm{O}_{3} \). Thus, \( Y \) is \( \mathrm{B}_{2} \mathrm{H}_{6} \).

Key Concepts

Reduction Reactions

Reduction reactions are chemical processes that involve the gain of electrons by a molecule, atom, or ion. In a typical reduction process, the reducing agent donates electrons and gets oxidized, while the substance being reduced gains those electrons. This concept is integral in many chemical reactions, including organic and inorganic transformations.

In the exercise presented, the compound \( X \) is reduced by \( \text{LiAlH}_4 \), a powerful reducing agent. Sodium tetrahydroaluminate, or \( \text{LiAlH}_4 \), is commonly used in laboratories to reduce esters, carboxylic acids, and ketones, among other compounds. It is particularly useful in converting molecules into alcohols by reducing the carbonyl group. The role of \( \text{LiAlH}_4 \) in the transformation of \( X \) effectively showcases the reduction process where electrons are transferred, and hydrogen is introduced to form new chemical species.

Understanding reduction reactions is key in synthetic chemistry, providing a pathway to manipulate and form different compound structures by selectively adding hydrogen atoms or gaining electrons from a reducing agent.

Hydride Compounds

Hydride compounds are a group of chemical compounds where hydrogen is bonded with a less electronegative element or group. These compounds can play diverse roles in chemical reactions, serving as reducing agents or chemical intermediates.

In the exercise example, hydride compounds are central to the identity of \( Y \), a hydride of boron. Diborane, \( \text{B}_2\text{H}_6 \), is identified as the compound \( Y \) due to its hydrogen content that closely matches the given percentage, \( 21.72\% \). This gaseous compound is known for its volatility and its ability to react violently with air, forming boron trioxide. This reaction produces a visible and explosive combustion, a characteristic behavior of many hydrides when exposed to oxygen.

The understanding of hydride chemistry extends beyond simple reactions. It has applications in energy storage, such as hydrogen fuel cells, and plays a critical role in materials science, especially in designing lightweight materials. The reactivity and versatility of hydrides make them a fascinating subject in inorganic chemistry and material applications.

Boron Chemistry

Boron chemistry is a specialized area in inorganic chemistry that deals with the properties and reactions of boron-containing compounds. Boron is a unique element, offering a diverse chemistry due to its ability to form stable covalent bonds and its electron-deficient nature.

The compound identified in the exercise, \( \text{B}_2\text{H}_6 \) (diborane), highlights key aspects of boron chemistry. Diborane is an example of a boron hydride, which features unusual bonding situations, such as the three-center two-electron bond. This feature is a hallmark of boron compounds, allowing boron to maintain stability despite having fewer than the typical octet of electrons.

Boron trioxide, \( \text{B}_2\text{O}_3 \), is another fundamental compound in boron chemistry, resulting from the complete oxidation of boron hydrides like \( \text{B}_2\text{H}_6 \). This robust oxide is primarily used in glass and ceramics industries, emphasizing the practical importance of boron compounds in industrial applications.

• Boron compounds display a broad range of properties and reactivities, making them valuable in catalysis, agriculture, and medicine.
• One of the noted applications is in the synthesis of boron-containing polymers and as reagents in organic transformations.

The ingenuity of boron chemistry paves the way for innovations in materials science and industrial processes, establishing boron as a pivotal element in modern chemistry.