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 \(Y\) 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 \( \text{Y} \) is both \( \text{B}_4\text{H}_{10} \) and \( \text{B}_2\text{H}_6 \) (option (d)).

Step 1: Understand the Role of LiAlH4

The compound \( \text{LiAlH}_4 \) is a strong reducing agent. It typically reduces compounds by adding hydrogen to them. In the problem, it reduces compound \( \text{X} \), resulting in the hydride \( \text{Y} \), which contains \( 21.72\% \) hydrogen.

Step 2: Calculate the Molecular Weight

The percentage composition of hydrogen in \( \text{Y} \) is given as \( 21.72\% \). Let the molar mass of \( \text{Y} \) be \( M \). The percentage of hydrogen allows us to set the equation: \( \frac{n \times 1}{M} \times 100 = 21.72 \), where \( n \) is the number of hydrogen atoms (each contributing \( 1 \) g/mol).

Step 3: Solve the Equation

Rearrange to find \( n = 21.72 \times \frac{M}{100} \). Assuming \( n = 10 \) based on option \( \text{a} \) and \( \text{b} \), the potential molecular weight \( M \) is approximately \( \frac{10}{21.72} \times 100 \approx 46 \) for \( \text{B}_4 \text{H}_{10} \), and it also fits for \( \text{B}_2 \text{H}_6 \) since \( \frac{6}{21.72} \times 100 \approx 27.6 \). Both correspond to the hydride proportion of boron hydrides.

Step 4: Analyze the Reaction with Air

\( \text{Y} \) reacts explosively with air to produce \( \text{B}_2\text{O}_3 \), indicating it is a boron hydride. This property is characteristic of \( \text{B}_4\text{H}_{10} \) and \( \text{B}_2\text{H}_6 \), as both react similarly under these conditions.

Step 5: Verify the Options Against Criteria

Both options (a) \( \text{B}_4\text{H}_{10} \) and (b) \( \text{B}_2\text{H}_6 \) satisfy all conditions, including the explosive reaction to form \( \text{B}_2\text{O}_3 \). Hence, the correct identification of \( \text{Y} \) based on the given options is (d) both (a) and (b).

Key Concepts

Reduction Reactions

Reduction reactions involve the gain of hydrogen or the loss of oxygen. In the context of boron hydrides, these reactions typically involve the use of strong reducing agents like lithium aluminum hydride ( LiAlH_4 ). LiAlH_4 acts by supplying hydrogen atoms, which then add to the compound being reduced. In the original exercise, compound X reacts with LiAlH_4 to form compound Y , a boron hydride that contains exactly 21.72% hydrogen. This specific transformation is a classic example of a reduction reaction because LiAlH_4 is reducing X by boosting its hydrogen content significantly. Such an increase in hydrogen saturation in the product is distinct to reduction processes. When interpreting results from a reduction reaction, it's important to consider the reactants' roles. Here, it's the reducing agent LiAlH_4 that is crucial for transforming X into Y , enabling it to attain the explosive reaction characteristics seen with Y when it subsequently interacts with air.

Percent Composition Calculation

Understanding percent composition is vital in chemistry to determine the proportion of each element within a compound. In the given exercise, you're working with the percent composition of hydrogen in compound Y to determine its identity.The percent composition formula is expressed as: \[ \text{Percent Composition of Element} = \left( \frac{\text{Mass of the Element in 1 Mole}}{\text{Molar Mass of the Compound}} \right) \times 100 \%\] This formula was used to verify Y, showing it includes 21.72% hydrogen. To calculate correctly, assume a value for the hydrogen atoms (say 10 for B_4H_{10}) and rearrange the equation to find the molar mass:\[ n = 21.72 \times \frac{M}{100} \]. Solving this allows you to check which molecular formulas for Y match your calculations, enabling determination if it's B_4H_{10} or B_2H_6.In essence, percent composition helps us connect molecular structures to their empirical formulas and properties of the compound.

Chemical Reactions with Air

Boron hydrides are known for their vigorous reactions with air. When analyzing compound Y , realizing its reactive nature with air to form boron trioxide ( B_2O_3 ) is key. When Y , either B_4H_{10} or B_2H_6 , is exposed to air, it reacts explosively. This explosive reaction signifies a rapid oxidation process, where hydrogen atoms in the hydride react with oxygen to produce water, while the boron components oxidize to form B_2O_3 . This characteristic is especially notable, reflecting the structural instability of boron hydrides in oxygen-rich environments. For students, it's crucial to recognize that the attractions between free oxygen in air and the electrons in boron hydrides drive these powerful reactions. Hence, understanding the behavior of Y upon exposure to air helps to identify it among other potential candidates, linking its chemical makeup to observable physical phenomena.