Problem 6 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

The correct statements among (a) to (d) are : [Main April 10, 2019 (II)] (i) saline hydrides produce \(\mathrm{H}_{2}\) gas when reacted with \(\mathrm{H}_{2} \mathrm{O}\). (ii) reaction of \(\mathrm{LiAlH}_{4}\) with \(\mathrm{BF}_{3}\) leads to \(\mathrm{B}_{2} \mathrm{H}_{6}\). (iii) \(\mathrm{PH}_{3}\) and \(\mathrm{CH}_{4}\) are electron \(-\) rich and electron \(-\) precise hydrides, respectively, (iv) \(\mathrm{HF}\) and \(\mathrm{CH}_{4}\) are called as molecular hydrides. (a) (i), (ii), (iii) and (iv) (b) (iii) and (iv) only (c) (i), (iii) and (iv) only (d) (i), (ii) and (iii) only

Step-by-Step Solution

Answer

Option (a) is correct: all statements (i), (ii), (iii), and (iv) are true.

Step 1: Analyzing Statement (i)

Statement (i) says that saline hydrides produce \( \mathrm{H}_{2} \) gas when reacted with \( \mathrm{H}_{2}\mathrm{O} \). Saline hydrides, such as \( \mathrm{NaH} \), react with water to form hydrogen gas: \( \mathrm{NaH} + \mathrm{H}_{2}\mathrm{O} \rightarrow \mathrm{NaOH} + \mathrm{H}_{2} \). This statement is correct.

Step 2: Analyzing Statement (ii)

Statement (ii) says that the reaction of \( \mathrm{LiAlH}_{4} \) with \( \mathrm{BF}_{3} \) leads to \( \mathrm{B}_{2} \mathrm{H}_{6} \). This reaction indeed produces diborane \( \mathrm{B}_{2} \mathrm{H}_{6} \). Thus, this statement is correct.

Step 3: Analyzing Statement (iii)

Statement (iii) claims that \( \mathrm{PH}_{3} \) and \( \mathrm{CH}_{4} \) are electron-rich and electron-precise hydrides, respectively. \( \mathrm{PH}_{3} \), with a lone pair, is electron-rich, and \( \mathrm{CH}_{4} \) has complete octet without a lone pair, making it electron-precise. Therefore, this statement is correct.

Step 4: Analyzing Statement (iv)

Statement (iv) asserts that \( \mathrm{HF} \) and \( \mathrm{CH}_{4} \) are molecular hydrides. Both \( \mathrm{HF} \) and \( \mathrm{CH}_{4} \) are indeed molecular hydrides because they consist of discrete molecules. This statement is correct.

Step 5: Conclusion

Since all individual statements (i), (ii), (iii), and (iv) are correct, the correct option according to the list given is option (a).

Key Concepts

Hydride Reactions

Hydride reactions are essential in chemistry as they involve compounds that contain hydrogen with more electronegative atoms. One common hydride reaction is the interaction of saline hydrides with water. For example, when sodium hydride \( \mathrm{NaH} \) is added to water, the reaction produces hydrogen gas \( \mathrm{H}_2 \) and sodium hydroxide. This can be represented by the equation:

• NaH + H\(_2\)O \( \rightarrow \) NaOH + H\(_2\)

Such reactions demonstrate the ability of hydrides, especially ionic ones like sodium hydride, to produce hydrogen gas. It's a significant reaction in industries where hydrogen production is required quickly and efficiently. This is due to the high reactivity of the hydride ions with water, leading to the release of hydrogen gas.

Electron-Rich Compounds

Electron-rich compounds are those that contain atoms with extra pairs of electrons, known as lone pairs. These compounds often act as nucleophiles in chemical reactions, seeking positively charged or electron-deficient counterparts. An example is phosphine (\( \mathrm{PH}_{3} \)), which is considered electron-rich because of its lone pair on the phosphorus atom.

• Electron-rich compounds can participate in various reactions, such as nucleophilic substitution and addition reactions.
• They play a crucial role in organic and inorganic chemistry, often seen in catalytic mechanisms.

Understanding the behavior of electron-rich compounds helps chemists manipulate chemical reactions for desired outcomes, such as in the creation of pharmaceuticals and polymers.

Molecular Hydrides

Molecular hydrides are compounds that feature hydrogen atoms bonded to more electronegative elements like carbon or fluorine. These hydrides are characterized by their existence as individual molecules, rather than a lattice or network. Examples include methane (\( \mathrm{CH}_{4} \)) and hydrogen fluoride (\( \mathrm{HF} \)).

• Methane is a simple hydrocarbon and a fundamental building block in organic chemistry.
• Hydrogen fluoride, though toxic, is used in processes like glass etching and in the synthesis of other chemicals.

Molecular hydrides are crucial in various chemical processes due to their stable molecular structures and are often used in research and industrial applications for their chemical properties.

Saline Hydrides

Saline hydrides are ionic compounds composed of metal cations and hydride anions (\( \mathrm{H}^{-} \)). A well-known example of a saline hydride is sodium hydride (\( \mathrm{NaH} \)). These hydrides are characterized by their ability to react vigorously with water to release hydrogen gas.

• The reaction of saline hydrides with water is useful for producing hydrogen gas efficiently.
• They serve as strong bases and are used in various chemical syntheses.

Saline hydrides are especially important in laboratory settings for deprotonation reactions and during the reduction of organic compounds, providing a versatile toolkit for chemists.

Diborane Synthesis

Diborane (\( \mathrm{B}_{2} \mathrm{H}_{6} \)) is a unique compound in boron chemistry, known for its electron-deficient nature. One of the methods to synthesize diborane involves the reaction of lithium aluminum hydride (\( \mathrm{LiAlH}_{4} \)) with boron trifluoride (\( \mathrm{BF}_{3} \)). This reaction is crucial as it yields diborane, an essential compound for many chemical processes.

• Diborane is used as a reducing agent and in the synthesis of other boron-containing compounds.
• Its electron-deficient structure leads to interesting bonding characteristics, making it a subject of study in various scientific research fields.

Understanding diborane synthesis is essential for those studying inorganic chemistry, as it highlights the unique properties and reactivities of boron hydrides.