Question

Diborane can't be obtained from (a) \(\mathrm{NaBH}_{4}+\mathrm{I}_{2}\) (b) \(\mathrm{Na}_{2} \mathrm{~B}_{4} \mathrm{O}_{7}+\mathrm{HCl}\) (c) \(\mathrm{BF}_{3}+\mathrm{NaH}\) (d) \(\mathrm{BF}_{3}^{2}+\mathrm{LiAlH}_{4}\)

Short answer

Diborane can't be obtained from (b) \(\mathrm{Na}_2 \mathrm{B}_4 \mathrm{O}_7 + \mathrm{HCl}\).

Step-by-step solution

Analyze the Reaction Pathways

To determine which option cannot produce diborane, we need to consider the chemical pathways and reactions involved in each choice. Diborane, \(\mathrm{B}_{2}\mathrm{H}_6\), is typically generated by reducing boron containing compounds using appropriate hydride reagents.

Review Option (a) - Sodium Borohydride and Iodine

In option (a), \(\mathrm{NaBH}_{4}\) is used with \(\mathrm{I}_{2}\). Here, the combination of sodium borohydride and iodine can lead to the formation of diborane because the iodide ion can oxidize hydrogen from \(\mathrm{NaBH}_{4}\), facilitating the formation of \(\mathrm{B}_2\mathrm{H}_6\).

Examine Option (b) - Borax and Hydrochloric Acid

In option (b), \(\mathrm{Na}_2 \mathrm{B}_4 \mathrm{O}_7\) with \(\mathrm{HCl}\) results in \(\mathrm{B}(\mathrm{OH})_3\), not diborane. This is because borax, when treated with an acid like HCl, forms boric acid, which does not lead to diborane.

Check Option (c) - Boron Trifluoride and Sodium Hydride

In option (c), \(\mathrm{BF}_3\) with \(\mathrm{NaH}\) can form diborane. The sodium hydride provides hydrogen which can reduce the boron trifluoride to form \(\mathrm{B}_2\mathrm{H}_6\).

Evaluate Option (d) - Boron Trifluoride and Lithium Aluminum Hydride

In option (d), the reaction between \(\mathrm{BF}_3\) and \(\mathrm{LiAlH}_4\) is a well-known method to synthesize diborane as the strong reducing agent \(\mathrm{LiAlH}_4\) provides sufficient hydrogen to reduce boron trifluoride, forming \(\mathrm{B}_2\mathrm{H}_6\).

Conclusion: Identify Which Option Fails to Produce Diborane

Summarizing the analysis, option (b), using \(\mathrm{Na}_2 \mathrm{B}_4 \mathrm{O}_7\) and \(\mathrm{HCl}\), is the method that cannot be used to produce diborane as it leads only to boric acid.

Key concepts

Sodium Borohydride Oxidation

Sodium borohydride (\(\mathrm{NaBH}_{4}\)) is commonly utilized as a powerful reducing agent in chemical reactions. It can donate hydride ions (\(\mathrm{H}^-\)) to other compounds, facilitating various transformations. In the context of diborane synthesis, \(\mathrm{NaBH}_{4}\) reacts with iodine (\(\mathrm{I}_2\)) to form boranes. This occurs because iodine acts as an oxidizing agent, capable of removing hydrogen atoms from sodium borohydride. The hydrogen atoms pair up and contribute to forming diborane (\(\mathrm{B}_2\mathrm{H}_6\)). This method highlights the resilience and versatility of \(\mathrm{NaBH}_{4}\) in generating boron-hydrogen compounds successfully.

• Sodium borohydride is a reductant.
• Reacts with iodine to produce diborane.
• Iodine's role as an oxidant is crucial.

By understanding this, students can grasp the underlying mechanics of diborane formation via this pathway.

Borax and Hydrochloric Acid Reaction

Borax, scientifically known as sodium tetraborate (\(\mathrm{Na}_2 \mathrm{B}_4 \mathrm{O}_7\)), has various applications, but producing diborane isn't one of them. When reacted with hydrochloric acid (\(\mathrm{HCl}\)), the outcome is boric acid (\(\mathrm{B(OH)_3}\)), not diborane. This specific reaction showcases how important the selection of reagents is when targeting a specific product.

• Borax reacts with acids to form boric acid.
• This pathway doesn't yield diborane.
• Understanding reagent compatibility is essential.

This reveals the critical aspect of chemical transformation expectations and reinforces that not all boron compounds directly lead to diborane.

Boron Trifluoride Reduction

Boron trifluoride (\(\mathrm{BF}_3\)) is a readily available boron compound. It is often involved in reactions to produce other boron-containing structures through reduction. When combined with sodium hydride (\(\mathrm{NaH}\)), a reducing agent, boron trifluoride can be converted into diborane. The reduction occurs as the hydride ion from sodium hydride provides essential hydrogen atoms needed for the creation of diborane.

• Boron trifluoride acts as a precursor for diborane.
• Naturally pairs with hydrides for reduction.
• Reveals the versatility of boron chemistry.

This process exemplifies how selecting the correct combination of reagents can efficiently route to desired boron-based products.

Lithium Aluminum Hydride as Reducing Agent

Lithium aluminum hydride (\(\mathrm{LiAlH}_4\)) is a well-known, potent reducing agent. It functions by providing reliable sources of hydride ions used in various reduction reactions. In synthesis involving boron trifluoride (\(\mathrm{BF}_3\)), \(\mathrm{LiAlH}_4\) facilitates the formation of diborane, by donating sufficient hydride ions required to reduce boron trifluoride effectively.

• Enhances the formation of boron-hydrogen bonds.
• Known for strong reducing capabilities.
• Works synergistically with boron trifluoride.

By recognizing these interactions, students realize the crucial impact of hydride agents in successful diborane synthesis.