/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 79 Which of the following are forme... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Which of the following are formed when \(\mathrm{BCl}_{3}\) is treated with \(\mathrm{NH}_{3}\). (a) \(\mathrm{B}\left(\mathrm{NH}_{2}\right)_{3}\) (b) \(\mathrm{B}_{2}\left(\mathrm{NH}_{3}\right)_{2}\) (c) \(\mathrm{BN}\) (d) All of these

Short Answer

Expert verified
(a) \(\mathrm{B\left(NH_2\right)_3}\) and (c) \(\mathrm{BN}\) are formed.

Step by step solution

01

Understanding the Reaction

When \(\mathrm{BCl}_3\) is treated with \(\mathrm{NH}_3\), \(\mathrm{BCl}_3\) acts as a Lewis acid (electron pair acceptor) and \(\mathrm{NH}_3\) acts as a Lewis base (electron pair donor). The reaction involves \(\mathrm{NH}_3\) donating a lone pair to \(\mathrm{BCl}_3\) to form an adduct.
02

Formation of Adduct

Initially, \(\mathrm{BCl}_3\) and \(\mathrm{NH}_3\) combine to form the adduct \(\mathrm{Cl}_3\mathrm{B}-\mathrm{NH}_3\). This complex can lead to further reactions, eventually leading to polymeric nitrogen-boron compounds.
03

Decomposition to Other Products

The interaction can lead to the loss of hydrogen chloride (\(\mathrm{HCl}\)), forming \(\mathrm{B\left(NH_2\right)_3}\) after elimination of \(\mathrm{HCl}\) molecules. Thermal decomposition of these adducts can sometimes yield \(\mathrm{BN}\) in a polymeric form.
04

Conclusion on Possible Products

Both \(\mathrm{B\left(NH_2\right)_3}\) and \(\mathrm{BN}\) are typically formed during the reaction process. There is no mention of \(\mathrm{B_2\left(NH_3\right)_2}\) as a common product without detailed conditions; thus it is not included in typical reaction predictions.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Lewis Acids and Bases
In the realm of inorganic chemistry, the terms "Lewis acids" and "Lewis bases" are crucial to understanding molecular interactions. A Lewis acid is defined as a substance that can accept a pair of electrons. On the other hand, a Lewis base is an entity that donates electron pairs.
  • Take \( \mathrm{BCl}_3 \) for example. It is a renowned Lewis acid because of its electron-deficient boron center.
  • Ammonia, \( \mathrm{NH}_3 \), serves as a Lewis base since it has a lone pair of electrons available for donation.
This interaction is not merely theoretical. It's exemplified when \( \mathrm{BCl}_3 \) comes into contact with \( \mathrm{NH}_3 \). The reaction is initiated as \( \mathrm{NH}_3 \) donates its lone pair to the vacant orbital of \( \mathrm{BCl}_3 \). This results in the formation of a new bond between boron and nitrogen. In practical chemistry, identifying which compounds serve as Lewis acids and bases helps in predicting reactions and understanding the nature of product formation.
Adduct Formation
Adduct formation is a fundamental concept in chemistry, particularly in reactions involving Lewis acids and bases. When talking about adduct formation, it refers to the creation of a complex where two or more distinct molecules come together.
  • Initially, when \( \mathrm{BCl}_3 \) and \( \mathrm{NH}_3 \) react, they form an adduct by combining into \( \mathrm{Cl}_3\mathrm{B}-\mathrm{NH}_3 \).
  • This complex manifests itself through the new covalent bond between boron and nitrogen, highlighting the transfer of an electron pair from nitrogen to boron.
Adduct formation is not just a straightforward endpoint but often a stepping stone to other chemical transformations. These new complexes can engage in further reactions or breakdown under suitable conditions to form various products, as seen when they undergo changes leading to polymeric compounds. This step is crucial because it showcases the dynamic nature of chemical reactions, where initial adducts can evolve into more complex structures.
Polymeric Compounds
Polymeric compounds, often resulting from chemical reactions involving simple molecules, have a wide array of applications. An interesting example is when the adduct \( \mathrm{Cl}_3\mathrm{B}-\mathrm{NH}_3 \) undergoes further reaction steps.
  • Through the release of hydrochloric acid (\( \mathrm{HCl} \)), one possible product is \( \mathrm{B\left(NH_2\right)_3} \). This can be considered a step towards polymerization.
  • At higher temperatures, these intermediates can decompose further into polymeric boron nitride (\( \mathrm{BN} \)).
Boron nitride bears significant industrial importance due to its unique properties similar to carbon-based polymers like graphite and diamond. It's notable for its electrical insulation and thermal conductivity, making it suitable for specialized material applications. Understanding the formation of polymeric compounds in such reactions opens the door to material science innovations, demonstrating how simple molecule interactions can lead to complex and useful materials.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In analogy to \(\mathrm{O}_{2}^{+}\left[\mathrm{PtF}_{6}\right]^{-}\)a compound \(\mathrm{N}_{2}^{+}\left[\mathrm{PtF}_{6}\right] \mathrm{Z}^{-}\), will not be formed because: (a) The ionization of \(\mathrm{N}_{2}\) gas is higher than that of \(\mathrm{N}\) atom (b) The ionization of \(\mathrm{N}_{2}\) gas is lower than that of \(\mathrm{O}_{2}\) gas (c) The ionization of \(\mathrm{N}_{2}^{2}\) gas is higher than that of \(\mathrm{O}_{2}\) gas (d) none of these

The number of P-O-P bridges in the structure of phosphorus pentoxide and phosphorus trioxide are respectively. (a) 5,5 (b) 5,6 (c) 6,5 (d) 6,6

A compound A of S, Cl and \(O\) has a vapour density of 67.5. It reacts with water to form two acids and reacts with \(\mathrm{KOH}\) to form two salts \(\mathrm{B}\) and \(\mathrm{C}\). while B gives white precipitate with \(\mathrm{AgNO}_{3}\) solution, \(\mathrm{C}\) gives white precipitate with \(\mathrm{BaCl}_{2}\) solution. Identify \(\mathrm{A}\) (a) \(\mathrm{SOCl}_{2}\) (b) \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) (c) \(\mathrm{SO}_{2} \mathrm{Cl}\) (d) none of these

Arrange the following acids (1) \(\mathrm{H}_{2} \mathrm{SO}_{3}\) (2) \(\mathrm{H}_{3} \mathrm{PO}_{3}\) (3) \(\mathrm{HClO}_{3}\), in the increasing order of acid strength. (a) \(2>3>1\) (b) \(3>2>1\) (c) \(1>3>2\) (d) \(1>2>3\)

\(\mathrm{B}(\mathrm{OH})_{3}+\mathrm{NaOH} \rightleftharpoons \mathrm{NaBO}_{2}+\mathrm{Na}\left[\mathrm{B}(\mathrm{OH})_{4}\right]+\mathrm{H}_{2} \mathrm{O}\) How can this reaction be made to proceed in forward direction? (a) addition of cis- 1,2 -diol (b) addition of borax (c) addition of trans- 1,2 -diol (d) addition of \(\mathrm{Na}_{2} \mathrm{HPO}_{4}\)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.