/*! 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 62 Which of the following statement... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Which of the following statements is correct? (a) Both \(\mathrm{AlCl}_{3}\) and \(\mathrm{BCl}_{3}\) are not Lewis acids. (b) Both \(\mathrm{AlCl}_{3}\) and \(\mathrm{BCl}_{3}\) are equally strong Lewis acids. (c) \(\mathrm{AlCl}_{3}\) is stronger Lewis acid than \(\mathrm{BCl}_{3}\). (d) \(\mathrm{BCl}_{3}\) is stronger Lewis acid than \(\mathrm{AlCl}_{3}\).

Short Answer

Expert verified
The correct statement is (d): µþ°ä±ô₃ is a stronger Lewis acid than ´¡±ô°ä±ô₃.

Step by step solution

01

Understanding Lewis Acids

A Lewis acid is defined as a compound or molecule that can accept a pair of electrons. Many metal halides, such as ´¡±ô°ä±ô₃ and µþ°ä±ô₃, exhibit Lewis acidity because they have an incomplete octet and can accept electron pairs to achieve a stable electron configuration.
02

Analyzing Aluminum Chloride (´¡±ô°ä±ô₃)

´¡±ô°ä±ô₃ is a common Lewis acid because aluminum has six valence electrons in its neutral state and can accept two more electrons to complete its octet. It readily accepts electron pairs from donors (Lewis bases).
03

Analyzing Boron Chloride (µþ°ä±ô₃)

µþ°ä±ô₃ is also a Lewis acid as boron has three valence electrons and an empty p-orbital, allowing it to accept a pair of electrons. Despite that, µþ°ä±ô₃ is known to be more acidic due to boron's higher electronegativity and stronger tendency to pull electron density towards itself.
04

Comparing Strength of Acidic Behavior

µþ°ä±ô₃ is considered a stronger Lewis acid than ´¡±ô°ä±ô₃. The greater electronegativity and smaller atomic radius of boron compared to aluminum allow µþ°ä±ô₃ to have a stronger affinity for electrons, enhancing its Lewis acidity.
05

Concluding the Correct Statement

Based on the analysis of Lewis acidity, statement (d) is correct: µþ°ä±ô₃ is a stronger Lewis acid than ´¡±ô°ä±ô₃ because of boron's smaller size and higher electronegativity.

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.

´¡±ô°ä±ô₃
Aluminum chloride, represented as \( ext{AlCl}_3\), is a prominent example of a Lewis acid. In chemistry, Lewis acids are those that can accept pairs of electrons. Aluminum in \( ext{AlCl}_3\) has only six valence electrons, which is not enough to complete its octet. This makes it eager to accept electron pairs from electron-rich species, known as Lewis bases.
´¡±ô°ä±ô₃ is often used in industrial processes, such as the Friedel-Crafts reaction, due to its strong ability to accept electrons. When \( ext{AlCl}_3\) accepts electron pairs, it forms a complex, achieving a stable electronic configuration. Understanding this behavior is crucial in predicting the reaction pathways in organic chemistry involving ´¡±ô°ä±ô₃.
µþ°ä±ô₃
Boron chloride, or \( ext{BCl}_3\), is another well-known Lewis acid. Boron is at the center and is surrounded by three chlorine atoms, creating vacancies for accepting electrons. Boron only has three valence electrons and an empty p-orbital, allowing \( ext{BCl}_3\) to accept an electron pair easily.
One interesting feature of \( ext{BCl}_3\) is that even though it resonates as stable through its resonance structures, it still has a significant electron deficiency. This deficiency leads to its desirable property of accepting electrons, making it a stronger Lewis acid compared to some other elements like aluminum, due partly to boron's higher electronegativity.
In chemical reactions, \( ext{BCl}_3\) acts aggressively to seek additional electrons, contributing to its inability to achieve an octet in its neutral state.
electronegativity
Electronegativity is a fundamental concept in understanding the strength of Lewis acids, such as \( ext{BCl}_3\) and \( ext{AlCl}_3\). It refers to the ability of an atom to attract or pull electrons towards itself. Generally, atoms with higher electronegativity values are better at attracting electrons.
In the case of \( ext{BCl}_3\) and \( ext{AlCl}_3\), boron is more electronegative than aluminum. This higher electronegativity allows \( ext{BCl}_3\) to act as a stronger Lewis acid, drawing in more electron density compared to \( ext{AlCl}_3\).
Understanding electronegativity helps explain why some substances have higher reactivity and acidity levels, as electron-withdrawing tendencies increase their capability to accept electron pairs.
octet rule
The octet rule is an essential principle in chemistry that refers to the preference of atoms to have eight electrons in their valence shell, resembling the electron configuration of noble gases. Most elements, including aluminum and boron, strive to achieve this stable configuration by forming bonds with other atoms.
For \( ext{AlCl}_3\), aluminum seeks to complete its octet by accepting electrons since it initially has only six valence electrons. This tendency classifies it as a Lewis acid. Similarly, \( ext{BCl}_3\) attempts to satisfy the octet rule by accepting electron pairs due to its three valence electrons and the subsequent electron deficiency.
Understanding the octet rule clarifies why elements exhibit certain behaviors in chemical reactions. It helps explain the electron-accepting nature of aluminum chloride and boron chloride, as both aim to complete their valence shells by attracting electron pairs.

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

Amorphous boron is extracted from borax by following steps: \(\operatorname{Borax} \stackrel{(A)}{\longrightarrow} \mathrm{H}_{3} \mathrm{BO}_{3} \stackrel{(B)}{\longrightarrow} \mathrm{B}_{2} \mathrm{O}_{3} \stackrel{(C)}{\longrightarrow}\) Boron (A) and \((C)\) are : (a) \(\mathrm{H}_{2} \mathrm{SO}_{4}, \mathrm{Al}\) (b) \(\mathrm{HCl}, \mathrm{C}\) (c) \(\mathrm{HCl}, \mathrm{Fe}\) (d) \(\mathrm{H}_{2} \mathrm{SO}_{4}, \mathrm{Na}\)

Ge(II) compounds are powerful reducing agents whereas \(\mathrm{Pb}(\mathrm{IV})\) compounds are strong oxidising agents. It'can be due to: (a) \(\mathrm{Pb}\) is more electropositive than Ge (b) ionisation potential of lead is less than that of Ge (c) ionic radii of \(\mathrm{Pb}^{2+}\) and \(\mathrm{Pb}^{4+}\) are larger than those of \(\mathrm{Ge}^{2+}\) and \(\mathrm{Ge}^{4+}\) (d) more pronounced inert pair effect in lead than in Ge

Which one of the following reactions does not occur? (a) \(\mathrm{F}_{2}+2 \mathrm{Cl}^{-} \longrightarrow 2 \mathrm{~F}+\mathrm{Cl}_{2}\) (b) \(\mathrm{Cl}_{2}+2 \mathrm{~F} \longrightarrow 2 \mathrm{Cl}^{-}+\mathrm{F}_{2}\) (c) \(\mathrm{Cl}_{2}+2 \mathrm{Br}^{-} \longrightarrow 2 \mathrm{Cl}^{-}+\mathrm{Br}_{2}\) (d) \(\mathrm{F}_{2}+2 \mathrm{Br}^{-} \longrightarrow 2 \mathrm{~F}^{-}+\mathrm{Br}_{2}\)

In the process \(X \mathrm{H}_{3}+\mathrm{H}^{+} \longrightarrow X \mathrm{H}_{4}^{+}\) ( \(X\) being \(\mathrm{N}, \mathrm{P}\), As or Sb), the maximum increase in the bond angle will be in the case of : (a) \(\mathrm{NH}_{3}\) (b) \(\mathrm{PH}_{3}\) (c) \(\mathrm{AsH}_{3}\) (d) \(\mathrm{SbH}_{3}\)

What is the basicity of phosphorus acid?

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.