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Which of the following compound react with \(\mathrm{BF}_{3} ?\) (a) \(\mathrm{Al}_{2} \mathrm{Cl}_{6}\) (b) \(\mathrm{NH}_{3}\) (c) Ethers (d) All of these

Short Answer

Expert verified
(b) \(\mathrm{NH_3}\) and (c) Ethers react with \(\mathrm{BF}_3\).

Step by step solution

01

Understanding the Problem

The problem requires us to identify which compound can react with boron trifluoride (\(\mathrm{BF}_3\)). Boron trifluoride is a Lewis acid, meaning it can accept an electron pair.
02

Identify Lewis Bases

To solve this problem, recognize that Lewis bases are compounds that donate electron pairs. Look for compounds that have lone pairs of electrons, as they are potential Lewis bases and can react with \(\mathrm{BF}_3\).
03

Analyze Ammonia \((\mathrm{NH_3})\)

\(\mathrm{NH}_3\) is a molecule with a nitrogen atom that has a lone pair of electrons. Thus, it behaves as a Lewis base and can donate these electrons to \(\mathrm{BF}_3\).
04

Consider Ethers

Ethers, characterized by an \(\mathrm{R-O-R'}\) structure, have oxygen atoms with lone pairs of electrons. These lone pairs can be donated, making ethers Lewis bases as well.
05

Evaluate \(\mathrm{Al_2Cl_6}\)

\(\mathrm{Al_2Cl_6}\) is an electron-deficient structure and doesn't have lone pairs to donate as a Lewis base, nor can it directly react with \(\mathrm{BF}_3\).
06

Conclude the Results

Since \(\mathrm{NH_3}\) and ethers can donate electron pairs, they can both react with \(\mathrm{BF}_3\). \(\mathrm{Al_2Cl_6}\) cannot. Therefore, the correct options are \((\mathrm{b})\) and \((\mathrm{c})\).

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Key Concepts

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

Boron trifluoride reactions
Boron trifluoride (\(\mathrm{BF}_3\)) is a classic example of a Lewis acid. A Lewis acid is any compound that can accept a pair of electrons. This ability is due to \(\mathrm{BF}_3\)'s electron-deficient structure. In the case of \(\mathrm{BF}_3\), the boron atom has an incomplete octet. This means it has only six valence electrons instead of the stable eight.

This deficiency makes \(\mathrm{BF}_3\) highly reactive toward Lewis bases. In a chemical reaction, \(\mathrm{BF}_3\) will accept an electron pair from other compounds that have lone pairs of electrons.

  • These compounds are typically referred to as Lewis bases.
  • When \(\mathrm{BF}_3\) reacts with a Lewis base, it forms a coordinate covalent bond.
  • This bond is different from a typical covalent bond because both electrons come from the Lewis base.
Lewis base identification
Identifying Lewis bases involves looking for compounds with lone pairs of electrons that can be donated. This is because a Lewis base is a compound that provides an electron pair to a Lewis acid in a chemical reaction. Here's how we can identify Lewis bases:

  • Compounds such as \(\mathrm{NH}_3\) (ammonia) have lone pairs. The nitrogen atom in \(\mathrm{NH}_3\) has a lone pair and can donate it to form a bond with \(\mathrm{BF}_3\).
  • Ethers, with the general structure \(\mathrm{R-O-R'}\), also have lone electron pairs on the oxygen atom. These lone pairs can be donated, making ethers good Lewis bases.
  • In contrast, compounds like \(\mathrm{Al}_2\mathrm{Cl}_6\), though also electron-deficient, do not have lone pairs for donation and thus do not function as Lewis bases.

Recognizing these characteristics is crucial in predicting which molecules can react as bases in reactions with Lewis acids like \(\mathrm{BF}_3\).
Electron pair donation
When it comes to electron pair donation, it's important to understand the role of lone pairs. Lone pairs are pairs of valence electrons that are not shared with another atom. They reside on atoms that can potentially participate in bonding.

  • For a Lewis base to donate an electron pair, the electrons must be readily available for bond formation. This typically means they are "lone" or unbonded electrons.
  • Taking \(\mathrm{NH}_3\) as an example, the molecule has a lone pair located on the nitrogen atom. When \(\mathrm{NH}_3\) reacts with \(\mathrm{BF}_3\), it donates this lone pair to the empty orbital of boron, allowing the formation of a bond.
  • Similarly, ethers have an oxygen atom with lone pairs that can be shared with a Lewis acid. The oxygen donates its lone pair to \(\mathrm{BF}_3\) or other suitable Lewis acids, forming a coordinate bond.

Understanding which molecules can share their electron pairs allows chemists to predict and control the outcomes of chemical reactions, especially those involving Lewis acid-base interactions.

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Most popular questions from this chapter

Which of the following statement is correct? (a) \(\mathrm{Al}\) forms \(\left[\mathrm{AlF}_{6}\right]^{3-}\) ion but \(\mathrm{B}\) does not form \(\left[\mathrm{BF}_{6}\right]^{3-}\) ion. (b) Boron and aluminium halides behave as Lewis acids. (c) The \(\mathrm{p} \pi-\mathrm{p} \pi\) back bonding occurs in the halides of boron and not in those of aluminium. (d) All of these

Maximum bond angle is present in case of (a) \(\mathrm{BCl}_{3}\) (b) \(\overrightarrow{\mathrm{BF}}_{3}\) (c) \(\mathrm{BBr}_{3}\) (d) same in all

Which of the following is not a Lewis acid? (a) \(\mathrm{AlCl}_{3} \cdot 6 \mathrm{H}_{2} \mathrm{O}\) (b) \(\mathrm{SnCl}_{4}\) (c) \(\mathrm{FeCl}_{3}\) (d) \(\mathrm{AlCl}_{3}\)

In the reaction: \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} .18 \mathrm{H}_{2} \mathrm{O} \frac{\text { heat }}{-18 \mathrm{H}_{2} \mathrm{O}}\) \(\mathrm{A} \quad{ }_{90^{\circ} \mathrm{C}}{\longrightarrow} \mathrm{B}+\mathrm{C}\). The product \(\mathrm{A}, \mathrm{B}\) and \(\mathrm{C}\) are respectively (a) \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}, \mathrm{Al}_{2} \mathrm{O}_{3}, \mathrm{SO}_{3}\) (b) \(\mathrm{Al}_{2} \mathrm{O}_{3}, \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}, \mathrm{SO}_{3}\) (c) \(\mathrm{Al}_{2} \mathrm{SO}_{4}, \mathrm{Al}_{2} \mathrm{O}_{3}, \mathrm{SO}_{3}\) (d) \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}, \mathrm{Al}_{2} \mathrm{O}_{3}, \mathrm{SO}_{2}\)

Aluminium chloride in acidified aqueous solution forms: (a) Octahedral \(\left[\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{+3}\) ion (b) Tetrahedral \(\left.[\mathrm{A}]\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{+3}\) ion (c) Octahedral \(\left[\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4}\right]^{+3}\) ion (d) Tetrahedral \(\left[\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)\right]^{+3}\) ion

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