Question

Which is strongest Lewis acid? (a) \(\mathrm{BBr}_{3}\) (b) \(\mathrm{BI}_{3}\) (c) \(\mathrm{BF}_{3}\) (d) \(\mathrm{BCl}_{3}\)

Short answer

BI3 is the strongest Lewis acid due to weaker backbonding.

Step-by-step solution

Understanding Lewis Acids

Lewis acids are chemical species that accept a pair of electrons. Their strength is determined by their ability to attract electrons, which can be influenced by factors such as electronegativity and the size of the atoms involved.

Analyzing the Halogen Elements

Consider the halogens in each Lewis acid: fluorine (F), chlorine (Cl), bromine (Br), and iodine (I). Electronegativity and size will affect how these halogens influence the electron-deficient boron atom.

Evaluating the Electronegativity Effect

Fluorine is the most electronegative element, which could make \(\mathrm{BF}_{3}\) a strong Lewis acid due to the electron-withdrawing ability enhancing the electron deficiency of boron.

Assessing the Electron Pair Acceptance Ability

Despite high electronegativity, \(\mathrm{BF}_{3}\) forms strong backbonding which compensates the boron's electron deficiency, thus weakening its Lewis acidity.

Evaluating the Effect of Atom Size and Backbonding

Analyze how the larger size of iodine in \(\mathrm{BI}_{3}\) leads to weaker backbonding than \(\mathrm{BF}_{3}\) and \(\mathrm{BCl}_{3}\), retaining boron's electron-accepting ability. As the atoms get larger from F to I, the overlap and backbonding decrease.

Determining the Strength based on Overall Factors

\(\mathrm{BI}_{3}\) has weaker backbonding compared to the others, particularly \(\mathrm{BF}_{3}\), making it the strongest Lewis acid despite bromine and iodine being less electronegative than fluorine and chlorine.

Key concepts

Electronegativity

Electronegativity refers to an atom's ability to attract electrons towards itself. In the context of Lewis acids, electronegativity is crucial because it affects how readily an element such as boron, paired with halogens, can accept electron pairs. Fluorine is the most electronegative element among the halogens, followed by chlorine, bromine, and iodine. This means that fluorine pulls electrons toward itself more effectively than the other halogens. When fluorine is bonded to boron in \(\mathrm{BF}_3\), it enhances the electron-deficient nature of boron, theoretically increasing its Lewis acidity. However, this needs to be carefully evaluated along with other factors like backbonding effects, which we will discuss further.

Electronegativity alone is not the sole determinant of a Lewis acid's strength, but it significantly influences electron distribution in a molecule. While \(\mathrm{BF}_3\) looks strong due to its high electronegativity factor, the interplay with backbonding changes the dynamics.

Backbonding

Backbonding is a unique interaction between boron and a nearby atom like fluorine. This involves a reverse flow of electrons from a filled orbital of an electronegative atom back into the empty orbital of boron. In \(\mathrm{BF}_3\), despite boron's electron deficiency, the strong backbonding with fluorine's lone pair reduces its ability to act as a strong Lewis acid, as it compensates for the electron vacancy in boron. This could seem counterintuitive given fluorine's high electronegativity.

When backbonding occurs, boron does not remain as electron-deficient. This is why compounds such as \(\mathrm{BF}_3\) are not as strong Lewis acids as one might expect just by considering the electronegativity of the bonded atoms. The deficiency is effectively compensated, making other candidates like \(\mathrm{BI}_3\) potentially stronger due to their lack of significant backbonding interactions.

Electron Deficiency

Electron deficiency is an important characteristic in Lewis acids, particularly in boron compounds. Electron-deficient species are those that do not have a full octet in their valence shell, making them eager to accept electron pairs. Boron, frequently bonded with halogens in Lewis acids like \(\mathrm{BF}_3\), \(\mathrm{BCl}_3\), \(\mathrm{BBr}_3\), and \(\mathrm{BI}_3\), inherently tends to be electron-deficient since it has an incomplete octet.

In the evaluation of Lewis acids, a compound with higher electron deficiency will typically be a stronger Lewis acid, as it will have a greater tendency to accept electrons. However, electron deficiency is reduced if effective backbonding occurs, which means fluorine, although electronegative and enhancing deficiency traits, also provides stabilization through backbonding in \(\mathrm{BF}_3\). In contrast, the larger atom sizes of iodine and bromine in \(\mathrm{BI}_3\) and \(\mathrm{BBr}_3\) produce weaker backbonding, allowing them to retain more electron deficiency, thus potentiating stronger acidity.

Halogens in Lewis Acids

Halogens play a pivotal role in determining the stability and strength of Lewis acids, primarily through their electronegativity and ability to form bonds with an electron-deficient central atom like boron. In the series \(\mathrm{BF}_3\), \(\mathrm{BCl}_3\), \(\mathrm{BBr}_3\), and \(\mathrm{BI}_3\), halogens affect both the electron deficiency of boron and the potential for backbonding. I'll break down how each halogen impacts Boron's properties:

• **Fluorine**: Its high electronegativity enhances boron's deficiency but simultaneously provides significant backbonding, stabilizing boron's electron need.
• **Chlorine**: Less electronegative than fluorine, providing moderate backbonding and thus moderate Lewis acidity.
• **Bromine and Iodine**: Being less electronegative and larger in size, they contribute less to electron pull and backbonding, meaning \(\mathrm{BBr}_3\) and \(\mathrm{BI}_3\) can potentially act as stronger acids due to higher sustained electron deficiency.

The differences in these properties across the halogens allow us to better understand why \(\mathrm{BI}_3\) emerges as the strongest Lewis acid among these compounds. The weaker backbonding keeps boron more electron-deficient and thus a better electron acceptor than in \(\mathrm{BF}_3\), \(\mathrm{BCl}_3\), or \(\mathrm{BBr}_3\).