Question

The tendency of \(BF_3, BCI_3 \, and \, BBr_3\) behave as Lewis acid decreases in the sequence

• A. \(BCI_3 > BF_3 > BBr_3\)
• B. \(BBr_3 > BCI_3 > BF_3\)
• C. \(BBr_3 > BF_3 > BCI_3\)
• D. \(BF_3 > BCI_3 > BBr_3\)

Answer 1

The Correct Option is B

To determine the tendency of \(BF_3\), \(BCl_3\), and \(BBr_3\) to behave as Lewis acids, we should analyze their ability to accept an electron pair. A Lewis acid is a chemical species that can accept a pair of electrons due to the presence of an empty orbital. In the case of boron halides, the central boron atom is electron-deficient and has an empty p-orbital, which makes these compounds potential Lewis acids.

In halides like \(BF_3\), \(BCl_3\), and \(BBr_3\), the ability to accept electrons generally depends on two factors:

1. Back-bonding: This occurs from the filled p-orbitals of the halogen atom to the empty p-orbitals on the boron atom. Fluorine, being highly electronegative and having good orbital overlap due to its small size, facilitates significant back-bonding in \(BF_3\). This decreases the electron deficiency of the boron, making it a weaker Lewis acid than expected.
2. Size of Halogen: As we move down the group from fluorine to bromine in the periodic table, the size of the halogen atom increases, decreasing the effectiveness of back-bonding due to less orbital overlap. Thus, the Lewis acidity increases because the electron deficiency at boron is less compensated.

Considering these factors, the order of Lewis acidity is:

\(BBr_3\) > \(BCl_3\) > \(BF_3\)

This order is because bromine is less effective at back-bonding due to its larger size compared to chlorine and fluorine, resulting in \(BBr_3\) having the greatest willingness to accept an electron pair, making it the strongest Lewis acid among the three.

Therefore, the correct answer is:

\(BBr_3\) > \(BCl_3\) > \(BF_3\)