91Ó°ÊÓ

The Lewis acids have empty orbital so that they can accept atleast one lone pair of electrons.

a. ${\mathrm{BBr}}_3$

The boron atom in the given molecule is the thirteen-group element. Therefore, it has 3 valence electrons present in the 2s orbital (two electrons) and 2p orbital (one electron).

Therefore, boron has an empty orbital in that it can accept lone pair of electrons. Therefore, ${\mathrm{BBr}}_3$acts as a Lewis acid.

b. ${\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{OH}$

The oxygen atom in the given molecule is the sixteenth-group element. Therefore, it has 6 valence electrons in which two are used to form bonds and 4 remain on oxygen.

Therefore, it cannot accept electrons. Therefore, ${\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{OH}$does not act as a Lewis acid.

c. ${\left({\mathrm{CH}}_3\right)}_3{\mathrm{C}}^{+}$

The carbon atom in the given molecule is an electron-deficient compound due to the positive charge on it. Therefore, it has an empty orbital to accept lone pair of electrons. Therefore, ${\left({\mathrm{CH}}_3\right)}_3{\mathrm{C}}^{+}$acts as a Lewis acid.

d. ${\mathrm{Br}}^{-}$

The bromide ion has extra electrons in its outermost shell. Therefore, it does not have an empty orbital to accept electrons. Therefore, it does not act as a Lewis base.