Answer:
The B atom in \[B{{F}_{3}}\] or \[BB{{r}_{3}}\]
has only six electrons in its valence shell and hence can accept a pair of
electrons to complete its octet. Therefore, both \[B{{F}_{3}}\] and \[BB{{r}_{3}}\]
act as Lewis acids. But in \[B{{F}_{3}}\], the sizes of empty 2p-orbital of B
and the 2p-orbital of F containing the lone pair of electrons are almost
identical and hence effective \[p\pi -p\pi \] bonding occurs. As a result, the
lone pair of F is donated to B atom and hence the electron-deficiency of boron
decreases. In contrast, in \[BB{{r}_{3}},\] the size of 4p-orbital of Br containing
the lone pair of electrons is much bigger than the empty 2p-orbital of B and
hence donation of lone pair of electrons of Br to B does not occur to any
significant extent. As a result, the electron- deficiency of B is much higher
in \[BB{{r}_{3}},\] than that in \[B{{F}_{3}},\] and hence \[BB{{r}_{3}},\] is
a stronger Lewis acid than \[B{{F}_{3}}\].
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