Question

Which of the following reactions is/are likely to be impossible :

• A. $(C{H}_3{)}_{2}O+B{F}_3$
• B. $(Si{H}_3{)}_{2}O+B{F}_3$
• C. $(Si{H}_3{)}_{3}N+B{F}_3$
• D. All are possible

Answer 1

Answer: (B), (C)

The correct options are
B $(Si{H}_3{)}_{2}O+B{F}_3$
C $(Si{H}_3{)}_{3}N+B{F}_3$

In $(C{H}_3{)}_{2}O$, $O$ can donate its lone pair to the empty $p$ orbital of $B{F}_3$, whereas it is not possible in the case of $(Si{H}_3{)}_{2}O$ and $(Si{H}_3{)}_{3}N$. This is because the lone pair of the central atom is already delocalised with $Si$ via $p\pi -d\pi$ backbonding. So, the lone pairs are not available for donation to the vacant p orbital of $B{F}_3$. Hence, $(Si{H}_3{)}_{2}O+B{F}_3$ and $(Si{H}_3{)}_{3}N+B{F}_3$ can not form lewis acid base pair adduct.