Question

The lewis acidity of $B{F}_3$ is less than $BC{l}_3$ even though fluorine is more electronegative than chlorine. It is due to:

• A. stronger $1p\left(B\right)-3p\left(Cl\right)\sigma -$ bonding
• B. stronger $2p\left(B\right)-3p\left(Cl\right)\pi -$ bonding
• C. stronger $2p\left(B\right)-2p\left(F\right)\sigma -$ bonding
• D. stronger $2p\left(B\right)-2p\left(F\right)\pi -$ bonding

Answer 1

The correct option is D stronger $2p\left(B\right)-2p\left(F\right)\pi -$ bonding

Boron and fluorine do not have $d-$orbitals. Hence, both of these participate in strong $2p\left(B\right)-2p\left(F\right)$ back $\pi$-bonding. On the other hand, due to large size and availability of vacant $d-$orbitals, $Cl$ does not participate in such type of back $\pi$-bonding. Hence, $B{F}_3$ is less acidic than $BC{l}_3$.