KVPY2018Chemistryp Block Elements (Group 13 & 14)
The tendency of X in BX 3   X = F ,   Cl ,   OMe ,   NMe 2 to form a π -bond with boron follows the order
Options
- ABCl 3 < BF 3 < B ( OMe ) 3 < B NMe 2 3
- BBF 3 < BCl 3 < B ( OMe ) 3 < B NMe 2 3
- CBCl 3 < B NMe 2 3 < B ( OMe ) 3 < BF 3
- DBCl 3 < BF 3 < B NMe 2 3 < B ( OMe ) 3
Correct answer
A. BCl 3 < BF 3 < B ( OMe ) 3 < B NMe 2 3
Step-by-step solution
Boron atom has an incomplete octet. It can accept a lone pair from X given in the question and can also donate electron to it, thus it can form a π -bond also known as back bonding. The extent of back bonding depends upon the electronegativity of X and on the size of the valence shell. Lesser is the electronegativity of X and similar is the size of valence shell of X and B , more will be the extend of back bonding. Cl and F are more electronegative than OMe and NMe group, so there extent of back bonding will b