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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

  1. ABCl 3 < BF 3 < B ( OMe ) 3 < B NMe 2 3
  2. BBF 3 < BCl 3 < B ( OMe ) 3 < B NMe 2 3
  3. CBCl 3 < B NMe 2 3 < B ( OMe ) 3 < BF 3
  4. 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

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