How to Draw the Lewis Structure and Determine the Resonance Exhibit for BCl₃

To draw the Lewis structure for BCl₃ (Boron Trichloride), follow these steps:

1. Count the total number of valence electrons: Boron (B) has 3 valence electrons, and each Chlorine (Cl) atom has 7 valence electrons. Since there are 3 Chlorine atoms, the total number of valence electrons is 3 (from Boron) + 3 × 7 (from Chlorine) = 24 valence electrons.
2. Place the least electronegative atom in the center: Boron is less electronegative than Chlorine, so it will be the central atom.
3. Connect the outer atoms to the central atom with single bonds: Draw single bonds between Boron and each Chlorine atom. This uses up 6 valence electrons (2 electrons per bond).
4. Distribute the remaining electrons: After forming the single bonds, you have 18 valence electrons left. Place these electrons around the Chlorine atoms to complete their octets. Each Chlorine atom will have 6 additional electrons, making a total of 8 electrons around each Chlorine atom.
5. Check the octet rule for the central atom: Boron only has 6 electrons around it (3 single bonds), which is less than an octet. However, Boron is an exception to the octet rule and can be stable with fewer than 8 electrons.

Resonance in BCl₃:

BCl₃ does not exhibit resonance. Resonance structures occur when there are multiple ways to arrange the electrons in a molecule, typically involving double or triple bonds. In BCl₃, all the bonds are single bonds, and there are no lone pairs on the central Boron atom that could be delocalized. Therefore, BCl₃ has only one valid Lewis structure and does not exhibit resonance.