This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: then, distribute the remaining e- in best compliance possible with the octet r u l e Example 1. Cl 2 step 1 . Total valence e- available = (2 Cl atoms) x (7 valence e- per Cl) = 14 valence e- step 2 . Cl Cl step 3 . Cl Cl one e- pair between atoms = bond 12 e- pairs left to distribute step 4 . Cl Cl Lewis dot structure of Cl 2 each Cl atom has or shares 8 valence e- each Cl atom "feels" a bit like a noble gas bonding pair of electrons lone pairs there are 6 lone pairs in Cl 2 there are 12 lone pair e- in Cl Chemistry 111 Fall, 2005 Prof. Simpson Example 2. CH 4 Example 3. NH 4 + Example 4. CH 3-Example 5. PF 5...
View Full Document