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Hybridization

Scheme for determining hybridization of the central atom of a species:
  1. Identify the central atom of the species.
  2. Write the outermost electronic configuration of the central atom.
  3. Determine oxidation state of the central atom.
  4. Excite the electrons (if necessary) to the orbitals of higher energy in order to make the number of unpaired electrons equal to the oxidation state of the central atom.
  5. Now start putting orbitals into your pocket, beginning from s-orbitals. The number of orbitals added to the pocket must have orbitals with unpaired electrons equal to number of other atoms of the species.
  6. All the orbitals added to the pocket (including s-orbital, whether it has paired or unpaired electrons) are now summed. If there are one s-orbital and one p-orbital in the pocket, then the hybridization is sp. If there are one s-orbital and two p-orbital in the pocket, it is sp2 hybridization, and so on.
  7. Each unpaired electron left (outside the pocket) will form a pi-bond. Each orbital with paired electrons in the pocket will exist as lone pair on the central atom.
Example
Find out hybridization of the central atom in ClO3 and draw its structure.
Solution
  1. The central atom is Cl.
  2. Outermost electron configuration of Cl = [Ne]3s23p5.
     
  3. Oxidation state of Cl is +5.
  4. Outermost electron configuration of Cl after excitation:
     
  5. Now start adding orbitals into your pocket beginning from s, and thereafter 3p orbitals. We will stop adding orbitals to the pocket after adding all three 3p-orbitals because then the orbitals with unpaired electrons in the pocket would become equal to the number of other atoms in ClO3 .
  6. So, the hybridization of Cl in ClO3  is sp3.
  7. Shape of ClO3  would be tetrahedral with one lone pair (pyramidal). Each unpaired electron left (outside the pocket) will form a π-bond and there will be one lone pair on Cl.
     
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