Question

Explain hybridisation and explain formation of H2O on the basis of hybridisation

Answer 1

Oxygen is sp3 hybridised in H2O molecule. Two hybrid orbitals are occupied by lone pairs and two are used in bonding with Hydrogen atoms. Since lone pairs does not contribute to the geometry of a molecule, therefore H2O has an angular geometry. The lone pair-lone pair repulsion is more than lone pair - bond pair or bond pair - bond pair repulsion, So the angle between H-O-H is 104.5° which is less than ideal tetrahedral angle of 109°28′.

H2O molecule has two lone pairs and two bond pairs, so the steric number of the central atom (O) is 2+2 as steric no = lone pairs + bond pairs except odd electron species and the stereochemically inactive lone pairs.

So O has steric number 4 and sp3 hybridization, tetrahedral arrangement of hybrid orbitals ( geometry) and angular shape (appearance to human eye when views practically by special techniques such as AFM). The bond angle HO^H = 104.5*
Total no of valence shell electron is 6+1+1 =8 , that is 4pairs . among the 4 pairs two r lone pair of oxygen and two pair bonded with hydrogen make the shape v shape.

Answer 2

Oxygen is sp3 hybridised in H2O molecule. Two hybrid orbitals are occupied by lone pairs and two are used in bonding with Hydrogen atoms. Since lone pairs does not contribute to the geometry of a molecule, therefore H2O has an angular geometry. The lone pair-lone pair repulsion is more than lone pair - bond pair or bond pair - bond pair repulsion, So the angle between H-O-H is 104.5° which is less than ideal tetrahedral angle of 109°28′.



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