What is hybridization? On the basis of hybridization of orbitals, deduce the shape of NH3, NH4, H2O.
The Correct Answer and Explanation is:
number of electron domains (bonding and lone pairs) around the central atom.
1. Ammonia (NH₃):
Nitrogen has five valence electrons and forms three sigma bonds with hydrogen atoms, leaving one lone pair. Thus, there are four electron domains around nitrogen. According to VSEPR theory, four electron domains suggest sp³ hybridization. The hybrid orbitals arrange in a tetrahedral geometry to minimize repulsion. However, due to the presence of a lone pair, which exerts greater repulsion than bonding pairs, the molecular shape is trigonal pyramidal, and the bond angle is approximately 107°.
2. Ammonium ion (NH₄⁺):
In the ammonium ion, nitrogen forms four sigma bonds with hydrogen atoms and has no lone pairs. With four bonding pairs, the nitrogen atom also undergoes sp³ hybridization. All hybrid orbitals are involved in bonding, resulting in a tetrahedral geometry with ideal bond angles of 109.5°.
3. Water (H₂O):
Oxygen has six valence electrons, two of which form sigma bonds with hydrogen atoms, and the remaining four electrons form two lone pairs. Therefore, there are four electron domains around oxygen, leading to sp³ hybridization. The electron domain geometry is tetrahedral, but due to the two lone pairs, the molecular shape is bent or V-shaped, with a bond angle of about 104.5°.
Thus, based on the concept of hybridization and VSEPR theory, NH₃ is trigonal pyramidal, NH₄⁺ is tetrahedral, and H₂O is bent.
