What is the order of increasing the bond angle ( H × H ) of the following? What is the theory involved CH 4 , H 2 O , NH 3

In CH 4 , the central atom carbon shows sp 3 -hybridisation and produces four sp 3 hybrid orbitals. These sp 3 - hybrid orbitals are arranged in a regular tetrahedral fashion. Since each sp 3 - hybrid orbital contains a half-filled electron, four hydrogen atoms (each containing a half-filled electron) overlap with it. As a result of which four sigma bonds are formed. Therefore, the shape of CH 4 molecule is tetrahedral and the bond angle is 109 0 28 ' . Although the central atom shows sp 3 - hybridisation, the shape of water is not tetrahedral. This is because of the greater lone pair-lone pair repulsion. Hence, the tetrahedral geometry of water distorted to 'V' shaped or bent. Thus, the geometry of water is 'V'-shaped and the bond angle is 104 0 27 1 or 104 . 5 0 . The electronic configuration of nitrogen Is 1 s 2 2 s 1 2 p x 1 2 p y 1 2 p z 1 . Here, the central atom, nitrogen shows sp 3 - hybridisation and forms four sp 3 - hybrid orbitals. Each of the three half-filled sp 3 -hybrid orbitals overlaps with 1 s -orbital of a hydrogen atom to form sigma bonds. A pair of electrons are present in the fourth sp 3 -orbital. Here, due to Ione pair-bond pair repulsion, according to VSEPR theory, the regular tetrahedral geometry is distorted. The shape of NH 3 molecule becomes pyramidal and the bond angle is reduced to 107 0 . Therefore, the order of increase in bond angle is H 2 O 104 . 5 0 < CH 4 107 2 < NH 3 109 . 5 0