Question

predict the shapes of H₂O and CH₄ using VSEPR theory

Answer 1

Dear student!
According to VSEPR theory,
In case of CH₄ the hybridization is obtained as:
No. of total valence electrons = 1 x 4 (for C)+ 4 x1 (forH) =8
Bond pair = 4
Electrons involved in octet/duet formation =4x 2 = 8
Hence no. of lone pair = 8-8 /2= 0
Thus, hybrid orbitals = 4 (4 lone pair + 0 bond pair)
Hybridization = sp³
Structure =Tetrahedral
Geometry or Shape =tetrahedral (as no lone pair)

In case of H₂O, the hybridization is obtained as:
No. of total valence electrons = 2 x 1(forH) +1 x6 (forO) =8
Bond pair = 2
Electrons involved in octet/duet formation =2x2 =4
Hence no. of lone pair = 8-4 /2= 2
Thus, hybrid orbitals = 4 (2lone pair +2 bond pair)
Hybridization = sp³
Structure =Tetrahedral
Geometry or Shape = bent or angular( due to two lone pair)