Answer: Bond angle for methane is 109.5°,Bond angle for ammonia is 107° and Bond angle for water is 104.5°
: \(Formula used :{\text{Number of electrons}} =\frac{1}{2}[V+N-C+A]\)
where, V = number of valence electrons present in central atom
N = number of monovalent atoms bonded to central atom
C = charge of cation
A = charge of anion
a. Methane:\(CH_4\)
In the given molecule, carbon is the central atom and there are 4 Hydrogens as monovalent atoms.
\(Formula used :{\text{Number of electrons}} =\frac{1}{2}[4+4-0+0]=4\)
The number of electrons is 4 that means the hybridization will be \(sp^3\) and geometry of the molecule will be tetrahedral.The bond angle for terahedral geometry is 109.5°.
b. Ammonia:\(NH_3\)
In the given molecule, nitrogen is the central atom and there are 3 Hydrogens as monovalent atoms.
\(Formula used :{\text{Number of electrons}} =\frac{1}{2}[5+3-0+0]=4\)
The number of electrons is 4 that means the hybridization will be \(sp^3\) and geometry of the molecule will be pyramidal as there are three bond pairs and one lone pair.The bond angle for pyramidal geometry is 107°.the bond angle reduces due to lone pair bond pair repulsions.
c. water :\(H_2O\)
In the given molecule, oxygen is the central atom and there are 2 Hydrogens as monovalent atoms.
\(Formula used :{\text{Number of electrons}} =\frac{1}{2}[6+2-0+0]=4\)
The number of electrons is 4 that means the hybridization will be \(sp^3\) and geometry of the molecule will be bent as there are two bond pairs and two lone pairs.The bond angle for pyramidal geometry is 104.5°.The bond angle reduces further due to greater lone pair-lone pair repulsions.
