Question

Display electronic distribution around the oxygen atom in water molecule and state shape of a molecules also write its H-O-H bond angle​

Answer 1

In a water molecule, the oxygen atom and hydrogen atoms share electrons in covalent bonds, but the sharing is not equal. In the covalent bond between oxygen and hydrogen, the oxygen atom attracts electrons a bit more strongly than the hydrogen atoms.

Answer 2

Answer : The shape of water molecule is bent or angular and the bond angle is $120^o$.

Explanation :

Lewis-dot structure : It shows the bonding between the atoms of a molecule and it also shows the unpaired electrons present in the molecule.

In the Lewis-dot structure the valance electrons are shown by 'dot'.

The given molecule is, $H_2O$

As we know that hydrogen has '1' valence electron and oxygen has '6' valence electrons.

Therefore, the total number of valence electrons in $H_2O$ = 2(1) + 6 = 8

According to Lewis-dot structure, there are 4 number of bonding electrons and 4 number of non-bonding electrons.

Formula used for hybridization :

$\text{Number of electron pair}=\frac{1}{2}[V+N-C+A]$

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

Now we have to determine the hybridization of the water molecules.

The given molecule is, $H_2O$

$\text{Number of electrons}=\frac{1}{2}\times [6+2]=4$

The number of electron pair are 4 that means the hybridization will be $sp^3$ and the electronic geometry of the molecule will be tetrahedral.

But as there are 2 atoms around the central oxygen atom, the third and fourth position will be occupied by lone pair of electrons. The repulsion between lone and bond pair of electrons is more and hence the molecular geometry will be bent or angular and the bond angle will be, $120^o$

Hence, the shape of water molecule is bent or angular and the bond angle is $120^o$.