Question

The bond angle in ${H_2}O$ is ${104.5^ \circ }$. This fact can be explained with the help of:
(A) Valence shell electron pair repulsion theory (VSEPR)
(B) Molecular orbital theory
(C) Presence of hydrogen bond
(D) Electronegativity difference between hydrogen and oxygen

Answer 1

Hint: Oxygen atom in water has $s{p^3}$ hybridization. Two lone pairs are present on the oxygen atom of the water molecule alongside two hydrogen atoms. The bond angle they are asking is H-O-H bond angle.

Complete step by step solution:
Let’s find out the hybridization of oxygen in ${H_2}O$.
Ground state configuration of O : $1{s^2}2{s^2}2{p_x}^22{p_y}^12{p_z}^1$
Now, both H-atoms will donate their one electron each to $2{p_y}$ and $2{p_z}$ orbital. An outermost orbit of oxygen contains one s-orbital and three p-orbitals. So, we can say that the hybridization of oxygen atoms in ${H_2}O$ is $s{p^3}$.

- It is evident that all the bonds have an angle of ${108^ \circ }28'$ between them. According to hybridization, bond angle in ${H_2}O$ should be ${108^ \circ }28'$ but it is not true. Why? Let’s see.

- We can see from the electronic configuration of oxygen atoms in ${H_2}O$ that oxygen has two filled orbitals in its outermost orbit. This filled orbital (filled with 2 electrons) is known as a lone pair of electrons. Oxygen has two of such lone pairs in this structure.

- These lone pairs repel each other very much because of their high electron density. Due to repulsion between them, the C-H bonds are also affected and they also feel the repulsion with the lone pair, so they also tend to be as far as possible with the lone pair. This is the reason why H-O-H bond angle is reduced from ${108^ \circ }28'$ to ${104.5^ \circ }$in water molecules.

- Now the theory that states that various effects of repulsion between orbitals of the valence shell of the central atom is called VSEPR theory. So, this explains the fact that H-O-H bond angle in water is ${104.5^ \circ }$.

- VSEPR stands for Valence shell electron pair repulsion theory.
So, the correct answer to this question is (A) Valence shell electron pair repulsion theory (VSEPR).

Note: Always remember that repulsion between two lone pairs is more than repulsion between a lone pair and a bond pair. While finding out the bond angles of any molecule, we should find the hybridization of the central atom only i.e. Here we are finding the hybridization of oxygen atoms.