Question

Compare the structures of ${{H}_{2}}O$ and ${{H}_{2}}{{O}_{2}}$ .

Answer 1

Hint : In both the structures there are only single bonds present. The bond angle has changed due to the presence of lone pairs. Due to their bonding, they can form a complicated structure.

Complete step by step solution :

Let us study the structure of ${{H}_{2}}O$.
In ${{H}_{2}}O$molecule, the oxygen atom is and hence has four $s{{p}^{3}}-hybridized$ orbitals. Two of these$s{{p}^{3}}-orbitals$ are half-filled and hence overlap with hydrogen to form two while the other two contain a lone pair of electrons each. Since the oxygen atom is $s{{p}^{3}}-hybridized$, therefore, the geometry of the molecule should be tetrahedral and the bond angle should be ${{109.5}^{\circ }}$. But experimentally, the oxygen atom is surrounded by two shared pairs, and two lone pairs of the electron. But according to VSEPR theory, lone pair-lone pair repulsion is greater than bond pair-bond pair repulsion. As a result, the angle decreases to${{104.5}^{\circ }}$.
The bond length is 95.7pm.

Now, the structure of ${{H}_{2}}{{O}_{2}}$
Hydrogen peroxide is a nonpolar molecule. The two oxygen atoms are linked to each other by a single covalent bond and each oxygen is further linked to a hydrogen atom by a single bond. The two bonds are, however, in different planes due to repulsion between different bonding and antibonding orbitals. The dihedral angle between the two planes being ${{111.5}^{\circ }}$and in the gas phase reduces to ${{90.2}^{\circ }}$. The bond length is 147.5pm and 95pm.

Note : Don’t get confused that if the hybridization is then the shape would be tetrahedral, lone pairs repulsion is a factor that should be considered. Water has a planar structure whereas hydrogen peroxide has a non-planar structure.