What is the difference between in the boiling points of \[{{\text{O}}_{\text{2}}}\] and \[{{\text{O}}_{\text{3}}}\] at standard pressure in terms of intermolecular forces?

Hint: Oxygen is one of the element periodic tables. The symbol of oxygen is \[{\text{O}}\]. The atomic number of oxygen is \[8\]. The symbol of oxygen molecule is \[{{\text{O}}_{\text{2}}}\] and it is a diatomic molecule, the number of atoms in the molecule is two. The symbol of ozone molecule is \[{{\text{O}}_{\text{3}}}\] and it is a triatomic molecule, the number of atoms in the molecule is three.

Complete step by step answer:
We must know that the boiling point of oxygen is \[ - {183^ \circ }C\] at standard temperature. The boiling point of Ozone is \[ - {112^ \circ }C\] at standard temperature.
We need to know that at standard pressure, the boiling point of ozone is greater than oxygen. The boiling point of difference in oxygen and ozone at standard pressure, because of the volatile nature of oxygen in the atmosphere. But this kind of volatility is absent in ozone. Ozone molecules are heavier than oxygen atoms. Heaviness of ozone molecules is due to the electron cloud. In the atmosphere, consequently greater dispersive forces in ozone increases the intermolecular forces.
The polarity of the molecule also affects the boiling point and melting point. The polarity of the oxygen is lesser than ozone molecules. The polarity of the oxygen is zero. The polarity of an ozone molecule depends on the formal positive charge and negative charges.

Hence, \[{60^ \circ }C\] is the difference between the boiling point of oxygen and ozone molecule.

Note: We also remember that ozone prevents over atmosphere from ultraviolet rays from sunlight. Oxygen is used for respiration in human beings. There are two types of forces between the molecules. There are intermolecular forces and intramolecular forces. The oxygen molecules having intramolecular forces in the atmosphere. But, ozone molecules have intermolecular forces in the atmosphere.