Question

Why is $PC{{l}_{5}}$ more covalent than $PC{{l}_{3}}$ ?

Answer 1

To find the more covalent compound, compare the oxidation states of the central atom. The oxidation state and covalent nature of a compound are directly proportional to each other. Higher the oxidation state of a compound is its polarizing power.

Complete answer:
As we understood that the covalent nature of a compound is dependent on the oxidation state of the central atom. Thus, the oxidation number of Phosphorus in Phosphorus pentachloride $PC{{l}_{5}}$ is +5, and in Phosphorus trichloride $PC{{l}_{3}}$ is +3.
The Oxidation state of an element in a compound is inversely proportional to the size of the cation. So, If the oxidation number increases the size of the cation decreases.
Also, the greater the oxidation number, the more is the polarizing power. Because the polarizing power is dependent on the charge on the cation and the radius of the atom. The charge on the cation is the Oxidation number of the element. Polarizing power is nothing but the high charge to the size of the cation ratio.
According to Fajan’s rule, the size of the cation indicates the polarizing power, hence, the polarizing power indicates the covalent nature of the compound. Therefore, an increase in the oxidation number of an element increases the covalent character and decreases the ionic character in a compound.

Note:
Remember that the covalent character is also reliant on the dipole moment of the bonds in the compound. $PC{{l}_{5}}$ is a nonpolar molecule as the individual dipole moments are cancelled out by the P-Cl bonds. However, $PC{{l}_{3}}$ is a polar molecule. Also, Phosphorus is a d-block element that eventually increases its covalency.