Question

What is the relationship between atomic radius and ionization energy?

Answer 1

Hint: This completely depends upon the trends of the modern periodic table and the relationships within themselves (here, atomic radius and ionization energy).
In general, we can say that this relationship is inversely proportional to each other i.e.
$$Atomic-radius\propto {\displaystyle \frac{1}{Ionization-energy}}$$

Complete answer:
Let us define the terms and relationship between them;
Atomic radius –
It is the measure of the size of an atom i.e. the distance between the nucleus of the atom to the electron of the outermost shell.
Ionization energy –
The energy required to move the electron from a gaseous atom i.e. to take the electron out of the attractive forces and make it free.
Now, we know that the nucleus of an atom is positive and the electrons are negatively charged entities. The attractive forces hold them all together. The ionization energy is the energy which overcomes the same and makes the electron free.
Therefore, smaller the atom more the ionization energy required; vice versa is also true.

Note:
Do note that the ionization energy is against the force of attraction the electrons hold towards the nucleus of an atom. More the force of attraction the electrons hold towards the respective nucleus, more will be the ionization energy required to overcome these forces and make the electron free from the same.