Question

Ionization energy is directly proportional to the atomic radius and the effective nuclear charge. (True/False)

• A. True
• B. False

Answer 1

The correct option is B

False

(a) Atomic radius: The values of ionization potential of an element decreases as its atomic radius increases. This is because the electrostatic force of attraction between the nucleus and the outermost electron decreases as the distance between them increases. So, the energy required for the removal of the electron will comparatively be less

Ionization potential $\alpha$ $\frac{1}{AtomicRadius}$

(b) Effective nuclear charge: The greater the effective charge on the nucleus of an atom, the more difficult it would be to remove an electron from the atom because electrostatic force of attraction between the nucleus and the outermost electron increases. So greater energy will be required to remove the electron

Ionization potential $\alpha$ Effective nuclear charge ($Z_{eff}$)

Therefore, you can see the

Ionization potential $\alpha$ $\frac{1}{AtomicRadius}$