Zr (Z=40) and Hf (Z=72) have similar atomic and ionic radii because of :

The correct option is lanthanoid contraction.

Step-by-step Explanation:

1. Position in the Periodic Table:

Zirconium (Zr, $Z=40$) belongs to the d-block, specifically the 4d transition series (5th period, Group 4).

Hafnium (Hf, $Z=72$) belongs to the 5d transition series (6th period, Group 4).

2. Expected Trend:
Generally, as we move down a group in the periodic table, the atomic and ionic radii increase because of the addition of a new electronic shell (principal energy level).

3. The Lanthanoid Contraction:

Between zirconium and hafnium, there lie 14 lanthanoid elements (from Cerium, $Z=58$ to Lutetium, $Z=71$). In these elements, the incoming electrons enter the inner 4f subshell.

The f-orbitals have a highly diffused shape, which results in very poor shielding (screening) of the outer electrons from the nuclear charge.

Consequently, as the nuclear charge increases by 32 units from Zr to Hf, the shielding is insufficient to offset the increased nuclear pull. The outer electrons are drawn closer to the nucleus, causing a contraction in the atomic and ionic sizes. This phenomenon is known as lanthanoid contraction.

4. Resulting Radii:
The decrease in size due to lanthanoid contraction almost exactly counterbalances the increase in size expected from adding a new principal energy shell. As a result, Zirconium and Hafnium have nearly identical atomic radii (Zr ≈ 160 pm, Hf ≈ 159 pm) and ionic radii (Zr⁴⁺ ≈ 72 pm, Hf⁴⁺ ≈ 71 pm).