Because of lanthanoid contraction, which of the following pairs of elements have nearly the same atomic…

$\mathrm{Zr} \left(40\right)$ and $\mathrm{Hf} \left(72\right)$: Because of the lanthanoid contraction, $\mathrm{Zr}$ (atomic radii $160$ pm) and $\mathrm{Hf}$ (atomic radii $158$ pm) have nearly the same atomic radii.
Lanthanoids include the elements from lanthanum $\mathrm{La} \left(\mathrm{Z}=57\right)$ to lutetium $\mathrm{Lu} \left(\mathrm{Z}=71\right)$. Zirconium $\mathrm{Zr} \left(40\right)$ belong to the second transition series $\left(4\mathrm{d}\right)$ and $\mathrm{Hf} \left(72\right)$ belongs to the third transition series $\left(5\mathrm{d}\right)$. Lanthanoid contraction is related to the intervention of the $4\mathrm{f}-$orbitals, which are filled before the $5\mathrm{d}-$series of elements start. The filling of $4\mathrm{f}$, if orbitals before $5\mathrm{d}-$orbitals result in a regular decrease in atomic radii, which compensates the expected increases in atomic size with increasing atomic number. As a result of lanthanoid contraction, the elements of second and third transition series have almost similar atomic radii.