Question

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

• A. $Zr$ (40) and $Hf$ (72)
• B. $Zr$ (40) and $Ta$ (73)
• C. $Ti$ (22) and $Zr$ (40)
• D. $Zr$ (40) and $Nb$ (41)

Answer 1

The correct option is A $Zr$ (40) and $Hf$ (72)

Lanthanides have the extra electron (also called differentiating electron) which enters in $(n-2)f$ orbital. Since $(n-2)f$
orbital lies comparatively deep within the inner to the penultimate shell, therefore the elements are also called Inner transition elements.
The lanthanoid have the general electronic configuration as $4f^{1-14}5d^{0-1}6s^2$, with some exception to fully-filled or half-filled electronic configuration, wherein the $4f$ electrons of lanthanides did not participate in bond formation. Lanthanoid atoms have similar configurations and similar physical and chemical properties, and the most common oxidation state are 3 and 4.
The atomic radii or the size of tri positive lanthanide ions decreases steadily from Lanthanum to Lutetium due to an increase in the nuclear charge and electrons entering in the inner $(n-2)f$ orbital.
The greater than expected decrease in ionic radii of the elements in the lanthanide series from atomic number 51 to atomic number 71, resulting in smaller atomic radii than expected for subsequent elements starting with atomic number 72 is known as lanthanide contraction.
Lanthanide contraction results due to poor shielding of the nuclear charge by $4f$ electrons due to which $6s$ electrons are drawn towards the nucleus thus leading to a decrease in atomic radius.

Element	Radius (pm)
$Zr$ (40)	145
$Hf$ (72)	144
$Ta$ (73)	134
$Ti$ (22)	132
$Nb$ (41)	134

So, it is clear from the table that Zr and Hf have nearly the same atomic numbers.

Therefore, the correct answer is option A.