f-Block Elements — Lanthanides & Actinides

Explanation:

Lanthanide contraction is caused by the poor shielding effect of $4f$ electrons. As electrons are sequentially added across the lanthanide series, they enter the highly diffuse $4f$ orbitals. These inner $f$-electrons shield the outer valence electrons from the increasing nuclear charge very ineffectively. This results in a steadily increasing effective nuclear charge ($Z_{\text{eff}}$) felt by the outer electrons, pulling them closer to the nucleus and causing a steady decrease in atomic and ionic radii across the series.

Answer: B) $\text{Zr}$ and $\text{Hf}$.

Explanation: Due to lanthanide contraction, Zirconium ($\text{Zr}$, from the 4d series in period 5) and Hafnium ($\text{Hf}$, from the 5d series in period 6) possess almost identical atomic radii ($\approx 160\text{ pm}$). Because their size and valence electron configurations are so similar, their chemical properties are nearly identical, making them famously known as "chemical twins" and extremely difficult to separate.

Explanation:

The ground state electronic configuration of Cerium is $[\text{Xe}]4f^{1}5d^{1}6s^2$. In the $+4$ oxidation state (${\text{Ce}}^{4+}$), Cerium loses all four of its outer valence electrons ($4f^1,5d^1,6s^2$), achieving a completely empty $4f$ shell ($4f^0$). This leaves it with the exceptionally stable noble gas core of Xenon ($[\text{Xe}]$). This tremendous thermodynamic stability provides a strong driving force for the $+4$ state, making Cerium the most prominent lanthanide to exhibit it.

Answer: C) All actinides occur naturally in the Earth's crust.

Explanation: This statement is incorrect. Only Thorium ($\text{Th}$), Protactinium ($\text{Pa}$), and Uranium ($\text{U}$) occur in significant quantities naturally. All elements beyond Uranium ($Z\ge 93$, known as the transuranium elements — $\text{Np, Pu, Am, Cm, Bk, Cf, Es, Fm, Md, No, Lr}$) are synthetic (man-made), produced only in nuclear reactors or particle accelerators.

Explanation:

Ground state configuration of $\text{Eu}$: $[\text{Xe}]4f^{7}6s^2$

When forming ${\text{Eu}}^{2+}$, it loses its two outermost $6s$ electrons, resulting in the configuration: $[\text{Xe}]4f^7$

The $4f^7$ configuration is exactly half-filled — all seven $f$ orbitals contain exactly one unpaired electron. According to Hund's rule, this yields maximum exchange energy and spin multiplicity, making ${\text{Eu}}^{2+}$ remarkably stable for a lanthanide ion. (Similarly, ${\text{Yb}}^{2+}$ with a $4f^{14}$ configuration is stabilised by a completely filled f-shell).