Consider the following two statements $Statement I:$ Among $\mathrm{Mg}, \mathrm{Al}, \mathrm{Mg}^{2+},…
Consider the following two statements
$Statement I:$ Among $\mathrm{Mg}, \mathrm{Al}, \mathrm{Mg}^{2+}, \mathrm{Al}^{3+}$ the one having smallest size is $\mathrm{Al}^{3+}$
$Statement II:$ Eu is having exceptionally high atomic radii among lanthanide elements
The correct answer is
Both statements I and II are correct
Both statements I and II are not correct
Statement I is correct but statement II not correct
Statement I is not correct but statement II is correct
Solution
$\mathrm{Al}$ atom is smaller than $\mathrm{Mg}$ atom as $\mathrm{Al}$ lies to the right of $\mathrm{Mg}$ and it has greater effective nuclear charge. A cation is smaller than its parent atom so $\mathrm{Al}^{3+}$ is smaller than $\mathrm{Al}$ and therefore $\mathrm{Al}^{3+}$ is smallest in size.
Thus, statement I is correct.
Among lanthanides there is a gradual decrease in the atomic and ionic radii as we move from left to right due to lanthanoid contraction.
As per the latest data from rsc.org; the atomic radii of Sm, Eu and Gd are:-
$
\begin{aligned}
& \mathrm{Sm}=236 \mathrm{pm}, 185 \mathrm{pm} \text { (covalent) } \\
& \mathrm{Eu}=235 \mathrm{pm}, 183 \mathrm{pm} \text { (covalent) } \\
& \mathrm{Gd}=234 \mathrm{pm}, 182 \mathrm{pm} \text { (covalent) }
\end{aligned}
$
Thus, the radii of lanthanides decreases gradually with europium following the same trend.
Thus, statement II is false