Consider the following two statements:Stateme | vedclass

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(C) $Al$ atom is smaller than $Mg$ atom because $Al$ lies to the right of $Mg$ in the same period and has a greater effective nuclear charge.$A$ catio

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Statement $I$ is correct but statement $II$ is not correct.

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(C) $Al$ atom is smaller than $Mg$ atom because $Al$ lies to the right of $Mg$ in the same period and has a greater effective nuclear charge.$A$ cation is smaller than its parent atom,so $Al^{3+}$ is smaller than $Al$. Therefore,$Al^{3+}$ is the smallest in size among the given species.Thus,$Statement-I$ is correct.Among lanthanides,there is a gradual decrease in atomic and ionic radii as we move from left to right due to lanthanoid contraction.The atomic radii of lanthanides generally decrease across the series. $Eu$ does not show an exceptionally high atomic radius compared to its neighbors in the context of the general trend of lanthanoid contraction.Thus,$Statement-II$ is false.

Consider the following two statements:
$Statement-I:$ Among $Mg, Al, Mg^{2+}, Al^{3+}$,the one having the smallest size is $Al^{3+}$.
$Statement-II:$ $Eu$ has an exceptionally high atomic radius among lanthanide elements.
The correct answer is:

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Consider the following two statements:$Statement-I:$ Among $Mg, Al, Mg^{2+}, Al^{3+}$,the one having the smallest size is $Al^{3+}$.$Statement-II:$ $Eu$ has an exceptionally high atomic radius among lanthanide elements.The correct answer is: