Question

The set representing the correct order of ionic radius is:
A) $\text{N}{{\text{a}}^{\text{+}}}\text{M}{{\text{g}}^{\text{2+}}}\text{A}{{\text{l}}^{\text{+3}}}\text{L}{{\text{i}}^{\text{+}}}\text{B}{{\text{e}}^{\text{+2}}}$
B) $\text{N}{{\text{a}}^{\text{+}}}\text{L}{{\text{i}}^{\text{+}}}\text{M}{{\text{g}}^{\text{2+}}}\text{A}{{\text{l}}^{\text{+3}}}\text{B}{{\text{e}}^{\text{+2}}}$
C) $\text{N}{{\text{a}}^{\text{+}}}\text{M}{{\text{g}}^{\text{2+}}}\text{L}{{\text{i}}^{\text{+}}}\text{A}{{\text{l}}^{\text{+3}}}\text{B}{{\text{e}}^{\text{+2}}}$
D) $\text{N}{{\text{a}}^{\text{+}}}\text{M}{{\text{g}}^{\text{2+}}}\text{L}{{\text{i}}^{\text{+}}}\text{B}{{\text{e}}^{\text{+2}}}$

Answer 1

To answer this question, the idea of ionic and atomic radius is required. Atomic radius is defined as the total distance of the outermost orbital of the atom from the centre or the nucleus of the atom. While ionic radii is defined as the distance of the nucleus to the outermost orbital in an ion. The atomic radius is greater than the cationic radii and smaller than the anionic radii.

Complete step by step solution:
In the periodic table, as we move down the group, the atomic radius and the ionic radius increases, while along the period, the atomic radius decreases. The elements asked for in the question are: lithium (group-1), sodium (group-1), beryllium (group-2), Magnesium (group-2), (Aluminium group-13). In $\text{N}{{\text{a}}^{\text{+}}}$ sodium has lost one electron to form the cation, similarly in $\text{L}{{\text{i}}^{\text{+}}}$ also, lithium has lost one electron to form the cation. As both sodium and lithium belong to the same group so $\text{N}{{\text{a}}^{\text{+}}}$ has larger radius than$\text{L}{{\text{i}}^{\text{+}}}$ because atomic and ionic radii increase down the group. Coming to $\text{B}{{\text{e}}^{\text{+2}}}$ and $\text{M}{{\text{g}}^{\text{+2}}}$, the former appears before the later in the periodic table hence the ionic radii of $\text{B}{{\text{e}}^{\text{+2}}}$ is smaller than that of $\text{M}{{\text{g}}^{\text{+2}}}$. Now, as $\text{B}{{\text{e}}^{\text{+2}}}$ appears after $\text{L}{{\text{i}}^{\text{+}}}$ in the periodic table so ionic radii of $\text{L}{{\text{i}}^{\text{+}}}$>$\text{B}{{\text{e}}^{\text{+2}}}$ and similarly,$\text{N}{{\text{a}}^{\text{+}}}$> $\text{M}{{\text{g}}^{\text{+2}}}$. Coming to $\text{A}{{\text{l}}^{\text{+3}}}$ as it appears after $\text{N}{{\text{a}}^{\text{+}}}$ and $\text{M}{{\text{g}}^{\text{+2}}}$ in the periodic table, so its ionic radii is smaller compared to both.
So the correct order of the ionic radii is: $\text{N}{{\text{a}}^{\text{+}}}\text{L}{{\text{i}}^{\text{+}}}\text{M}{{\text{g}}^{\text{2+}}}\text{A}{{\text{l}}^{\text{+3}}}\text{B}{{\text{e}}^{\text{+2}}}$ and the correct answer is option B.
Note:
Among the alkali metals like sodium and lithium as only one electron is lost to form the cation but the number of protons in the nucleus remain the same, so the effective nuclear charge increases but this increase is more when two electrons are lost as in the case of beryllium and magnesium and finally in case of aluminium it is highest as three electrons are lost.