Identify the species

We are comparing the ionic radii of: $Na^+,\; Mg^{2+},\; Al^{3+}$

Check electronic configuration

• $Na$ has $Z=11$, so: $Na^+ : 1s^2 2s^2 2p^6$
• $Mg$ has $Z=12$, so: $Mg^{2+} : 1s^2 2s^2 2p^6$
• $Al$ has $Z=13$, so: $Al^{3+} : 1s^2 2s^2 2p^6$

Thus, all three ions are isoelectronic; each has 10 electrons.

Use the isoelectronic trend

For an isoelectronic series, ionic radius decreases as nuclear charge increases, because the same number of electrons is pulled more strongly by more protons.

Nuclear charges are: $Na^+ : 11, \quad Mg^{2+} : 12, \quad Al^{3+} : 13$

Therefore, $r(Na^+) > r(Mg^{2+}) > r(Al^{3+})$

Use the given value

Given: $r(Na^+) = 1.02\,\text{Å}$

So both $Mg^{2+}$ and $Al^{3+}$ must have radii less than $1.02\,\text{Å}$, and also: $r(Mg^{2+}) > r(Al^{3+})$

Check the options

• A: $1.05$ and $0.99$

  Here $Mg^{2+} = 1.05\,\text{Å} > Na^+$, which is impossible. So A is incorrect.

• B: $0.72$ and $0.54$

  Both are less than $1.02$, and: $0.72 > 0.54$ This matches the trend. So B is correct.

• C: $0.85$ and $0.99$

  Here $Al^{3+}$ is larger than $Mg^{2+}$, which is wrong for an isoelectronic series. So C is incorrect.

• D: $0.68$ and $0.72$

  Again $Al^{3+}$ is larger than $Mg^{2+}$, which is incorrect. So D is incorrect.

Final answer

$r(Mg^{2+}) = 0.72\,\text{Å}, \quad r(Al^{3+}) = 0.54\,\text{Å}$

Hence, the correct option is B.