Question

The correct increasing order of the ionic radii of the isoelectronic species \(O^{2-}, N^{3-}, F^{-}, Mg^{2+}, Na^{+}\) and \(Al^{3+}\) is}

• A. \(O^{2-}<N^{3-}<F^{-}<Mg^{2+}<Na^{+}<Al^{3+}\)
• B. \(Al^{3+}<N^{3-}< F^{-}<Mg^{2+}<Na^{+}
• C. \(Al^{3+}<Mg^{2+}<Na^{+}<F^{-}<O^{2-}<N^{3-}\)
• D. \(F^{-}<O^{2-}<N^{3-}<Al^{3+}<Mg^{2+}<Na^{+}\)
• E. \(O^{2-}<N^{3-}<F^{-}<Mg^{2+}<Al^{3+}

Hint:

For isoelectronic species, compare only nuclear charge: more protons means smaller radius.

Answer 1

The Correct Option is C

All these species are isoelectronic, that is, each has 10 electrons. For isoelectronic species: \[ \text{Higher nuclear charge} \Rightarrow \text{smaller ionic radius} \] Now compare their atomic numbers: \[ Al(13)>Mg(12)>Na(11)>F(9)>O(8)>N(7) \] So the ion with the highest positive charge and greatest nuclear attraction is smallest: \[ Al^{3+} \] And the one with the least nuclear charge is largest: \[ N^{3-} \] Therefore, increasing order is: \[ Al^{3+}<Mg^{2+}<Na^+<F^-<O^{2-}<N^{3-} \]
Hence, the correct answer is: \[ \boxed{(C)} \]