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The ionic size of Na+, F-, Mg2+ and Al3+ varies as

  • A.\(\text{Al}^{3+} \gt \text{Mg}^{2+} \gt \text{Na}^{+} \gt \text{F}^{-}\)
  • B.\(\text{F}^{-} \gt \text{Na}^{+} \gt \text{Mg}^{2+} \gt \text{Al}^{3+}\)
  • C.\(\text{Al}^{3+} \gt \text{Mg}^{2+} \gt \text{F}^{-} \gt \text{Na}^{+}\)
  • D.\(\text{Na}^{+} \gt \text{F}^{-} \gt \text{Mg}^{2+} \gt \text{Al}^{3+}\)

Answer: B

The size order is \(\text{F}^{-} \gt \text{Na}^{+} \gt \text{Mg}^{2+} \gt \text{Al}^{3+}\).

All four ions are isoelectronic, with 10 electrons each (the neon configuration). For isoelectronic ions, size falls as nuclear charge rises, because the same electron cloud is pulled in more strongly. The nuclear charges are F 9, Na 11, Mg 12, Al 13, so F- is largest and Al3+ smallest.

The orders starting with Al3+ reverse the trend, and the order putting Na+ before F- ignores that fluoride has two fewer protons for the same ten electrons.

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