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NaF, KF, MgO and CaO are crystalline solids. They have NaCl structure. Their lattice energies vary in the order

  • A.NaF \(\lt\) KF \(\lt\) MgO \(\lt\) CaO
  • B.KF \(\lt\) NaF \(\lt\) CaO \(\lt\) MgO
  • C.MgO \(\lt\) CaO \(\lt\) NaF \(\lt\) KF
  • D.CaO \(\lt\) MgO \(\lt\) KF \(\lt\) NaF

Answer: B

The lattice energies increase as KF < NaF < CaO < MgO.

For compounds with the same (NaCl) structure, lattice energy varies roughly as \(z^{+}z^{-}/(r^{+} + r^{-})\). The oxides combine 2+ and 2− ions, so their charge product is four times that of the fluorides, and they lie far above them: MgO and CaO have lattice energies of several thousand kJ mol−1, NaF and KF under one thousand.

Within each pair the smaller cation wins: Na+ is smaller than K+, and Mg2+ smaller than Ca2+. That gives KF < NaF and CaO < MgO. Options putting a fluoride above an oxide, or reversing the size order, fail.

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