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Which has higher lattice energy: MgO or LiF ?

  • A compound has higher lattice energy if it's ions have smaller size and greater charge

    1. Li cation is smaller than Mg cation => +1 for (LiF)
    2. Fluoride anion is smaller than oxide anion => +1 for (LiF)
    3. Mg cation has greater charge than Li cation => +1 for (MgO)
    4. Oxide anion has greater charge than fluoride anion => +1 for (MgO)

Therefore, 'charge' factor favour MgO while 'size' factor favour LiF.

So, which one will have higher lattice energy?

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$\ce {MgO}$ ($\approx 3800$ $\pu{kJ mol^{-1}}$) has higher lattice energy than $\ce {LiF}$ ($\approx 1045$ $\pu{kJ mol^{-1}}$) mainly because of the greater charge on $\ce{Mg^2^{+}}$ ion and $\ce {O^{2-}}$ as lattice energy is directly proportional to the charges of the combing atoms.

Your reasoning isn't incorrect but remember that $\ce{Li}$ and $\ce{Mg}$ show diagonal relationship so the size of their ions has only slight difference in magnitude. Similarly, oxygen and fluorine are placed next to each other in the periodic table so their anions don't have a large difference in size.

Thus, the "charge factor" dominates.

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according to me MgO has more lattice energy that LiF.

The lattice energy of an ionic compound is the enthalpy change which occurs when one mole of an ionic compound dissociates into its ions in gaseous state.

More ionic is a compound, stronger would be the ionic bond and more would be the lattice enthalpy (Ref).

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