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In largely ionic compounds (e.g. NaBr and NaI), it seems to be generally true that, the greater the differences of electronegativity between the forming substances (i.e. The compound has more "ionic character" or has lower "covalent character"), the stronger its bond would be. (e.g. melting points: NaF>NaCl>NaBr>NaI with all of them having the same crystal structure. Pradyot, Patnaik (2003). Handbook of Inorganic Chemicals. The McGraw-Hill Companies, Inc. ISBN 978-0-07-049439-8.)

What I do not understand is that, as covalent bonds are not necessarily weaker than ionic bonds (For example, diamond and silicon both has very strong covalent bonds), why is it proper to claim that "more ionic bonds have greater strength" for such salts?

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    $\begingroup$ "It is often said that": Please edit, adding one or more sources for this claim. $\endgroup$
    – Karsten
    Commented Oct 21, 2023 at 21:33
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    $\begingroup$ When you say 'stronger than' you should qualify it by comparing with other molecules, it may not always be true. You can check by finding lists of bond formation/dissociation energy. $\endgroup$
    – porphyrin
    Commented Oct 22, 2023 at 8:26

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If a bond is ionic, the strength depends on the charge of the bonding partners and the distance of the ions. Strength here refers to the energy required to increase the distance between the two ions to infinity (in a vacuum).

For a covalent bond, the term "strength" often refers to the (homolytic) bond dissociation energy. In trying to rationalize bond dissociation energies, it does make sense to consider differences in electronegativity as one aspect among many others.

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    $\begingroup$ I think it is useful to distinguish between bonding in an ion pair, and bonding in a macroscopic lump of an ionic material. In the former there is a clear pairwise bond. In the latter each ion sits in the potential due to the whole of the rest of the lattice, and to my mind there is no simple pair bond. $\endgroup$
    – Ian Bush
    Commented Oct 22, 2023 at 21:35
  • $\begingroup$ @IanBush Yes, and you need the Madelung constant to figure out the lattice energy, not an easy calculation. $\endgroup$
    – Karsten
    Commented Oct 23, 2023 at 0:34

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