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    ankara55t's Avatar
    ankara55t Posts: 53, Reputation: 2
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    #1

    Dec 29, 2009, 06:36 AM
    Boiling points
    Could we say that the reason that NaF has a higher boiling point than NaI is that the bonds of NaF are stronger than those of NaI and that more heat is required to break them than for NaI? Are there any other major reasons that the boiling points of these differ?
    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
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    #2

    Dec 29, 2009, 09:48 AM

    Well, yes that's right.

    If it were to be compared to another another compound for example magnesium oxide, MgO, then you could add that magnesium has a charge of 2+ and oxygen a charge of 2-, hence the bond is even stringer as the electrostatic attraction is stronger.
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    dihydrogenoxide Posts: 9, Reputation: 2
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    #3

    Jan 2, 2010, 03:02 PM

    Also consider how electronegative Fluorine is, creating stronger van der waals forces, or london dispersion forces. Both compounds are straight line structure. However, Fluorine is more electronegative than iodine. So in NaF, the side with the Fluorine is slightly negatively charged, which creates a higher attraction to the Na side of the other NaF molecules. Therefore, it is harder to separate the molecules, giving it a higher boiling point. This phenomenon can also be seen in NaI, but because Iodine is less electronegative than Fluorine, the london dispersion forces aren't as strong, so it has a lower boiling point than NaF.
    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
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    #4

    Jan 4, 2010, 12:24 AM
    Quote Originally Posted by dihydrogenoxide View Post
    Also consider how electronegative Fluorine is, creating stronger van der waals forces, or london dispersion forces. Both compounds are straight line structure. However, Fluorine is more electronegative than iodine. So in NaF, the side with the Fluorine is slightly negatively charged, which creates a higher attraction to the Na side of the other NaF molecules. Therefore, it is harder to seperate the molecules, giving it a higher boiling point. This phenomenon can also be seen in NaI, but because Iodine is less electronegative than Fluorine, the london dispersion forces aren't as strong, so it has a lower boiling point than NaF.
    Hmm, not all completely true...

    Van der Waals occur in covalent compounds. NaF is ionic (the covalent character is neglible)

    Both compounds are not 'straight'. They are arranged in an ionic lattice. A giant ionic lattice.

    However, your post made me think of something. Yes, fluorine is more electronegative, but that doesn't explain the high boiling point. Fluorine, having the same charge as iodine, we can say that both have the same force of attraction regarding that only. But fluorine is also smaller in size. Because of this, the ions are closer to each other and the bond is stronger. Here, we talk about charge density.

    Charge density depends on the charge of an ion and its size.
    The smaller, the greater charge density. The stronger the charge, the greater the charge density.

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