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Why does iodine have a higher boiling point than chlorine? Watch

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    i have narrowed it down to 2 options:

    the covalent bonds between iodine atoms are stronger.

    and

    the forces between iodine molecules are stronger.

    I think it's the second one??
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    It’s actually neither.

    When you evaporate something you aren’r breaking the intramolecular bonds (bonds between atoms in a molecule), but the intermolecular forces between the individual molecules themselves.

    The only intermolecular force present in Cl and I are London Forces/Van der waals/dispersion forces (whichever you like to call them). These are the instantaneous dipoles in molecules which in turn create an induced dipole in nearby molecules, simulating a very weak electrostatic attraction.

    The reason Iodine’s boiling point is higher is because it has more electrons than Chlorine therefore there is a higher chance for an instantaneous dipole to be made, and therefore more induced dipoles in other molecules.

    Because there are more London Forces in Iodine than Chlorine, it takes more energy to break it all up.

    Hopefully I have explained it clear enough (I’m in year 13 studying chemistry so it might not be explained all that well!) but hope you get what I’m trying to say.
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    Iodine has a higher atomic number, so more protons, and therefore more electrons, than chlorine.
    When you boil a halogen, you break the intermolecular forces. In this case, as all halogens are bi molecular, they have no permanent dipoles, or hydrogen bonds, therefore the only intermolecuar force left are London forces. London forces increase with the number of electrons, as the instantaneous dipole increases in strength, and therefore can induce another dipole more easily. Therefore because iodine has more electrons, it has stronger London forces between molecules of I2. This means Iodine has a higher melting/boiling point.
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    Thank you both!
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