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Quick question regarding thermal stability of group 2 ions watch

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    I never seem to be far from this board but you are all so great its hard to resist!

    Anyway, I would like to understand why in a compound like Mg(Co3) the physical size of the cation determines the polarity of the bond which determines its thermal stability.

    Thanks!
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    (Original post by lefneosan)
    I never seem to be far from this board but you are all so great its hard to resist!

    Anyway, I would like to understand why in a compound like Mg(Co3) the physical size of the cation determines the polarity of the bond which determines its thermal stability.

    Thanks!
    it might be best if you could draw a hess cycle for this,
    ie

    MCO3 ----> MO + CO2

    combine them with lattice energy. evaluate the term for delta H for this reaction. If delta H is negative, the reaction is most likely to be spontaneous, MCO3 will decompose under heat, and vice versa.

    Or to make life simple, read the following page very carefully.
    http://www.chemguide.co.uk/inorganic/group2/thermstab.html
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    What happens is, the Cation exerts a pull on the Carbonate anion, drawing electrons towards the Cation. Now, every metal in Group 2 produces cations with a +2 charge; however, the size of the Cation obviously increases down the group due to the larger mass of the atom, which means, down the group, the charge density decreases. So a lower charge density means less of a pull on the Carbonate ion, so less distortion of the Carbonate ion - the more distorted it is, the more easily it can be broken down by heat, hence down the group less distortion means more thermal stability, as more and more heat is being required to break down the Carbonate.
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    Mg2+ compared to Ba2+ has a greater charge density, therefore more electron-withdrawing hence more polarising. Basically charge stays the same, ionic radii increases down the group.
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    Thanks you are all legends!!
    p.s If I have a question later about lattice energies can I ask here or should I start a new thread?
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    (Original post by lefneosan)
    Thanks you are all legends!!
    p.s If I have a question later about lattice energies can I ask here or should I start a new thread?
    might be best to just start here
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    cool
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    Its not lattice energies but I doubt it matters!
    When it says "no rotation around rigid CC double bond" what do they mean?
    Really appreciate some help on this one!
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    Imagine the two carbon atoms are two balls of plasticine and you link them together by a straw, which represents a single bond. You could easily twist one of the plasticine balls around without pulling straw out, just as a C-C single bond can freely rotate around the bond without breaking it. But say you link the two balls with two straws, now you can't rotate either of them without pulling one of the straw out. This is essentially the same as a C=C bond which cannot rotate unless you break the double bond (i.e. pull one of the straws out). This is what is meant by restricted rotation.
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    Thanks kyri, but then how can functional groups e.g. a methyl group swap around with a hydrogen on one side of this rigid bond and not switch the whole thing around?
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    (Original post by lefneosan)
    Thanks kyri, but then how can functional groups e.g. a methyl group swap around with a hydrogen on one side of this rigid bond and not switch the whole thing around?
    there are two different things, conformation and configuration.
    for alkenes, let say, we have cis and trans, because these are different configurations - rotation about the double bond is not possible.

    conformation involves rotation about single bonds. Imagine a butane group, but think of it as (Me)H2C- CH2(Me), the Me groups are bigger than the hydrogen, so the conformation will try to minimize the energy of this structure by free rotation about single bond(possible) as E required is about 3-4 kJmol-1, favourable compared to the unfavourable interaction due to sterics.
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    I'm not sure what you mean by that. If I understood correctly, you do switch the whole thing around. A methyl can't just "swap" with one other group. You have to move all 3 groups the same way. I've drawn a hypothetical molecule to show this.

    Name:  Free rotation.gif
Views: 54
Size:  4.0 KB

    The C-C bond is twisted by 120 degrees so all the substituents on the nearby carbon atom are all shifted around by 120 degrees. This would not be allowed with a C=C double bond due to its rigidness (high energy barrier for rotation).
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    Thanks a lot for all this. When I have more time, I will give you all some of these mysteriously desirable reps!
 
 
 
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