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    I have 6 questions, and would be grateful if anyone could answer them. I have basically gone through all of A2, and highlighted the bits i dont understand, so could really do with some help.

    1) On an acyl chlrodie group, why does the presecne of the C=O bond make the C-Cl bond more reactive?

    2) Why is scandium not a transition metal?

    3) Why can the O in the C=O bonds in edta ligand not attach to the central metal ion?

    4) Why does execess ammonia only react with Cu(OH)2, and not with Fe(OH)2?

    5) Why do variable oxidation states help catalysts?

    6) How do you determine the feasability of electrode potential reactions?

    I have looked all these questions up on wiki/ chemguide, and only get poor answers.

    I would REALLY appreciate any help. Thanks again
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    1. The O on the C=O is electronegative and will pull a lot of the electron density from the the carbon atom. The Cl attached is therefore a good leaving group because of the lack of electron density on the carbon. The Cl may leave the molecule more readily therefore which implies a greater degree of reactivity.

    2. Sc is not a transition metal as, upon ionisation, it loses its 1 3d electron. Transition metals are definied as having partially filled d shells upon ionisation.

    3. C=O has not got a lone pair of electrons to donate to a central metal cation whilst the C-O^- attached to the same carbon does.

    4. Im not 100% sure but it's prob something to do with stability of the complexes.

    5. Catalysis usually involves the multichange of oxidation numbers through intermediate compounds so catalysts that can change oxidation number easily can lower the Ea of a reaction.

    6. If an overall EMF is positive in a reaction, the reaction will proceed. You work out the overall EMF of a reaction by finding the sum of the EMF for the oxidation and reduction reactions involved.

    Im sure the other guys on here can give you more details answers than me so hold up for them.
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    (Original post by vinny221)
    1. The O on the C=O is electronegative and will pull a lot of the electron density from the the carbon atom. The Cl attached is therefore a good leaving group because of the lack of electron density on the carbon. The Cl may leave the molecule more readily therefore which implies a greater degree of reactivity.

    2. Sc is not a transition metal as, upon ionisation, it loses its 1 3d electron. Transition metals are definied as having partially filled d shells upon ionisation.

    3. C=O has not got a lone pair of electrons to donate to a central metal cation whilst the C-O^- attached to the same carbon does.

    4. Im not 100% sure but it's prob something to do with stability of the complexes.

    5. Catalysis usually involves the multichange of oxidation numbers through intermediate compounds so catalysts that can change oxidation number easily can lower the Ea of a reaction.

    6. If an overall EMF is negative in a reaction, the reaction will proceed. You work out the overall EMF of a reaction by finding the sum of the EMF for the oxidation and reduction reactions involved.
    I think that the Cl leaving group abiliy has little to do with the C=O group. Chloride tends to be a good leaving group, but this is to do with the stability of the chloride ion. Are you sure you didn't mean to ask why the presence of the Cl increases the reactivity of the Carbonyl group?

    Also, CO does have a pair of electrons to donate from the C-O sigma star orbital. The reason the metal bonds to the carbon and not the oxygen is because the coefficient of this orbital is greater on the oxygen atom and so there is better overlap.
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    (Original post by Fly.)
    I think that the Cl leaving group abiliy has little to do with the C=O group. Chloride tends to be a good leaving group, but this is to do with the stability of the chloride ion. Are you sure you didn't mean to ask why the presence of the Cl increases the reactivity of the Carbonyl group?

    Also, CO does have a pair of electrons to donate from the C-O sigma star orbital. The reason the metal bonds to the carbon and not the oxygen is because the coefficient of this orbital is greater on the oxygen atom and so there is better overlap.
    Yea that makes sense.
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    1) On an acyl chlrodie group, why does the presecne of the C=O bond make the C-Cl bond more reactive?
    The O atom is highly elecronegative and thus shifts the electron density so that it is less concentrated upon Cl, making it reactive.
    2) Why is scandium not a transition metal?
    It cannot form ions having partially filled d-orbitals

    3) Why can the O in the C=O bonds in edta ligand not attach to the central metal ion?

    Already answered by others.
    4) Why does execess ammonia only react with Cu(OH)2, and not with Fe(OH)2?
    The stability constant of Fe(NH3)4(aq) is less than that of Fe(OH)2, while the stability constant of Cu(NH3)4 is greater than Cu(OH)2
    5) Why do variable oxidation states help catalysts?
    Already answered by others.

    6) How do you determine the feasability of electrode potential reactions?

    Add the cell potentials of both half-cell reactions. If the result is positive, then the forward reaction is feasible.
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    (Original post by vinny221)
    1. The O on the C=O is electronegative and will pull a lot of the electron density from the the carbon atom. The Cl attached is therefore a good leaving group because of the lack of electron density on the carbon. The Cl may leave the molecule more readily therefore which implies a greater degree of reactivity.
    That's not exactly true. Chlorine is more electronegative than oxygen, so a great degree of electron density is actually pulled away from the carbon towards the chlorine. But that doesn't answer the question.

    The presence of the carbonyl C=O group is there not to substitute for pulling away electron density but is there because it has a double bond, which is stronger than the C-Cl bond. Hence, creating the greater possibility that the Cl group will come off, rather than the oxygen.
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    (Original post by aiman)
    6) How do you determine the feasability of electrode potential reactions?[/B]
    Add the cell potentials of both half-cell reactions. If the result is positive, then the forward reaction is feasible.
    Not necessarily.

    You need to apply the anti-clockwise rule before you know whether the reaction is feasible or not. A positive/negative e.m.f just dictates where the electron flow is going to in each electrode.
 
 
 
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