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Exothermic lattice energy Watch

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    Regarding 13a, what factors determine how exothermic the lattice energy is?

    Regarding 13b), I know that some compounds will show a difference between the experimental lattice energy and the purely ionic model due to the fact that the compounds have a small degree of covalent character. This degree of covalent character is due to the polarisation of the ionic bond which is in turn due to the attraction of the cation for the outer electrons of the anion..

    The higher the charge density of the cation, the stronger its polarising power.
    The larger the anion, the more polarisable it is..

    I eliminated C from the answers because the O2- ion in CaO is smaller than the S2- ion in CaS.. Therefore D would show more covalent character (and so it cannot be C)..

    But then I get stuck. How do I find the correct answer out of A, B and D?

    Cheers
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    http://en.wikipedia.org/wiki/Kapustinskii_equation

    this is the simplest thoery/equation for explaining this
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    13a) the higher the charges and the small the ions, the greater the strength of the bonding and so the more exothermic the lattice energy :yep:

    13b) Yeah you can rule out C with that reasoning In the same way you can rule out A since F- is smaller than O2-

    So now the choice is between B and C, which have the same anion. It's the charge density that counts for the cation - so Ca2+ is about 30% larger than Li+ but has double the charge.
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    Lol this alone is putting me off chemistry in the sixth form!
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    (Original post by Coke1)
    Lol this alone is putting me off chemistry in the sixth form!
    It's bound to be confusing if you've not been taught it :p:
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    (Original post by EierVonSatan)
    13a) the higher the charges and the small the ions, the greater the strength of the bonding and so the more exothermic the lattice energy :yep:

    13b) Yeah you can rule out C with that reasoning In the same way you can rule out A since F- is smaller than O2-

    So now the choice is between B and C, which have the same anion. It's the charge density that counts for the cation - so Ca2+ is about 30% larger than Li+ but has double the charge.
    In Q13a), both A and B have a total charge of -1 (since both their cations are +1, and their anions are -1; the product of which = -1).. however A's anion (F-) is smaller than B's anion (Cl-) so we know A will have a larger exothermic lattice energy (so we can eliminate B).

    C and D both have a total charge of -4, however C's anion (O2-) is smaller than D's anion (S2-), so we know C will have a larger exothermic lattice energy (and so we can eliminate D).. Now we are left with only A and C..

    A has a total charge of -1 but a small ionic radius, C has a total charge of -4 but a large ionic radius. How do we determine the answer?

    Thanks a lot !
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    (Original post by sabre2th1)
    In Q13a), both A and B have a total charge of -1 (since both their cations are +1, and their anions are -1; the product of which = -1).. however A's anion (F-) is smaller than B's anion (Cl-) so we know A will have a larger exothermic lattice energy (so we can eliminate B).

    C and D both have a total charge of -4, however C's anion (O2-) is smaller than D's anion (S2-), so we know C will have a larger exothermic lattice energy (and so we can eliminate D).. Now we are left with only A and C..

    A has a total charge of -1 but a small ionic radius, C has a total charge of -4 but a large ionic radius. How do we determine the answer?

    Thanks a lot !
    The size of C isn't much larger than A, but C has double the charge. So you'd expect C to have the larger lattice energy
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    (Original post by EierVonSatan)
    The size of C isn't much larger than A, but C has double the charge. So you'd expect C to have the larger lattice energy
    Ah I see.. but isn't it Isn't it quadruple the charge?

    Also regarding this;

    13b) Yeah you can rule out C with that reasoning In the same way you can rule out A since F- is smaller than O2-

    So now the choice is between B and C, which have the same anion. It's the charge density that counts for the cation - so Ca2+ is about 30% larger than Li+ but has double the charge.
    Don't the parts in bold go against each other? Since first you have said rule C out, but then you said ''the choice is between B and C''

    Also B is Li2O, doesn't the diatomic lithium molecule affect the charge density etc?

    Thanks!
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    (Original post by sabre2th1)
    Ah I see.. but isn't it Isn't it quadruple the charge?
    I meant double per ion, sorry.

    Don't the parts in bold go against each other? Since first you have said rule C out, but then you said ''the choice is between B and C''

    Also B is Li2O, doesn't the diatomic lithium molecule affect the charge density etc?

    Thanks!
    Confusing myself :lol: So yes, we can rule out A and C.

    Then decide between B and D, which is actually easier. S2- is bigger i.e. more polarisable than O2- and Ca2+. The previous argument between the two cations still applies

    When you have multiple of ions like in B, just treat them as they're singular :yep:
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    (Original post by EierVonSatan)
    I meant double per ion, sorry.



    Confusing myself :lol: So yes, we can rule out A and C.

    Then decide between B and D, which is actually easier. S2- is bigger i.e. more polarisable than O2- and Ca2+. The previous argument between the two cations still applies

    When you have multiple of ions like in B, just treat them as they're singular :yep:
    Ah.. Thanks a lot! You have helped a lot over the past few days, hopefully I will utilise some of the knowledge (you passed on) tomorrow !
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    (Original post by JMaydom)
    http://en.wikipedia.org/wiki/Kapustinskii_equation

    this is the simplest thoery/equation for explaining this
    Thanks but its quite complicated for an AS student lol, but I understand now anyways so no worries
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    (Original post by sabre2th1)
    Ah.. Thanks a lot! You have helped a lot over the past few days, hopefully I will utilise some of the knowledge (you passed on) tomorrow !
    Good luck :gah:
 
 
 
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