Results are out! Find what you need...fast. Get quick advice or join the chat
x

Unlock these great extras with your FREE membership

  • One-on-one advice about results day and Clearing
  • Free access to our personal statement wizard
  • Customise TSR to suit how you want to use it

Exothermic lattice energy

Announcements Posted on
Rate your uni — help us build a league table based on real student views 19-08-2015
  1. Offline

    ReputationRep:


    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
  2. Offline

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

    this is the simplest thoery/equation for explaining this
  3. Offline

    ReputationRep:
    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.
  4. Offline

    ReputationRep:
    Lol this alone is putting me off chemistry in the sixth form!
  5. Offline

    ReputationRep:
    (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:
  6. Offline

    ReputationRep:
    (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 !
  7. Offline

    ReputationRep:
    (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
  8. Offline

    ReputationRep:
    (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!
  9. Offline

    ReputationRep:
    (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:
  10. Offline

    ReputationRep:
    (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 !
  11. Offline

    ReputationRep:
    (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
  12. Offline

    ReputationRep:
    (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:

Reply

Submit reply

Register

Thanks for posting! You just need to create an account in order to submit the post
  1. this can't be left blank
    that username has been taken, please choose another Forgotten your password?
  2. this can't be left blank
    this email is already registered. Forgotten your password?
  3. this can't be left blank

    6 characters or longer with both numbers and letters is safer

  4. this can't be left empty
    your full birthday is required
  1. By joining you agree to our Ts and Cs, privacy policy and site rules

  2. Slide to join now Processing…

Updated: May 14, 2012
TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

Today on TSR

Win a mini-fridge

Don't miss our Freshers competition!

Poll
Do you prefer exams or coursework?
Study resources
x

Looking for some help?

Ask our friendly student community a question

Ask a question now

Or get help from our smart tools and guides

GCSE help A-level help Uni application help Everything else
Quick reply
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.