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### Topic: Diff. between value calculated from Born-Haber cycle and that from charge/ radii  (Read 4446 times)

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#### Charismaztex

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##### Diff. between value calculated from Born-Haber cycle and that from charge/ radii
« on: February 08, 2010, 07:23:50 AM »
Q. Why is there generally a difference between the lattice enthalpy of a compound calculated from a Born-Haber cycle and that calculated from the charge numbers and radii of its ions? [2 marks]

(Is it because of the average values we take from measuring enthalpy changes, especially affected by the surroundings?)

Any help would be great,
Charismaztex

#### Angelant99

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##### Re: Diff. between value calculated from Born-Haber cycle and that from charge/ radii
« Reply #1 on: February 09, 2010, 09:15:35 AM »
Theoretical lattice enthalpies (calculated from charge numbers and radii) are generally different from experimental values (calculated through Born-Haber cycles) for a very good reason: ionic polarisation.

If you assume that the ions in an ionic compound are nice and spherical, you can work out what the LE (lattice enthalpy) should be, but your answer is likely to be slightly off.

Why? Well, positive and negative ions are not usually spherical - positive ions will polarise neighbouring negative ions to different extents, and this increase in polarisation makes the bonding more covalent (and distorts the spherical shape.)

Naturally therefore, for compounds in which there is a higher degree of covalency, there will be a greater difference between the experimental and theoretical LE values.
(There's actually a rule that dictates the likelihood that this will happen: it's called Fajan's rule.)
Basically, a compound will have a higher degree of covalency if the positive ion is small (and has a high charge density); the negative ion is large.

E.g.  BeF2, beryllium fluoride - the theoretical LE value is -3150kJ/mol  and the experimental value is -3505kJ/mol

That's quite a bit of difference. The reason for this is because a beryllium ion has a 2+ charge, but is relatively small; despite the 2 fluoride ions being only slightly larger themselves, there is quite a lot of covalency in the compound.

Counter e.g.
If we look at the opposite - compounds with more ionic bonding - like, say K2O, potassium oxide,  the difference between the 2 LE values is, according to my Nuffield book of data, only 6kJ/mol less. (-2232 vs -2238)
That's because this time, we've got fairly large potassium ions with only +1 charges each, next to a rather miniscule by comparison, oxide ion: so the 2 LE values end up being very close together.

Quick note:
Positive ion(s) large, negative ion(s) small and both have small charges? Likely to be ionic.

Positive ion(s) small, negative ion(s) large and greater difference in charges? Likely to be more covalent.

Hope any of that was useful

Chemistry. Enough said