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Why is an LiI anion more polarised than a LiF anion?

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Reply 1
Compare the electronegativity of I and F.
Original post by Claisen
Compare the electronegativity of I and F.

F is more electronegative then shouldn't the LiF ion be more polarised as a result?
Reply 3
I'm looking at this from two perspectives. The difference in electronegativities between the two atoms is greatest in LiF, so by this definition, LiF is a more polar molecule (Pauling). We can also look at it from a Hard/Soft Acid/Base (HSAB based on Fajan's rule) perspective. Li+ and F- are both 'hard' species and so neither have an overwhelming power to polarise to the other. This is in comparison to I- which is a 'soft' species and so is very polarisable.


So, we can use two different theories and they give different answers... The answer you're looking for obviously uses the HSAB/Fajan explanation above but typically I think it is fair to say more chemists follow the Pauling method.
Original post by Anonymous1502
Question above^

The iodide ion is MUCH bigger than the fluoride ion. This means that it holds onto its outer electrons more loosely. This allows the extremely small (high charge density) lithium ion to polarise the iodide charge cloud affording a degree of covalent character to the ionic compound.

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