Does this have something to do with the inert pair effect?
1 Answer
Ahh, Thallium is a very nice element. Because it's weird.
The +3 oxidation states are favorable except for the heavier elements, such as Tl, which prefer the +1 oxidation state due to its stability; this is known as the inert pair effect. Chemwiki.UCDavis
As an example in group 13 the +1 oxidation state of Tl is the most stable and Tl(III) compounds are comparatively rare. Inert pair effect, Wikipedia
So, why?
Inert pair effect was introduced to describe why the facts above are correct. However, the original explanation, provided by Sidgwick, relied on the fact that the $s$ subshell has an "inert" pair of electrons. This can't be correct, and Wikipedia uses the trend in ionization potentials of "$s$" electrons to prove it wrong: $$\ce{In < Al < Tl < Ga}$$
A better explanation lies within the lower bond enthalpies of $\ce{MX}$ where M is one of the p-block heavy elements.
To reach a high oxidation number, elements will need to have a strong bond. They wanna bond in order to lose energy (i.e. become more stable) after all! If the bond is weak, this energy isn't provided. Thus, a heavy element is better off with being less oxidised.
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$\begingroup$ -1, inert pair effect is definitely a real thing; it's due to the lanthanide contraction for the 6p elements + relativistic 6s contraction (the "inert pair"). Also, the thallium halides aren't molecular so invoking bond enthalpies doesn't give the full picture either. $\endgroup$– J. LSCommented May 4, 2015 at 15:57
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$\begingroup$ [email protected] I never said it's not real. Just that the description isn't very authentic. (Or rather, the first description) How do you explain the trend in ionization potentials? $\endgroup$– M.A.R.Commented May 4, 2015 at 16:01
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$\begingroup$ @J.LS Figures out, the pattern in ionization potentials is due to the relativistic effects you mention. $\endgroup$– M.A.R.Commented May 4, 2015 at 16:24
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$\begingroup$ No, the pattern in ionization potentials is due to the transition metal and lanthanide contractions and the associated decrease in shielding for Ga/Tl respectively; relativistic effects are only important for Tl and probably are of secondary importance. $\endgroup$– J. LSCommented May 4, 2015 at 16:51