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Why is it that only Promethium (Pm) is the only radioactive lanthanide?

I was trying to figure out what percentage of all radioactive elements in the Periodic Table the f-block accounts for, when it struck me that Promethium was the only radioactive lanthanide shown. I found this a bit odd, since I expected to see a couple of other radioactive elements there as well, especially since Promethium is (somewhat) near the middle of the series.

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    $\begingroup$ Nuclear properties have absolutely nothing to do with chemistry (f-block and everything). Promethium being radioactive is just bad luck. Technetium is another such case. $\endgroup$
    – Ivan Neretin
    Sep 6, 2016 at 17:13
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    $\begingroup$ Glen Seaborg might disagree a little bit on it being chemistry or not ;), but @IvanNeretin is generally correct. The issue isn't f-electrons so much as nuclear structure, where it turns out that the Samarium nucleus is more tightly bound (see nuclear shell model), so flipping one of Promethium's neutrons to a proton results in a lower-energy nucleus (not that it stops there in general, but...). $\endgroup$
    – Jon Custer
    Sep 6, 2016 at 17:20
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    $\begingroup$ It is not the only radioactive lanthanide. It is the only one with no stable isotopes. Other stable lanthanides also have naturally occurring radioactive isotopes (samarium is a good example). $\endgroup$
    – Gimelist
    Sep 7, 2016 at 5:51
  • $\begingroup$ @IvanNeretin I would advise against making such a sweeping absolutist statement. It is well-established that nuclear properties can affect chemical properties. Granted, it's not typical, but it's nevertheless incorrect to take the hard-line stance that it doesn't happen at all. See, for instance: tandfonline.com/doi/abs/10.1080/00222348808212315?src=recsys $\endgroup$
    – theorist
    Apr 1, 2020 at 5:58
  • $\begingroup$ @theorist See, in natural sciences there is always an exception, or a subtlety, or a tiny deviation in nearly any rule and law. If we start with exceptions, the students never get to know the rules in the first place. Here I stand, and I can do no other. $\endgroup$
    – Ivan Neretin
    Apr 1, 2020 at 6:06

1 Answer 1

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Lanthanoids are not generally supposed to be radioactive, with exception of mentioned very long lived radioisotopes.

The longest half-life radioactive nuclides of lanthanoids ( in years )

$\ce{^{150}Nd}:\ \pu{6.7e18}$
$\ce{^{151}Eu}:\ \pu{5e18}$
$\ce{^{148}Sm}:\ \pu{7e15}$
$\ce{^{144}Nd}:\ \pu{2.29e15}$
$\ce{^{152}Gd}:\ \pu{ 108e12}$
$\ce{^{147}Sm}:\ \pu{106e9}$
$\ce{^{138}La}:\ \pu{ 102e9}$
$\ce{^{176}Lu}:\ \pu{ 38.5e9}$

Promethium is rather a victim of circumstances there was no free position to have at least one stable isotope. It is related also to the Liquid drop model.

There are few semi-empirical rules, related to proton and neutron fermionic pairing within the nucleus model:

Elements with the odd proton number have maximally 2 stable isotopes.

2 Isotopes of adjacent elements with the same nucleon number, like $\ce{^N_{A}X, ^N_{A+1}Y}$, are not both stable. ( But some of these unstable ones have very long half-life as seen above.)

Isotopes not close enough to the Valley of stability are not stable, as they have too little or too many neutrons.

The stability of isotopes of similar nucleon number generally decreases in order

For Pm and Tc, both have odd proton numbers what gives them disadvantage.

All their possible stable isotope candidates are already taken by their neighbors as energetically preferred variants. Therefore all isotopes of given 2 elements beta decay to nuclei of adjacent elements with less energy.

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