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So neutrons are neutral in terms of charge, and adding neutrons to an atom affects its atomic mass. But when neutrons are added to the nucleus, the nuclear radius would be affected. Couldn't that affect how the charge of the nucleus is distributed within the nucleus?

With an atom large enough that, say, 40 neutrons could be added (and still be stable), I would imagine that the nucleus would be slightly rearranged to accommodate this (so that neutrons and protons are distributed throughout the nucleus and there isn't a shell of 40 neutrons surrounding what is the pre-existing nucleus).

I imagine that this redistribution (if it does occur) would dilute the charge coming off of the nucleus (as the protons are farther apart and the nucleus is less dense in terms of charge), which would lead the electron shells to be farther from the nucleus. This would increase atomic radius, affecting a wide-range of chemical characteristics. I could equally see an opposite effect, where the nuclear charge is strengthened (which would decrease atomic radius).

Is this an actual reaction to the adding of nuclear charges? If so, does it occur in large enough quantities to play a significant role in chemistry? Or have I just made a false assumption?

Mithoron
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Jean Valjean
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  • This is an interesting question, but one that may be better posted to Physics.SE, at least in terms of whether such changes actually would occur to the nucleus. – hBy2Py Apr 03 '16 at 21:31
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    You've just made a false assumption. A charge is a charge; you can't dilute it with neutral matter (or rather, you can, but it would make no difference). From the electron's point of view, a nucleus is but a dot, no matter how many neutrons are there.

    Sure enough, isotopes would differ in chemical and physical properties (that's why it is possible to separate them), but only very slightly (that's why it is so difficult).

     – Ivan Neretin Apr 03 '16 at 21:34
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    Some quantum chemistry programs effectively let you simulate this, by letting you specify various models for the nuclear charge (e.g. point, Gaussian, or something else, and with variable parameters). See, for example, the ADF program's docs on the subject. – Aesin Apr 03 '16 at 22:21
  • @Aesin, to be sure I understand, the nuclear charge model affects reactivity negligibly? It only shows up in certain subtle spectroscopic measurements? – hBy2Py Apr 04 '16 at 01:45
  • The proton to neutron ratio does effect the stability of the nucleus itself. Unstability in this context meaning that the nucleus would undergo radioactive decay. – MaxW Apr 04 '16 at 02:39
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    @Brian: That's my understanding, yes. Or more specifically, spectroscopic properties involving orbitals with a significant amplitude inside the nucleus. – Aesin Apr 04 '16 at 15:37
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    Possible duplicate of Chemical properties of isotopes – Mithoron Jun 18 '17 at 12:25

2 Answers2

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Even if such a phenomenon were to occur, and the resulting nuclei were sufficiently stable to participate in chemistry (they would not be sufficiently stable, in reality), the "inflation" of the charge of the nucleus would almost certainly have a negligible effect on the chemistry. The nucleus is really really really tiny with respect to the length scales of bonds and the electron cloud.

To give an idea of the scales involved: the largest stable element, uranium, has a nuclear radius of about fifteen femtometers $\left(15~\mathrm{fm}\right)$, per Wikipedia, or $1.5\times 10^{-14}~\mathrm m$. Bonds and electron clouds are measured in Angstroms $\left(10^{-10}~\mathrm m\right)$. Thus, the nuclear radius would have to expand by well over $\sim\!\!100$-fold in order for the assumption of a point-charge nucleus to begin to fail.

It is well accepted that the effect of isotopic substitutions on chemistry primarily is due to the change in mass of the nucleus, leading to the kinetic isotope effect. Another potential isotope effect on reactivity involves the change in nuclear spin, leading to a "magnetic isotope effect" only relevant for certain classes of reactions between species with unpaired electrons (e.g., organic radicals and/or various inorganic species).

Isotopic substitution does affect the behavior of nuclei in certain spectroscopies, however, leading to isotopic shifts. The isotopic mass change can have noticeable effects on the vibrational spectrum, due to the effect on the local distribution of mass. As described in the link to ADF's documentation provided by Aesin in a comment to the question, these effects can also be seen in the hyperfine structure of atomic spectra. As well, the NMR isotopic shift of light elements, especially hydrogen, can sometimes be observed.

hBy2Py
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    ... and to expand the radius by 100-fold would require $100^3$-fold increase in the number of nucleons since the nucleus is approximately spherical. (Not entirely true if the nucleus has angular momentum. But such complication is a but much for this question.) Now a million nucleons is around 4000-times as many as we find in heavy atoms, like Uranium, so this has never been observed. – Eric Towers Apr 04 '16 at 01:04
  • @EricTowers Precisely! :-) – hBy2Py Apr 04 '16 at 01:17
  • But isn't there a chance that the neutrons could have some sort of slight shielding effect? – SendersReagent Apr 04 '16 at 01:56
  • @DGS Shielding effect of what type? – hBy2Py Apr 04 '16 at 02:46
  • @Brian Maybe due to polarizability within a neutron, there can be random neutralizing or enhancing effects? – SendersReagent Apr 04 '16 at 07:02
  • You can't shield a charge (or rather, you can shield it with an opposite charge, but that's another story). – Ivan Neretin Apr 04 '16 at 07:26
  • @DGS Agree w/Ivan, the mere presence of a neutron can do nothing to shield the positive charge of protons. Even if a neutron could undergo a local polarization analogous to the induced dipole of a London interaction (another good question for Phys.SE, not here!), on the Angstrom length scale of the surrounding electron cloud a 'polarized neutron' would still appear just as a zero-charge constituent of a ~point nucleus. – hBy2Py Apr 04 '16 at 11:02
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The hydrogen atom (a single proton and electron) has the greatest effect from an added neutron, turning it into deuterium. From http://en.wikipedia.org/wiki/Deuterium: "Chemically, there are differences in bond energy and length for compounds of heavy hydrogen isotopes compared to normal hydrogen, which are larger than the isotopic differences in any other element."

Winton Baker
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