How is the size of $\ce{Li+}$ ($\pu{0.68}$) larger than $\ce{Ga^3+}$ ($0.62$)?

This is very difficult for me to understand why they say that size of Li+ with 3 protons and just 2 electrons should be considerably more than that of $\ce{Ga^3+}$. I understand the poor shielding effect of $\ce{3d^{10}}$ electrons but it isn't obvious to me why it will make its size smaller than $\ce{Li+}$ which has just one shell with 2 electrons and more effective nuclear charge. Where is my logic wrong?

  • 3
    $\begingroup$ Because removing three electrons from Ga leaves a much stronger net positive charge, pulling in the orbitals, than pulling one from Li. $\endgroup$ – DrMoishe Pippik Jun 11 '18 at 18:02
  • $\begingroup$ We removed remove the 1 electron from 4p1 and then remove 2 electrons from 4s2 these three electrons where the ones on which the nucleus pulled on with more force but now they are not there. $\endgroup$ – Jasmine Jun 11 '18 at 18:55
  • $\begingroup$ @drmoishepippik but what about Ga3+ containing 3 shells and Li only one? $\endgroup$ – Jasmine Jun 11 '18 at 18:57
  • $\begingroup$ Could you link the source of the image? $\endgroup$ – HolgerFiedler Jun 11 '18 at 20:24
  • $\begingroup$ @holgerfielder chemistry.stackexchange.com/questions/98166/… $\endgroup$ – Jasmine Jun 12 '18 at 2:50

That is impressive. What you may need to bear in mind is that as the nuclear charge increases, all the electron orbitals move in closer to the nucleus. You may have been assuming, consciously or unconsciously, that the electron shells appear at a relatively fixed distance from the nucleus -- that is, that the n=1 shell is always at about the same distance from the nucleus, the n=2 shell is always at about the same additional distance, and so on.

Not the case! For example, if you take a look at the quantum mechanical model of the hydrogen atom and solve it for a "hydrogen-like" atom with atomic number Z and one electron (so, He+, Li++, Be+3,....,U+92 ha ha) then you will find that the average orbital radius of the electron scales like 1/Z. So if we ignored electron-electron repulsion entirely, the radius of each shell would decrease like 1/Z as Z increases. That's quite a contraction. It would mean that the n=1 shell in carbon, say, would be 1/6 the size of what it is in hydrogen. In the case of Ga, Z=31, the n=1 shell would be 1/31 the size it is in H! Each new shell would contract similarly.

So in the absence of electron-electron repulsion we would expect the size of atoms to shrink drastically as Z increases, going across the Periodic Table, and bumping up slightly at the beginning of each row with the addition of a new shell. Of course, real atoms do have electron-electron repulsion, lots of it, so the effect is considerably muted.

Nevertheless, the size of atoms does shrink considerably going from left to right in the Periodic Table, and it's worth remembering this is because all of the electron shells shrink, not just the valence shell.

| improve this answer | |
  • $\begingroup$ This surely helps. I didnt know that the size radius by 1/Z and i think it may even lead to repulsion between the electrons in 2 consecutive shells $\endgroup$ – Jasmine Jun 11 '18 at 19:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.