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Apparently the of last four, $\ce{Mg^2+}$ is closest in radius to $\ce{Li+}$. Is this true, and if so, why would a whole larger shell ($\ce{Mg^2+}$) be closer in radius to $\ce{Li+}$ than its isoelectronic neighbor ($\ce{Be^2+}$) with just one extra proton?

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    $\begingroup$ It's a common misconception that a 2s orbital in lithium is the same size as a 2s orbital in, say, fluorine. It's not true at all. That "just one extra proton" has quite a large impact. To directly answer the question, en.wikipedia.org/wiki/Diagonal_relationship Second paragraph, second sentence: "On moving across a period of the periodic table, the size of the atoms decreases, and on moving down a group the size of the atoms increases." $\endgroup$ – orthocresol Apr 20 '16 at 22:42
  • $\begingroup$ Consider the size of the first s-orbital of $\ce{Mg}$. The nuclear charge is 12+. Therefore the first electrons, close to the nucleus, are attracted by a relatively large force. These first electrons the 'shield' the nucleus from some of the charge when adding more electrons to the orbitals, these latter electrons experience a smaller charge than the first. The result of this is that nuclear radius does not change according to number of electrons (only). $\endgroup$ – Eljee Nov 26 '16 at 16:25
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Many people may think that orbitals, or ‘shells’ essentially work like planet orbits around the sun; i.e. no matter how many electrons populate ‘Mercury’, ‘Venus’ and ‘Earth’ or how many protons are in the ‘Sun’ Venus’ orbit will always have the same dimension.

This is very much not so when studing atomic orbitals. The principle of hydrogen-like orbitals (which are the only orbitals that can be exactly calculated with the Born-Oppenheimer–approximation) is that we are dealing with an electron free to rotate and translate on a potential energy surface which is formed by the electrostatic potential of a proton or a nucleus with a charge of $z+$. Commonly, taking the 1s orbital as an example, there is a distance from said nucleus where the probability of locating the electron in the sphere defined by that radius is larger than a certian cutoff value, e.g. $90~\%$. If we go from a hydrogen-like system to a $\ce{He+}$ cation, we are doubling the charge thus strongly increasing the attractive force between nucleus and electron. As such, the $90~\%$-radius is much closer to the nucleus than before.

This holds true with limitations when no longer observing hydrogen-like atoms but actual multi-electron atoms: the more protons in the nucleus the more contracted the orbitals become. While that effect in itself is rather small, the second one is substantial: the higher an ion’s charge is, the more contracted its orbitals become. Therefore, $\ce{Be^2+}$, which is even considered to be the smallest atom or atomic ion known (yes, smaller than hydrogen) is substantially smaller than lithium.

When directly comparing two atom sizes of elements that are above and below each other in the periodic table, naturally under otherwise identical conditions the lower one will be larger; even though it has at least 2 (lithium) more protons it has also started populating an additional shell which is further removed from the nucleus. Similarly, if you compare $\ce{Li+}$ and $\ce{Mg+}$, magnesium will still be substantially larger. (In this case, the additional remaining electron in magnesium’s 3s orbital is the deal breaker but note that the same argument applies to $\ce{C+}$ and $\ce{P+}$ where no shell is completely empty.

As noted before, however, if we substract another electron, i.e. if we compare the isolobal $\ce{Li+}$ and $\ce{Mg^2+}$, the latter’s orbitals shrink substantially. By empirical observations scientists have discovered what is known as the diagonal relationship: two ions that are top-left bottom-right diagonally placed in the periodic table, whereby the bottom-right one’s charge is greater by one, have approximately the same radius.

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