I am currently studying the structures of ionic compounds and how we can use radius ratios to predict the possible arrangement of ions in the lattice. My book mentions that the cation to anion radius ratio in caesium chloride is about 0.93, and since this number lies between 0.732 and 1, it suggests a cubic arrangement with a coordination number of 8.

But caesium is in the 6th period of the periodic table whereas chlorine is in the 3rd. A caesium cation would be similar to xenon, which is in the 5th period. And a chloride anion would be similar to argon, a period 3 element.This means that there is a difference of two shells between the two atoms. How is the ionic radius of chloride still larger? In general, how many "shells of difference" are required for a cation to be larger than an anion?

  • $\begingroup$ I think it may be simply nuclear charge. The effective nuclear charge for an electron in a 5p Cs atom is 13.651. On the other hand, the effective nuclear charge for an electron in a 3s Cl atom is 6.1161. knowledgedoor.com/2/elements_handbook/… I do not think there is an effect similar to lanthanide contraction for cesium. $\endgroup$ Dec 12, 2016 at 6:45

1 Answer 1


This is something students often forget when comparing radii: shells are not simply additive.

If you move along eight elements to go from lithium to sodium, you are not only filling the second shell and putting one electron into the third, you are also adding a proton to the nucleus with every additional electron. The larger the nuclear charge is, the greater the electrostatic potential the electrons experience. And the greater the potential the stronger the force that draws them towards the nucleus, thus the more compact an orbital will become. Thus, while each new period will enlargen the atomic radius each added proton will reduce it. Therefore, the atoms do not become infinitely larger in size — although a sodium atom will always be larger than an electronically comparable lithium atom.

A similar discussion can be made for charges. Adding an electron to the mix increases the electron-electron repulsion (more negative charge) while removing one decreases it. Thus, with each positive charge a cation will get smaller while an anion will get larger with each negative charge. All of this results in rather complex calculations; one should always remember that an atomic radius cannot be easily guessed unless it is a straightforward up/down or left/right movement in the periodic table.

As for your second question: an anion may be smaller than a cation if the anion has very few shells and the cation very many. Unfortunately, the Wikipedia article on ionic radii does not quote a size of the hydride anion which would otherwise be a strong contestant. Instead, the smallest anion in the tables is $\ce{F-}$ with a crystal ionic radius of $119~\mathrm{pm}$ in hexacoordinate state. Searching for a cation that is larger quickly gives potassium with $152~\mathrm{pm}$ crystal ionic radius. In effective ionic radius terms (the second table), fluoride is mentioned as having $133~\mathrm{pm}$ while $\ce{K+}$ has $138~\mathrm{pm}$. Thus, by both sets of values the cation in $\ce{KF}$ is larger than the anion.

In crystal ionic radii, chloride is smaller than caesium and fracium while the latter two’s effective ionic radii are larger than chloride’s.

As you can see, there is no hard and fast rule. Fluoride to potassium is a period and two elements difference while chloride to eka-francium is four periods and two elements.

  • $\begingroup$ What's the difference between effective ionic radius and crystal ionic radius? $\endgroup$
    – Newton
    Dec 12, 2016 at 16:51
  • $\begingroup$ @Kalyan I must refer you to the Wikipedia article. Something about the effective being closer to values Pauling used, I think. $\endgroup$
    – Jan
    Dec 12, 2016 at 16:52

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