I saw in the book Chemistry of the Elements by Greenwood & Earnshaw, 2nd ed., p 66, that the distance between calcium atoms decrease when it's forming an ionic structure.

The closest $\ce{M-M}$ approach in these compounds [metal hydrides] is often less than for the metal itself: this should offer no surprise since this is a common feature of many compounds in which there is substantial separation of charge. For example, the shortest $\ce{Ca-Ca}$ interatomic distance is $\pu{393 pm}$ in calcium metal, $\pu{360 pm}$ in $\ce{CaH2}$, $\pu{380 pm}$ in $\ce{CaF2}$, and only $\pu{340 pm}$ in $\ce{CaO}$ (why?). [sic]

Why does this happen?

  • $\begingroup$ You understand the passage correctly. It is referring to Ca-Ca distances not Ca-X. $\endgroup$ Mar 29, 2017 at 18:08

1 Answer 1


Simply put, the distance is measure from the center of the atoms. In the metal, the uncharged Ca atoms are kept at the van der waals contact distance.

In the ionic compounds the atomic radii of the Ca2+ ions have shrunk due to the loss of the outer shell electrons. Furthermore, negative ions cancel out the repelling charges and allow the Ca2+ ions to pack even closer together.


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