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Having the same atomic radii of 145 pm, should one mole of tin and one mole of magnesium take up the same volume? Since tin is much heavier, that means tin is much denser than magnesium. Is this true, or is there an error in my thought process.

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  • $\begingroup$ You have to specify conditions, and if we are speaking of crystalline matter, what polymorphs you are interested in. It's very unlikely that both are going to have the exact same molar volume. $\endgroup$ – andselisk May 29 '19 at 16:22
  • $\begingroup$ I think this is a superb question! The measure of atomic sizes is done by many methods, all of which have value. So when you see a number like atomic radius = 145 pm, is that definitive? No, because the packing of atoms is not always the same, as noted below. But the contrast between densities and molar volumes is instructive. Tin is denser ("heavier") (7.31 g/cc) than Mg (1.731), but the molar volumes are fairly close (Sn: 16.29 cc) vs Mg (14.00 cc). So the numbers of atoms in a given volume are similar, but tin weighs 118.7, while Mg weighs a paltry 24.3. $\endgroup$ – James Gaidis May 30 '19 at 14:58
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    $\begingroup$ I do not think this is a homework question. In fact it offers an opportunity to clarify some interesting issues. $\endgroup$ – matt_black May 30 '19 at 17:21
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No, that is not how it works.

Atomic radius means the radius of an atom. A single atom. It does not in and by itself mean that a solid chunk of a material has a particular density. A solid chunk of material has many atoms, and they are arranged in certain ways and sometimes more than one way of this arrangement is possible.

Tin and Magnesium are very different. Magnesium is arranged in a hexagonal close packed structure, basically just repeating units of skewed (rombohedric) prisms. This can be hard to imagine, but imagine just stacking perfect spheres so they rest on each other and likely you will end up at a close packing arrangement. (there are only two*, try to find them. Hint: one has three layers before it repeats the bottom, the other has only two layers then it repeats)

*: There are actually 4, but you'll need a lot of layers and a lot of spheres to find those.

Tin has two different ones, so already here your question has to be no'ed without further investigation, but lets anyway. The first one, grey tin (${\alpha}$) has a diamond like structure. I am finding it hard to describe them in a meaningful way, but they have examples over at the excellent Wikipedia. The other one, tin (${\beta}$) which is white has a body centered tetragonal - which is basically stack 4 - put 1 in the middle and then stack 4 on top of that again (yes, the spheres will roll off, but there are so many of them per layer that hardly any of them do.) The white one is amusing, if you bend it it screams like you are hurting it. Sounds absolutely disgusting, like nails on a chalkboard. This is related to it's structure.

Summary: When you put a whole mole (ha... ha...) of atoms together, they organize in structures, and it is the number of atoms per volume that you have to multiply with the atomic weight to get the density of the chunk. Fortunately for most kinds of structures, you don't have to do this, someone already has done the work, by actually measuring the densities of crystals and made tables. Just look it up.

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    $\begingroup$ One should note that there are more than 2 close packed structures (see en.wikipedia.org/wiki/Periodic_table_(crystal_structure)). The main one, DHCP, found in several Lanthanides/Actinides, is ABACABAC. Samarium is particularly weird. In theory, any periodic arrangement of (A,B,C) would be close packing. In reality, the longer the period the less likely it will be found in nature. $\endgroup$ – Jon Custer May 29 '19 at 17:57
  • $\begingroup$ @JonCuster Right you are. I was drawing from memory, didn't remember those weird ones. I am still leaning towards keeping it simple for the OP though. Should I edit what you suggest in or... ? $\endgroup$ – Stian Yttervik May 29 '19 at 20:02
  • $\begingroup$ Up to you. A parenthetical comment added would be more than enough. $\endgroup$ – Jon Custer May 29 '19 at 20:52

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