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Calculate the radius of iridium atom, given that Ir has an FFC crystal structure, a density of $22.4\text{g/cm}^3$. and atomic weight of $192.2\text{g/mol}$.

Everyone whom I asked this question is giving different answers. By the way, I was going to ask something about this. Why are they finding that the atomic radius of iridium atom is $1.34\text{pm}, 1.35\text{pm}, \text{1.36pm}$? According to textbook solution, it seems $1.36\text{pm}$. However, they said that all answers correct. Thereupon, I wanted to check their answer from wikipedia, that gives anohter answer too. What would you say? Are all answers correct?

Kindest Regards!

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  • $\begingroup$ atoms are not spheres with precisely defined radii. It should not be surprising that different sources give slightly different numbers. In any case this belongs on chemistry.stackexchange.com $\endgroup$ – agentp Dec 10 '17 at 19:12
  • $\begingroup$ So, are those answers correct? Maybe different? $\endgroup$ – Mascular Dec 10 '17 at 19:27
  • $\begingroup$ Probably the 1.36pm is the most correct (this is the result of the calculation), but measuring it with other methods will give a different result. I think your question is a good one, but it should have been asked on the chemistry.stackexchange.com . $\endgroup$ – peterh - Reinstate Monica Dec 10 '17 at 21:26
  • $\begingroup$ I just wanna know what is correct. $\endgroup$ – Mascular Dec 11 '17 at 13:56
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The radius $r$ can be calculated purely from geometry. Consider the FCC lattice. Its lattice parameter $a$ is related to radius through the arrangement of atoms. What is the close-packed direction in the FCC lattice? Along the face diagonal, because two corner atoms touch the face-centered atom. The length of the face diagonal is $\sqrt{a}$ by the pythagorean theorem. That length is also four $r$. You can draw the lattice and the atoms of a face to convince yourself this is true. So $4r=\sqrt{a}$.

Now it remains to calculate the lattice parameter. How many atoms per unit cell? Again you can draw the lattice and atoms of the unit cell to convince yourself there are 4 atoms per unit cell (remember corners count for 1/8 atoms and faces 1/2). How many unit cells per mole? Well there are $6.022\times 10^{23}$ atoms per mol, divided by atoms per unit cell gives unit cells per mole. How big is the unit cell? Of course $a^3$ cm cubed per unit cell. Multiply by unit cells per mole to get cm cubed per mole. Now multiply by density (grams per cm cubed) to get grams per mole. Since those are units of atomic weight we can set what we have equal to the value given. Solve for $a$, then substitute into the earlier expression to obtain $r$.

The probable reason there are different values is that different methods may have been used to calculate the atomic radius. What we just did is one method, and accurate enough for bulk calculation purposes in an engineering context. Another method is to use Density Functional Theory to model the electrostatic force given lattice parameters and electron orbitals and other properties. It is also possible that the values are for Ir present in various alloys or compounds, or at different temperatures. Atomic radii vary for the same atom when bonded to others, and for different temperatures due to thermal expansion.

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  • $\begingroup$ So are all answers correct? I wanna know it. $\endgroup$ – Mascular Dec 10 '17 at 18:54
  • $\begingroup$ Unfortunately I have no way of knowing that. For the purposes of a homework problem the above calculation is correct enough. $\endgroup$ – wwarriner Dec 10 '17 at 19:29
  • $\begingroup$ Just share your thinkings with me. Everyone finds anohter answer and I doubt it. $\endgroup$ – Mascular Dec 10 '17 at 20:26
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    $\begingroup$ Some have already shared with you - now you need to do the calculations yourself and understand the assumptions. $\endgroup$ – Solar Mike Dec 10 '17 at 22:11
  • $\begingroup$ I just wanna know what is correct. So I mean are all answers correct? $\endgroup$ – Mascular Dec 11 '17 at 13:57

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