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The ionic radius of iodine in the high temperature form can be determined once the unit cell is known, but how can I calculate the radius of $\ce{I}$ given that the cubic lattice parameter is 6.10 Å.

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Considering by cubic lattice you mean a Simple Cubic structure, you could do the following:

(could not put it here because of copyright)

In a cubic lattice the atoms touch each other along the edges(see image1 and image2), hence the edge length/parameter will be equal to twice the radius (see image2). $$\therefore\ \ce{Edge\ Length(a)=2\times radius(r)}$$ $$\therefore\ a \ = \ 2\times r$$

This method can be adopted for any lattice, you only need to find the axis along which the atoms touch each other(image1 bottom row).


If they exhibited an fcc close packing structure, you would notice that the atoms touch each other along a face-diagonal.

$$ \begin{align} \therefore\ce{Length\ of\ face\ diagonal&\ =\ radius\ of\ top\ left\ atom +\\ &\quad\ diameter\ of\ atom\ at\ face-center + \\ &\quad\ radius\ of\ bottom\ left\ atom}\\ \therefore\ce{Length\ of\ face\ diagonal&\ =\ r\ + ( 2\times r ) + r}\\ \therefore\ce{\sqrt{2}\times a&\ =\ 4\times r}\\ \therefore\ce{a&\ = 2\sqrt{2}\times r} \end{align} $$


Similarly for bcc structure the atoms touch each other along the body-diagonal $$\therefore\ce{\sqrt{3}\times a\ =\ 4\times r}$$

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  • $\begingroup$ So, the answer must be in terms of the diagonal unit of the cube. $\endgroup$ Jan 16, 2013 at 14:35
  • $\begingroup$ @ordinarychemistrystudent I thought you stated it followed a cubic lattice, in which case it would be the first section, that is $a\ =\ 2\times r$ where $a$ is edge length $\endgroup$ Jan 17, 2013 at 14:42

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