If a face-centered cubic unit cell has a volume of 1.447 $\times$ 10⁸ pm³ and the ions at the corners touch the ion on the face, what must be the ion's radius? A)186 pm B)388 pm C)4,243 pm D)125 pm E)1,050 pm

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Quizplus · Accepted Answer

The radius of an ion in a face-centered cubic (FCC) unit cell can be determined using the formula for the edge length of the unit cell, $a = 2\sqrt{2}r$, where $r$ is the radius of the ion. Given the volume of the unit cell is $1.447 \times 10^{18} pm^3$, we can find the edge length $a$ by taking the cube root of the volume, which gives $a = 1129 pm$. Substituting $a$ into the formula for the FCC unit cell gives $r = \frac{a}{2\sqrt{2}} = \frac{1129 pm}{2\sqrt{2}} \approx 186 pm$.

If a Face-Centered Cubic Unit Cell Has a Volume of 1.447