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Strontium oxide has a face-centered cubic unit cell of oxide ions with the strontium in octahedral holes. If the radius of Sr²⁺ is 127 pm and the density of SrO is 4.7 g/cm³, what is the radius of the oxide ion? (100 cm = 1 × 10¹² pm)

a. 176 pm
b. 132 pm
c. 127 pm
d. 137 pm
e. 273 pm

1 Answer

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Final answer:

c. 127 pm. The radius of the oxide ion in strontium oxide cannot be accurately determined from the given information without additional data such as the edge length of the unit cell or a known relationship between the ionic radii.

Step-by-step explanation:

To find the radius of the oxide ion in a face-centered cubic unit cell of strontium oxide (SrO), one must consider the density of the compound and the radius of the strontium ion. The density of SrO is given as 4.7 g/cm³.

With strontium (Sr²⁺) having a radius of 127 pm, we can use the ratio of ionic sizes in a face-centered cubic lattice, where the cation fits into the octahedral hole formed by the anions.

The size of the octahedral hole is related to the size of the anions in the lattice. Since the strontium ion fits exactly into the octahedral hole, we know that the radius of the oxide ion should be larger than that of the strontium ion to accommodate the ionic sizes appropriately within the lattice structure.

However, without additional information such as the edge length of the unit cell or a specific relationship between the ionic radii, we cannot calculate the exact radius of the oxide ion. The correct way to determine the oxide ion radius would involve calculations based on the unit cell dimensions or a known empirical relationship.

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