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Chapter 9: Problem 71

Cobalt fluoride crystallizes in a closest packed array of fluoride ions with the cobalt ions filling one-half of the octahedral holes. What is the formula of this compound?

Short Answer

Expert verified
The compound cobalt fluoride crystallizes in a closest packed array of fluoride ions with cobalt ions filling half of the octahedral holes. The ratio of cobalt ions to fluoride ions is 1:2, which leads to the empirical formula of the compound as \(CoF_2\).

Step by step solution

01

Understand the closest packed array structure with octahedral holes

In a closest packed array, the arrangement of ions is such that it occupies the maximum volume possible, creating the most efficient packing. In our case, fluoride ions form the closest packed array, and cobalt ions are present in octahedral holes. In this type of lattice, there are two types of holes formed: tetrahedral and octahedral holes. The number of tetrahedral holes is twice the number of atoms in the closest packed array, and the number of octahedral holes is equal to the number of atoms in that array. Since we are interested in octahedral holes, we will focus on the statement that the number of octahedral holes is equal to the number of fluoride ions in the closest packed array.
02

Calculate the ratio of cobalt ions to fluoride ions

Now, as we know that cobalt ions fill one-half of the octahedral holes, we can determine the ratio of cobalt ions (Co) to fluoride ions (F): Let 'x' represent the number of fluoride ions. Number of octahedral holes = x (since it's equal to the number of fluoride ions) Half of the octahedral holes filled by cobalt ions = x/2 So the ratio of cobalt ions to fluoride ions can be represented as: Co : F = x/2 : x
03

Deduce the empirical formula of the compound

To get the simplest whole number ratio of cobalt ions to fluoride ions, we can divide both sides of the ratio by the smaller number, in this case x/2. (Co : F) = (x/2 : x) = (1 : 2) This means that there is one cobalt ion for every two fluoride ions in the crystal lattice. Therefore, the empirical formula of this compound is CoF2.

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Most popular questions from this chapter

You and a friend each synthesize a compound with the formula \(\mathrm{XeCl}_{2} \mathrm{F}_{2}\). Your compound is a liquid and your friend's compound is a gas (at the same conditions of temperature and pressure). Explain how the two compounds with the same formulas can exist in different phases at the same conditions of pressure and temperature.

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