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Chapter 10: Problem 76

The structure of manganese fluoride can be described as a simple cubic array of manganese ions with fluoride ions at the center of each edge of the cubic unit cell. What is the charge of the manganese ions in this compound?

Short Answer

Expert verified
The given structure has 1 Mn ion and 3 F ions within the unit cell. Fluoride ions carry a -1 charge each. Using charge balance, let x represent the charge of the Mn ion: \( 1 * x + 3 * (-1) = 0 \). Solving for x, the charge of the manganese ions is +3. Thus, the compound is MnF₃ with manganese ions having a +3 charge.

Step by step solution

01

Understand the arrangement of ions in the unit cell

In the given structure, Mn ions are arranged in a simple cubic array, and F ions are at the center of each edge of the cubic unit cell. In a simple cubic unit cell, there is one ion at each corner, which makes a total of 8 ions. There are 12 edges in a simple cubic unit cell, and since F ions are at the center of each edge, there are 12 F ions.
02

Count Mn and F ions in the unit cell

In the simple cubic arrangement of Mn ions, we have 1 Mn ion per corner. Since there are 8 corners, there are 8 * 1 = 8 Mn ions in the unit cell. Each Mn ion, however, is shared by 8 adjacent cubes, so it contributes 1/8 of an Mn ion per unit cell: \( Mn_{contributed} = 8 * \frac{1}{8} = 1 \) For the F ions, there is 1 F ion at the center of each of the 12 edges. Each F ion is shared by 4 adjacent cubes, so they contribute 1/4 of an F ion per unit cell: \( F_{contributed} = 12 * \frac{1}{4} = 3 \) So, in the unit cell, we have a total of 1 Mn ion and 3 F ions.
03

Use charge balance to determine the charge of Mn ions

In order to determine the charge of Mn ions, we need to consider charge balance in the compound. The overall charge of the compound should be zero. Fluoride ions carry a charge of -1 each. Let x be the charge of the Mn ion. Then, considering the charge balance: \( (1 Mn) * x + (3 F) * (-1) = 0 \) Solving for x, we get: \( x = +3 \) Therefore, the charge of the manganese ions in this compound is +3. The compound formula would be MnF₃.

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