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Chapter 10: Problem 74

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 ratio of manganese ions to fluoride ions in the compound is 1:6. Since each fluoride ion has a charge of -1, the charge of the manganese ions must be +6 (Mn⁶⁺) to balance the negative charge from the fluoride ions.

Step by step solution

01

Determine the number of manganese and fluoride ions in a unit cell

We are given that the manganese ions form a simple cubic array, which means there is one manganese ion at each corner of the unit cell. There are a total of 8 corners in a cubic unit cell, so there are 8 manganese ions/cell. However, since each corner contributes 1/8 of an ion to the cell (because they are shared by 8 adjacent cells), there is effectively 8 * (1/8) = 1 manganese ion/cell. For fluoride ions, they are located at the center of each edge of the cubic unit cell. Each unit cell has 12 edges, and since each edge contributes 1/2 of an ion to the cell (because they are shared by 2 adjacent cells), there are effectively 12 * (1/2) = 6 fluoride ions/cell.
02

Find the ratio of manganese ions to fluoride ions

To determine the ratio of manganese ions to fluoride ions in the compound, we will use the number of ions per unit cell. Mn : F 1 : 6 The ratio of manganese ions to fluoride ions in the compound is 1 : 6.
03

Determine the charge of the manganese ions

Fluoride ions have a charge of -1 (F⁻). Since there are six fluoride ions in the unit cell, the total negative charge contributed by fluoride ions is -6. To balance the charges in the compound, the positive charge from the manganese ions must equal the negative charge from the fluoride ions. Let x be the charge of the manganese ion (Mn ^x+). Since there is 1 manganese ion in the unit cell, the equation for the balancing of charges is: 1 * x = 6 * (-1) Solve for x: x = -6 Therefore the charge of the manganese ions in this compound is +6 (Mn⁶⁺).

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Most popular questions from this chapter

Explain how a p-n junction makes an excellent rectifier.

The Group \(3 \mathrm{~A} /\) Group 5 A semiconductors are composed of equal amounts of atoms from Group \(3 \mathrm{~A}\) and Group \(5 \mathrm{~A}-\) for example, InP and GaAs. These types of semiconductors are used in lightemitting diodes and solid-state lasers. What would you add to make a p-type semiconductor from pure GaAs? How would you dope pure GaAs to make an n-type semiconductor?

Consider the following enthalpy changes: $$ \begin{aligned} \mathrm{F}^{-}+\mathrm{HF} \longrightarrow \mathrm{FHF}^{-} \quad \Delta H=-155 \mathrm{~kJ} / \mathrm{mol} \\ \left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}=\mathrm{O}+\mathrm{HF} \longrightarrow\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}=\mathrm{O}-\mathrm{-HF} \\ \Delta H=-46 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{HOH}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}--\mathrm{HOH}(\text { in ice }) \\ \Delta H=-21 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$ How do the strengths of hydrogen bonds vary with the electronegativity of the element to which hydrogen is bonded? Where in the preceding series would you expect hydrogen bonds of the following type to fall?

Atoms are assumed to touch in closest packed structures, yet every closest packed unit cell contains a significant amount of empty space. Why?

Rationalize the difference in boiling points for each of the following pairs of substances: a. \(n\) -pentane \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\) \(36.2^{\circ} \mathrm{C}\) b. \(\mathrm{HF} \quad 20^{\circ} \mathrm{C}\) \(\mathrm{HCl} \quad-85^{\circ} \mathrm{C}\) c. \(\mathrm{HCl} \quad-85^{\circ} \mathrm{C}\) \(\mathrm{LiCl} \quad 1360^{\circ} \mathrm{C}\) d. \(n\) -pentane \(\quad \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\) \(36.2^{\circ} \mathrm{C}\) \(n\) -hexane \(\quad \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3} \quad 69^{\circ} \mathrm{C}\)
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