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Chapter 21: Problem 67

Iron is present in the earth's crust in many types of minerals. The iron oxide minerals are hematite \(\left(\mathrm{Fe}_{2} \mathrm{O}_{3}\right)\) and magnetite \(\left(\mathrm{Fe}_{3} \mathrm{O}_{4}\right) .\) What is the oxidation state of iron in each mineral? The iron ions in magnetite are a mixture of \(\mathrm{Fe}^{2+}\) and \(\mathrm{Fe}^{3+}\) ions. What is the ratio of \(\mathrm{Fe}^{3+}\) to \(\mathrm{Fe}^{2+}\) ions in magnetite? The formula for magnetite is often written as \(\mathrm{FeO} \cdot \mathrm{Fe}_{2} \mathrm{O}_{3}\). Does this make sense? Explain.

Short Answer

Expert verified
In hematite (Fe₂O₃), the oxidation state of iron is +3. In magnetite (Fe₃O₄), the ratio of Fe³⁺ to Fe²⁺ ions is 2 : 1, with two Fe³⁺ ions and one Fe²⁺ ion present. The formula FeO·Fe₂O₃ makes sense for magnetite, as it represents the combination of one Fe²⁺ ion from FeO and two Fe³⁺ ions from Fe₂O₃ to form magnetite's formula, Fe₃O₄.

Step by step solution

01

1. Determine oxidation state of Iron in hematite (Fe₂O₃)

For hematite (Fe₂O₃), we let x be the oxidation state of iron (Fe). Since oxygen has an oxidation state of -2, we can write these down in an equation: $$2x + 3(-2) = 0$$ Solve this equation to find the value of x, which is the oxidation state of iron in hematite.
02

2. Determine oxidation state of Iron in magnetite (Fe₃O₄)

For magnetite (Fe₃O₄), the oxidation state of iron is a mixture of Fe³⁺ and Fe²⁺ ions, which means it has more than one value. Let's call the amount of Fe³⁺ ions a and the amount of Fe²⁺ ions b. Since there are 3 iron atoms in total, we have a + b = 3. Next, since oxygen has an oxidation state of -2, we can write these down in another equation involving the oxidation states: $$3a + 2b + 4(-2) = 0$$ We now have two equations with two unknowns, a and b: $$a + b = 3$$ $$3a + 2b = 8$$
03

3. Solve the system of equations to find the ratio of Fe³⁺ to Fe²⁺ ions in magnetite

To solve these equations, we can first multiply the first equation by -2 and then add the two equations: $$-2(a + b) = -6$$ $$-2a - 2b = -6$$ $$3a + 2b - 2a - 2b = 8 - 6$$ $$a = 2$$ Now we can plug the value of a back into the first equation to find the value of b: $$2 + b = 3$$ $$b = 1$$ Since a = 2 and b = 1, we now know there are two Fe³⁺ ions and one Fe²⁺ ion in magnetite. So, the ratio of Fe³⁺ to Fe²⁺ ions in magnetite is 2 : 1.
04

4. Analyzing the formula FeO·Fe₂O₃ for Magnetite

We know magnetite can be written as Fe₃O₄, now let's analyze whether the formula FeO·Fe₂O₃ makes sense or not. If we split the two compounds within the formula, we have: - FeO, which contains 1 Fe²⁺ ion and 1 O²⁻ ion - Fe₂O₃, which contains 2 Fe³⁺ ions and 3 O²⁻ ions As seen, the FeO part of the formula corresponds to the Fe²⁺ ions, while the Fe₂O₃ part corresponds to the Fe³⁺ ions. Adding up the ions in these two compounds, we have: Fe₂O₃: 2 Fe³⁺ ions + 3 O²⁻ ions FeO: 1 Fe²⁺ ion + 1 O²⁻ ion Combining both gives us the formula Fe₃O₄, which is the formula for magnetite. Therefore, the given formula FeO·Fe₂O₃ does make sense for magnetite.

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Most popular questions from this chapter

Draw all the geometrical isomers of \(\mathrm{Cr}(\mathrm{en})\left(\mathrm{NH}_{3}\right)_{2} \mathrm{BrCl}^{+} .\) Which of these isomers also have an optical isomer? Draw the various isomers.

Draw all geometrical isomers of \(\mathrm{Pt}(\mathrm{CN})_{2} \mathrm{Br}_{2}\left(\mathrm{H}_{2} \mathrm{O}\right)_{2} .\) Which of these isomers has an optical isomer? Draw the various optical isomers.

Which of the following crystal field diagram(s) is(are) correct for the complex given? a. \(\mathrm{Zn}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\) (tetrahedral) b. \(\mathrm{Mn}(\mathrm{CN})_{6}{ }^{3-}\) (strong field) c. \(\mathrm{Ni}(\mathrm{CN})_{4}{ }^{2-}\) (square planar, diamagnetic)

Name the following complex ions. a. \(\mathrm{Ru}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}^{2+}\) b. \(\mathrm{Fe}(\mathrm{CN})_{6}{ }^{4-}\) c. \(\mathrm{Mn}\left(\mathrm{NH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}\right)_{3}{ }^{2+}\) d. \(\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{NO}_{2}{ }^{2+}\) e. \(\mathrm{Ni}(\mathrm{CN})_{4}{ }^{2-}\) f. \(\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}^{+}\) g. \(\mathrm{Fe}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}{ }^{3-}\) h. \(\mathrm{Co}(\mathrm{SCN})_{2}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4}^{+}\)

What is the electron configuration for the transition metal ion in each of the following compounds? a. \(\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]\) b. \(\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right] \mathrm{Cl}\) c. \(\left[\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Br}_{2}\) d. \(\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4}\left(\mathrm{NO}_{2}\right)_{2}\right] \mathrm{I}\)
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