Transition Elements · Practice

All questions with answers · Transition Elements

1. Write the balanced equation for the preparation of iron (II) oxide from iron (II) oxalate, with the condition over the arrow.

   Answer: FeC2O4(s) →[Δ][in absence of air] FeO(s) + CO(g) + CO2(g)$\mathrm{FeC}{}_2\mathrm{O}{}_4(\mathrm{s})\underset{\text{in absence of air}}{\overset{\mathrm{\Delta }}{\to }}\mathrm{FeO}(\mathrm{s})+\mathrm{CO}(\mathrm{g})+\mathrm{CO}{}_2(\mathrm{g})$

   Heating iron (II) oxalate breaks it down into FeO, carbon monoxide and carbon dioxide. The book draws the oxalate as two COO groups joined to Fe. Air must be absent. Otherwise hot air oxidizes the FeO to Fe2O3$\mathrm{Fe}{}_2\mathrm{O}{}_3$. Most likely slip: leaving out CO, or the condition 'in absence of air'.

2. Write the balanced equation for heating iron (II) sulphate strongly.

   Answer: 2FeSO4(s) →[Δ] Fe2O3(s) + SO2(g) + SO3(g)$2\mathrm{FeSO}{}_4(\mathrm{s})\stackrel{\mathrm{\Delta }}{\to }\mathrm{Fe}{}_2\mathrm{O}{}_3(\mathrm{s})+\mathrm{SO}{}_2(\mathrm{g})+\mathrm{SO}{}_3(\mathrm{g})$

   Heating iron (II) sulphate gives iron (III) oxide, sulphur dioxide and sulphur trioxide. Iron is oxidized from +2 to +3. One S is reduced from +6 to +4 in SO2$\mathrm{SO}{}_2$; the other stays +6 in SO3$\mathrm{SO}{}_3$. Most likely wrong answer: FeO and SO3$\mathrm{SO}{}_3$ only.

3. Magnetic iron oxide is heated with concentrated sulphuric acid. The products are ...

   • B. Fe2(SO4)3$\mathrm{Fe}{}_2(\mathrm{SO}{}_4){}_3$ and water only
   • A. FeSO4$\mathrm{FeSO}{}_4$, Fe2(SO4)3$\mathrm{Fe}{}_2(\mathrm{SO}{}_4){}_3$ and water
   • C. FeSO4$\mathrm{FeSO}{}_4$ and water only
   • D. FeSO4$\mathrm{FeSO}{}_4$, Fe2(SO4)3$\mathrm{Fe}{}_2(\mathrm{SO}{}_4){}_3$, SO2$\mathrm{SO}{}_2$ and water

   Answer: FeSO4$\mathrm{FeSO}{}_4$, Fe2(SO4)3$\mathrm{Fe}{}_2(\mathrm{SO}{}_4){}_3$ and water

   Fe3O4$\mathrm{Fe}{}_3\mathrm{O}{}_4$ is a mixed iron (II) and iron (III) oxide. With hot concentrated acids, it gives iron (II) and iron (III) salts and water: Fe3O4 + 4H2SO4 → FeSO4 + Fe2(SO4)3 + 4H2O$\mathrm{Fe}{}_3\mathrm{O}{}_4+4\mathrm{H}{}_2\mathrm{SO}{}_4\to \mathrm{FeSO}{}_4+\mathrm{Fe}{}_2(\mathrm{SO}{}_4){}_3+4\mathrm{H}{}_2\mathrm{O}$. No gas forms. Iron metal, not its oxide, gives SO2$\mathrm{SO}{}_2$ with the hot concentrated acid.

4. Write the balanced equations for the conversions Fe → Fe3O4$\mathrm{Fe}{}_3\mathrm{O}{}_4$ → FeO. Use water vapour in the first step and carbon monoxide in the second. Write the condition over each arrow.

   Answer: (1) 3Fe + 4H2O(v) →[500\,^C] Fe3O4 + 4H2$3\mathrm{Fe}+4\mathrm{H}{}_2\mathrm{O}(\mathrm{v})\stackrel{500{}_{}^{}{}_{}^{\circ }\mathrm{C}}{\to }\mathrm{Fe}{}_3\mathrm{O}{}_4+4\mathrm{H}{}_2$. (2) Fe3O4 + CO →[400-700\,^C] 3FeO + CO2$\mathrm{Fe}{}_3\mathrm{O}{}_4+\mathrm{CO}\stackrel{400-700{}_{}^{}{}_{}^{\circ }\mathrm{C}}{\to }3\mathrm{FeO}+\mathrm{CO}{}_2$.

   Step 1: red hot iron (500 °C) reacts with water vapour to give magnetic iron oxide and hydrogen. Step 2: higher oxides are reduced by CO (or H2$\mathrm{H}{}_2$) at 400–700 °C to FeO. The 3 Fe in Fe3O4$\mathrm{Fe}{}_3\mathrm{O}{}_4$ give 3 FeO; the fourth O joins CO as CO2$\mathrm{CO}{}_2$. Most likely slip: iron as the step 2 product. That needs higher than 700 °C.

5. Give reasons: Iron (II) oxalate is heated in the absence of air to prepare iron (II) oxide.

   Answer: Key points: (1) FeO is easily oxidized in hot air. (2) In air it would change to iron (III) oxide: 4FeO + O2 →[Δ] 2Fe2O3$4\mathrm{FeO}+\mathrm{O}{}_2\stackrel{\mathrm{\Delta }}{\to }2\mathrm{Fe}{}_2\mathrm{O}{}_3$. (3) So air is kept out to obtain FeO.

   Heating iron (II) oxalate gives FeO, CO and CO2$\mathrm{CO}{}_2$. FeO is easily oxidized in hot air to Fe2O3$\mathrm{Fe}{}_2\mathrm{O}{}_3$. Heating in absence of air keeps the product as FeO. Key words: easily oxidized in hot air. Most likely wrong answer: 'the oxalate does not decompose in air'. It does; the product is the problem.

6. An iron oxide is found in nature as hematite and is used in red pigments in paints. Which statement about this oxide is correct?

   • D. It is FeO; it is easily oxidized in hot air.
   • A. It is Fe2O3$\mathrm{Fe}{}_2\mathrm{O}{}_3$; it is insoluble in water and gives an iron (III) salt and water with hot concentrated acids.
   • C. It is Fe2O3$\mathrm{Fe}{}_2\mathrm{O}{}_3$; it gives an iron (II) salt and water with hot concentrated acids.
   • B. It is Fe3O4$\mathrm{Fe}{}_3\mathrm{O}{}_4$; it is a strong magnet and gives two salts with hot concentrated acids.

   Answer: It is Fe2O3$\mathrm{Fe}{}_2\mathrm{O}{}_3$; it is insoluble in water and gives an iron (III) salt and water with hot concentrated acids.

   Hematite is iron (III) oxide, Fe2O3$\mathrm{Fe}{}_2\mathrm{O}{}_3$. It is used in red pigments in paints. It is insoluble in water. With hot concentrated acids it gives iron (III) salts and water: Fe2O3 + 3H2SO4 → Fe2(SO4)3 + 3H2O$\mathrm{Fe}{}_2\mathrm{O}{}_3+3\mathrm{H}{}_2\mathrm{SO}{}_4\to \mathrm{Fe}{}_2(\mathrm{SO}{}_4){}_3+3\mathrm{H}{}_2\mathrm{O}$. Iron stays +3.

7. Give reasons: Magnetic iron oxide, Fe3O4$\mathrm{Fe}{}_3\mathrm{O}{}_4$, is considered a mixed oxide.

   Answer: Key points: (1) It can be considered as mixed iron (II) and iron (III) oxides (FeO + Fe2O3$\mathrm{Fe}{}_2\mathrm{O}{}_3$). (2) It reacts with hot concentrated acids to give iron (II) and iron (III) salts: Fe3O4 + 4H2SO4 → FeSO4 + Fe2(SO4)3 + 4H2O$\mathrm{Fe}{}_3\mathrm{O}{}_4+4\mathrm{H}{}_2\mathrm{SO}{}_4\to \mathrm{FeSO}{}_4+\mathrm{Fe}{}_2(\mathrm{SO}{}_4){}_3+4\mathrm{H}{}_2\mathrm{O}$. (3) Two salts prove two oxidation states of iron.

   An oxide with one oxidation state gives one kind of salt. Fe3O4$\mathrm{Fe}{}_3\mathrm{O}{}_4$ gives two: an iron (II) salt and an iron (III) salt. So it contains iron (II) and iron (III) together. Key words: gives iron (II) and iron (III) salts, proving it is a mixed oxide. Most likely wrong answer: 'because it is magnetic'. Magnetism does not show two oxidation states.

8. Write the balanced equations for the conversions FeO → Fe2O3$\mathrm{Fe}{}_2\mathrm{O}{}_3$ → Fe, where the second step takes place in the Midrex furnace. Write the condition over each arrow.

   Answer: (1) 4FeO + O2 →[Δ] 2Fe2O3$4\mathrm{FeO}+\mathrm{O}{}_2\stackrel{\mathrm{\Delta }}{\to }2\mathrm{Fe}{}_2\mathrm{O}{}_3$ (heating in hot air). (2) 2Fe2O3 + 3CO + 3H2 →[Δ] 4Fe + 3CO2 + 3H2O$2\mathrm{Fe}{}_2\mathrm{O}{}_3+3\mathrm{CO}+3\mathrm{H}{}_2\stackrel{\mathrm{\Delta }}{\to }4\mathrm{Fe}+3\mathrm{CO}{}_2+3\mathrm{H}{}_2\mathrm{O}$ (water gas from natural gas).

   Step 1: FeO is easily oxidized in hot air to Fe2O3$\mathrm{Fe}{}_2\mathrm{O}{}_3$. Step 2: in the Midrex furnace, water gas (CO + H2$\mathrm{H}{}_2$) from natural gas reduces Fe2O3$\mathrm{Fe}{}_2\mathrm{O}{}_3$ to iron. CO becomes CO2$\mathrm{CO}{}_2$, and H2$\mathrm{H}{}_2$ becomes H2O$\mathrm{H}{}_2\mathrm{O}$. Most likely slip: using CO alone, as in the blast furnace. The Midrex reducing agent is a mixture of CO and H2$\mathrm{H}{}_2$.

9. Which statement about iron (II) oxide, FeO, is correct?

   • B. It is a red powder used in red pigments in paints.
   • C. It is a black powder that gives an iron (III) salt and water with dilute acids.
   • D. It is a black powder that gives an iron (II) salt and hydrogen gas with dilute acids.
   • A. It is a black powder, insoluble in water, and gives an iron (II) salt and water with dilute acids.

   Answer: It is a black powder, insoluble in water, and gives an iron (II) salt and water with dilute acids.

   FeO is a black powder, insoluble in water. It is easily oxidized in hot air. With dilute acids it gives an iron (II) salt and water: FeO + H2SO4 → FeSO4 + H2O$\mathrm{FeO}+\mathrm{H}{}_2\mathrm{SO}{}_4\to \mathrm{FeSO}{}_4+\mathrm{H}{}_2\mathrm{O}$. Iron stays +2, and no gas forms.

10. FeO is added to dilute sulphuric acid, and Fe2O3$\mathrm{Fe}{}_2\mathrm{O}{}_3$ is heated with concentrated sulphuric acid. Which statement compares the two reactions correctly?

    • A. FeO gives iron (II) sulphate and Fe2O3$\mathrm{Fe}{}_2\mathrm{O}{}_3$ gives iron (III) sulphate; both also give water and no gas.
    • C. Both give a salt and hydrogen gas.
    • D. FeO gives iron (III) sulphate because it is easily oxidized; Fe2O3$\mathrm{Fe}{}_2\mathrm{O}{}_3$ gives iron (III) sulphate too.
    • B. Both give iron (III) sulphate, because acids oxidize iron.

    Answer: FeO gives iron (II) sulphate and Fe2O3$\mathrm{Fe}{}_2\mathrm{O}{}_3$ gives iron (III) sulphate; both also give water and no gas.

    An oxide and an acid give a salt and water, with no gas. Iron keeps its oxidation state. FeO (+2) gives FeSO4$\mathrm{FeSO}{}_4$. Fe2O3$\mathrm{Fe}{}_2\mathrm{O}{}_3$ (+3) gives Fe2(SO4)3$\mathrm{Fe}{}_2(\mathrm{SO}{}_4){}_3$ with the hot concentrated acid. FeO is easily oxidized by hot air, not by a dilute acid.

11. Carbon monoxide is passed over heated iron (III) oxide at 250 °C. Write the balanced equation of the reaction, with the condition over the arrow.

    Answer: 3Fe2O3(s) + CO(g) →[230-300\,^C] 2Fe3O4(s) + CO2(g)$3\mathrm{Fe}{}_2\mathrm{O}{}_3(\mathrm{s})+\mathrm{CO}(\mathrm{g})\stackrel{230-300{}_{}^{}{}_{}^{\circ }\mathrm{C}}{\to }2\mathrm{Fe}{}_3\mathrm{O}{}_4(\mathrm{s})+\mathrm{CO}{}_2(\mathrm{g})$

    At 230–300 °C, CO removes only a little oxygen: Fe2O3$\mathrm{Fe}{}_2\mathrm{O}{}_3$ becomes Fe3O4$\mathrm{Fe}{}_3\mathrm{O}{}_4$. 250 °C lies in this range, so write 230–300 °C (or 250 °C). 3 Fe2O3$\mathrm{Fe}{}_2\mathrm{O}{}_3$ hold 6 Fe and 9 O. 2 Fe3O4$\mathrm{Fe}{}_3\mathrm{O}{}_4$ hold 6 Fe and 8 O. The ninth O joins CO as CO2$\mathrm{CO}{}_2$. Most likely wrong answer: iron as the product. That needs higher than 700 °C.

12. Carbon monoxide is passed over heated Fe2O3$\mathrm{Fe}{}_2\mathrm{O}{}_3$, and the solid left is FeO. In which temperature range was the reaction carried out?

    • B. 230–300 °C
    • D. Any temperature, because CO always gives the same product
    • C. Higher than 700 °C
    • A. 400–700 °C

    Answer: 400–700 °C

    With CO, the temperature decides how much oxygen is removed. 230–300 °C: Fe2O3$\mathrm{Fe}{}_2\mathrm{O}{}_3$ → Fe3O4$\mathrm{Fe}{}_3\mathrm{O}{}_4$. 400–700 °C: higher oxides → FeO. Higher than 700 °C: Fe2O3$\mathrm{Fe}{}_2\mathrm{O}{}_3$ → Fe. So FeO points to 400–700 °C.

13. Iron (III) oxide is reduced by carbon monoxide at 400–700 °C to iron (II) oxide. Write the balanced equation, with the condition over the arrow.

    Answer: Fe2O3(s) + CO(g) →[400-700\,^C] 2FeO(s) + CO2(g)$\mathrm{Fe}{}_2\mathrm{O}{}_3(\mathrm{s})+\mathrm{CO}(\mathrm{g})\stackrel{400-700{}_{}^{}{}_{}^{\circ }\mathrm{C}}{\to }2\mathrm{FeO}(\mathrm{s})+\mathrm{CO}{}_2(\mathrm{g})$

    The book prepares FeO by reducing higher oxides with hydrogen or carbon monoxide at 400–700 °C. Fe2O3$\mathrm{Fe}{}_2\mathrm{O}{}_3$ holds 2 Fe, so 2 FeO form. 2 FeO keep 2 O; the third O joins CO to give CO2$\mathrm{CO}{}_2$. Most likely wrong answer: iron as the product. That needs CO at higher than 700 °C.

14. FeO is heated in air. The solid product is then heated with concentrated sulphuric acid. What is the iron compound in the final solution, and what is iron's oxidation state in it?

    • C. FeSO4$\mathrm{FeSO}{}_4$ and Fe2(SO4)3$\mathrm{Fe}{}_2(\mathrm{SO}{}_4){}_3$; +2 and +3
    • A. Fe2(SO4)3$\mathrm{Fe}{}_2(\mathrm{SO}{}_4){}_3$; +3
    • D. FeSO4$\mathrm{FeSO}{}_4$; +2, because the acid reduces iron (III)
    • B. FeSO4$\mathrm{FeSO}{}_4$; +2

    Answer: Fe2(SO4)3$\mathrm{Fe}{}_2(\mathrm{SO}{}_4){}_3$; +3

    Step 1: FeO is easily oxidized in hot air: 4FeO + O2 →[Δ] 2Fe2O3$4\mathrm{FeO}+\mathrm{O}{}_2\stackrel{\mathrm{\Delta }}{\to }2\mathrm{Fe}{}_2\mathrm{O}{}_3$. Step 2: Fe2O3$\mathrm{Fe}{}_2\mathrm{O}{}_3$ with hot concentrated H2SO4$\mathrm{H}{}_2\mathrm{SO}{}_4$ gives iron (III) sulphate and water: Fe2O3 + 3H2SO4 → Fe2(SO4)3 + 3H2O$\mathrm{Fe}{}_2\mathrm{O}{}_3+3\mathrm{H}{}_2\mathrm{SO}{}_4\to \mathrm{Fe}{}_2(\mathrm{SO}{}_4){}_3+3\mathrm{H}{}_2\mathrm{O}$. Iron is +3 in the final salt.

15. Consider the following statements about magnetic iron oxide, Fe3O4$\mathrm{Fe}{}_3\mathrm{O}{}_4$: (I) It is a strong magnet. (II) When heated in air, it is oxidized to iron (III) oxide. (III) It is known naturally as hematite. Which statements are correct?

    • D. (I), (II) and (III)
    • C. (II) and (III) only
    • A. (I) and (II) only
    • B. (I) and (III) only

    Answer: (I) and (II) only

    (I) is correct: Fe3O4$\mathrm{Fe}{}_3\mathrm{O}{}_4$ is a strong magnet. (II) is correct: 2Fe3O4 + 1/2O2 →[Δ] 3Fe2O3$2\mathrm{Fe}{}_3\mathrm{O}{}_4+\frac{1}{2}\mathrm{O}{}_2\stackrel{\mathrm{\Delta }}{\to }3\mathrm{Fe}{}_2\mathrm{O}{}_3$; its iron (II) part is oxidized to iron (III). (III) is wrong: Fe3O4$\mathrm{Fe}{}_3\mathrm{O}{}_4$ is known naturally as magnetite. Hematite is Fe2O3$\mathrm{Fe}{}_2\mathrm{O}{}_3$.

16. 1. Fe → X ((1) + Cl2$\mathrm{Cl}{}_2$, heat)
    2. X → Y ((2) + NH4OH$\mathrm{NH}{}_4\mathrm{OH}$(aq))
    3. Y → Fe2O3$\mathrm{Fe}{}_2\mathrm{O}{}_3$ ((3) above 200 °C)
    4. Fe2O3$\mathrm{Fe}{}_2\mathrm{O}{}_3$ → Z ((4) + H2$\mathrm{H}{}_2$, 400–700 °C)

    Study the diagram. Identify substances X, Y and Z.

    • D. X = FeCl3$\mathrm{FeCl}{}_3$, Y = Fe(OH)3$\mathrm{Fe}(\mathrm{OH}){}_3$, Z = Fe3O4$\mathrm{Fe}{}_3\mathrm{O}{}_4$
    • C. X = FeCl3$\mathrm{FeCl}{}_3$, Y = Fe(OH)3$\mathrm{Fe}(\mathrm{OH}){}_3$, Z = Fe
    • B. X = FeCl2$\mathrm{FeCl}{}_2$, Y = Fe(OH)2$\mathrm{Fe}(\mathrm{OH}){}_2$, Z = FeO
    • A. X = FeCl3$\mathrm{FeCl}{}_3$, Y = Fe(OH)3$\mathrm{Fe}(\mathrm{OH}){}_3$, Z = FeO

    Answer: X = FeCl3$\mathrm{FeCl}{}_3$, Y = Fe(OH)3$\mathrm{Fe}(\mathrm{OH}){}_3$, Z = FeO

    (1) 2Fe + 3Cl2 → 2FeCl3$2\mathrm{Fe}+3\mathrm{Cl}{}_2\to 2\mathrm{FeCl}{}_3$: X is iron (III) chloride. (2) An alkali precipitates reddish brown iron (III) hydroxide: Y is Fe(OH)3$\mathrm{Fe}(\mathrm{OH}){}_3$. (3) Above 200 °C, 2Fe(OH)3 → Fe2O3 + 3H2O$2\mathrm{Fe}(\mathrm{OH}){}_3\to \mathrm{Fe}{}_2\mathrm{O}{}_3+3\mathrm{H}{}_2\mathrm{O}$. (4) Fe2O3 + H2 →[400-700\,^C] 2FeO + H2O$\mathrm{Fe}{}_2\mathrm{O}{}_3+\mathrm{H}{}_2\stackrel{400-700{}_{}^{}{}_{}^{\circ }\mathrm{C}}{\to }2\mathrm{FeO}+\mathrm{H}{}_2\mathrm{O}$: Z is FeO.

17. Compare the three iron oxides: classify each description under the oxide it describes.

    Answer: FeO: Prepared by heating iron (II) oxalate in the absence of air, Gives only an iron (II) salt and water with dilute acids · Fe2O3$\mathrm{Fe}{}_2\mathrm{O}{}_3$: Used in red pigments in paints, Prepared by heating iron (II) sulphate · Fe3O4$\mathrm{Fe}{}_3\mathrm{O}{}_4$: A strong magnet, known naturally as magnetite, Gives iron (II) and iron (III) salts with hot concentrated acids, Prepared by reducing Fe2O3$\mathrm{Fe}{}_2\mathrm{O}{}_3$ with CO at 230–300 °C

    FeO: prepared from iron (II) oxalate in absence of air. With dilute acids, it gives only an iron (II) salt and water. Fe2O3$\mathrm{Fe}{}_2\mathrm{O}{}_3$: red pigments in paints; prepared by heating iron (II) sulphate. Fe3O4$\mathrm{Fe}{}_3\mathrm{O}{}_4$: a strong magnet (magnetite); gives two salts with hot concentrated acids; made from Fe2O3$\mathrm{Fe}{}_2\mathrm{O}{}_3$ and CO at 230–300 °C.

18. 1. Fe3O4$\mathrm{Fe}{}_3\mathrm{O}{}_4$ → Fe2O3$\mathrm{Fe}{}_2\mathrm{O}{}_3$ (A)
    2. Fe2O3$\mathrm{Fe}{}_2\mathrm{O}{}_3$ → Fe (B)

    Study the diagram. Which choice gives conditions A and B?

    • B. A: H2$\mathrm{H}{}_2$ at 400–700 °C; B: CO, higher than 700 °C
    • C. A: heating in air; B: CO at 230–300 °C
    • A. A: heating in air; B: CO, higher than 700 °C
    • D. A: heating in absence of air; B: CO, higher than 700 °C

    Answer: A: heating in air; B: CO, higher than 700 °C

    Step A adds oxygen, so it is an oxidation: 2Fe3O4 + 1/2O2 →[Δ] 3Fe2O3$2\mathrm{Fe}{}_3\mathrm{O}{}_4+\frac{1}{2}\mathrm{O}{}_2\stackrel{\mathrm{\Delta }}{\to }3\mathrm{Fe}{}_2\mathrm{O}{}_3$, by heating in air. Step B removes all the oxygen: Fe2O3 + 3CO →[higher than 700\,^C] 2Fe + 3CO2$\mathrm{Fe}{}_2\mathrm{O}{}_3+3\mathrm{CO}\stackrel{\text{higher than }700{}_{}^{}{}_{}^{\circ }\mathrm{C}}{\to }2\mathrm{Fe}+3\mathrm{CO}{}_2$. Lower temperatures with CO give Fe3O4$\mathrm{Fe}{}_3\mathrm{O}{}_4$ or FeO, not iron.

19. What happens when iron reacts with dilute sulphuric acid, and the iron salt formed is then dried and heated strongly? Write the equations.

    Answer: Key points: (1) Fe + H2SO4 →[(dil)] FeSO4 + H2$\mathrm{Fe}+\mathrm{H}{}_2\mathrm{SO}{}_4\stackrel{(\mathrm{dil})}{\to }\mathrm{FeSO}{}_4+\mathrm{H}{}_2$: iron (II) sulphate and hydrogen gas. (2) 2FeSO4 →[Δ] Fe2O3 + SO2 + SO3$2\mathrm{FeSO}{}_4\stackrel{\mathrm{\Delta }}{\to }\mathrm{Fe}{}_2\mathrm{O}{}_3+\mathrm{SO}{}_2+\mathrm{SO}{}_3$: iron (III) oxide, sulphur dioxide and sulphur trioxide.

    Iron displaces hydrogen from the dilute acid and gives iron (II) sulphate. The produced hydrogen keeps iron at +2. Heating iron (II) sulphate strongly gives red iron (III) oxide, SO2$\mathrm{SO}{}_2$ and SO3$\mathrm{SO}{}_3$. Most likely slip: writing Fe2(SO4)3$\mathrm{Fe}{}_2(\mathrm{SO}{}_4){}_3$ in step 1.

20. Arrange the following oxides in ascending order according to the magnetic moment of their metal ions (smallest first). (₂₆${}_{26}$Fe, ₂₉${}_{29}$Cu, ₃₀${}_{30}$Zn)

    Answer: ZnO → CuO → FeO → Fe2O3$\mathrm{Fe}{}_2\mathrm{O}{}_3$

    Count the unpaired electrons in each metal ion. ZnO: Zn²⁺$\mathrm{Zn}{}^{2+}$ is [Ar] 3d¹⁰$3d^{10}$, 0. CuO: Cu²⁺$\mathrm{Cu}{}^{2+}$ is [Ar] 3d⁹$3d^9$, 1. FeO: Fe²⁺$\mathrm{Fe}{}^{2+}$ is [Ar] 3d⁶$3d^6$, 4. Fe2O3$\mathrm{Fe}{}_2\mathrm{O}{}_3$: Fe³⁺$\mathrm{Fe}{}^{3+}$ is [Ar] 3d⁵$3d^5$, 5. Order: ZnO < CuO < FeO < Fe2O3$\mathrm{Fe}{}_2\mathrm{O}{}_3$. Common slip: putting CuO last because it has the most 3d electrons.

21. Write the equations for roasting siderite in air until iron (III) oxide is obtained, and explain why the final product is not FeO.

    Answer: Key points: (1) FeCO3 →[Δ] FeO + CO2$\mathrm{FeCO}{}_3\stackrel{\mathrm{\Delta }}{\to }\mathrm{FeO}+\mathrm{CO}{}_2$. (2) 2FeO + 1/2O2 →[Δ] Fe2O3$2\mathrm{FeO}+\frac{1}{2}\mathrm{O}{}_2\stackrel{\mathrm{\Delta }}{\to }\mathrm{Fe}{}_2\mathrm{O}{}_3$. (3) Roasting is strong heating in air, and FeO is easily oxidized in hot air, so it changes to Fe2O3$\mathrm{Fe}{}_2\mathrm{O}{}_3$.

    Siderite, iron (II) carbonate, decomposes on heating into FeO and CO2$\mathrm{CO}{}_2$. Roasting is strong heating in air. FeO is easily oxidized in hot air, so it becomes Fe2O3$\mathrm{Fe}{}_2\mathrm{O}{}_3$. Most likely slip: stopping at FeO and forgetting the air.