Transition Elements · Practice

All questions with answers · Transition Elements

1. Element X is in the first transition series. Its ion X³⁺$\mathrm{X}{}^{3+}$ has a half-filled 3d sublevel. Which choice identifies X and gives one of its uses? (₂₄${}_{24}$Cr, ₂₅${}_{25}$Mn, ₂₆${}_{26}$Fe, ₂₇${}_{27}$Co)

   • C. Chromium; metal plating and leather tanning
   • A. Iron; a catalyst in the manufacture of ammonia by the Haber-Bosch method
   • B. Manganese; ferromanganese alloy for railway tracks
   • D. Cobalt; cobalt-60 for food preserving

   Answer: Iron; a catalyst in the manufacture of ammonia by the Haber-Bosch method

   A half-filled 3d sublevel is 3d⁵$3d^5$. X³⁺$\mathrm{X}{}^{3+}$ has lost three electrons, so the atom has 8 electrons beyond [Ar]: [Ar] 4s²\,3d⁶$4s^{2}3d^6$. That is iron (₂₆${}_{26}$Fe). Iron is used as a catalyst in the Haber-Bosch method. It is also used in the Fischer-Tropsch method.

2. Match each transition element in column (A) with one of its uses in column (B).

   Answer: Scandium → A light, very hard alloy with aluminum for MiG fighter jets · Titanium → Dental implants and artificial joints · Nickel → A catalyst in the hydrogenation of oils · Copper → Electric cables · Zinc → Galvanizing other metals to protect them from rusting

   Scandium with aluminum gives a light, very hard alloy for MiG fighter jets. Titanium is used in dental implants and artificial joints, because the body does not reject it. Finely divided nickel is a catalyst in the hydrogenation of oils. Copper is a good conductor, so it is used in electric cables. Zinc is used in galvanizing other metals. Common slip: swapping nickel and zinc, since both protect other metals.

3. Classify each element, using the configurations given, as a transition element or not a transition element.

   Answer: Transition element: ₂₁${}_{21}$Sc: atom [Ar] 4s²\,3d¹$4s^{2}3d^1$; Sc³⁺$\mathrm{Sc}{}^{3+}$: [Ar], ₂₉${}_{29}$Cu: atom [Ar] 4s¹\,3d¹⁰$4s^{1}3d^{10}$; Cu²⁺$\mathrm{Cu}{}^{2+}$ in CuSO4$\mathrm{CuSO}{}_4$: [Ar] 3d⁹$3d^9$, ₇₉${}_{79}$Au: atom [Xe] 4f¹⁴\,6s¹\,5d¹⁰$4f^{14}6s^{1}5d^{10}$; Au³⁺$\mathrm{Au}{}^{3+}$: [Xe] 4f¹⁴\,5d⁸$4f^{14}5d^8$ · Not a transition element: ₃₀${}_{30}$Zn: atom [Ar] 4s²\,3d¹⁰$4s^{2}3d^{10}$; Zn²⁺$\mathrm{Zn}{}^{2+}$: [Ar] 3d¹⁰$3d^{10}$, ₃₁${}_{31}$Ga: atom [Ar] 4s²\,3d¹⁰\,4p¹$4s^{2}3d^{10}4p^1$; Ga³⁺$\mathrm{Ga}{}^{3+}$: [Ar] 3d¹⁰$3d^{10}$

   The book's definition: the d or f orbitals contain electrons but are not complete, in the atom or in one of its oxidation states. Sc: atom 3d¹$3d^1$, incomplete. Cu: Cu²⁺$\mathrm{Cu}{}^{2+}$ is 3d⁹$3d^9$. Au: Au³⁺$\mathrm{Au}{}^{3+}$ is 5d⁸$5d^8$. Zn and Ga: the d sublevel is d¹⁰$d^{10}$ in the atom and in the ion. So they are not transition elements.

4. Consider the following statements about iron (₂₆${}_{26}$Fe: [Ar] 4s²\,3d⁶$4s^{2}3d^6$): (I) Iron does not show the oxidation state +8, which would match the loss of all its 4s and 3d electrons. (II) Its oxidation states above +3 are not important. (III) In FeS, formed from iron and sulphur, iron is +3. Which statements are correct?

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

   Answer: (I) and (II) only

   (I) is correct: iron differs from the elements before it; it does not reach +8. (II) is correct: the book says the states above +3 are not important. Iron's important states are +2 and +3. (III) is wrong: FeS is iron (II) sulphide. Sulphur is −2, so iron is +2.

5. Element	Atomic mass	Atomic radius	Density (g/cm³)	Melting point (°C)
   Mn$\mathrm{Mn}$	54.9	1.17	7.21	1247
   Fe$\mathrm{Fe}$	55.9	1.16	7.87	1538
   Co$\mathrm{Co}$	58.9	1.16	8.7	1490
   Ni$\mathrm{Ni}$	58.7	1.15	8.9	1492

   Use the table. Which statement is correct?

   • C. The element with the largest atomic radius has the highest density.
   • B. From Mn to Ni, the melting point increases with the atomic number.
   • A. From Mn to Ni, the density increases while the atomic radius stays nearly constant.
   • D. Iron has both the highest melting point and the highest density of the four.

   Answer: From Mn to Ni, the density increases while the atomic radius stays nearly constant.

   Density: 7.21 → 7.87 → 8.70 → 8.90, so it increases. Radius: 1.17 → 1.16 → 1.16 → 1.15, nearly constant. The melting point does not rise steadily: iron (1538 °C) is the highest; cobalt (1490 °C) and nickel (1492 °C) are lower. Mn has the largest radius but the lowest density. The mass generally rises while the radius barely changes, so the density rises.

6. Arrange the following ions in descending order according to their magnetic moment (largest first). (₂₂${}_{22}$Ti, ₂₄${}_{24}$Cr, ₂₅${}_{25}$Mn, ₂₉${}_{29}$Cu)

   Answer: Mn²⁺$\mathrm{Mn}{}^{2+}$ → Cr³⁺$\mathrm{Cr}{}^{3+}$ → Ti³⁺$\mathrm{Ti}{}^{3+}$ → Cu+$\mathrm{Cu}{}^{+}$

   Mn²⁺: [Ar] 3d⁵$3d^5$, 5 unpaired. Cr³⁺: [Ar] 3d³$3d^3$, 3. Ti³⁺: [Ar] 3d¹$3d^1$, 1. Cu⁺: [Ar] 3d¹⁰$3d^{10}$, 0, so it is diamagnetic. Descending order: Mn²⁺ > Cr³⁺ > Ti³⁺ > Cu⁺. Common slip: counting the atom's electrons instead of the ion's.

7. Give reasons: Finely divided iron is used as a catalyst in the preparation of ammonia by the Haber-Bosch method.

   Answer: Key points: (1) Iron is a transition element with 4s and 3d electrons. (2) These electrons form bonds between the atoms of the iron surface and the reacting molecules (N2$\mathrm{N}{}_2$ and H2$\mathrm{H}{}_2$). (3) This increases the concentration of the reacting molecules on the surface and weakens the bonds in them. (4) So the activation energy decreases and the speed of the reaction increases: N2 + 3H2 →[Fe] 2NH3$\mathrm{N}{}_2+3\mathrm{H}{}_2\stackrel{\mathrm{Fe}}{\to }2\mathrm{NH}{}_3$.

   The book links catalytic activity to the 4s and 3d electrons. On the iron surface, they form bonds with N2$\mathrm{N}{}_2$ and H2$\mathrm{H}{}_2$ molecules. More molecules gather on the surface, and their bonds weaken. The activation energy falls, so ammonia forms faster. Finely divided iron has a larger surface. Most likely wrong answer: 'iron reacts with nitrogen and is used up'. A catalyst is not used up.

8. Which choice compares the reduction of iron ore in the blast furnace and in the Midrex furnace correctly?

   • D. Both furnaces use coke; the Midrex furnace adds natural gas only as a fuel.
   • A. Blast furnace: CO from coke, higher than 700 °C. Midrex: a mixture of CO and H2$\mathrm{H}{}_2$ from natural gas.
   • B. Blast furnace: H2$\mathrm{H}{}_2$ from coke. Midrex: CO only, from methane.
   • C. Blast furnace: CO from coke, at 230–300 °C. Midrex: a mixture of CO and H2$\mathrm{H}{}_2$ from natural gas.

   Answer: Blast furnace: CO from coke, higher than 700 °C. Midrex: a mixture of CO and H2$\mathrm{H}{}_2$ from natural gas.

   Blast furnace: coke gives CO (C + O2 → CO2$\mathrm{C}+\mathrm{O}{}_2\to \mathrm{CO}{}_2$, then CO2 + C → 2CO$\mathrm{CO}{}_2+\mathrm{C}\to 2\mathrm{CO}$). CO reduces Fe2O3$\mathrm{Fe}{}_2\mathrm{O}{}_3$ to iron at higher than 700 °C. Midrex: natural gas gives water gas, a mixture of CO and H2$\mathrm{H}{}_2$. Both gases reduce the ore. At 230–300 °C, CO would give only Fe3O4$\mathrm{Fe}{}_3\mathrm{O}{}_4$.

9. Classify the following alloys according to their type.

   Answer: Intermetallic alloy: Ni3Al$\mathrm{Ni}{}_3\mathrm{Al}$, a compound whose formula valency cannot explain, Duralumin (aluminum-copper) · Interstitial alloy: Iron with a little boron, whose atoms are much smaller than iron atoms (no fixed formula), Steel · Substitutional alloy: An alloy of cobalt (radius 1.16) and nickel (radius 1.15), metals with the same crystal lattice and similar chemical properties, Ferrochrome in stainless steel

   Interstitial: steel, and iron with boron, because much smaller atoms fit in the spaces. Substitutional: ferrochrome, and cobalt-nickel, because the atoms have nearly the same radius, lattice and properties. Intermetallic: duralumin and Ni3Al$\mathrm{Ni}{}_3\mathrm{Al}$, because the elements combine chemically and the formula disobeys valency.

10. An iron nail is dipped in concentrated nitric acid, then washed. It is rubbed with sandpaper and put in dilute sulphuric acid. Which products form?

    • A. FeSO4$\mathrm{FeSO}{}_4$ and hydrogen gas
    • D. FeSO4$\mathrm{FeSO}{}_4$, Fe2(SO4)3$\mathrm{Fe}{}_2(\mathrm{SO}{}_4){}_3$ and sulphur dioxide
    • C. Fe2(SO4)3$\mathrm{Fe}{}_2(\mathrm{SO}{}_4){}_3$ and hydrogen gas
    • B. No products: the nail stays passive

    Answer: FeSO4$\mathrm{FeSO}{}_4$ and hydrogen gas

    Concentrated nitric acid made the nail passive: a thin oxide layer formed. Abrasion with sandpaper removed the layer. Fresh iron then reacts with dilute sulphuric acid: Fe + H2SO4 → FeSO4 + H2$\mathrm{Fe}+\mathrm{H}{}_2\mathrm{SO}{}_4\to \mathrm{FeSO}{}_4+\mathrm{H}{}_2$. The produced hydrogen keeps iron at +2.

11. 1. Fe → A ((1) water vapour, 500 °C)
    2. A → B ((2) CO, 400–700 °C)
    3. B → C ((3) hot air)
    4. C → D ((4) hot conc. H2SO4$\mathrm{H}{}_2\mathrm{SO}{}_4$)

    Study the diagram. Identify substances A, B, C and D, then write the balanced equations for steps (1) and (4).

    Answer: A = Fe3O4$\mathrm{Fe}{}_3\mathrm{O}{}_4$, B = FeO, C = Fe2O3$\mathrm{Fe}{}_2\mathrm{O}{}_3$, D = Fe2(SO4)3$\mathrm{Fe}{}_2(\mathrm{SO}{}_4){}_3$ (with water). (1) 3Fe + 4H2O →[500\,^C] Fe3O4 + 4H2$3\mathrm{Fe}+4\mathrm{H}{}_2\mathrm{O}\stackrel{500{}_{}^{}{}_{}^{\circ }\mathrm{C}}{\to }\mathrm{Fe}{}_3\mathrm{O}{}_4+4\mathrm{H}{}_2$. (4) Fe2O3 + 3H2SO4 →[conc.,\ Δ] Fe2(SO4)3 + 3H2O$\mathrm{Fe}{}_2\mathrm{O}{}_3+3\mathrm{H}{}_2\mathrm{SO}{}_4\stackrel{\text{conc.},\mathrm{\Delta }}{\to }\mathrm{Fe}{}_2(\mathrm{SO}{}_4){}_3+3\mathrm{H}{}_2\mathrm{O}$.

    (1) Red hot iron and water vapour give magnetic iron oxide and hydrogen. (2) CO at 400–700 °C reduces it to FeO. (3) FeO is easily oxidized in hot air to Fe2O3$\mathrm{Fe}{}_2\mathrm{O}{}_3$. (4) Fe2O3$\mathrm{Fe}{}_2\mathrm{O}{}_3$ and the hot concentrated acid give iron (III) sulphate and water. Most likely slip: two salts for D. Of the iron oxides, only Fe3O4$\mathrm{Fe}{}_3\mathrm{O}{}_4$ gives two salts. Fe2O3$\mathrm{Fe}{}_2\mathrm{O}{}_3$ holds iron only at +3, so D is iron (III) sulphate alone.

12. Three iron oxides X, Y and Z are tested. X, a black powder, gives only iron (II) sulphate and water with dilute H2SO4$\mathrm{H}{}_2\mathrm{SO}{}_4$. Y is strongly attracted by a magnet and gives two iron salts with hot concentrated H2SO4$\mathrm{H}{}_2\mathrm{SO}{}_4$. Z, a red solid, gives only iron (III) sulphate and water with hot concentrated H2SO4$\mathrm{H}{}_2\mathrm{SO}{}_4$. Identify X, Y and Z.

    • D. X = Fe3O4$\mathrm{Fe}{}_3\mathrm{O}{}_4$, Y = FeO, Z = Fe2O3$\mathrm{Fe}{}_2\mathrm{O}{}_3$
    • B. X = Fe2O3$\mathrm{Fe}{}_2\mathrm{O}{}_3$, Y = Fe3O4$\mathrm{Fe}{}_3\mathrm{O}{}_4$, Z = FeO
    • A. X = FeO, Y = Fe3O4$\mathrm{Fe}{}_3\mathrm{O}{}_4$, Z = Fe2O3$\mathrm{Fe}{}_2\mathrm{O}{}_3$
    • C. X = FeO, Y = Fe2O3$\mathrm{Fe}{}_2\mathrm{O}{}_3$, Z = Fe3O4$\mathrm{Fe}{}_3\mathrm{O}{}_4$

    Answer: X = FeO, Y = Fe3O4$\mathrm{Fe}{}_3\mathrm{O}{}_4$, Z = Fe2O3$\mathrm{Fe}{}_2\mathrm{O}{}_3$

    X gives only an iron (II) salt, so iron is +2: X is FeO, the black powder. Y is a strong magnet and gives two salts: the mixed oxide Fe3O4$\mathrm{Fe}{}_3\mathrm{O}{}_4$. Z is red and gives only an iron (III) salt: Fe2O3$\mathrm{Fe}{}_2\mathrm{O}{}_3$. Use the oxidation state of iron in each salt to decide.