Transition Elements · Practice

All questions with answers · Transition Elements

1. Which statement describes an electric current?

   • D. A flow of metal atoms along the wire
   • A. A flow of electric charges through a conductor or a solution
   • C. A flow of heat along a metal wire
   • B. The energy stored inside a battery

   Answer: A flow of electric charges through a conductor or a solution

   An electric current is a flow of electric charges. In a metal wire, free electrons move. In a solution, ions move. A source such as a battery pushes the charges round a closed path. Common slip: thinking the current is the energy stored in the battery.

2. Two metal plates dip into a solution that contains copper ions (Cu²⁺$\mathrm{Cu}{}^{2+}$) and zinc ions (Zn²⁺$\mathrm{Zn}{}^{2+}$). The plates are joined to a battery. Towards which plate do the metal ions move?

   • A. Both kinds of ion move towards the plate joined to the negative terminal.
   • C. They do not move; only electrons carry a current in a solution.
   • D. Copper ions move to one plate and zinc ions to the other.
   • B. Both kinds of ion move towards the plate joined to the positive terminal.

   Answer: Both kinds of ion move towards the plate joined to the negative terminal.

   In a solution, ions carry the current. Opposite charges attract. Both ions are positive, so both move towards the plate joined to the negative terminal. The kind of metal does not decide the direction; the charge does. This is why two metals can deposit together on one object.

3. A steel car part is to be coated with chromium by electroplating. Which solution should be used?

   • B. A solution that contains iron ions
   • D. Water with chromium metal powder stirred into it
   • C. Pure water
   • A. A solution that contains chromium ions

   Answer: A solution that contains chromium ions

   In electroplating, the coat is made of the metal whose ions are in the solution. The object is joined to the negative terminal. The positive chromium ions move to it. There they gain electrons and become chromium atoms. So the solution must contain chromium ions.

4. A student wants to coat a spoon with copper. The spoon is joined to the positive terminal and a copper plate to the negative terminal, in copper (II) sulphate solution. After ten minutes, the spoon has no copper coat. What is the student's error?

   • D. The terminals do not matter; the student should simply wait longer.
   • A. The spoon should be joined to the negative terminal, where copper ions gain electrons.
   • C. The copper plate is not needed; the spoon alone should be in the solution.
   • B. The solution should contain ions of the spoon's metal, not copper ions.

   Answer: The spoon should be joined to the negative terminal, where copper ions gain electrons.

   In electroplating, the object must be the cathode: it is joined to the negative terminal. Copper ions are positive, so they are attracted to it. Each copper ion gains two electrons there and becomes a copper atom: Cu²⁺ + 2e- → Cu$\mathrm{Cu}{}^{2+}+2\mathrm{e}{}^{-}\to \mathrm{Cu}$. Here the spoon was positive. It repelled the copper ions, so no coat formed.

5. Write the scientific term: A material usually formed from two metals or more, which may also consist of a metal and a non-metal.

   Answer: Alloy. Key points: two metals or more (such as ferrochromium or ferromanganese), or a metal and a non-metal, such as carbon in steel.

   This is the book's definition of an alloy. Most alloys contain two metals or more, such as ferrochromium, ferromanganese, ferrovanadium and ferronickel. Some contain a non-metal: steel is iron with carbon. Common slip: answering 'compound'. An alloy is not always a chemical compound.

6. Element	Kind	Atomic radius
   W	metal	1.24
   X	metal	1.25
   Y	non-metal	0.77
   Z	metal	1.6

   The table gives the atomic radius of four elements (made-up values). Which pair is most likely to form an interstitial alloy?

   • A. W with Y: Y atoms are small enough to enter the spaces of the W lattice.
   • D. X with Z: an alloy must be made of two metals.
   • C. W with Z: the large Z atoms fill the spaces better.
   • B. W with X: their atoms have almost the same radius.

   Answer: W with Y: Y atoms are small enough to enter the spaces of the W lattice.

   In an interstitial alloy, atoms with a small atomic size enter the spaces of a metal's crystal lattice. Y (0.77) is much smaller than W (1.24), so it fits in the spaces. W and X have nearly equal radii: they suit a substitutional alloy. Steel is the book's example: small carbon atoms in the spaces of iron.

7. Metal M has an atomic radius of 1.24 (made-up value). Which of the following can replace some M atoms in its crystal lattice to form a substitutional alloy?

   • B. Metal Q: atomic radius 1.25, the same crystal lattice as M, very different chemical properties
   • A. Metal P: atomic radius 1.25, the same crystal lattice as M, similar chemical properties
   • C. Element R: atomic radius 0.75, the same crystal lattice as M, similar chemical properties
   • D. Metal T: atomic radius 1.60, a different crystal lattice, similar chemical properties

   Answer: Metal P: atomic radius 1.25, the same crystal lattice as M, similar chemical properties

   In a substitutional alloy, atoms of another metal replace some atoms of the pure metal. The book's three conditions: the same atomic radius, crystal lattice and chemical properties. Only metal P meets all three. Q fails on chemical properties. R is small enough to enter the spaces instead. T is too large and has another lattice.

8. Cementite (Fe3C$\mathrm{Fe}{}_3\mathrm{C}$) and the lead-gold alloy (Au2Pb$\mathrm{Au}{}_2\mathrm{Pb}$) belong to the same type of alloy. What do they have in common?

   • C. Atoms of one element replace atoms of the other because they have the same atomic radius.
   • B. Small atoms enter the spaces of the main metal's crystal lattice.
   • A. Their elements combine chemically to form a compound whose formula disobeys the valence law.
   • D. They are physical mixtures with no fixed ratio of atoms.

   Answer: Their elements combine chemically to form a compound whose formula disobeys the valence law.

   Both are intermetallic alloys. Their elements combine chemically to form a chemical compound. The formulas Fe3C$\mathrm{Fe}{}_3\mathrm{C}$ and Au2Pb$\mathrm{Au}{}_2\mathrm{Pb}$ disobey the valence law. Steel is also iron and carbon, but there the carbon atoms only sit in the spaces. So steel is interstitial, while cementite is intermetallic.

9. Which of the following processes prepares an alloy?

   • B. A current deposits copper on a plate from a solution of copper ions. Then the plate is moved to a solution of tin ions, and tin deposits on top.
   • A. A current passes through one solution of copper ions and tin ions, and both metals deposit on a plate at the same time.
   • C. A current passes through a solution of tin ions only, and tin deposits on a copper plate.
   • D. Copper powder and tin powder are shaken together in a bottle at room temperature.

   Answer: A current passes through one solution of copper ions and tin ions, and both metals deposit on a plate at the same time.

   The book gives two methods. By melting: the metals are melted together and left to cool gradually. By electro-deposition: two metals or more deposit at the same time from a solution of their ions. Only one solution holding both ions deposits copper and tin together. Two layers made one after the other do not give an alloy. One metal deposited on another does not either.

10. Give reasons: Introducing atoms of an element with a small atomic size into the crystal lattice of a pure metal increases its hardness.

    Answer: Key points: (1) In a pure metal, atoms are arranged in compact closed rows; on hammering, a layer of atoms can slip over the other layers. (2) The small atoms enter the spaces between the metal atoms. (3) They prevent the movement of the metal layers, so the hardness increases (example: steel).

    A pure metal has its atoms in compact close rows. On hammering, one layer can slip over the others, so the metal is relatively soft. Small atoms sit in the spaces of the lattice. They block the layers from slipping, so the metal becomes harder. Key words: spaces of the crystal lattice, prevent the movement of the layers. Most likely wrong answer: 'the small atoms replace metal atoms'. They replace none; they fill the spaces.

11. Classify the following alloys according to their type. (Atomic radius: Fe 1.16, Cr 1.17, Ni 1.15)

    Answer: Intermetallic alloy: Cementite, Fe3C$\mathrm{Fe}{}_3\mathrm{C}$, Au2Pb$\mathrm{Au}{}_2\mathrm{Pb}$ · Interstitial alloy: Steel (iron and carbon) · Substitutional alloy: Ferrochrome (Fe-Cr) in stainless steel, Copper-gold alloy, Ferronickel (Fe-Ni)

    Substitutional: ferrochrome, copper-gold and ferronickel. Their atoms have nearly the same radius, crystal lattice and chemical properties. Interstitial: steel, where small carbon atoms sit in the spaces of the iron lattice. Intermetallic: cementite, Fe3C$\mathrm{Fe}{}_3\mathrm{C}$, and Au2Pb$\mathrm{Au}{}_2\mathrm{Pb}$. Their elements combine chemically, and their formulas disobey the valence law. Common slip: putting steel and cementite together because both are Fe + C.

12. A made-up metal Q has valency 3 only. In an alloy, Q and carbon (valency 4) are found as the compound Q4C$\mathrm{Q}{}_4\mathrm{C}$. What formula does the valence law predict for a compound of Q and carbon, and what type of alloy is this?

    • C. Q4C3$\mathrm{Q}{}_4\mathrm{C}{}_3$; an interstitial alloy
    • D. Q3C4$\mathrm{Q}{}_3\mathrm{C}{}_4$; an interstitial alloy
    • A. Q4C3$\mathrm{Q}{}_4\mathrm{C}{}_3$; an intermetallic alloy
    • B. Q3C4$\mathrm{Q}{}_3\mathrm{C}{}_4$; an intermetallic alloy

    Answer: Q4C3$\mathrm{Q}{}_4\mathrm{C}{}_3$; an intermetallic alloy

    Swap the valencies: Q (3) and C (4) give Q4C3$\mathrm{Q}{}_4\mathrm{C}{}_3$. The real formula, Q4C$\mathrm{Q}{}_4\mathrm{C}$, disobeys the valence law. Such a formula is a sign of an intermetallic alloy: the elements combine chemically into a compound. Cementite, Fe3C$\mathrm{Fe}{}_3\mathrm{C}$, is the book's example with carbon.

13. Iron handles are coated with an alloy X by electro-deposition, from one solution that contains copper ions and zinc ions. What is X, and to which terminal are the handles joined?

    • C. X is brass; the handles are joined to the positive terminal.
    • A. X is brass; the handles are joined to the negative terminal.
    • B. X is bronze; the handles are joined to the negative terminal.
    • D. X is bronze; the handles are joined to the positive terminal.

    Answer: X is brass; the handles are joined to the negative terminal.

    Brass is the alloy of copper and zinc (Cu + Zn). Bronze is copper with tin. The handles must receive both metals, so they are joined to the negative terminal. Copper ions and zinc ions are positive, so both move to the handles. They deposit there at the same time and form a brass coat.

14. An engineer wants to harden a metal by putting atoms of another element into the spaces of its crystal lattice, without removing any of its own atoms. Which property must the added element have?

    • B. Its atoms must have the same atomic radius as the metal atoms.
    • C. It must have the same crystal lattice and chemical properties as the metal.
    • D. It must combine chemically with the metal to form a compound with a fixed formula.
    • A. Its atoms must be much smaller than the atoms of the metal.

    Answer: Its atoms must be much smaller than the atoms of the metal.

    The spaces between metal atoms are small. Only atoms of an element with a small atomic size can enter them. This gives an interstitial alloy, such as steel (carbon in iron). Same-size atoms with the same lattice and chemical properties give a substitutional alloy. Chemical combination gives an intermetallic alloy.

15. Consider the following statements about substitutional alloys: (I) Some atoms of the pure metal are replaced by atoms of another metal. (II) The added atoms must be much smaller than the atoms of the pure metal. (III) Ferrochrome in stainless steel is an example. Which statements are correct?

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

    Answer: (I) and (III) only

    (I) is the book's description of a substitutional alloy. (III) is correct: ferrochrome in stainless steel is a book example. (II) is wrong: the atoms must have the same atomic radius, not a much smaller one. Much smaller atoms enter the spaces and form an interstitial alloy.

16. Compare the three types of alloys: classify each description under the type it describes.

    Answer: Intermetallic alloy: The elements combine chemically to form a chemical compound., The chemical formula disobeys the valence law., The metals are not in the same group of the periodic table. · Interstitial alloy: Atoms of an element with a small atomic size enter the spaces of the lattice., Steel is the book's example. · Substitutional alloy: Some atoms of the pure metal are replaced by atoms of another metal., The metals have the same atomic radius, crystal lattice and chemical properties.

    Interstitial: atoms with a small atomic size enter the spaces of the lattice; steel is the example. Substitutional: atoms of another metal replace some atoms of the pure metal. They need the same atomic radius, crystal lattice and chemical properties. Intermetallic: the elements combine chemically into a compound. Its formula disobeys the valence law, and the metals are not in the same group.

17. Write the scientific term: An alloy formed by introducing atoms of an element with a small atomic size into the spaces of the crystal lattice of a pure metal.

    Answer: Interstitial alloy (example: steel, iron and carbon).

    In an interstitial alloy, small atoms fill the spaces between the metal atoms. They prevent the layers from slipping, so the hardness increases. Steel, iron with carbon, is the book's example. Common slip: 'substitutional alloy'. There, atoms of another metal replace some atoms of the pure metal.

18. The book's table (for illustration only) gives these atomic radii: Sc 1.44, Ti 1.32, Fe 1.16, Ni 1.15. Consider the atomic radius only. Which metal is most likely to replace some iron atoms, and what type of alloy does it form with iron?

    • C. Sc; a substitutional alloy
    • D. Ti; an interstitial alloy
    • B. Ni; an interstitial alloy
    • A. Ni; a substitutional alloy

    Answer: Ni; a substitutional alloy

    A substitutional alloy needs the same atomic radius (and the same crystal lattice and chemical properties). Ni (1.15) and Fe (1.16) are nearly equal, so Ni atoms can take the places of some Fe atoms. Ferronickel is the book's example. Sc and Ti atoms are much larger than Fe atoms. Only atoms much smaller than Fe can enter the spaces.

19. Duralumin alloys are formed from aluminum with nickel, or aluminum with copper. Which choice gives their type and a correct statement about them?

    • C. Intermetallic; the elements combine chemically, and the formula obeys the valence law like any compound.
    • D. Interstitial; aluminum atoms enter the spaces of the copper lattice.
    • B. Substitutional; Al and Cu atoms take each other's places because they are in the same group.
    • A. Intermetallic; the elements combine chemically, and the formula disobeys the valence law.

    Answer: Intermetallic; the elements combine chemically, and the formula disobeys the valence law.

    The book calls the Al-Ni and Al-Cu alloys duralumin and lists them as intermetallic alloys. The elements combine chemically to form a compound. Its formula disobeys the valence law. The metals are not in the same group: aluminum is a main-group metal, while nickel and copper are transition metals.

20. Cobalt (atomic radius 1.16) and nickel (atomic radius 1.15) can form a substitutional alloy because their radii are almost equal. Which statement explains why the radius changes so little from Co to Ni?

    • A. The increasing effective nuclear charge is nearly balanced by the increasing repulsion between 3d electrons.
    • C. Both atoms have the same number of 3d electrons.
    • B. The effective nuclear charge stays the same from Co to Ni.
    • D. Nickel has five stable isotopes, which makes its atom larger than expected.

    Answer: The increasing effective nuclear charge is nearly balanced by the increasing repulsion between 3d electrons.

    Two opposite factors act from Cr to Cu. The effective nuclear charge increases, which pulls the electrons in and decreases the radius. The number of 3d electrons increases, so their repulsion increases, which increases the radius. The two nearly balance, so the radius is nearly constant. The book says this explains the use of these metals in making alloys.

21. Explain the role of carbon monoxide in the blast furnace. Support your answer with an equation.

    Answer: Key points: (1) CO is the reducing agent: it reduces iron (III) oxide to iron. (2) 3CO + Fe2O3 →[higher than 700\,^C] 2Fe + 3CO2$3\mathrm{CO}+\mathrm{Fe}{}_2\mathrm{O}{}_3\stackrel{\text{higher than }700{}_{}^{}{}_{}^{\circ }\mathrm{C}}{\to }2\mathrm{Fe}+3\mathrm{CO}{}_2$. (3) CO gains oxygen and is oxidized to CO2$\mathrm{CO}{}_2$. (4) CO comes from coke: C + O2 → CO2$\mathrm{C}+\mathrm{O}{}_2\to \mathrm{CO}{}_2$, then CO2 + C → 2CO$\mathrm{CO}{}_2+\mathrm{C}\to 2\mathrm{CO}$.

    In the blast furnace, coke burns to CO2$\mathrm{CO}{}_2$, which meets hot coke and becomes CO. CO then takes oxygen from the iron ore. So CO is the reducing agent, and iron is set free above 700 °C. CO itself becomes CO2$\mathrm{CO}{}_2$. Most likely wrong answer: 'CO only heats the furnace'. Its key role is reduction.

22. Ferromanganese alloy is used in railway tracks. Which statement is correct?

    • B. It is softer than steel, so the tracks bend without breaking.
    • C. Pure manganese is used for tracks because it is harder than any of its alloys.
    • D. It is an alloy of manganese with aluminum, used because it is light.
    • A. It is harder than steel, while pure manganese is not used because it is brittle.

    Answer: It is harder than steel, while pure manganese is not used because it is brittle.

    Ferromanganese is an alloy of iron and manganese. It is used in railway tracks because it is harder than steel. Manganese is not used as a pure metal because it is brittle. So it is always used as alloys or compounds. The aluminum-manganese alloy is used for soft drink cans because it resists corrosion.