Transition elements

1. Overview

The transition elements are a collection of metals located in the central block of the Periodic Table, between Groups II and III. These elements are vital in construction and the chemical industry due to their high strength and ability to act as catalysts to speed up industrial reactions.

Key Definitions

• Transition Element: A metal found in the central block of the Periodic Table that forms at least one stable ion with a partially filled d-subshell (often simplified at IGCSE as "metals in the middle of the table").
• Catalyst: A substance that increases the rate of a chemical reaction by providing an alternative pathway with lower activation energy, without being chemically changed at the end of the reaction.
• Oxidation Number: A value assigned to an atom in a compound that represents the number of electrons lost or gained (e.g., in FeCl₂, Iron has an oxidation number of +2).

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Core Content

Transition elements are distinct from Group I (alkali metals) and Group II (alkaline earth metals) because of their specific physical and chemical properties.

Physical Properties

1. High Densities: They are much "heavier" for their size compared to Group I metals. For example, Iron (Fe) has a density of 7.87 g/cm³, whereas Sodium (Na) is only 0.97 g/cm³.
2. High Melting Points: They remain solid at much higher temperatures. For example, Iron melts at 1538°C, while Sodium melts at only 98°C.

Chemical Properties

1. Form Coloured Compounds: Unlike Group I metals which form white compounds, transition metals form brilliantly coloured salts.
   • Copper(II) sulfate: Blue
   • Iron(II) salts: Pale green
   • Iron(III) salts: Orange/Brown/Yellow
2. Act as Catalysts: Both the pure elements and their compounds are used to speed up reactions.
   • Iron (Fe): Used in the Haber Process to make ammonia.
   • Manganese(IV) oxide (MnO₂): Used to decompose hydrogen peroxide.

Example Reaction: Decomposition of Hydrogen Peroxide

• Word Equation: Hydrogen peroxide $\rightarrow$ water + oxygen (using Manganese(IV) oxide catalyst)
• Symbol Equation: $2H_2O_2(aq) \rightarrow 2H_2O(l) + O_2(g)$

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Extended Content (Extended Only)

Variable Oxidation Numbers

Transition elements do not have a fixed number of electrons in their outer shell that they lose. Instead, they can form multiple stable ions with different charges. This is referred to as having variable oxidation numbers.

Common Examples:

• Iron (Fe):
  • Iron(II) ions: $Fe^{2+}$ (forms green compounds)
  • Iron(III) ions: $Fe^{3+}$ (forms reddish-brown compounds)
• Copper (Cu):
  • Copper(I) ions: $Cu^{+}$
  • Copper(II) ions: $Cu^{2+}$

Reactivity with Sodium Hydroxide (aq): This property is often tested by adding aqueous sodium hydroxide to identify the specific ion present via a precipitation reaction:

• Iron(II) reaction: Iron(II) chloride(aq) + Sodium hydroxide(aq) $\rightarrow$ Iron(II) hydroxide(s) + Sodium chloride(aq) $FeCl_2(aq) + 2NaOH(aq) \rightarrow Fe(OH)_2(s) + 2NaCl(aq)$ (Result: Green precipitate)

• Iron(III) reaction: Iron(III) chloride(aq) + Sodium hydroxide(aq) $\rightarrow$ Iron(III) hydroxide(s) + Sodium chloride(aq) $FeCl_3(aq) + 3NaOH(aq) \rightarrow Fe(OH)_3(s) + 3NaCl(aq)$ (Result: Red-brown precipitate)

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Key Equations

Reaction	Balanced Symbol Equation	Notes
Haber Process	$N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$	Uses Iron (Fe) catalyst
Contact Process	$2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)$	Uses Vanadium(V) oxide ($V_2O_5$) catalyst
Peroxide Breakdown	$2H_2O_2(aq) \rightarrow 2H_2O(l) + O_2(g)$	Uses Manganese(IV) oxide ($MnO_2$) catalyst

Key to Symbols:

• $N_2$ = Nitrogen gas
• $H_2$ = Hydrogen gas
• $NH_3$ = Ammonia gas
• $\rightleftharpoons$ = Reversible reaction
• $(aq)$ = Aqueous (dissolved in water)
• $(s)$ = Solid

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Common Mistakes to Avoid

• ❌ Wrong: Saying transition metals are "reactive" like Group I.
  • ✓ Right: Transition metals are generally much less reactive than Group I and II metals (e.g., Gold and Platinum are very unreactive).
• ❌ Wrong: Describing the metals themselves as coloured.
  • ✓ Right: While some are (Copper is pinkish-brown), the key property is that they form coloured compounds.
• ❌ Wrong: Forgetting the Roman numerals in names.
  • ✓ Right: Always write Iron(II) or Iron(III) to indicate the oxidation state, as "Iron chloride" is ambiguous.

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Exam Tips

• Command Word "State": This topic has appeared 29 times as a "State" question. If asked to "state three properties of transition metals," simply list: High density, high melting point, and form coloured compounds.
• Identification Questions: If a question mentions a "blue solution" or "green precipitate," immediately think of Transition Elements (Copper and Iron respectively).
• Catalyst Contexts: Expect questions on the Haber Process. Remember that the catalyst (Iron) does not appear in the chemical equation itself but is written over the arrow.
• Typical Values: In calculation questions, be ready to see atomic masses such as 56.0 (Fe), 63.5 (Cu), or 65.0 (Zn).
• Frequency: With 44 past paper appearances, this is a high-yield topic. Ensure you can distinguish transition metals from Group I metals (who have low density and low melting points).

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Exam-Style Questions

Practice these original exam-style questions to test your understanding. Each question mirrors the style, structure, and mark allocation of real Cambridge 0620 Theory papers.

Exam-Style Question 1 — Short Answer [6 marks]

Question:

Iron is a transition element widely used in industry.

(a) State two typical physical properties of transition elements, other than being metallic. [2]

(b) Iron is used as a catalyst in the Haber process. State the meaning of the term catalyst. [1]

(c) Describe one advantage of using a catalyst in an industrial process. [2]

(d) Iron(II) chloride ($FeCl_2$) is a compound of iron. State the oxidation number of iron in $FeCl_2$. [1]

Worked Solution:

(a)

1. High density Transition elements generally have high densities.

2. High melting point Transition elements generally have high melting points.

How to earn full marks:

• State either high density or high melting point for 1 mark each.
• Do not state "hard" - this is a general property of metals, not specific to transition elements.

(b)

1. A substance that speeds up a chemical reaction without being used up itself. This is the definition of a catalyst.

How to earn full marks:

• Must include both "speeds up a reaction" and "not used up" for the mark.

(c)

1. Increases the rate of reaction, so less energy is needed to achieve desired production rate. This is a key advantage.
2. Can lower the activation energy of the reaction. Lower activation energy leads to faster reaction rates.

How to earn full marks:

• Mentioning that a catalyst speeds up the reaction is worth 1 mark.
• Mentioning that this reduces energy costs or lowers activation energy is worth the second mark.

(d)

1. +2 The oxidation number of chlorine is -1, and there are two chlorine atoms, so the iron must be +2 to balance the charge.

How to earn full marks:

• Correct sign (+ or -) must be included.

Common Pitfall: Remember that transition elements have properties that are specific to them. Stating a general property of metals, like being "shiny", will not earn you the mark. Also, make sure you include the + or - sign when stating oxidation numbers.

Exam-Style Question 2 — Short Answer [5 marks]

Question:

Copper is a transition element that forms coloured compounds.

(a) State two colours that compounds of copper can be. [2]

(b) Copper(II) sulfate ($CuSO_4$) is a soluble salt. Describe what you would observe when aqueous ammonia is added dropwise to a solution of copper(II) sulfate, until excess. [3]

Worked Solution:

(a)

1. Blue Many copper(II) compounds are blue.

2. Green Some copper(II) compounds, especially those containing chloride ions, are green.

How to earn full marks:

• Award 1 mark for each correct colour stated.
• Accept "blue" or "green".

(b)

1. Initially, a light blue precipitate forms. Copper(II) hydroxide is initially precipitated.
2. With excess ammonia, the precipitate dissolves. A complex ion is formed.
3. A dark blue solution is formed. The tetraamminecopper(II) complex is a dark blue solution.

How to earn full marks:

• 1 mark for mentioning "light blue precipitate".
• 1 mark for stating the precipitate "dissolves" or "disappears".
• 1 mark for stating the final solution is "dark blue" or "deep blue".

Common Pitfall: When describing colour changes, be precise. Saying "blue" is not enough when the question asks for the change in colour. Also, remember to mention the precipitate dissolves in excess ammonia, not just that the solution changes colour.

Exam-Style Question 3 — Extended Response [9 marks]

Question:

Manganese is a transition element that can exist in several oxidation states. Manganese(IV) oxide, $MnO_2$, is used as a catalyst in the decomposition of hydrogen peroxide ($H_2O_2$).

(a) Write the balanced chemical equation for the decomposition of hydrogen peroxide into water and oxygen. [2]

(b) Describe how you would demonstrate experimentally that $MnO_2$ acts as a catalyst in this reaction. Your answer should include: - the apparatus and procedure - the expected observations [5]

(c) Manganese also forms $MnCl_2$. Calculate the relative formula mass, $M_r$, of $MnCl_2$. [2] $[A_r: Mn = 55, Cl = 35.5]$

Worked Solution:

(a)

1. $2H_2O_2(aq) \rightarrow 2H_2O(l) + O_2(g)$ This shows the correct formulas and balancing.

How to earn full marks:

• 1 mark for correct formulas of reactants and products.
• 1 mark for correct balancing. State symbols are not required.

(b)

1. Apparatus: Test tube, delivery tube, measuring cylinder, gas syringe (or inverted measuring cylinder over water).
   📊Test tube containing hydrogen peroxide solution with a delivery tube leading to a gas syringe to collect oxygen gas. The gas syringe is marked with volume measurements.
   A diagram is helpful but not essential.
2. Procedure: Add a small amount of $MnO_2$ to hydrogen peroxide solution in the test tube. Collect the gas produced in the gas syringe/measuring cylinder. Measure the volume of gas produced over a set time. Repeat the experiment without $MnO_2$. This describes the experiment.
3. Observation: The rate of oxygen production is much faster with $MnO_2$ than without it. This identifies the catalytic effect.
4. Additional observation: The $MnO_2$ remains unchanged at the end of the reaction. This confirms the $MnO_2$ is a catalyst.
5. Control: Weigh the $MnO_2$ before and after the experiment to show its mass doesn't change. This adds further evidence of the catalytic role.

How to earn full marks:

• 1 mark for describing suitable apparatus.
• 2 marks for describing the procedure clearly and accurately.
• 1 mark for stating the rate of reaction is faster with $MnO_2$.
• 1 mark for indicating the $MnO_2$ remains unchanged (mass or appearance).

(c)

1. $M_r(MnCl_2) = 55 + (2 \times 35.5)$ This is the correct calculation.
2. $M_r(MnCl_2) = 126$ This is the final answer.

How to earn full marks:

• 1 mark for the correct calculation.
• 1 mark for the correct answer.
• $M_r = \boxed{126}$

Common Pitfall: In experimental design questions, be specific about your measurements and controls. Don't just say "measure the rate"; explain how you will measure the rate. Also, remember that a catalyst is not consumed in the reaction, so you need to show how you would confirm that the $MnO_2$ remains unchanged.

Exam-Style Question 4 — Extended Response [7 marks]

Question:

A student investigates the reaction between iron(II) sulfate solution and potassium manganate(VII) solution. Potassium manganate(VII) is a strong oxidising agent.

(a) State the colours of:

(i) iron(II) sulfate solution [1]

(ii) potassium manganate(VII) solution [1]

(b) In the reaction, iron(II) ions ($Fe^{2+}$) are oxidised to iron(III) ions ($Fe^{3+}$). State the change in oxidation number of manganese in this reaction, given that manganate(VII) ions ($MnO_4^−$) are reduced to manganese(II) ions ($Mn^{2+}$). [2]

(c) In a titration, the student finds that $20.0 cm^3$ of $0.020 mol/dm^3$ potassium manganate(VII) solution reacts completely with $25.0 cm^3$ of iron(II) sulfate solution. The equation for the reaction is:

$MnO_4^−(aq) + 5Fe^{2+}(aq) + 8H^+(aq) \rightarrow Mn^{2+}(aq) + 5Fe^{3+}(aq) + 4H_2O(l)$

Calculate the concentration of the iron(II) sulfate solution in $mol/dm^3$. [3]

Worked Solution:

(a) (i) Pale green Iron(II) solutions are typically pale green.

(ii) Purple Potassium manganate(VII) solutions are purple.

How to earn full marks:

• Accept "light green" for iron(II) sulfate.
• Accept "pink" for potassium manganate(VII), although purple is more accurate.

(b)

1. From +7 to +2 This is the oxidation number change.

How to earn full marks:

• 1 mark for stating "+7".
• 1 mark for stating "+2".

(c)

1. Moles of $MnO_4^− = concentration \times volume = 0.020\ mol/dm^3 \times (20.0/1000)\ dm^3 = 0.00040\ mol$ Calculating the moles of potassium manganate(VII).

2. Moles of $Fe^{2+} = 5 \times moles\ of\ MnO_4^− = 5 \times 0.00040\ mol = 0.0020\ mol$ Using the stoichiometry of the equation to find the moles of iron(II) ions.

3. Concentration of $Fe^{2+} = moles / volume = 0.0020\ mol / (25.0/1000)\ dm^3 = 0.080\ mol/dm^3$ Calculating the concentration of the iron(II) sulfate solution.

How to earn full marks:

• 1 mark for calculating the moles of $MnO_4^−$ correctly.
• 1 mark for using the stoichiometry to find the moles of $Fe^{2+}$.
• 1 mark for calculating the concentration of $Fe^{2+}$ correctly.
• Concentration = $\boxed{0.080\ mol/dm^3}$

Common Pitfall: Remember to convert volumes from $cm^3$ to $dm^3$ before using them in calculations. Also, pay close attention to the stoichiometry of the balanced equation to correctly relate the moles of reactants. Forgetting to multiply by 5 in this case is a common mistake.