Chemistry · Unit 18

Transition Metals

The d-block elements

The d-block behaves differently enough from the rest of the table to need its own unit, and the reason is the d-orbitals sitting close in energy to the outer s.

The unit covers transition metal properties, variable oxidation states, complex ions, why their compounds are coloured, and their catalytic behaviour.

This unit breaks down into 20 short steps and 120 questions, starting at difficulty 1 and building to 3. Below you can see exactly what it covers, how the path is structured, and worked examples with explanations.

Steps: 20
Questions: 120
Difficulty: 1-3

What this unit covers

Transition Metal Properties
Variable Oxidation States
Complex Ions
Coloured Compounds
Transition Metal Catalysis

Where this fits

The most advanced unit in the Chemistry track. Needs Chemical Bonding and Redox Reactions.

Where people slip

Not every d-block element is a transition metal. The definition requires an incomplete d subshell in a stable ion, which excludes zinc and scandium on the usual reading.

How the unit is structured

Transition Metals runs as 20 short steps that unlock in order. 15 are practice rounds and 5 are challenge rounds that pull together everything before them. Questions start at difficulty 1 and climb to 3 as you progress.

Step 1 · easierStep 20 · harder

Challenge rounds

Example questions

30 real questions from this unit, with the answer and the reason behind it, grouped by what they practise. There are 120 in the unit altogether.

Coloured Compounds

Fill the blankLevel 2
1. Iron(II) compounds in solution are usually ____ in colour.
- greencorrect
- purple
- orange-brown
- colourless
Iron(II) (Fe2+) compounds are typically pale green, which helps tell them apart from orange-brown iron(III).
Multiple choiceLevel 2
2. Iron(III) compounds in solution are typically what colour?
- Orange-browncorrect
- Bright green
- Deep blue
- Pink
Iron(III) (Fe3+) compounds are usually orange-brown or yellow-brown, unlike the green of iron(II).
Odd one outLevel 2
3. Which of these compounds is white or colourless rather than coloured?
- Zinc sulfatecorrect
- Copper(II) sulfate
- Iron(III) chloride
- Potassium manganate(VII)
Zinc has a full d sub-shell, so zinc sulfate is white; the others are all coloured transition metal compounds.
Picture questionLevel 2
4. 🧪 This flask holds a solution of potassium manganate(VII), KMnO4. What colour is it?
- Purplecorrect
- Blue
- Green
- Colourless
Potassium manganate(VII) solution is a strong purple colour.
Fact or fibLevel 3
5. Since the splitting energy equals h times the frequency absorbed, a larger d-orbital splitting means the complex absorbs light of a higher frequency (shorter wavelength).
Answer: True
Because delta E = h f, a bigger splitting gap requires a higher-frequency, shorter-wavelength photon to be absorbed.
Match the pairsLevel 3
6. Match each ion to the colour of its aqueous solution.
Answer: [Cu(H2O)6]2+ = blue; MnO4- = purple; Cr2O7 2- = orange; [Ni(H2O)6]2+ = green
Common colours: hydrated copper(II) is blue, manganate(VII) is purple, dichromate(VI) is orange and hydrated nickel(II) is green.

Complex Ions

Build the sentenceLevel 3
7. Build the rule that decides spin state in a complex.
Answer: strong field ligands give low spin complexes
Strong-field ligands cause a large splitting, so electrons pair up before occupying the upper orbitals, giving low-spin complexes.
Fact or fibLevel 3
8. The binding of oxygen to the iron in haemoglobin is a reversible ligand substitution, letting oxygen be picked up in the lungs and released in the tissues.
Answer: True
O2 acts as a ligand that reversibly coordinates to the haem iron, so it can be loaded and unloaded as the body needs.
Fill the blankLevel 3
9. A molecule or ion that donates a lone pair of electrons to a central metal ion is called a ____.
- ligandcorrect
- catalyst
- isotope
- monomer
A ligand donates a lone pair to the central metal ion, forming a coordinate (dative) bond in a complex ion.
Guess the numberLevel 3
10. How many water molecules surround the central ion in the complex [Cu(H2O)6]2+ (its coordination number)?
Answer: 6 ligands
Six water ligands surround the copper ion, so the coordination number is 6.
Multiple choiceLevel 3
11. In the ion [Cu(H2O)6]2+, what is the role of the water molecules?
- They act as ligands, bonding to the central copper ion
- They act as a catalyst for the reaction
- They are simply the solvent and are not bonded
- They are negative counter-ions
The six water molecules are ligands, each donating a lone pair to the central Cu2+ ion to form a complex ion.
Picture questionLevel 3
12. 💊 Cisplatin, [Pt(NH3)2Cl2], has a coordination number of 4 and is used as an anti-cancer drug. What is its shape?
- Square planarcorrect
- Octahedral
- Trigonal planar
- Linear
Cisplatin is square planar; the cis arrangement of the two chloride ligands is what makes it medically active.

Transition Metal Catalysis

Build the sentenceLevel 2
13. Build the sentence about the Haber process catalyst.
Answer: Iron is the catalyst in the Haber process
Iron is the catalyst used in the Haber process to make ammonia.
Fill the blankLevel 2
14. A substance that speeds up a reaction without being used up is called a ____.
- catalystcorrect
- reactant
- product
- solvent
A catalyst increases the rate of a reaction but is not used up, so it can be reused.
Multiple choiceLevel 2
15. Which transition metal is used as the catalyst in the Haber process?
- Ironcorrect
- Nickel
- Copper
- Zinc
The Haber process, which makes ammonia, uses an iron catalyst.
Match the pairsLevel 3
16. Match each industrial process to the catalyst it uses.
Answer: Haber process = Iron; Contact process = Vanadium(V) oxide; Hydrogenation of oils = Nickel; Decomposition of hydrogen peroxide = Manganese(IV) oxide
Haber uses iron, the Contact process uses vanadium(V) oxide, hydrogenation uses nickel and hydrogen peroxide decomposition uses manganese(IV) oxide.
Put in orderLevel 3
17. Put the three stages of heterogeneous catalysis on a solid surface into the correct order.
Answer: Reactants adsorb onto the catalyst surface -> Bonds weaken and the reaction takes place -> Products desorb from the surface
Reactants first adsorb onto the surface, then react as bonds weaken, and finally the products desorb.
Tap the pairsLevel 3
18. Match each transition metal catalyst to the industrial process it is used in.
Answer: Iron = Haber process; Vanadium(V) oxide = Contact process; Nickel = Hydrogenation of alkenes; Platinum = Catalytic converter
Iron catalyses the Haber process, V2O5 the Contact process, nickel the hydrogenation of alkenes and platinum the catalytic converter.

Transition Metal Properties

Multiple choiceLevel 1
19. Where in the periodic table are the transition metals found?
- In the central d-block, between groups 2 and 3
- On the far left, in group 1
- In the top-right corner with the non-metals
- In the two detached rows at the bottom
The transition metals make up the central d-block of the periodic table, between groups 2 and 3.
Choose all that applyLevel 2
20. Which of these are typical properties of transition metals? (Select all that apply.)
- High densitycorrect
- High melting pointcorrect
- Very low density, like group 1 metals
- React violently with cold water
Transition metals are typically dense, hard and high-melting, and much less reactive than group 1 metals.
Fact or fibLevel 2
21. Transition metals are generally less reactive than group 1 metals.
Answer: True
Unlike group 1 metals, transition metals react slowly or not at all with water, so they are much less reactive.
Fill the blankLevel 2
22. The transition metals occupy the central ____ of the periodic table.
- d-blockcorrect
- s-block
- p-block
- f-block
Transition metals are d-block elements, filling their 3d sub-shell across the central block.
Match the pairsLevel 2
23. Match each transition metal to its chemical symbol.
Answer: Iron = Fe; Copper = Cu; Zinc = Zn; Nickel = Ni
Iron is Fe, copper is Cu, zinc is Zn and nickel is Ni.
Guess the numberLevel 3
24. How many unpaired electrons are there in an Fe3+ ion, which has the configuration [Ar] 3d5?
Answer: 5 unpaired electrons
In 3d5 each of the five d orbitals holds one electron, so all five electrons are unpaired.

Variable Oxidation States

Multiple choiceLevel 2
25. Iron commonly forms two different ions. What are they?
- Fe2+ and Fe3+correct
- Fe+ and Fe2+
- Fe2+ and Fe4+
- Fe3+ and Fe6+
Iron shows variable oxidation states, most commonly forming Fe2+ (iron(II)) and Fe3+ (iron(III)).
Build the sentenceLevel 3
26. Build the balanced reduction half-equation for the manganate(VII) ion in acid solution.
Answer: MnO4- + 8H+ + 5e- -> Mn2+ + 4H2O
MnO4- + 8H+ + 5e- -> Mn2+ + 4H2O balances charge and atoms as manganese is reduced from +7 to +2.
Choose all that applyLevel 3
27. Which of these are genuine oxidation states shown by manganese in its compounds?
- +2, as in Mn2+
- +4, as in MnO2
- +7, as in MnO4-
- +8, as in MnO4
Manganese ranges widely: +2 (Mn2+), +4 (MnO2) and +7 (MnO4-) are all real; +8 and +1 are not.
Fact or fibLevel 3
28. Transition metals show variable oxidation states because the 3d and 4s subshells are very close in energy, so different numbers of electrons can be removed using similar amounts of energy.
Answer: True
The small energy gap between 3d and 4s lets varying numbers of electrons be removed, giving several accessible oxidation states.
Guess the numberLevel 3
29. What is the oxidation state of manganese in potassium manganate(VII), KMnO4?
Answer: 7 oxidation state
In KMnO4, K is +1 and each O is -2 (total -8), so manganese must be +7 to balance.
Match the pairsLevel 3
30. Match each compound to the oxidation state of its transition metal.
Answer: Cu2O = +1; FeO = +2; Fe2O3 = +3; MnO2 = +4
Oxide ions are -2, so the metal states are Cu +1 in Cu2O, Fe +2 in FeO, Fe +3 in Fe2O3 and Mn +4 in MnO2.

Where these questions come from. Each unit starts as a plan of the concepts it should cover and the difficulty it should span. Questions are written against that plan with AI assistance, then checked by a validator that rejects anything without a single defensible answer, an explanation, or plausible wrong options. How we write questions sets out the whole process, and corrections are fixed in the bank and reach the site and the app the same day.

How you practise

This unit mixes 17 different question formats, so you are recalling and applying rather than recognising the same layout every time.

Build the sentence
Choose all that apply
Fact or fib
Fill the blank
Guess the number
Listen and choose
Match the pairs
Multiple choice
Odd one out
Picture question
Put in order
Sequence recall
Sort into groups
Spell it
Tap the pairs
True or false
Type the answer

Practise Transition Metals

120 questions across 20 steps. Start with step one and crawl at your own pace.

Play this unit

More units in Chemistry

See all 18 units in Chemistry →