Transition elements: A Level Chemistry questions with answers
Free Transition elements questions for A Level Chemistry (9701), part of Inorganic & organic. Each question has the answer and the marking points an examiner would look for.
This question is about complexes of copper and platinum.
(a)
Define the term transition element.
[1]
(b)
Concentrated HCl is added to aqueous copper(II) sulfate. Describe the colour change and the change in shape of the complex.
[2]
(c)
Explain why [Cu(H₂O)₆]²⁺ is coloured.
[3]
(d)
Cisplatin, Pt(NH₃)₂Cl₂, is square planar. Explain why it shows geometrical isomerism.
[2]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
d-block element forming a stable ion with partially filled d subshell
an element that forms one or more stable ions with an incomplete d subshell B1
1
(b)
blue to yellow-green; octahedral to tetrahedral
blue to yellow / green B1 [Cu(H₂O)₆]²⁺ octahedral to [CuCl₄]²⁻ tetrahedral B1
2
(c)
d orbital splitting; absorbs visible light
ligands split the d orbitals into two energy levels M1 an electron is promoted, absorbing light of a particular frequency in the visible range M1 the complementary colour is seen M1
3
(d)
cis and trans arrangements
square planar with two pairs of different ligands M1 the like ligands can be adjacent (cis) or opposite (trans) M1
2
Question 2Structured · 12 marksA* challenge
An iron tablet of mass 2.00 g is dissolved in dilute sulfuric acid and made up to 250 cm3. 25.0 cm3 of this solution requires 25.60 cm3 of 0.0100 mol dm−3 KMnO4 for complete reaction. MnO4− + 8H+ + 5Fe2+ → Mn2+ + 5Fe3+ + 4H2O
(a)
Calculate the percentage by mass of iron in the tablet.
[4]
(b)
State why no indicator is needed, and describe the colour change at the end-point.
[2]
(c)
Explain why transition metal complexes are coloured.
[3]
(d)
[Cu(H2O)6]2+ reacts with excess concentrated HCl. Give the formula, shape and colour of the new complex, and the type of reaction.
[3]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
35.7%
mol MnO4−C1 × 5 for Fe2+C1 × 10 for 250 cm3C1 35.7 A1
4
(b)
MnO4− is self-indicating; colourless to pale pink (permanent)
self-indicating B1 to (permanent) pale pink B1
2
(c)
ligands split d orbitals into two energy levels; electrons absorb light to move to higher level; complementary colour transmitted
d-orbital splitting by ligands B1 electron promotion absorbing visible light B1 remaining frequencies seen B1
This question is about complexes of cobalt and copper.
(a)
Define the terms ligand and coordination number.
[2]
(b)
Excess concentrated hydrochloric acid is added to aqueous [Co(H₂O)₆]²⁺. Write the equation, state the colour change and the shape of each complex.
[4]
(c)
Explain why the formation of [Cu(EDTA)]²⁻ from [Cu(H₂O)₆]²⁺ is favourable, in terms of entropy.
[3]
(d)
Explain why different ligands give complexes of the same metal ion different colours.
[2]
(e)
Draw the two geometrical isomers of the square planar complex [Pt(NH₃)₂Cl₂] and state which is used as an anticancer drug.
[3]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
species donating a lone pair to a central metal ion; number of dative bonds to central ion
ligand: a species that uses a lone pair to form a dative bond to a central metal ion B1 coordination number: the number of dative bonds formed with the central ion B1
2
(b)
[Co(H₂O)₆]²⁺ + 4Cl⁻ ⇌ [CoCl₄]²⁻ + 6H₂O; pink to blue; octahedral to tetrahedral
equation correct B1 pink to blue B1 octahedral → tetrahedral B1 Cl⁻ larger so only four fit B1
4
(c)
1 EDTA replaces 6 H₂O; number of particles increases; ΔS positive (chelate effect)
one hexadentate EDTA⁴⁻ replaces six water molecules B1 the number of particles in solution increases from 2 to 7 B1 so ΔS is large and positive and the complex is very stable (chelate effect) B1
3
(d)
different ligands split d orbitals by different amounts; different ΔE, different frequency absorbed
different ligands split the d orbitals by different energies (ΔE) B1 so different frequencies of light are absorbed B1
2
(e)
cis and trans; cisplatin (cis)
cis isomer drawn B1 trans isomer drawn B1 cis-platin is the anticancer drug B1
3
Question 4Structured · 14 marks
This question is about transition elements and their complexes.
(a)
Define a transition element.
[1]
(b)
Give the electronic configuration of Fe²⁺ and explain why Fe³⁺ is more stable than Fe²⁺.
[2]
(c)
Explain why aqueous Cu²⁺ ions are coloured.
[3]
(d)
Write an equation for the reaction of [Cu(H₂O)₆]²⁺ with excess concentrated hydrochloric acid, and state the colour change and the type of reaction.
[3]
(e)
2.50 g of an iron supplement is dissolved in acid and made up to 250 cm3. 25.0 cm3 portions react with 24.80 cm3 of 0.0200 mol dm−3 KMnO₄. Calculate the percentage by mass of iron in the supplement. MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O
[4]
(f)
Explain why no indicator is needed in this titration.
[1]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
d-block element forming one or more stable ions with incomplete d subshell
a d-block element that forms one or more stable ions with an incomplete d subshell B1
1
(b)
[Ar]3d⁶; Fe³⁺ 3d⁵ half-filled, reduced repulsion
[Ar] 3d6B1 Fe³⁺ has a half-filled 3d5 subshell with no paired electrons, so less repulsion B1
2
(c)
d orbitals split by ligands; electron absorbs visible light to promote; complementary colour transmitted
ligands split the 3d orbitals into two energy levels B1 an electron absorbs a frequency of visible light and is promoted B1 the remaining (complementary) colour is seen B1
3
(d)
[Cu(H₂O)₆]²⁺ + 4Cl⁻ → [CuCl₄]²⁻ + 6H₂O; blue to yellow; ligand exchange
[Cu(H₂O)₆]²⁺ + 4Cl⁻ ⇌ [CuCl₄]²⁻ + 6H₂O B1 blue to yellow / green B1 ligand exchange (shape octahedral → tetrahedral) B1
ligands split the d-orbitals into two energy levels B1 electrons absorb light to move to higher level B1 remaining (complementary) colour transmitted B1
3
(b)
blue ppt then dark blue solution
pale blue precipitate B1 dissolves to give a deep blue solution B1
2
Question 6Structured · 12 marks
An alloy containing copper is dissolved in nitric acid; the excess acid is removed and the solution neutralised. Excess potassium iodide is added: 2Cu²⁺ + 4I⁻ → 2CuI + I₂. The iodine produced requires 18.40 cm3 of 0.0500 mol dm−3 sodium thiosulfate: I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻. The mass of alloy used was 0.200 g.
(a)
State the electronic configuration of Cu, Cu⁺ and Cu²⁺.
[3]
(b)
Explain why CuI is white whereas Cu²⁺ compounds are coloured.
[2]
(c)
Calculate the percentage by mass of copper in the alloy.
[4]
(d)
State the indicator used and the colour at the end-point, and explain why it is added near the end-point only.
Cu⁺ has a full 3d subshell B1 no electron can be promoted between split d orbitals, so no visible light is absorbed B1
2
(c)
29.2%
moles S₂O₃²⁻ = 9.20 × 10−4M1 moles I₂ = 4.60 × 10−4, moles Cu²⁺ = 9.20 × 10−4M1 mass Cu = 0.0584 g M1 29.2% A1
4
(d)
starch; blue-black to colourless (white ppt); added late to avoid starch–iodine complex not releasing iodine
starch B1 blue-black to colourless (white suspension of CuI) B1 added when the solution is pale yellow, because at high [I₂] a large amount of iodine is bound in the starch complex and released only slowly, so the end-point is not sharp B1
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Original Cambridge-style practice questions written for mentorriyaz.in. These are not Cambridge past paper questions, and this site is not affiliated with or endorsed by Cambridge University Press & Assessment. Updated 2026-11-04. Paper generator · mentorriyaz.in