Unit 1 · Inorganic Chemistry

Transition Metals

Electronic configuration, characteristic properties, complex ions, isomerism, chemistry of individual transition metals, and ion identification.

1.1

Definition & Electronic Configuration

Transition Metal — Definition A transition metal is an element that forms at least one stable ion with an incompletely filled d sub-shell.

The d-block elements occupy Groups 3–12 of the Periodic Table (Period 4: Sc to Zn).
Note: Sc and Zn do not satisfy this definition strictly (see Section 1.3).

Writing Electronic Configurations

The 3d sub-shell fills after the 4s sub-shell (Aufbau principle), but the 3d electrons are lost first when forming cations (since 4s has higher energy in multi-electron ions).

General pattern for Period 4 d-block: [Ar] 3d^n 4s^2 (n = 1 to 10) Exceptions (half-filled and fully-filled d-shells are extra stable): Cr: [Ar] 3d^5 4s^1 (NOT [Ar] 3d^4 4s^2) Cu: [Ar] 3d^10 4s^1 (NOT [Ar] 3d^9 4s^2) When forming cations: 4s electrons lost FIRST Fe: [Ar] 3d^6 4s^2 Fe²⁺:[Ar] 3d^6 (both 4s removed) Fe³⁺:[Ar] 3d^5 (one 3d also removed)
ElementSymbolZConfig (atom)Common ionsIon config
ScandiumSc21[Ar] 3d¹ 4s²Sc³⁺[Ar] (empty d)
TitaniumTi22[Ar] 3d² 4s²Ti²⁺, Ti⁴⁺[Ar] 3d²; [Ar]
VanadiumV23[Ar] 3d³ 4s²V²⁺ to V⁵⁺[Ar] 3d³ etc.
ChromiumCr24[Ar] 3d⁵ 4s¹Cr²⁺, Cr³⁺[Ar] 3d⁴; [Ar] 3d³
ManganeseMn25[Ar] 3d⁵ 4s²Mn²⁺, Mn⁴⁺, Mn⁷⁺[Ar] 3d⁵; [Ar] 3d³
IronFe26[Ar] 3d⁶ 4s²Fe²⁺, Fe³⁺[Ar] 3d⁶; [Ar] 3d⁵
CobaltCo27[Ar] 3d⁷ 4s²Co²⁺, Co³⁺[Ar] 3d⁷; [Ar] 3d⁶
NickelNi28[Ar] 3d⁸ 4s²Ni²⁺[Ar] 3d⁸
CopperCu29[Ar] 3d¹⁰ 4s¹Cu⁺, Cu²⁺[Ar] 3d¹⁰; [Ar] 3d⁹
ZincZn30[Ar] 3d¹⁰ 4s²Zn²⁺[Ar] 3d¹⁰ (full d)
1.2

Properties of Transition Metals

1. Variable Oxidation States

Transition metals exhibit multiple oxidation states because the energy difference between 3d and 4s orbitals is small — varying numbers of d and s electrons can be involved in bonding.

MetalOxidation statesMost commonExample compound
Ti+2, +3, +4+4TiO₂ (titanium dioxide)
V+2, +3, +4, +5+5V₂O⁵ (vanadium pentoxide)
Cr+2, +3, +6+3Cr₂O₃ (green); CrO₄²⁻ (yellow)
Mn+2, +3, +4, +6, +7+2, +7MnSO₄; KMnO₄
Fe+2, +3+2, +3FeCl₂; FeCl₃
Co+2, +3+2CoCl₂; Co₂O₃
Cu+1, +2+2Cu₂O; CuSO₄

2. Formation of Coloured Ions

Many transition metal ions are coloured in solution because of d-d transitions — electrons absorb photons of visible light and jump to higher-energy d-orbitals. The colour observed is the complementary colour of the light absorbed.

This only occurs when the d sub-shell is partially filled (Sc³⁺ and Zn²⁺ are colourless because d is empty and full respectively).

IonColourIonColour
Ti³⁺PurpleCr³⁺Green (dark)
V²⁺VioletCrO₄²⁻Yellow
V³⁺GreenCr₂O⁷²⁻Orange
Mn²⁺Very pale pinkMnO₄⁻Purple
Fe²⁺Pale greenFe³⁺Yellow-brown
Co²⁺Pink (aq); blue (anhydrous)Ni²⁺Green
Cu²⁺Blue (aq)Zn²⁺Colourless

3. Catalytic Activity

Transition metals and their compounds are excellent catalysts because:

  • They can adopt variable oxidation states, acting as electron transfer agents.
  • Their surfaces can adsorb reactant molecules, weakening bonds (heterogeneous catalysis).
CatalystReactionProcess
FeN₂ + 3H₂ → 2NH₃Haber process
V₂O⁵2SO₂ + O₂ → 2SO₃Contact process
Pt or PdC=C + H₂ → C–CHydrogenation of alkenes
NiVegetable oils + H₂Hardening of fats (margarine)
MnO₂2H₂O₂ → 2H₂O + O₂Decomposition of H₂O₂
Fe³⁺Electron transfer reactionsBiological (enzymes)

4. Magnetic Properties

Paramagnetism: Substances with unpaired electrons are attracted to a magnetic field. All transition metals with unpaired d-electrons are paramagnetic.
Ferromagnetism: Fe, Co, and Ni are ferromagnetic — they can be permanently magnetised because of cooperative alignment of electron spins in domains.
Diamagnetism: Substances with all electrons paired are weakly repelled (Sc³⁺, Zn²⁺, Cu⁺ are diamagnetic).

5. Formation of Complex Ions

Transition metals form complex ions by accepting lone pairs from ligands (Lewis bases) into empty d/s/p orbitals. This is explained in detail in Section 1.4.

6. Physical Properties

Transition metals generally have high melting points, high densities, and are good conductors of heat and electricity. These arise from strong metallic bonding involving both 3d and 4s electrons.

MetalMP (°C)Density (g/cm³)Atomic radius (pm)
Sc15412.99162
Ti16684.51147
Cr19077.19128
Mn12467.43127
Fe15387.87126
Co14958.90125
Ni14558.91124
Cu10858.96128
Zn4207.13134

Note how atomic radius stays nearly constant across the series (128–162 pm) — the increasing nuclear charge is offset by increasing d-electron shielding.

1.3

Anomalous Properties of Zinc and Scandium

⚠️
Why Sc and Zn are not strictly transition metals By IUPAC definition: a transition metal forms at least one ion with an incomplete d sub-shell. Sc and Zn do not satisfy this — yet they are always studied alongside transition metals as they occupy the same d-block.

Scandium (Sc)

Sc only forms Sc³⁺ ions, which have configuration [Ar] — an empty d sub-shell. Therefore Sc has no partially filled d-shell in any of its ions.

Anomalies:

  • Sc³⁺ is colourless (no d-d transitions possible)
  • No catalytic activity through variable oxidation states
  • Not paramagnetic in its common ion
  • Only one oxidation state (+3)

Zinc (Zn)

Zn only forms Zn²⁺ ions, which have configuration [Ar] 3d¹⁰ — a completely filled d sub-shell. Therefore Zn has no partially filled d-shell in any of its ions.

Anomalies:

  • Zn²⁺ is colourless (full d — no d-d transitions)
  • Low melting point (420°C) compared to other transition metals
  • Less dense; softer
  • Only one oxidation state (+2)
  • Diamagnetic (all electrons paired)
  • Low reactivity but forms complexes (e.g. [Zn(NH₃)₄]²⁺)
1.4

Complex Ions and Isomerism

Complex Ion A complex ion consists of a central metal atom/ion bonded to a number of ligands by coordinate (dative) bonds. The ligands donate lone pairs into empty orbitals on the metal.

Coordination number = total number of coordinate bonds from ligands to the metal.

Ligands

Monodentate ligands — donate one lone pair per ligand:

H₂O (water) NH₃ (ammonia) Cl⁻ (chloride) CN⁻ (cyanide) OH⁻ (hydroxide) CO (carbon monoxide)

Bidentate ligands — donate two lone pairs per ligand:

en (ethylenediamine, H₂N–CH₂–CH₂–NH₂) ox²⁻ (oxalate, C₂O₄²⁻)

Polydentate (chelating) ligands — donate multiple lone pairs:

EDTA⁴⁻ (ethylenediaminetetraacetate) — hexadentate; forms very stable complexes

Naming Complex Ions

Rules: 1. Ligands named BEFORE the metal. 2. Anionic ligands: replace –ide with –o (e.g. chloride → chloro; hydroxide → hydroxo) Neutral ligands: use special names: H₂O = aqua; NH₃ = ammine; CO = carbonyl 3. Number of ligands: mono, di, tri, tetra, penta, hexa (or bis, tris, tetrakis for complex ligand names) 4. Metal name: in CATION or neutral → English name + oxidation state (Roman) in ANION → Latin root + –ate + oxidation state (Fe → ferrate; Cu → cuprate; Co → cobaltate; Cr → chromate; Mn → manganate) Examples: [Cu(H₂O)₆]²⁺ hexaaquacopper(II) [Fe(CN)₆]⁴⁻ hexacyanoferrate(II) [CoCl₂(NH₃)₄]⁺ tetraamminedichlorocobalt(III) [Cr(en)₃]³⁺ tris(ethylenediamine)chromium(III)

Shapes of Complex Ions

CNShapeExample
2Linear[Ag(NH₃)₂]⁺ (diamminesilver(I))
4Tetrahedral[CoCl₄]²⁻ (tetrachlorocobaltate(II))
4Square planar[Pt(NH₃)₂Cl₂] (cisplatin)
6Octahedral[Fe(H₂O)₆]²⁺ (hexaaquairon(II))
OCTAHEDRAL COMPLEX [M(L)₆] — Coordination Number 6 M L L L L L L Bond angles: 90° and 180° | M–L distances equal | Symmetrical arrangement

Isomerism in Complex Ions

Structural Isomerism

  • Ionisation isomers: [CoCl(NH₃)₅]SO₄ vs [Co(SO₄)(NH₃)₅]Cl — different ions in solution.
  • Hydrate isomers: [Cr(H₂O)₆]Cl₃ (violet) vs [CrCl(H₂O)₅]Cl₂·H₂O (pale green) — different number of water molecules in coordination sphere.
  • Linkage isomers: ambidentate ligands (e.g. NO₂⁻) can bond through N or O. [Co(NO₂)(NH₃)₅]²⁺ (nitro, N-bonded) vs [Co(ONO)(NH₃)₅]²⁺ (nitrito, O-bonded).

Geometric (Stereochemical) Isomerism

Occurs in square planar (4-coordinate) and octahedral (6-coordinate) complexes when the same ligands can be adjacent (cis) or opposite (trans):

Square planar example: [Pt(NH₃)₂Cl₂] cis (Cl atoms adjacent, 90°): cisplatin — used as anticancer drug trans (Cl atoms opposite, 180°): transplatin — no anticancer activity Octahedral example: [CoCl₂(NH₃)₄]⁺ cis: two Cl atoms at 90° to each other trans: two Cl atoms at 180° to each other (opposite vertices)

Optical Isomerism

Octahedral complexes with three bidentate ligands (e.g. [Cr(en)₃]³⁺ or [Co(ox)₃]³⁻) can exist as non-superimposable mirror images called enantiomers (labelled Λ and Δ). They rotate polarised light in opposite directions.

Example 1

Naming and Working Out Charges

(a) Name [Fe(H₂O)₅(OH)]²⁺. (b) What is the oxidation state of Co in [CoCl₄]²⁻? (c) What is the charge on [Ni(CN)₄]^n if Ni is +2?

a
Ligands: 5 aqua (H₂O) + 1 hydroxo (OH⁻). Metal: Fe, charge +2. Ion is cationic → use English name: pentaaquahydroxoiron(II) ion.
b
4×(−1) + Co = −2. Co + (−4) = −2. Co = +2.
c
Ni = +2; CN⁻ contributes 4×(−1) = −4. Total charge = +2 + (−4) = −2. Ion is [Ni(CN)₄]²⁻ (tetracyanonickelate(II)).
1.5

Chemistry of Individual Transition Metals

Iron (Fe)

Oxidation States and Reactions

Fe²⁺ (pale green): [Fe(H₂O)₆]²⁺ — oxidised easily to Fe³⁺ Fe³⁺ (yellow-brown): [Fe(H₂O)₆]³⁺ — stronger oxidant/Lewis acid Fe²⁺ → Fe³⁺: 4Fe(OH)₂ + O₂ + 2H₂O → 4Fe(OH)₃ (rusting mechanism) Fe²⁺ + ½Cl₂ → Fe³⁺ + Cl⁻ Fe³⁺ + I⁻ → Fe²⁺ + ½I₂ (I⁻ reduces Fe³⁺) Key complexes: [Fe(CN)₆]⁴⁻ — hexacyanoferrate(II), yellow solution [Fe(CN)₆]³⁻ — hexacyanoferrate(III), red-brown solution Fe³⁺ + SCN⁻ → [Fe(SCN)(H₂O)₅]²⁺ — blood red (test for Fe³⁺)

Copper (Cu)

Oxidation States and Reactions

Cu⁺ (colourless): unstable in aq. → disproportionates: 2Cu⁺ → Cu + Cu²⁺ Cu²⁺ (blue in water): [Cu(H₂O)₆]²⁺ Cu²⁺ + 4NH₃ → [Cu(NH₃)₄(H₂O)₂]²⁺ + 2H₂O (deep blue — test for Cu²⁺) Cu²⁺ + 2OH⁻ → Cu(OH)₂ (blue ppt) → CuO (black) on heating Cu is below H in reactivity series: Cu + 2H₂SO₄(conc.) → CuSO₄ + SO₂ + 2H₂O Cu + 4HNO₃(conc.) → Cu(NO₃)₂ + 2NO₂ + 2H₂O Cu¶Sn⁴ — bronze (alloy) Cu₂Zn — brass

Chromium (Cr)

Oxidation States and Reactions

Cr²⁺ (blue): strong reducing agent Cr³⁺ (green/violet): most stable; [Cr(H₂O)₆]³⁺ Cr⁶⁺: strong oxidant (toxic, carcinogenic) Key equilibrium (pH-dependent): CrO₄²⁻ (yellow, alkaline) ⇌ Cr₂O⁷²⁻ (orange, acidic) 2CrO₄²⁻ + 2H⁺ ⇌ Cr₂O⁷²⁻ + H₂O Reactions of Cr₂O⁷²⁻ as oxidant: Cr₂O⁷²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O (E° = +1.33 V) K₂Cr₂O⁷ oxidises alcohols: primary → carboxylic acid; secondary → ketone Amphoteric behaviour of Cr(OH)₃: Cr(OH)₃ + 3H⁺ → Cr³⁺ + 3H₂O (acid) Cr(OH)₃ + OH⁻ → [Cr(OH)₄]⁻ (base)

Manganese (Mn)

Oxidation States and Reactions

Mn shows most oxidation states of any transition metal: Mn²⁺ (pale pink): MnSO₄ — most stable in acidic solution Mn⁴⁺: MnO₂ (brown/black solid) — catalyst for H₂O₂ decomposition Mn⁷⁺: MnO₄⁻ (permanganate, purple) — powerful oxidant KMnO₄ reactions: In acid: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O (purple → colourless) In alkali: MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻ (purple → brown ppt) Uses: KMnO₄ as oxidant in volumetric analysis (self-indicating, Section 11) MnO₂ in dry cells (depolariser)
1.6

Identification of Transition Metal Ions

Reactions with NaOH(aq)

Adding NaOH solution to a transition metal salt solution typically forms a coloured metal hydroxide precipitate. Some are amphoteric and dissolve in excess NaOH.

IonPrecipitate + colourExcess NaOH
Fe²⁺Fe(OH)₂ — green pptInsoluble; oxidises in air to brown Fe(OH)₃
Fe³⁺Fe(OH)₃ — red-brown pptInsoluble
Cu²⁺Cu(OH)₂ — blue pptInsoluble
Cr³⁺Cr(OH)₃ — green pptDissolves (amphoteric): [Cr(OH)₄]⁻ (dark green soln)
Co²⁺Co(OH)₂ — pink/blue pptInsoluble
Ni²⁺Ni(OH)₂ — green pptInsoluble
Mn²⁺Mn(OH)₂ — white/pale pink pptInsoluble; oxidises in air → MnO₂ (brown)
Zn²⁺Zn(OH)₂ — white pptDissolves: [Zn(OH)₄]²⁻ (colourless soln)

Reactions with NH₃(aq)

IonWith excess NH₃Result
Cu²⁺Dissolves in excess[Cu(NH₃)₄(H₂O)₂]²⁺ — deep blue soln
Co²⁺Dissolves in excess[Co(NH₃)₆]²⁺ — yellow/brown
Ni²⁺Dissolves in excess[Ni(NH₃)₆]²⁺ — blue/violet
Cr³⁺Dissolves in excess[Cr(NH₃)₆]³⁺ — purple
Fe²⁺, Fe³⁺Insoluble in excessHydroxide ppt remains
Zn²⁺Dissolves in excess[Zn(NH₃)₄]²⁺ — colourless

Other Specific Tests

TestReagentObservationIon confirmed
SCN⁻ testKSCN(aq)Blood-red solution [Fe(SCN)(H₂O)₅]²⁺Fe³⁺
Prussian blueK₃[Fe(CN)₆] addedDark blue/Prussian blue pptFe²⁺
Turnbull blueK₄[Fe(CN)₆] addedDark blue pptFe³⁺
Deep blueExcess NH₃[Cu(NH₃)₄]²⁺ deep blueCu²⁺
Lilac/purpleAcidify + KMnO₄Decolourises KMnO₄Reducing agents (Fe²⁺, Mn²⁺)
Example 2

Ion Identification Flow

An aqueous solution gives a blue precipitate with NaOH that does not dissolve in excess NaOH but dissolves in excess NH₃ to give a deep blue solution. Identify the metal ion.

1
Blue ppt with NaOH → Cu(OH)₂: consistent with Cu²⁺.
2
Insoluble in excess NaOH: rules out Cr³⁺, Zn²⁺ (which would dissolve). Still consistent with Cu²⁺.
3
Dissolves in excess NH₃ → deep blue [Cu(NH₃)₄(H₂O)₂]²⁺: confirms Cu²⁺.
✏️

Exercises

🧠

Interactive Quiz

Unit 1 Quiz — Transition Metals

25 Questions
Q1

A transition metal is defined as an element that:

The IUPAC definition: forms at least one stable ion with an incompletely filled d sub-shell. This excludes Sc (Sc³⁺ has empty d) and Zn (Zn²⁺ has full d). Having d electrons or variable states alone is not sufficient.
Q2

The electronic configuration of Cr is:

Cr is [Ar] 3d⁵ 4s¹ — an exception to the Aufbau principle. A half-filled 3d sub-shell (3d⁵) gives extra stability, so one 4s electron is promoted to give 3d⁵ 4s¹ rather than 3d⁴ 4s².
Q3

The colour of Fe³⁺ in aqueous solution is:

Fe³⁺ is yellow-brown in solution as [Fe(H₂O)₆]³⁺. Pale green is Fe²⁺; blue is Cu²⁺; purple is MnO₄⁻ or Ti³⁺. The d⁵ configuration of Fe³⁺ allows d-d transitions absorbing in the blue/green region.
Q4

Zn²⁺ ions are colourless because:

Zn²⁺ has [Ar] 3d¹⁰ — all 10 d-electrons fill all 5 d-orbitals. There is no vacancy for an excited electron to jump into (no empty d orbital), so no d-d absorption of visible light occurs → colourless.
Q5

The coordination number of the complex [CoCl₄]²⁻ is:

Coordination number = total number of coordinate bonds from ligands to the metal. [CoCl₄]²⁻ has 4 Cl⁻ ligands, each donating one lone pair → CN = 4. Shape is tetrahedral.
Q6

Which catalyst is used in the Contact process (manufacture of H₂SO₄)?

V₂O⁵ (vanadium pentoxide) catalyses 2SO₂ + O₂ → 2SO₃ in the Contact process. Fe is the Haber process catalyst; Ni is used for hydrogenation of oils; Pt is used in catalytic converters and Ostwald process.
Q7

The oxidation state of Mn in KMnO₄ is:

K = +1; each O = −2; so: +1 + Mn + 4(−2) = 0 → Mn = +7. KMnO₄ is potassium permanganate — Mn is in its highest common oxidation state of +7.
Q8

Adding excess NH₃(aq) to Cu²⁺ solution gives:

Cu(OH)₂ blue ppt first forms, then dissolves in excess NH₃ to form the tetraamminecopper(II) complex [Cu(NH₃)₄(H₂O)₂]²⁺ — deep blue/royal blue. This is the characteristic test for Cu²⁺.
Q9

Cisplatin is an example of which type of isomerism?

Cisplatin [Pt(NH₃)₂Cl₂] is the cis isomer — Cl atoms at 90° to each other in a square planar complex. The trans isomer has Cl atoms at 180°. This is geometric isomerism. Cisplatin is an anticancer drug; transplatin has no anticancer activity.
Q10

The test reagent that gives a blood-red colour with Fe³⁺ is:

Fe³⁺ + SCN⁻ → [Fe(SCN)(H₂O)₅]²⁺ — blood-red. KSCN (potassium thiocyanate) is the confirmatory test for Fe³⁺. NaOH gives a red-brown ppt; NH₃ also gives a ppt; KI reduces Fe³⁺ to Fe²⁺.
Q11

The electronic configuration of Fe³⁺ is:

Fe: [Ar] 3d⁶ 4s². Forming Fe³⁺: lose 4s² first → [Ar] 3d⁶ then lose one 3d → [Ar] 3d⁵. This is the half-filled d⁵ configuration — same as Mn²⁺. This partly explains the stability of Fe³⁺.
Q12

The chromate/dichromate equilibrium CrO₄²⁻ ⇌ Cr₂O⁷²⁻ shifts to the right when:

2CrO₄²⁻ + 2H⁺ ⇌ Cr₂O⁷²⁻ + H₂O. Adding acid (H⁺) drives equilibrium to the right → orange Cr₂O⁷²⁻ forms. Adding alkali (OH⁻) drives it left → yellow CrO₄²⁻ forms. Colour is therefore pH-dependent.
Q13

Transition metals are good catalysts partly because they:

The key catalytic feature is variable oxidation states — the metal can be oxidised and then reduced back in a catalytic cycle, transferring electrons between reactants. Additionally, transition metal surfaces can adsorb reactants (heterogeneous catalysis), providing an alternative lower-energy pathway.
Q14

Which of the following gives a white precipitate that dissolves in excess NaOH?

Zn²⁺ + 2OH⁻ → Zn(OH)₂↓ (white ppt). Excess NaOH: Zn(OH)₂ + 2OH⁻ → [Zn(OH)₄]²⁻ (dissolves). Zn(OH)₂ is amphoteric. Cr³⁺ gives green ppt that also dissolves. Fe²⁺ green ppt stays; Cu²⁺ blue ppt stays; Mn²⁺ pale ppt stays.
Q15

The shape of [Fe(H₂O)₆]²⁺ is:

CN = 6 → octahedral. Six H₂O ligands point to the six corners of an octahedron around the central Fe²⁺ ion. All bond angles are 90° (adjacent) or 180° (opposite). This is the most common coordination geometry for transition metal ions in water.
Q16

Why does atomic radius change very little across the first-row transition metals (Sc to Zn)?

As Z increases across the series, nuclear charge increases (pulls electrons in) but each new electron enters the 3d sub-shell, which shields effectively. The two effects nearly cancel, so atomic radius changes little (roughly 162 pm at Sc to 124–134 pm at Ni/Cu/Zn).
Q17

An optical isomer of a complex ion is its non-superimposable mirror image. This occurs for:

[Cr(en)₃]³⁺ has three bidentate (en) ligands arranged octahedrally, giving a chiral (propeller-like) structure that exists as Λ and Δ enantiomers — non-superimposable mirror images. Square planar, tetrahedral with all-same ligands, and linear complexes do not show optical isomerism.
Q18

EDTA is described as a hexadentate ligand. This means it:

Hexadentate means 6 donor atoms — EDTA⁴⁻ has 2 N atoms and 4 O atoms that each donate a lone pair to the metal, forming 6 coordinate bonds from a single ligand. This makes EDTA complexes very stable (chelate effect — entropy gain from releasing 6 water molecules).
Q19

Prussian blue is formed when K₃[Fe(CN)₆] (potassium hexacyanoferrate(III)) is added to a solution containing:

K₃[Fe(CN)₆] + Fe²⁺ → Prussian blue (dark blue precipitate). This is the test for Fe²⁺. Conversely, K₄[Fe(CN)₆] + Fe³⁺ also gives Prussian blue (Turnbull's blue). The KSCN test is more selective for Fe³⁺.
Q20

The disproportionation of Cu⁺ in water gives:

Disproportionation: 2Cu⁺(aq) → Cu(s) + Cu²⁺(aq). One Cu⁺ is oxidised to Cu²⁺; another is reduced to Cu(0). Cu⁺ is unstable in aqueous solution and spontaneously disproportionates, which is why Cu²⁺ is much more common than Cu⁺ in water.
Q21

Which transition metal complex is used as an anticancer drug?

Cisplatin cis-[Pt(NH₃)₂Cl₂] is used to treat cancers (e.g. testicular, ovarian). It works by binding to DNA, preventing replication of cancer cells. The trans isomer has no anticancer activity — demonstrating how geometry affects biological function.
Q22

When KMnO₄ reacts in acidic solution, the half-equation is: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O. The change in oxidation state of Mn is:

In KMnO₄, Mn is +7. In MnSO₄ (Mn²⁺), Mn is +2. Change = +7 → +2 = decrease of 5, consistent with gaining 5 electrons (5e⁻ in the half-equation). The solution turns from purple (MnO₄⁻) to nearly colourless (Mn²⁺).
Q23

The correct IUPAC name for [Ag(NH₃)₂]⁺ is:

IUPAC: ligands first, then metal. NH₃ is the neutral ligand named ammine (double m). Two of them = diammine. Metal Ag in cation = silver with oxidation state (I). Full name: diamminesilver(I). This complex is Tollens' reagent.
Q24

Why is Fe ferromagnetic but Cu is not?

Fe has 4 unpaired 3d electrons per atom. In iron metal, these spins align cooperatively in magnetic domains, creating ferromagnetism. Cu has [Ar] 3d¹⁰ 4s¹ — the 3d is full, and the 4s electron is delocalized in the metal. No unpaired d-electrons → no ferromagnetism. Cu is weakly diamagnetic.
Q25

A solution of Cr³⁺ gives a green precipitate with NaOH. On adding excess NaOH, the precipitate dissolves. On then adding H₂SO₄(aq), a green precipitate re-forms. This sequence demonstrates:

Cr(OH)₃ dissolves in excess NaOH: Cr(OH)₃ + OH⁻ → [Cr(OH)₄]⁻ (acts as acid). Re-adding H₂SO₄ reprotonates and re-precipitates Cr(OH)₃. This sequence demonstrates that Cr(OH)₃ is amphoteric — it reacts with both acids and bases.
📝

Unit Test

ℹ️
Instructions Total: 50 marks  |  Time: 55 minutes  |  Show all working and write ionic equations where appropriate.

Section A — Short Answer

30 marks
Q1 [5 marks]

(a) Write the full electronic configuration of Cr and Cu, explaining why each is an exception to the Aufbau principle. [3]
(b) Write the configuration of Fe²⁺ and Fe³⁺. Explain which is lost first when Fe is ionised. [2]

(a) Cr: [Ar] 3d⁵ 4s¹ (NOT 3d⁴ 4s²). A half-filled 3d sub-shell (5 electrons in 5 orbitals, each singly occupied) has extra stability through exchange energy. One 4s electron is promoted to 3d to achieve this.
Cu: [Ar] 3d¹⁰ 4s¹ (NOT 3d⁹ 4s²). A completely filled 3d sub-shell also has extra stability. Again, one 4s electron is promoted to complete the 3d.
(b) Fe: [Ar] 3d⁶ 4s². The 4s sub-shell is lost first when forming cations (4s is higher in energy in multi-electron atoms than 3d once core charge builds up).
Fe²⁺: [Ar] 3d⁶ (both 4s electrons removed).
Fe³⁺: [Ar] 3d⁵ (additionally one 3d electron removed).
Q2 [5 marks]

(a) Name the complex [CoCl₂(en)₂]⁺. State its coordination number and oxidation state of Co. [3]
(b) Draw the two geometric isomers of [CoCl₂(en)₂]⁺ and state which shows optical isomerism. [2]

(a) en = ethylenediamine (bidentate, counts as 2 coordinate bonds each). Ligands: 2×Cl⁻ (dichloro) + 2×en (bis(ethylenediamine)). Metal: Co, positive charge complex = +1 overall. Charge: 2(−1) + Co + 0 + 0 = +1 → Co = +3.
Name: dichloridobis(ethylenediamine)cobalt(III). CN = 6 (2×Cl + 2×2 from en). Oxidation state = +3.
(b) cis isomer: both Cl atoms adjacent (90°) — this cis isomer has a chiral axis and exists as Λ and Δ optical isomers.
trans isomer: Cl atoms opposite (180°) — has a plane of symmetry; no optical isomers.
Q3 [5 marks]

Give the observations and write ionic equations for the reactions of the following with NaOH(aq) and then with excess NaOH: (a) FeSO₄ solution [2] (b) CrCl₃ solution [2] (c) ZnSO₄ solution [1]

(a) FeSO₄: Fe²⁺(aq) + 2OH⁻(aq) → Fe(OH)₂(s)↓ — green precipitate. Insoluble in excess NaOH. In air, slowly turns red-brown (oxidised to Fe(OH)₃).

(b) CrCl₃: Cr³⁺(aq) + 3OH⁻(aq) → Cr(OH)₃(s)↓ — green precipitate. In excess NaOH: Cr(OH)₃(s) + OH⁻(aq) → [Cr(OH)₄]⁻(aq) — dark green solution (amphoteric).

(c) ZnSO₄: Zn²⁺(aq) + 2OH⁻(aq) → Zn(OH)₂(s)↓ — white precipitate. In excess NaOH: Zn(OH)₂(s) + 2OH⁻(aq) → [Zn(OH)₄]²⁻(aq) — colourless solution (amphoteric).
Q4 [5 marks]

Manganese exhibits oxidation states +2, +4, and +7 in its common compounds.
(a) Give one compound/ion for each oxidation state and state its colour. [3]
(b) Write the half-equation for the reduction of MnO₄⁻ in acidic solution and in alkaline solution. State the product in each case. [2]

(a) +2: MnSO₄ (Mn²⁺ in solution) — very pale pink
+4: MnO₂ — black/dark brown solid
+7: KMnO₄ (MnO₄⁻) — purple/violet solution

(b) In acidic solution: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O. Product: Mn²⁺ (pale pink/colourless). Solution decolourises from purple.
In alkaline solution: MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻. Product: MnO₂ (brown precipitate). Purple solution forms brown ppt.
Q5 [5 marks]

Explain why transition metals are good catalysts with reference to BOTH: (a) their variable oxidation states (give an example) [2]; (b) the properties of their surfaces (heterogeneous catalysis) [3].

(a) Transition metals can change oxidation state by accepting/donating electrons, allowing them to act as electron transfer intermediates. Example: Fe³⁺ + e⁻ ↔ Fe²⁺. In the reaction between I⁻ and S₂O⁸²⁻ (thiosulfate), Fe³⁺ first oxidises I⁻ to I₂ (Fe³⁺ → Fe²⁺), then Fe²⁺ reduces S₂O⁸²⁻ back (Fe²⁺ → Fe³⁺). Overall rate is much faster via this two-step route than the direct (very slow) reaction.

(b) Heterogeneous catalysis (e.g. Fe in Haber process, Ni in hydrogenation): reactant molecules are adsorbed onto the catalyst surface. This:
• Weakens bonds within the adsorbed molecules (lower activation energy)
• Holds reactants close together in the correct orientation
• Provides an alternative reaction pathway with lower E₂₃
• Products are then desorbed from the surface, regenerating the catalyst
The partially filled d-orbitals of transition metal surfaces form suitable bonds with adsorbed molecules — neither too strong (which would poison the catalyst) nor too weak (which would not activate the molecules).
Q6 [5 marks]

Describe a chemical test to distinguish between each of the following pairs. In each case state the reagent used, the observation with each ion, and write ionic equations.
(a) Fe²⁺ from Fe³⁺ [3]
(b) Cu²⁺ from Zn²⁺ [2]

(a) Reagent: KSCN(aq).
Fe²⁺: no significant red colour formed (very pale or no colour).
Fe³⁺: blood-red colour → [Fe(SCN)(H₂O)₅]²⁺ formed.
Equation: Fe³⁺(aq) + SCN⁻(aq) → [Fe(SCN)]²⁺(aq) (blood-red)

Alternatively: use NaOH(aq).
Fe²⁺: green precipitate Fe(OH)₂↓; Fe³⁺: red-brown precipitate Fe(OH)₃↓.

(b) Reagent: excess NH₃(aq).
Cu²⁺: pale blue precipitate Cu(OH)₂ forms first, then dissolves in excess NH₃ to give deep blue [Cu(NH₃)₄(H₂O)₂]²⁺.
Zn²⁺: white precipitate Zn(OH)₂ forms, dissolves in excess NH₃ to give colourless [Zn(NH₃)₄]²⁺.
The deep blue colour of the Cu²⁺ ammonia complex distinguishes them.

Section B — Extended Response

20 marks
Q7 [10 marks]

(a) State five characteristic properties of transition metals and explain the electronic origin of each property. [5 marks]

(b) Explain why Sc and Zn are anomalous members of the d-block in terms of the IUPAC definition. Compare their properties with those of a typical transition metal (e.g. Fe or Cu), addressing: colour of ions, number of oxidation states, catalytic activity, and magnetic properties. [5 marks]

(a) Five characteristic properties:
1. Variable oxidation states: arise because the energy difference between 3d and 4s electrons is small, so different numbers of electrons can be involved in bonding. E.g. Fe: +2 (loses 4s²) and +3 (loses 4s² + one 3d).
2. Coloured ions: partially filled d-orbitals split in energy in a ligand field. Electrons absorb visible light photons to make d-d transitions to higher d-orbitals. The complementary colour is observed. E.g. Cu²⁺ absorbs red → appears blue.
3. Catalytic activity: variable oxidation states allow electron transfer; d-orbital surfaces can adsorb reactants. E.g. V₂O⁵ in Contact process cycles between V⁵⁺ and V⁴⁺.
4. Complex ion formation: available d, s, p orbitals accept lone pairs from ligands. High charge density of small ions attracts ligands. E.g. [Fe(H₂O)₆]²⁺.
5. Paramagnetic/ferromagnetic behaviour: unpaired d-electrons create magnetic moments. Fe, Co, Ni show ferromagnetism due to cooperative domain alignment.

(b) Sc and Zn anomalies:
IUPAC definition: a transition metal must form at least one stable ion with an incompletely filled d sub-shell.
Sc³⁺ = [Ar] (empty d); Zn²⁺ = [Ar] 3d¹⁰ (full d). Neither satisfies the definition.

Comparison with Fe/Cu:
• Colour: Sc³⁺ and Zn²⁺ are colourless (no d-d transitions possible: no vacancy/no electrons to promote). Fe²⁺ is pale green; Cu²⁺ is blue.
• Oxidation states: Sc only has +3; Zn only has +2. Very limited. Fe has +2, +3; Cu has +1, +2; Mn has +2 to +7.
• Catalytic activity: Sc and Zn cannot cycle between oxidation states → no catalytic activity via electron transfer. Fe (Haber) and V (Contact) are important industrial catalysts.
• Magnetic properties: Sc³⁺ (empty d, diamagnetic) and Zn²⁺ (full d, all paired, diamagnetic). Fe has 4 unpaired d-electrons → paramagnetic and ferromagnetic in the metal. Cu (metal, 3d¹⁰ 4s¹) has one unpaired electron → weakly paramagnetic.
Despite not fitting the IUPAC definition, Sc and Zn are always studied with transition metals as they complete the d-block and have some similar physical properties (metallic conductors, hard, relatively high MP except Zn).
Q8 [10 marks]

(a) Describe the types of isomerism that can occur in transition metal complexes. Give a named example of each type and state its significance. [6 marks]

(b) The complex [CoCl₂(NH₃)₄]⁺ exists as two geometric isomers. (i) State their names and sketch the two structures, labelling the Co and all ligands. (ii) State which isomer could also exhibit optical isomerism and explain why. [4 marks]

(a) Types of isomerism:
1. Geometric (cis-trans) isomerism: occurs in square planar (4-coordinate) and octahedral complexes when identical ligands can be adjacent (cis, 90°) or opposite (trans, 180°). Example: [Pt(NH₃)₂Cl₂] — cisplatin (cis) is an anticancer drug; transplatin is inactive, showing that geometry determines biological activity.

2. Optical isomerism: non-superimposable mirror image (enantiomer) pairs. Occur in octahedral complexes with 3 bidentate ligands or 2 bidentate + 2 different monodentate ligands in cis arrangement. Example: [Cr(en)₃]³⁺ exists as Λ and Δ forms; the two enantiomers rotate plane-polarised light in opposite directions. Significance: drug chirality (enantiomers can have different pharmacological effects).

3. Linkage isomerism: ambidentate ligands (e.g. NO₂⁻, SCN⁻) can bond through different atoms. Example: [Co(NO₂)(NH₃)₅]²⁺ (nitro, N-bonded) vs [Co(ONO)(NH₃)₅]²⁺ (nitrito, O-bonded). Different spectroscopic and chemical properties.

4. Ionisation isomerism: different ions in solution due to ligands swapping in/out of coordination sphere. Example: [CoCl(NH₃)₅]SO₄ (gives SO₄²⁻ ions in solution) vs [Co(SO₄)(NH₃)₅]Cl (gives Cl⁻ ions). Different conductivities in solution.

5. Hydrate isomerism: water molecules inside vs outside the coordination sphere. Example: [Cr(H₂O)₆]Cl₃ (violet) vs [CrCl(H₂O)₅]Cl₂·H₂O (pale green). Different colour and number of ions in solution.

(b) [CoCl₂(NH₃)₄]⁺ is octahedral (CN = 6: 2 Cl + 4 NH₃).
(i) Two geometric isomers:
cis-[CoCl₂(NH₃)₄]⁺: the two Cl⁻ ligands are 90° apart (adjacent). Violet/purple colour.
trans-[CoCl₂(NH₃)₄]⁺: the two Cl⁻ ligands are 180° apart (opposite). Green colour.
[Sketches: octahedral diagrams with Co at centre, showing Cl and NH₃ positions]

(ii) The cis isomer can show optical isomerism. In the cis isomer, the two Cl⁻ ligands and the four NH₃ ligands create an arrangement with no plane of symmetry — the molecule has a non-superimposable mirror image. The two non-superimposable forms (enantiomers) rotate plane-polarised light in opposite directions. The trans isomer has a C₂ axis and planes of symmetry → superimposable on its mirror image → no optical isomers.