Chemistry (9701) | General characteristic chemical properties of the first set of transition elements, titanium to copper

A-Level Chemistry 9701: Transition Elements (Ti to Cu) - Study Notes

Hello future Chemists! This chapter takes us away from the predictable world of Group 1 and 2 elements and into the fascinating realm of the d-block. Transition elements are often called the "chameleons of Chemistry" because they display amazing colours, variable behaviours, and act as brilliant catalysts. Don't worry if these concepts seem complex—we'll break down the key characteristics of the first row (Titanium to Copper) into easy, manageable steps!

Why is this important? Transition metal chemistry is essential for industry (think catalysts in the Haber process) and biology (iron in haemoglobin). Mastering these characteristics is key to achieving top grades in your Inorganic Chemistry papers.

1. Defining the Transition Elements

1.1 The Formal Definition (Syllabus 28.1.1)

A transition element is defined as a d-block element which forms one or more stable ions with incomplete d orbitals (i.e., $d^1$ to $d^9$).

• Crucial Note: Zinc (Zn) is in the d-block, but it is not considered a transition element because its only stable ion, $Zn^{2+}$, has the electronic configuration $[Ar] 3d^{10}$. The d orbital is complete.
• Similarly, Scandium (Sc) is often excluded as its common ion, $Sc^{3+}$, has configuration $[Ar] 3d^0$.

1.2 Electronic Structure: Why They Are Special (Syllabus 28.1.4)

The unique chemistry of transition elements stems from the small energy difference between the 3d and 4s sub-shells.

• During ionisation (forming cations), electrons are always removed from the 4s sub-shell first, even though the 3d is filled first when writing the ground state configuration.
• Example: Iron (Fe) is $[Ar] 3d^6 4s^2$.
  $Fe^{2+}$ is $[Ar] 3d^6$. (Two 4s electrons removed).

Key Takeaway: The similarity in energy between the 3d and 4s sub-shells allows transition elements to lose a variable number of electrons easily, leading directly to their characteristic properties.

2. The General Characteristic Properties (The Big Four)

Transition metals are characterised by four main properties (Syllabus 28.1.3):

1. Variable Oxidation States
2. Catalytic Behaviour
3. Formation of Complex Ions
4. Formation of Coloured Compounds

2.1 Variable Oxidation States (Syllabus 28.1.3a, 28.1.4)

Most transition metals can exist in several different stable oxidation states.

• The Explanation: Because the 3d and 4s orbitals are very similar in energy, the atom can lose electrons not just from the 4s orbital, but also a varying number from the 3d orbital without requiring massive amounts of energy.
• Analogy: Think of 4s and 3d orbitals as two drawers in a filing cabinet placed right next to each other. It’s almost as easy to grab files (electrons) from the 3d drawer as it is from the 4s drawer.
• Example: Vanadium has oxidation states +2, +3, +4, and +5.

2.2 Catalytic Behaviour (Syllabus 28.1.3b, 28.1.5)

Transition metals and their compounds are excellent catalysts, both in homogeneous and heterogeneous reactions.

• The Explanation: Catalytic activity is linked to:
  1. Having multiple stable oxidation states: The metal can cycle through these states, providing an alternative reaction pathway with a lower activation energy ($E_a$).
     Example (Homogeneous): $Fe^{2+}$ ions catalysing the reaction between $\mathrm{S_2O_8^{2-}}$ and $\mathrm{I^-}$.
  2. Having vacant d orbitals: This allows them to form temporary dative bonds with reactant molecules (ligands), concentrating the reactants on the surface (in heterogeneous catalysis) or activating them (in homogeneous catalysis).

3. Complex Ions and Ligands (Syllabus 28.2.1-28.2.7)

3.1 Definitions (Syllabus 28.2.2, 28.2.4, 28.2.6a)

• Complex Ion (or Complex): A molecule or ion formed by a central metal atom/ion surrounded by one or more ligands.
• Ligand: A species (ion or molecule) that contains a lone pair of electrons that forms a dative covalent bond (coordinate bond) to a central metal atom or ion.
• Coordination Number: The total number of dative bonds formed between the ligands and the central metal ion.

Analogy: The central metal ion is like a busy person with empty hands (vacant orbitals), and the ligands are friendly people offering food (lone pair of electrons) to be held via dative bonds.

3.2 Types of Ligands (Syllabus 28.2.3)

Ligands are classified by the number of lone pairs they can donate:

• Monodentate: Forms one dative bond. Examples: Water ($\mathrm{H_2O}$), Ammonia ($\mathrm{NH_3}$), Chloride ($\mathrm{Cl^-}$), Cyanide ($\mathrm{CN^-}$).
• Bidentate: Forms two dative bonds (using two lone pairs from different atoms within the same molecule). Examples: 1,2-diaminoethane (en, $\mathrm{H_2NCH_2CH_2NH_2}$), ethanedioate ion ($\mathrm{C_2O_4^{2-}}$).
• Polydentate: Forms multiple dative bonds (more than two). Example: $\mathrm{EDTA^{4-}}$ (forms six dative bonds, making it hexadentate).

3.3 Shapes (Geometry) of Complexes (Syllabus 28.2.5)

The shape depends primarily on the coordination number (C.N.).

• C.N. = 4:
  • Tetrahedral (109.5°): Common for complexes with large ligands like $\mathrm{Cl^-}$, e.g., $[\mathrm{CuCl_4}]^{2-}$
  • Square Planar (90°): Common for $d^8$ ions (like $Pt^{2+}$ or $Ni^{2+}$), e.g., $[\mathrm{Pt(NH_3)_2Cl_2}]$ (cisplatin).
• C.N. = 6:
  • Octahedral (90°): Most common, especially with small ligands like $\mathrm{H_2O}$ or $\mathrm{NH_3}$. E.g., $[\mathrm{Cu(H_2O)_6}]^{2+}$.
• C.N. = 2:
  • Linear (180°): Rare, often seen with $Ag^+$ ions.

3.4 Ligand Exchange Reactions (Syllabus 28.2.7)

Ligands in a complex can be replaced by other ligands. This often results in a colour change, making these reactions very easy to observe.

We focus on Copper(II) and Cobalt(II) ions:

1. Copper(II) Ions: Start as pale blue $[\mathrm{Cu(H_2O)_6}]^{2+}$.
   • With Ammonia ($\mathrm{NH_3}$): Water ligands are replaced by ammonia.
     $ \mathrm{[Cu(H_2O)_6]^{2+}} + \mathrm{4NH_3} \rightleftharpoons \mathrm{[Cu(NH_3)_4(H_2O)_2]^{2+}} + \mathrm{4H_2O} $
     Observation: Pale blue solution (or precipitate) changes to a dark blue solution.
   • With Concentrated Chloride ($\mathrm{Cl^-}$): The small $\mathrm{H_2O}$ ligands are replaced by the much larger $\mathrm{Cl^-}$ ions, changing the coordination number from 6 to 4 and the shape from octahedral to tetrahedral.
     $ \mathrm{[Cu(H_2O)_6]^{2+}} + \mathrm{4Cl^-} \rightleftharpoons \mathrm{[CuCl_4]^{2-}} + \mathrm{6H_2O} $
     Observation: Blue solution changes to a yellow/green solution.
2. Cobalt(II) Ions: Start as pink $[\mathrm{Co(H_2O)_6}]^{2+}$.
   • With Concentrated Chloride ($\mathrm{Cl^-}$):
     $ \mathrm{[Co(H_2O)_6]^{2+}} + \mathrm{4Cl^-} \rightleftharpoons \mathrm{[CoCl_4]^{2-}} + \mathrm{6H_2O} $
     Observation: Pink solution changes to a deep blue solution (tetrahedral complex).

Key Takeaway: Ligand exchange is an equilibrium process, often driven by the concentration of the new ligand, and is usually accompanied by a dramatic colour change.

4. Colour of Complexes (Syllabus 28.1.3d, 28.3)

4.1 The Role of d-Orbitals (Syllabus 28.3.1, 28.3.2)

Transition metal ions are colourful because they have partially filled d orbitals ($d^1$ to $d^9$).

In an isolated, gaseous transition metal ion, all five d orbitals are of equal energy—they are degenerate.

When ligands approach the central ion (forming a complex), they repel the electrons in the d orbitals. This repulsion causes the five d orbitals to split into two sets of non-degenerate orbitals:

• Octahedral Complexes (C.N. 6): The d orbitals split into three lower-energy orbitals and two higher-energy orbitals.
• Tetrahedral Complexes (C.N. 4): The d orbitals split into two lower-energy orbitals and three higher-energy orbitals.

4.2 How Colour is Produced (Syllabus 28.3.3)

Colour arises because electrons in the lower energy d orbitals can be promoted to the higher energy d orbitals ($\Delta E$) by absorbing a specific frequency of visible light.

1. The complex absorbs a specific colour/frequency of light ($\nu$). $$ \Delta E = h \nu $$
2. The remaining frequencies of light are transmitted or reflected.
3. The colour observed is the complementary colour to the colour absorbed.

Example: If a complex absorbs red light, we observe green light (the complementary colour).

4.3 Factors Affecting Colour (Syllabus 28.3.4, 28.3.5)

The magnitude of the energy gap, $\Delta E$, determines the frequency of light absorbed, and thus the colour observed.

• Nature of the Ligand: Different ligands cause different amounts of splitting ($\Delta E$). Stronger ligands (e.g., $\mathrm{CN^-}$, $\mathrm{NH_3}$) cause a larger $\Delta E$, meaning they absorb higher energy light (shorter wavelength, often blue/violet) and transmit lower energy light (red/yellow). We see a shift in colour when ligands are exchanged (e.g., from pale blue $\mathrm{H_2O}$ complexes to deep blue $\mathrm{NH_3}$ complexes).
• Oxidation State of the Metal: Higher oxidation states generally lead to greater splitting ($\Delta E$).
• Coordination Number/Geometry: Octahedral complexes usually show larger splitting than tetrahedral complexes.

Common Mistake to Avoid: Don't say the complex is the colour it absorbs! It is the complementary colour (what is reflected/transmitted).

5. Stability and Stereoisomerism

5.1 Stability Constants ($K_{stab}$) (Syllabus 28.5)

$K_{stab}$ measures the position of equilibrium for the formation of a complex ion in a solvent. It essentially tells us how stable the complex is.

For the overall reaction: $$ \mathrm{M^{m+}} + \mathrm{nL} \rightleftharpoons \mathrm{[ML_n]^{m+}} $$ The stability constant $K_{stab}$ is: $$ K_{stab} = \frac{\mathrm{[ML_n]^{m+}}}{\mathrm{[M^{m+}] [L]^n}} $$

• Interpretation: A large numerical value of $K_{stab}$ means the equilibrium lies far to the right, indicating a very stable complex ion.
• Ligand Exchange & $K_{stab}$: Ligand exchange reactions occur when a new complex can be formed that has a higher $K_{stab}$ than the original complex.

5.2 The Chelate Effect

When polydentate (chelating) ligands replace monodentate ligands, the resulting complex is usually much more stable (much larger $K_{stab}$). This phenomenon is the chelate effect.

• Explanation: When a monodentate ligand (like $\mathrm{H_2O}$) is replaced by a bidentate ligand (like 'en'), the total number of particles (molecules/ions) increases when the reaction proceeds forward. This increase in the number of particles leads to a large positive entropy change ($\Delta S$), making the reaction much more thermodynamically feasible ($\Delta G = \Delta H - T\Delta S$).
• Analogy: Trying to replace 6 single marbles (monodentate ligands) with 3 double-marbles (bidentate ligands) is easier and more orderly than doing the reverse, increasing the randomness (entropy).

5.3 Stereoisomerism (Syllabus 28.4)

Complexes can exhibit stereoisomerism (different spatial arrangements). We look at two types:

A. Geometrical (cis/trans) Isomerism

This occurs when ligands can be arranged differently around the central ion, particularly in Square Planar and Octahedral complexes.

• cis-isomer: Identical ligands are adjacent (side-by-side). Example: cis-$[\mathrm{Pt(NH_3)_2Cl_2}]$ (Cisplatin, used in chemotherapy).
• trans-isomer: Identical ligands are opposite (180° apart).
• Polarity: The cis-isomer is often polar, while the trans-isomer is usually non-polar (or less polar) because the dipoles cancel out.

B. Optical Isomerism

This occurs when the complex ion cannot be superimposed on its mirror image (it is chiral). This requires the use of bidentate ligands in octahedral complexes.

• Example: $[\mathrm{Ni(H_2NCH_2CH_2NH_2)_3}]^{2+}$ (the tris(ethylenediamine)nickel(II) ion). The two non-superimposable mirror images rotate plane-polarised light in opposite directions.

Key Takeaway: The large $K_{stab}$ seen in polydentate ligands is due to a favourable increase in entropy ($\Delta S$). Stereoisomers are physically distinct forms, affecting properties like polarity and biological activity.

6. Redox Reactions of Transition Metals (Syllabus 28.2.8, 28.2.9)

Since transition elements have variable oxidation states, they are excellent oxidising and reducing agents. We can predict the feasibility of these reactions using standard electrode potentials ($E^\circ$ values).

6.1 Predicting Feasibility (Syllabus 28.2.8)

A reaction is feasible if the Standard Cell Potential ($E^\circ_{cell}$) is positive. $$ E^\circ_{cell} = E^\circ_{\text{reduction reaction}} - E^\circ_{\text{oxidation reaction}} $$

• Rule Reminder: The species with the more positive $E^\circ$ value will undergo reduction (act as the oxidising agent).

6.2 Essential Redox Examples (Syllabus 28.2.9)

You must be able to describe and perform calculations involving the following reactions in acid solution:

A. Permanganate ($\mathrm{MnO_4^-}$) / Ethanedioate ($\mathrm{C_2O_4^{2-}}$)

• Context: This is used in titrations (usually heated because the reaction is slow, though $Mn^{2+}$ acts as an autocatalyst).
• Reduction (Oxidising Agent): $\mathrm{MnO_4^-}$ (Manganese changes from +7 to +2, decolourisation occurs). $$ \mathrm{MnO_4^- (aq) + 8H^+ (aq) + 5e^- \rightleftharpoons Mn^{2+} (aq) + 4H_2O (l)} $$
• Oxidation (Reducing Agent): $\mathrm{C_2O_4^{2-}}$ (Carbon changes from +3 to +4). $$ \mathrm{C_2O_4^{2-} (aq) \rightleftharpoons 2CO_2 (g) + 2e^-} $$

B. Permanganate ($\mathrm{MnO_4^-}$) / Iron(II) ($\mathrm{Fe^{2+}}$)

• Context: A standard titration to find the concentration of $\mathrm{Fe^{2+}}$ ions.
• Reduction (Oxidising Agent): $\mathrm{MnO_4^-}$ (Manganese +7 to +2). (See equation above).
• Oxidation (Reducing Agent): $\mathrm{Fe^{2+}}$ (Iron changes from +2 to +3). $$ \mathrm{Fe^{2+} (aq) \rightleftharpoons Fe^{3+} (aq) + e^-} $$

C. Copper(II) ($\mathrm{Cu^{2+}}$) / Iodide ($\mathrm{I^-}$)

• Context: Quantitative determination of copper.
• Reduction (Oxidising Agent): $\mathrm{Cu^{2+}}$ (Copper changes from +2 to +1, forming a white precipitate). $$ \mathrm{2Cu^{2+} (aq) + 2e^- \rightleftharpoons 2CuI (s) + I_2 (aq)} \quad \text{NOTE: The $\mathrm{Cu(I)}$ is precipitated as $\mathrm{CuI}$.} $$
• Oxidation (Reducing Agent): $\mathrm{I^-}$ (Iodine changes from -1 to 0, forming brown iodine). $$ \mathrm{2I^- (aq) \rightleftharpoons I_2 (aq) + 2e^-} $$

Did you know? In the $\mathrm{Cu^{2+}}/\mathrm{I^-}$ reaction, the white $\mathrm{CuI}$ precipitate is often masked by the brown colour of the iodine produced, so the final solution appears brown/dark yellow before being titrated against thiosulfate.

Final Key Takeaway: Transition metals are redox workhorses because they can easily switch oxidation states, allowing them to accept or donate electrons in catalytic cycles or quantitative reactions.