Transition Metals
In IB Chemistry, a transition metal is defined as:
Definition
Transition metal
An element that forms at least one ion with a partially filled d-subshell.
This means that, in at least one of its common ions, the d-orbitals are not completely full.
Note
- For the first-row transition metals (Sc → Zn), the 4s orbital is filled before the 3d orbitals.
- When these atoms form positive ions (cations), the 4s electrons are lost first, then the 3d electrons.
- Example:
- Fe atom: [Ar] 4s² 3d⁶
- Fe²⁺ ion: [Ar] 3d⁶ (4s electrons lost)
Note
- By the strict IB definition, Sc → Zn are the first-row transition elements, but Sc³⁺ and Zn²⁺ themselves do not have partially-filled d-subshells.
- IB specifically notes Zn as a borderline case and usually does not treat it as a “typical” transition metal.
Why Are Transition Metals Different?
Transition metals show some special patterns in their chemistry that most other metals do not. Two key ones for MYP/early IB are:
- Variable oxidation states
- Complex ion formation
Later in IB you will also study their coloured compounds and catalytic behaviour.
Variable Oxidation State
- Many transition metals can form more than one stable ion (more than one oxidation state).
- This happens because they can lose different combinations of their 4s and 3d electrons.
- This makes transition metals more flexible when bonding with other elements and forming compounds.
| Metal | Common ions |
|---|---|
| $\text{Fe}$ | $\text{Fe}^{2+}$, $\text{Fe}^{3+}$ |
| $\text{Cu}$ | $\text{Cu}^{+}$, $\text{Cu}^{2+}$ |
| $\text{Cr}$ | $\text{Cr}^{2+}$, $\text{Cr}^{3+}$ |
| $\text{Mn}$ | $\text{Mn}^{2+}$, $\text{Mn}^{4+}$, $\text{Mn}^{7+}$ |
Hint
- Chromium has the electron configuration: [Ar] 3d⁵ 4s¹
- It can lose:
- 2 electrons → Cr²⁺ : [Ar] 3d⁴
- 3 electrons → Cr³⁺ : [Ar] 3d³
- In compounds such as Cr⁶⁺ (e.g. CrO₄²⁻), more 3d electrons are effectively removed.
- Copper has the electron configuration: [Ar] 3d¹⁰ 4s¹
- It can lose:
- 1 electron (from 4s) → Cu⁺ : [Ar] 3d¹⁰
- 2 electrons (4s and one 3d) → Cu²⁺ : [Ar] 3d⁹
Complex Formation
- Transition metals often form complex ions.
- A complex ion consists of:
- a central metal ion (often a transition metal), surrounded by
- one or more ligands that are bonded to the metal.
Definition
Ligand
A ligand is a species (ion or molecule) that donates a lone pair of electrons to the central metal ion, forming a coordinate (dative) covalent bond.
In a coordinate (dative) bond, both electrons in the shared pair come from one atom (the ligand).
Note
- The metal ion is called the central metal ion.
- The ligand is the species that donates the lone pair of electrons.
Types of ligands
Ligands can be classified by how many bonds they form with the central metal ion:
- Monodentate ligand: can form 1 coordinate bond
- Bidentate ligand: can form 2 coordinate bonds
- Multidentate (polydentate) ligand: can form more than 2 coordinate bonds
When forming a complex ion , the coordination number refers to the number of bonds a central atom forms.
Example
- Monodentate ligands:
- H₂O (aqua)
- NH₃ (ammine)
- Cl⁻ (chloro)
- CN⁻ (cyano)
- OH⁻ (hydroxo)
- Bidentate ligands:
- Ethanedioate (oxalate), C₂O₄²⁻
- Ethanediamine (en), H₂N–CH₂–CH₂–NH₂
- Multidentate ligands:
- EDTA⁴⁻, a ligand with six donor sites (hexadentate), often used to bind metal ions very strongly.
Coordination Number
Definition
Coordination number
The number of coordinate bonds formed between the central metal ion and the ligands.
This is not always the same as the number of ligands (for example, a bidentate ligand forms 2 bonds but counts as 1 ligand).
Example
- Ammonia (NH₃) is a neutral monodentate ligand.
- It can form a complex with the cobalt(III) ion: $$[Co(NH_3)_6]^{3+}$$
Here:
- The central metal ion is Co³⁺.
- There are 6 monodentate ligands (NH₃).
- The coordination number is 6.
Why Are Transition Metals Colourful?
- Many transition metal ions and their complexes are brightly coloured.
- This is mainly due to d–d electron transitions.
- When ligands surround a transition metal ion and form a complex ion, the five d-orbitals of the metal ion are no longer all at the same energy.
- The electric field produced by the ligands causes the d-orbitals to split into two energy levels.
- Electrons in the lower-energy d-orbitals can absorb a specific amount of energy (a photon of light)
- They are then promoted to the higher-energy d-orbitals
- If the energy difference between these two levels corresponds to the energy of visible light, the complex will absorb certain colours and appear the complementary colour.
Note
- The colour you see is not the colour absorbed, but the mixture of the remaining wavelengths that are transmitted or reflected to your eye.
- If a complex absorbs red light, it may appear green.
- If it absorbs yellow–orange, it can appear blue–violet.
- The observed colour is complementary to the absorbed colour.
Why Does This Happen?
Note
- The explanation below about why the transitional metals are colourful is optional but good for the context.
- Certain aspects of that will be covered in IB Chemistry.
- When a transition metal complex ion is formed, the arrangement of ligands around the metal ion (its geometry) causes the d-orbitals to split.
- In an octahedral complex (6 ligands around the metal, e.g. $\left[\mathrm{Cu}\left(\mathrm{H}_2 \mathrm{O}\right)_{\mathrm{e}}\right]^{2+}$ ):
- The five d-orbitals split into:
- A higher-energy set (often written as $e_g$ ) - typically $d_{z^2}$ and $d_{x^2-y^2}$
- A lower-energy set (often written as $t_{2 g}$ ) - typically $d_{x y}, d_{x z}, d_{y z}$
- The energy gap between them is called $\mathbf{\Delta}_o$
- Electrons in the lower-energy set can absorb light of energy $\boldsymbol{\Delta} \boldsymbol{E}$ and move to the higher-energy set: $$\Delta E=h \nu=\frac{h c}{\lambda}$$
- If $\lambda$ (the wavelength) is in the visible region, the compound will appear coloured.
Note
The size of the splitting energy $\Delta$ determines which wavelength of light is absorbed, and therefore which colour the complex appears.
Example
- When potassium dichromate, K₂Cr₂O₇, dissolves in water, it forms dichromate ions, Cr₂O₇²⁻.
- These ions absorb particular wavelengths of visible light and the solution appears orange.
- Although the detailed explanation involves more than simple d–d transitions (charge-transfer effects are important), it is still a good example of a coloured transition metal compound.
What Factors Affect Color
The position of the d–d transition (which wavelength is absorbed) and the intensity of colour depend on several interrelated factors:
- Oxidation state of the metal ion
- A higher oxidation state usually means a stronger attraction between the metal ion and the ligands.
- This often leads to a larger splitting energy $\Delta$ between the d-orbitals.
- A larger $\Delta$ means higher-energy (shorter wavelength) light is absorbed.
- So changing the oxidation state can change the colour of the complex.
- Nature (strength) of the ligands
- Different ligands cause different amounts of splitting of the d-orbitals.
- This is described by the spectrochemical series, which orders ligands from strong field (large splitting) to weak field (small splitting).
- A simplified spectrochemical series is: $$\mathrm{CN}^{-}>\mathrm{NH}_3>\mathrm{H}_2 \mathrm{O}>\mathrm{OH}^{-}>\mathrm{Cl}^{-}>\mathrm{Br}^{-}>\mathrm{I}^{-}$$
- Stronger-field ligands (such as CN⁻, NH₃) cause a larger splitting, changing the wavelength of light absorbed and therefore the colour.
- Geometry of the complex
- Octahedral and tetrahedral complexes have different splitting patterns (different size and arrangement of energy levels).
- As a result, octahedral and tetrahedral complexes of the same metal and ligands can absorb different wavelengths, leading to different colours.
Why Are Transition Metals Good Catalysts?
Transition metals and their compounds often act as excellent catalysts, both:
- Heterogeneous catalysts – in a different phase from the reactants (e.g. solid catalyst, gaseous reactants)
- Homogeneous catalysts – in the same phase as the reactants (e.g. all in solution)
Two key reasons:
- Variable oxidation states
- Transition metals can change oxidation state easily by gaining or losing electrons.
- This allows them to form temporary intermediate compounds with reactants.
- These steps often have lower activation energies, so the overall reaction is faster.
- Partially filled d-orbitals and adsorption
- In heterogeneous catalysis, transition metal surfaces can adsorb reactant molecules.
- This weakens bonds in the reactant molecules and holds them in a favourable position for reaction.
Example
- Haber Process (ammonia synthesis):$$
\mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_3(\mathrm{~g})
$$ - Iron (Fe) acts as a heterogeneous catalyst.
- Contact process (sulfuric acid manufacture): $$2 \mathrm{SO}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_3(\mathrm{~g})
$$ - Vanadium(V) oxide, $\mathbf{V}_{\mathbf{2}} \mathbf{O}_{\mathbf{5}}$, is the catalyst.
- Hydrogenation of alkenes: $$
\mathrm{C}=\mathrm{C}+\mathrm{H}_2 \xrightarrow{\mathrm{Ni}} \mathrm{C}-\mathrm{C}
$$ - Nickel (Ni) is used as a solid catalyst in margarine and oil hardening.
Example
Consider the question where you are asked why the complex ion $\left[\mathrm{Cu}\left(\mathrm{NH}_3\right)_4\left(\mathrm{H}_2 \mathrm{O}\right)_2\right]^{2+}$ appears blue in solution. For answering that, you can use the following facts.
- Ligands cause d-orbital splitting
- The Cu²⁺ ion is surrounded by ligands (four NH₃ and two H₂O molecules) in an approximately octahedral arrangement.
- This creates a crystal field, splitting the copper ion’s d-orbitals into two energy levels.
- Electron transitions
- Electrons in the lower-energy d-orbitals can absorb energy from incoming white light.
- They are promoted to the higher-energy d-orbitals. This is a d–d transition.
- Complementary colour
- The complex absorbs certain wavelengths in the red–orange region of the spectrum.
- These absorbed wavelengths are removed from the white light.
- The remaining transmitted/reflected light is mainly in the blue region, so the complex appears blue.
Self review
- What property of transition metals allows them to act as effective catalysts in reactions?
- How does the strength of a ligand affect the colour of a transition metal complex?
- Describe the difference between a monodentate ligand and a multidentate ligand. Give one example of each.
- Explain in simple terms why a transition metal complex can absorb one colour of light and appear as another.
