Chapter 21: Transition Elements

Chapter 21: Transition Elements

Chapter 21
Transition Elements

This chapter is intertwined with several other topics that we have learnt previously, such as Atomic Structure, Chemical Bonding, Redox, Electrochemistry, Reaction Kinetics, Chemical Equilibria, etc.

Physical Properties

Worked Example 1
(2013 A Level Paper 2 Qn 4)

The table below gives data about some physical properties of the elements calcium, iron and copper.

Property	Calcium	Iron	Copper
Relative atomic mass	40.1	55.8	63.5
Atomic radius (metallic)/nm	0.197	0.126	0.128
Ionic radius (2+)/nm	0.099	0.076	0.069
Melting point/K	1112	1808	1358
density/g cm-3	1.54	7.86	8.92
Electrical conductivity/10⁶ S cm⁻¹	0.298	0.100	0.596

Property
Relative atomic mass
Calcium
40.1
Iron
55.8
Copper
63.5

Property
Atomic radius (metallic)/nm
Calcium
0.197
Iron
0.126
Copper
0.128

Property
Ionic radius (2+)/nm
Calcium
0.099
Iron
0.076
Copper
0.069

Property
Melting point/K
Calcium
1112
Iron
1808
Copper
1358

Property
density/g cm-3
Calcium
1.54
Iron
7.86
Copper
8.92

Property
Electrical conductivity/106 S cm-1
Calcium
0.298
Iron
0.100
Copper
0.596

(a) (i) Explain why the atomic radii of iron and copper are both less than that of calcium

• Answer: Fe and Cu both have a higher number of protons, thus larger nuclear charge than Ca. However valence electrons in the 4s orbital of Fe and Cu experience less effective shielding from the 3d orbital electrons than the valence electrons of Ca. Hence, effective nuclear charge of Fe and Cu are higher than that of Ca, thus they have a smaller atomic radii than calcium as their valence electrons are more closely attracted to the nucleus.

• (ii) Use relevant data from the table to explain why the densities of iron and copper are significantly greater than that of calcium ( no calculations are required)

• Answer: Both Cu and Fe have smaller atomic radii at 0.128 nm and 0.126nm than Ca, at 0.197nm. Cu and Fe have larger Ar at 63.5 and 55.8 than Ca, at 40.1. Thus the ions of Cu and Fe can be packed more closely together in a giant metallic structure. Since the mass of each Cu and Fe atom is larger than that of Ca, Cu and Fe have higher mass per unit volume, hence higher densities than Ca.

(b) The high electrical conductivity of copper makes it a very useful element for making electrical components

• (i) Apart from it’s high cost, suggest why copper is not usually used for overhead electrical cables.

• The density of copper, 8.92 g cm⁻³ is too high, hence copper wires are too heavy to be used as overhead cables.

• (ii) The high conductivity of copper is a consequence of its electronic configuration.
  Complete the electronic configuration of:

• Cu    : 1s²2s²2p⁶3s²3p⁶3d¹⁰4s¹
• Cu²⁺ : 1s²2s²2p⁶3s²3p⁶3d⁹

• (iii) The copper used for electrical wiring must be very pure. Outline how impure copper is purified industrially.

Answer: (Recall Chapter 20: Electrochemistry)
Electrolysis. The impure copper is placed at the anode and pure copper at the cathode.
Electrolyte is CuSO₄ (aq). Regulate the operating voltage such that it is just sufficient for Cu to be oxidised to Cu²⁺ in the solution. It will be then reduced at the cathode to Cu metal. Less reactive impurities will drop to the bottom as anode sludge, while more reactive impurities will remain as ions in the solution. Complexes and Ligands
(Related to Chemical Bonding)

Firstly, you need to know the difference between complexes and ligands:

Complexes	Ligands
Contains a central metal atom or ion datively bonded to one or more surrounding molecules or anions called ligands If the complex contains an overall charge, it is known to be a complex ion	An ion of molecule containing at least one atom bearing a lone pair of electrons, which can be donated to a low-lying vacant orbital of central metal atom or ion to form a dative covalent bond In other words, they can act as Lewis Bases Note: Only ONE lone pair of electrons per atom is used to form dative bond
Examples of neutral complexes: Ni(CO)₄, Fe(OH)₃ Examples of complex ions: [Al(H₂O)₆]³⁺, [Ag(NH₃)₂]⁺ (Tollen’s Reagent)	Examples of ligands commonly found in Data Booklet: H₂O, Br⁻, CN⁻, OH⁻, NH₃

Complexes
Contains a central metal atom or ion datively bonded to one or more surrounding molecules or anions called ligands If the complex contains an overall charge, it is known to be a complex ion
Examples of neutral complexes: Ni(CO)₄, Fe(OH)₃ Examples of complex ions: [Al(H₂O)₆]³⁺, [Ag(NH₃)₂]⁺ (Tollen’s Reagent)

Ligands
An ion of molecule containing at least one atom bearing a lone pair of electrons, which can be donated to a low-lying vacant orbital of central metal atom or ion to form a dative covalent bond In other words, they can act as Lewis Bases Note: Only ONE lone pair of electrons per atom is used to form dative bond
Examples of ligands commonly found in Data Booklet: H₂O, Br⁻, CN⁻, OH⁻, NH₃

Secondly, we need to know the types of ligands, determined by the number of dative bonds they are able to form with central atom / ion:

Types of Ligands	Number of Dative Bonds able to form with central atom / ion
Monodentate ligands (E.g. NH₃ in [Ag(NH₃)₂]⁺)	1
Bidentate ligands	2
Hexadentate ligands	6
Polydentate ligands	>1

Types of Ligands
Monodentate ligands (E.g. NH₃ in [Ag(NH₃)₂]⁺)
Bidentate ligands
Hexadentate ligands
Polydentate ligands

Number of Dative Bonds able to form with central atom / ion
1
2
6
>1

Now, we need to know the criteria for the formation of complexes:

• 1. Presence of ligands to donate electron pairs to form dative bonds
• 2. Presence of a central metal ion (can be transition or non-transition) that can
  • – Accommodate lone pair of electrons in its low-lying vacant orbitals to form a
    dative bond
  • – Possess a high charge density

Note: The central metal ion is a Lewis Acid and the ligand is a Lewis Base, and both form a complex ion.

Ligand Exchange

This process occurs when a stronger ligand displaces a weaker ligand from the metal complex.

[Ni(H₂O)6]²⁺ (aq) +6NH₃(aq)

⇌

[Ni (NH₃)₆]²⁺(aq) + 6H₂O(I )

Let’s look at a sample worked example below:

Describe and explain what is seen when concentrated HCl is added to a solution of Cu²⁺ ions in water with the aid of a balanced equation.

Solution:
Equation:
[Cu(H₂O)₆]²⁺ (aq) + 4Cl⁻ (aq) ⇌ [CuCl₄]²⁻ (aq) + 6H₂O (l) Observation:
The blue solution turns green (mixture of blue [Cu(H₂O)₆]²⁺ and yellow [CuCl₄]²⁻) then yellow. Explanation:
Ligand exchange has occurred where Cl⁻ displaces H₂O ligands.

Coloured compounds and ions

The fun thing about transition metals is their tendency to form coloured compounds, both in the solid and aqueous states. The not-so-fun thing however, is that memorising the different compounds and their respective colours can be overwhelming (and sadly, eventually necessary).

The table below summarises some of the colours students are required to know for common ions (with H₂O as the ligand)

Aqua complex ion	Electronic configuration of the cation	Colour
[Sc(H₂O)₆]³⁺	[Ar]3d⁰	Colourless
[Ti(H₂O)₆]³⁺	[Ar]3d¹	Violet
[V(H₂O)₆]³⁺	[Ar]3d²	Green
[V(H₂O)₆]²⁺	[Ar]3d³	Violet
[Cr(H₂O)₆]³⁺	[Ar]3d³	Violet / Green
[Cr(H₂O)₆]²⁺	[Ar]3d⁴	Blue
[Mn(H₂O)₆]³⁺	[Ar]3d⁴	Violet
[Mn(H₂O)₆]²⁺	[Ar]3d⁵	Faint pink
[Fe(H₂O)₆]³⁺	[Ar]3d⁵	Pale violet / Yellow
[Fe(H₂O)₆]²⁺	[Ar]3d⁶	Pale green
[Co(H₂O)₆]²⁺	[Ar]3d⁷	Pink
[Ni(H₂O)₆]³⁺	[Ar]3d⁸	Green
[Cu(H₂O)₆]²⁺	[Ar]3d⁹	Blue
[Zn(H₂O)₆]²⁺	[Ar]3d¹⁰	Colorless

The property of forming coloured compounds can be attributed to a phenomenon known as d-orbital splitting.
The five d-orbitals of a gaseous transition metal, Mⁿ⁺, are usually at the same energy level. But in octahedral complexes, upon the approach of ligands, the 3d orbitals are split into two different sets of energy levels. The d_𝑧² and d_x² − _y² orbitals point directly at the x,y and z axis. Due to stronger inter-electronic repulsion from direct interaction with the ligand’s lone pair of electrons, the energy level of the d_z² and d_x² – _y² orbitals are raised by a greater extent.
Hence, the d_yz, d_xz , and d_xy orbitals are at a lower energy level.

But in tetrahedral complexes, the four ligands approach the central metal atom or ion in between the axes. Hence, the d_yz, d_xz, and dxy orbitals now experience stronger inter-electronic repulsion from direct interaction with the ligand’s lone pair of electrons, the energy level of d_yz, d_xz, and dxy orbitals are raised by a greater extent. The d_z² and d_x²- _y² orbitals are at a lower energy level.

However, It is unnecessary to state in your answers which is of the higher energy level. Writing that the d-orbitals are split into two different energy levels is sufficient.

Since the d-orbitals are split into two different energy levels, there is an energy gap, ΔE, which corresponds to the energy of the visible region on the electromagnetic spectrum. Transition metal complexes will absorb the light of wavelength with that energy, causing an electron to be promoted from the d-orbital of a lower energy to another d-orbital of a higher energy level. This is called the d-d transition.

Since light of that energy is absorbed, the complementary colour ( the colour opposite on the colour wheel) would be reflected into our eyes. For example, Cu²⁺ (aq) appears blue since light of orange wavelength is absorbed.

The colour of complexes is directly affected by the energy gap, ΔE, since the light absorbed has energy corresponding to it. Since energy of light is E=hf where h is Plank’s constant and f is frequency of the light, and v=fλ where v is the speed and λ is the wavelength of light, rearranging the first equation, E=h  ᴠ  λ .

Hence, we can conclude that E is inversely proportional to λ.

Hence we can conclude that a bigger ΔE leads to light of a smaller wavelength being absorbed. Thus the colour of the compound reflected into our eyes is that of a larger wavelength, such as red.
The opposite can be said for a smaller ΔE as well. Light of a larger wavelength would be absorbed. The colour reflected into our eyes would be that of a smaller wavelength, such as blue.

ΔE is affected by two factors,

• 1. Strength of ligand
  A stronger ligand would result in a larger ΔE. Hence compounds with stronger ligands are more likely to appear yellow.

I – < Br⁻ < CI ⁻ < F⁻ < OH⁻ < CH₃COO⁻ < H₂O < NH₃ < CN⁻ ― ∆E increases → I – < Br⁻ < CI ⁻ < F⁻ < OH⁻ < CH₃COO⁻ < H₂O < NH₃ < CN⁻ ― ∆E increases →

• 2. Oxidation state of the cation
  Different numbers of electrons in the d-orbitals of the transition metal ion would result in varied interactions with the electrons as the ligands approach the central ion. Hence, ΔE will vary.

Next, here is a table listing the common reactions undergone by transition metal ions

Reaction	Explanation
Precipitation reactions	Common reagents used include NH₃ (aq) and NaOH (aq) Explanation is usually couched in terms of ionic product exceeding Kₛₚ Cu²⁺(aq) + 2OH⁻ (aq) blue solution NH₃(aq) ⟶ Cu(OH)₂(s) pale blue precipitate
Ligand exchange reactions	Prediction of whether such a reaction occurs is usually based on the given relative stability constants of the complex ions or the given observations (ie colour changes, dissolution of precipitate etc.) Explanation is usually couched in terms of the relative stabilities of the complex or the relative strengths of the ligands [Cu(H₂O)₆]²⁺(aq) + 4NH₃(aq) blue solution excess NH₃(aq) ⟶ [Cu(NH₃)₄(H₂O)₂]²⁺(aq) + 2H₂O(I ) pale blue precipitate [Cu(H₂O)₆]²⁺(aq) + 4Cl ⁻ (aq) blue solution conc. HCl ⟶ [CuC(I )₄²⁻ + 6H₂O(I) yellow solution [Fe(H₂O)₆]³⁺(aq) + SCN⁻(aq) blue solution KSCN(aq) ⟶ [Fe(SCN)(H₂O)₅]²⁺ (aq) + H₂O(I ) blood red solution [Fe(H2O)6]3+(aq) + 3C2O42-(aq) yellow solution Na2C2O4(aq) ⟶ [Fe(C2O4)3]3-(aq) + 6H2O(I ) colourless solution [Co(H2O6]2+(aq) + 4CI –(aq) pink solution conc. HCI ⟶ [CoCI 4]2-(aq) + 6H2O(I ) blue solution
Redox reactions	Justification of whether a redox reaction can occur is usually based on E° values 2Fe2+(aq) + Cl2(aq) pale green solution ⟶ 2Fe3+(aq) + 2Cl –(aq) yellow solution 2Fe3+(aq) + 2I– ⟶ 2Fe2+(aq) + I2(aq) 2C2+(aq) + 4I–(aq) ⟶ 2CuI(s) + I2(aq) cream ppt.  brown solution Cu2+(aq) + Zn(s) ⟶ Cu(s) + Zn2+(aq) pink solid
Acid-base reactions	[Cr(H2O6]3+ (aq) + H2O(l ) ⇆ [ Cr(OH) ( H2O)5]2+(aq) + H3O+(aq) hydrolysis of aqua complex ion reaction with sodium carbonate to give CO2 2CrO42-(aq)+2H+(aq) yellow solution ⟶ Cr2O72-(aq) + H2O(I ) orange solution Cr2O72-(aq) + 2OH–(aq) orange solution ⟶ 2CrO42-(aq) + H2O(I ) yellow solution

Reaction: Precipitation reactions
Explanation: Common reagents used include NH₃ (aq) and NaOH (aq) Explanation is usually couched in terms of ionic product exceeding Ksp Cu2+(aq) + 2OH⁻ (aq) blue solution NH3(aq) ⟶ Cu(OH)2(s) pale blue precipitate

Reaction: Ligand exchange reactions

Explanation: Prediction of whether such a reaction occurs is usually based on the given relative stability constants of the complex ions or the given observations (ie colour changes, dissolution of precipitate etc.) Explanation is usually couched in terms of the relative stabilities of the complex or the relative strengths of the ligands [Cu(H₂O)₆]²⁺(aq) + 4NH₃(aq) blue solution excess NH₃(aq) ⟶ [Cu(NH₃)₄(H₂O)₂]²⁺(aq) + 2H₂O(I ) pale blue precipitate [Cu(H₂O)₆]²⁺(aq) + 4Cl ⁻ (aq) blue solution conc. HCl ⟶ [CuC(I )₄²⁻ + 6H₂O(I) yellow solution [Fe(H₂O)₆]³⁺(aq) + SCN⁻(aq) blue solution KSCN(aq) ⟶ [Fe(SCN)(H₂O)₅]²⁺ (aq) + H₂O(I ) blood red solution [Fe(H2O)6]3+(aq) + 3C2O42-(aq) yellow solution Na2C2O4(aq) ⟶ [Fe(C2O4)3]3-(aq) + 6H2O(I ) colourless solution [Co(H2O6]2+(aq) + 4CI –(aq) pink solution conc. HCI ⟶ [CoCI 4]2-(aq) + 6H2O(I ) blue solution

Reaction: Redox reactions
Explanation: Justification of whether a redox reaction can occur is usually based on E° values 2Fe2+(aq) + Cl2(aq) pale green solution ⟶ 2Fe3+(aq) + 2Cl –(aq) yellow solution 2Fe3+(aq) + 2I– ⟶ 2Fe2+(aq) + I2(aq) 2C2+(aq) + 4I–(aq) ⟶ 2CuI(s) + I2(aq) cream ppt.  brown solution Cu2+(aq) + Zn(s) ⟶ Cu(s) + Zn2+(aq) pink solid

Reaction: Ligand exchange reactions
Explanation: [Cr(H2O6]3+ (aq) + H2O(l ) ⇆ [ Cr(OH) ( H2O)5]2+(aq) + H3O+(aq) hydrolysis of aqua complex ion reaction with sodium carbonate to give CO2 2CrO42-(aq)+2H+(aq) yellow solution ⟶ Cr2O72-(aq) + H2O(I ) orange solution Cr2O72-(aq) + 2OH–(aq) orange solution ⟶ 2CrO42-(aq) + H2O(I ) yellow solution