Exhaustive Guide: Properties of Transition Elements | Class 12 Chemistry

Transition Elements: Properties & Reactivity

Module 2 | CBSE Class 12 Chemistry | d- & f-Block Chapter

Table of Contents

• 1. Standard Electrode Potentials (E°)
  • 1.1 The Anomaly of Copper
  • 1.2 Trends in the M³⁺/M²⁺ Standard Potentials
• 2. Magnetic Properties
  • 2.1 Calculation of Magnetic Moment (μ)
• 3. Formation of Coloured Ions
• 4. Formation of Complex Compounds
• 5. Catalytic Properties
• 6. Formation of Interstitial Compounds
• 7. Alloy Formation
• 8. NCERT Solved Examples
• 9. Previous Year Questions (PYQs)

1. Standard Electrode Potentials (E°)

The stability of a particular oxidation state in aqueous solution is determined by the standard electrode potential (E°) of the metal. The total energy change involved in the conversion of a solid metal M to its aqueous ion M²⁺(aq) relies on three thermochemical processes:

1. Enthalpy of Atomization (Δ_aH): Energy required to convert solid metal into gaseous atoms (Endothermic).
2. Ionization Enthalpy (IE): Energy required to remove electrons from the isolated gaseous atom (Endothermic).
3. Hydration Enthalpy (Δ_hydH): Energy released when gaseous ions are surrounded by water molecules (Exothermic).

The overall standard potential depends on the net sum of these three enthalpy changes. Generally, moving left to right across the 3d series, E° values become less negative (meaning the elements become less reactive/less easily oxidized). This is primarily due to the steady increase in ionization enthalpies.

1.1 The Anomaly of Copper (Highly Tested Concept)

Why is the E°(M²⁺/M) value for Copper positive (+0.34 V)?
Copper is unique in the 3d series because it is the only metal with a positive standard reduction potential. This indicates its inability to liberate hydrogen gas from dilute acids.
Reason: The high energy required to transform Cu(s) to Cu²⁺(g) (i.e., its high enthalpy of atomization and high ionization enthalpy) is not balanced by its hydration enthalpy.

1.2 Trends in the M³⁺/M²⁺ Standard Potentials

An examination of E°(M³⁺/M²⁺) values explains the stability of higher oxidation states:

• Scandium: E°(Sc³⁺/Sc²⁺) is very low because Sc³⁺ achieves a highly stable noble gas configuration.
• Zinc: The value for Zn is extremely high because removing an electron from the stable 3d¹⁰ core of Zn²⁺ requires massive energy.
• Manganese: E°(Mn³⁺/Mn²⁺) is highly positive. This is because Mn²⁺ has a highly stable half-filled d-orbital (3d⁵). Hence, Mn³⁺ readily accepts an electron (acting as an oxidizing agent) to revert to the stable Mn²⁺ state.

2. Magnetic Properties

When a magnetic field is applied to substances, mainly two types of magnetic behavior are observed: diamagnetism and paramagnetism.

• Diamagnetic substances: Repelled by the applied field. They have all their electrons paired.
• Paramagnetic substances: Attracted by the applied field. They contain one or more unpaired electrons.
• Ferromagnetic substances: Very strongly attracted by the applied field (extreme form of paramagnetism). Examples: Fe, Co, Ni.

Since transition metal ions frequently have incompletely filled d-orbitals, most of their compounds are paramagnetic.

2.1 Calculation of Magnetic Moment (μ)

The magnetic moment of transition metal compounds arises mainly from the spin of the unpaired electrons. The orbital angular momentum contribution is usually "quenched" by the surrounding environment. Therefore, the magnetic moment is calculated using the "spin-only" formula:

μ = √[n(n+2)] B.M.

Where:
n = Number of unpaired electrons.
B.M. = Bohr Magneton (the unit of magnetic moment). 1 B.M. = eh / 4πmc = 9.27 × 10^-24 A m².

Shortcut to remember: A single unpaired electron (n=1) gives μ = √3 ≈ 1.73 B.M. For n=2, μ ≈ 2.83 B.M. For n=3, μ ≈ 3.87 B.M. The value before the decimal point roughly corresponds to the number of unpaired electrons.

3. Formation of Coloured Ions

One of the most distinctive properties of transition metal compounds is their brilliant colours, both in the solid state and in aqueous solutions.

Cause of Colour (d-d Transition): When an electron from a lower energy d-orbital is excited to a higher energy d-orbital, the energy of excitation corresponds to the frequency of light absorbed from the visible region. The colour observed is the complementary colour of the light absorbed.

When transition metal ions form complexes, the ligands cause the five degenerate d-orbitals to split into two sets of different energies (crystal field splitting, which you will study in Coordination Compounds). An electron can jump between these split d-orbitals. This is called a d-d transition.

Exceptions (Colourless Ions): Ions with completely empty d-orbitals (d⁰) or completely filled d-orbitals (d¹⁰) are colourless because d-d transitions are impossible.
Examples of Colourless ions: Sc³⁺(d⁰), Ti⁴⁺(d⁰), Zn²⁺(d¹⁰), Cu⁺(d¹⁰).

4. Formation of Complex Compounds

Complex compounds are those in which the metal ions bind a number of anions or neutral molecules (ligands) to give complex species with characteristic properties (e.g., [Fe(CN)₆]^3-, [Cu(NH₃)₄]²⁺).

The transition metals form a large number of complex compounds due to three main reasons:

1. Comparatively smaller sizes of the metal ions.
2. High ionic charges (high charge/radius ratio).
3. The availability of vacant d-orbitals of suitable energy to accept lone pairs of electrons donated by ligands.

5. Catalytic Properties

The transition metals and their compounds are known for their exceptional catalytic activity. Many industrial processes utilize them:

• Vanadium(V) oxide (V₂O₅): Used in the Contact Process for manufacturing sulphuric acid.
• Finely divided iron (Fe): Used in Haber’s Process for synthesizing ammonia.
• Nickel (Ni): Used in the catalytic hydrogenation of oils to fats.

Why are they good catalysts?
1. Variable Oxidation States: They can easily switch between oxidation states, forming unstable intermediate compounds which provide an alternate, lower activation energy pathway. (e.g., Iron(III) catalyzing the reaction between iodide and persulphate).
2. Surface Area: In solid form, they provide a large surface area for reactant molecules to adsorb, bringing them closer and weakening their bonds.

6. Formation of Interstitial Compounds

Interstitial compounds are those which are formed when small atoms like H, C, or N are trapped inside the crystal lattices of metals. They are usually non-stoichiometric and are neither typically ionic nor covalent (e.g., TiC, Mn₄N, Fe₃H, VH_0.56).

Principal Characteristics of Interstitial Compounds:

• They have very high melting points, higher than those of pure metals.
• They are very hard (some borides approach diamond in hardness).
• They retain metallic conductivity.
• They are chemically inert.

7. Alloy Formation

An alloy is a blend of metals prepared by mixing the components. Alloys may be homogeneous solid solutions in which the atoms of one metal are distributed randomly among the atoms of the other.

Hume-Rothery Rules for Alloy Formation: Alloys are readily formed by transition metals because their atomic radii are very similar. Because the atomic radii of transition metals differ by less than 15%, atoms of one metal can easily replace those of another in the crystal lattice.

The alloys so formed are hard and generally have high melting points.
Classic Examples:
1. Brass: Alloy of Copper (Cu) and Zinc (Zn).
2. Bronze: Alloy of Copper (Cu) and Tin (Sn).
3. Ferrous alloys containing chromium, vanadium, tungsten, molybdenum, and manganese are used for the production of a variety of steels and stainless steel.

8. NCERT Solved Examples (Step-by-Step)

NCERT Example 8.8: Calculate the magnetic moment of a divalent ion in aqueous solution if its atomic number is 25.

Solution:
Step 1: Write the electronic configuration
Atomic number Z = 25 refers to Manganese (Mn).
Electronic configuration of Mn atom: [Ar] 3d⁵ 4s².
For a divalent ion (Mn²⁺), it loses two electrons from the outermost 4s shell.
Configuration of Mn²⁺: [Ar] 3d⁵.

Step 2: Determine unpaired electrons (n)
The 3d⁵ configuration has 5 unpaired electrons (one in each of the five d-orbitals according to Hund's Rule). So, n = 5.

Step 3: Calculate magnetic moment (μ)
Using the spin-only formula: μ = √[n(n+2)] B.M.
μ = √[5(5+2)] = √[5(7)] = √35
μ ≈ 5.92 B.M.

NCERT Example 8.4: Why is E°(Mn³⁺/Mn²⁺) couple much more positive than that for E°(Cr³⁺/Cr²⁺) or E°(Fe³⁺/Fe²⁺)?

Solution:
A large positive E° value indicates a strong tendency to gain an electron and undergo reduction (from M³⁺ to M²⁺).
Mn²⁺ has a 3d⁵ configuration, which is exactly half-filled and therefore exceptionally stable. Thus, Mn³⁺ has a very strong tendency to gain an electron to achieve this stable 3d⁵ state, making the E°(Mn³⁺/Mn²⁺) highly positive.
In contrast, Cr³⁺ (d³, which is stable in an octahedral field) and Fe³⁺ (d⁵, already stable) do not have the same immense driving force to form their +2 states.

9. Previous Year Questions (PYQs) & Exhaustive Question Bank

Part A: Conceptual (1-2 Marks)

[CBSE 2018, 2020]

Q1. Why are most transition metal compounds coloured?

Answer: The colour of transition metal compounds is due to the presence of incomplete d-orbitals. In the presence of ligands, the d-orbitals split into two sets of different energy levels. When white light falls on these compounds, electrons absorb specific wavelengths to jump from a lower d-orbital to a higher d-orbital (d-d transition). The transmitted or reflected light corresponds to the complementary colour, which is what we see.

[CBSE 2017, 2019]

Q2. Why is Cu⁺ ion colourless while Cu²⁺ ion is coloured in aqueous solutions?

Answer: The electronic configuration of Cu⁺ is [Ar] 3d¹⁰. Since the d-subshell is completely filled, there is no space for d-d transitions to occur, making it colourless. On the other hand, the configuration of Cu²⁺ is [Ar] 3d⁹, which contains one unpaired electron. This permits d-d transitions, imparting colour (blue) to the solution.

Part B: Assertion-Reason Type (1 Mark)

[CBSE Sample Paper 2024]

Q3. Assertion (A): Transition metals form a large number of complex compounds.
Reason (R): Transition metals have small size, high charge density, and vacant d-orbitals of suitable energy.

Answer: Both Assertion and Reason are correct, and Reason is the correct explanation for Assertion. The high charge/size ratio intensely attracts ligands (anions or neutral molecules), and the vacant d-orbitals accommodate the lone pairs donated by these ligands, facilitating the formation of complex ions.

Part C: Application Based (2-3 Marks)

[CBSE 2015, 2022]

Q4. Explain the following observations:
(i) Transition metals and their many compounds act as good catalysts.
(ii) Interstitial compounds are well known for transition metals.

Answer:
(i) Transition metals are good catalysts due to their ability to exhibit variable oxidation states and their propensity to form complexes. This allows them to form unstable reaction intermediates with reactants, providing a new path with lower activation energy. Additionally, they provide a large surface area for reactants to adsorb.
(ii) Transition metals possess large crystal lattices with microscopic voids (interstices). Small non-metal atoms like Hydrogen, Carbon, and Nitrogen can easily slip into these voids and get trapped, forming interstitial compounds without altering the core metallic bonding.

[CBSE 2016, 2018]

Q5. Calculate the magnetic moment of a divalent ion in aqueous solution if its atomic number is 26.

Answer:
Atomic number Z = 26 corresponds to Iron (Fe).
Ground state electronic configuration of Fe: [Ar] 3d⁶ 4s².
Divalent ion (Fe²⁺) configuration: [Ar] 3d⁶.
According to Hund's rule, filling 6 electrons in the 5 d-orbitals results in 1 paired set and 4 unpaired electrons. (n = 4).
Using the spin-only formula: μ = √[n(n+2)]
μ = √[4(4+2)] = √24
μ ≈ 4.90 B.M.

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This module is strictly mapped to the latest rationalised NCERT syllabus for Class 12 Chemistry.
Coming up in Module 3: Important Compounds (Potassium Dichromate & Potassium Permanganate) and the f-Block Elements.

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