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Exhaustive Guide: d-Block Elements & Properties | Class 12 Chemistry

Exhaustive Guide: d-Block Elements & Properties | Class 12 Chemistry | ChemCA

d-Block Elements & General Properties

Module 1 | CBSE Class 12 Chemistry | d- & f-Block Chapter

1. Position in the Periodic Table

The d-block of the periodic table contains the elements of the groups 3 to 12 in which the d orbitals are progressively filled in each of the four long periods.

The f-block consists of elements in which 4f and 5f orbitals are progressively filled. They are placed in a separate panel at the bottom of the periodic table.

Transition Metals (IUPAC Definition): A transition element is defined as an element which has incompletely filled d orbitals in its ground state or in any one of its common oxidation states.
Why are Zn, Cd, and Hg not considered transition metals?
Zinc, cadmium, and mercury (Group 12) have completely filled d orbitals (d10 configuration) in their ground state as well as in their common oxidation states (e.g., Zn2+ is 3d10). Hence, they are not regarded as transition elements. However, being the end members of the 3d, 4d, and 5d series, their chemistry is studied along with the chemistry of the transition elements.

2. Electronic Configurations of the d-Block Elements

The general electronic configuration of d-block elements is:

(n-1)d1-10 ns1-2

The (n-1) stands for the inner d orbitals which may have one to ten electrons and the outermost ns orbital may have one or two electrons.

2.1 Exceptions in the 3d Series: Chromium (Cr) and Copper (Cu)

The electronic configuration of elements in a series usually follows a regular filling pattern (Aufbau principle). However, half-filled and completely filled subshells are exceptionally stable due to symmetrical electron distribution and maximum exchange energy.

Element Atomic Number (Z) Expected Configuration Actual Configuration Reason
Chromium (Cr) 24 [Ar] 3d4 4s2 [Ar] 3d5 4s1 Exactly half-filled d-orbitals (3d5) provide extra stability.
Copper (Cu) 29 [Ar] 3d9 4s2 [Ar] 3d10 4s1 Completely filled d-orbitals (3d10) provide extra stability.
NCERT Concept: Why is Copper (Z=29) considered a transition metal if its ground state is 3d10?
Answer: Copper exhibits a +2 oxidation state (Cu2+) in many of its compounds. In the +2 state, its configuration is [Ar] 3d9, which represents an incompletely filled d-orbital. Hence, it is a transition metal.

3. General Physical Properties

Transition metals display typical metallic properties such as high tensile strength, ductility, malleability, high thermal and electrical conductivity, and metallic lustre. Except for Zn, Cd, Hg, and Mn, they have one or more typical metallic structures at normal temperatures.

3.1 Enthalpy of Atomization

Transition metals have very high melting and boiling points. They also have high enthalpies of atomization.

Reason for High Enthalpy of Atomization: The high melting points and enthalpies of atomization are attributed to the involvement of a large number of electrons from the (n-1)d orbitals in addition to the ns electrons in the interatomic metallic bonding.

Rule of Thumb: Greater the number of unpaired d-electrons, stronger is the metallic bonding.

Trends to observe:

  • In any row, the maximum enthalpy of atomization is near the middle (e.g., V, Cr, W) because they have the maximum number of unpaired electrons (d5).
  • There is a distinct dip in the curve at Manganese (Mn) and Technetium (Tc). This is because their completely half-filled d5 configuration is highly stable and tightly held by the nucleus, making the d-electrons less available for metallic bonding.
  • Metals of the 4d and 5d series have much higher enthalpies of atomization than the corresponding elements of the 3d series. This explains why heavy transition metals (like Tungsten, W) have exceptionally high melting points and are used in applications requiring extreme heat resistance.

4. Variation in Atomic and Ionic Radii

Along a period: The atomic radii of transition metals progressively decrease with an increase in atomic number initially. This is because the new electron enters a d orbital, and the shielding effect of a d electron is not very effective. Thus, the increased nuclear charge pulls the electrons closer.

Towards the end of the series, the atomic radii remain almost constant (Fe, Co, Ni) and then slightly increase (Cu, Zn) due to increased electron-electron repulsion among the filled d orbitals overriding the increased nuclear charge.

4.1 The Lanthanoid Contraction

When we move down a group (from 3d to 4d series), the atomic radii increase as expected due to the addition of a new electron shell. However, the atomic radii of the 5d series are virtually the same as those of the corresponding members of the 4d series.

Lanthanoid Contraction: The steady decrease in atomic and ionic radii of lanthanoids with increasing atomic number due to the poor shielding effect of 4f electrons is called the lanthanoid contraction.

Consequence: Because the 4f subshell is filled before the 5d series begins (after Lanthanum), the poor shielding of the 4f electrons causes the outer electrons to be pulled strongly by the nucleus. This contraction perfectly cancels out the expected increase in size from adding a new principal shell.
Classic NCERT Example: Zirconium (Zr, 4d series) and Hafnium (Hf, 5d series) have almost identical atomic radii (Zr = 160 pm, Hf = 159 pm). This makes their physical and chemical properties almost identical, making them very difficult to separate.

5. Ionisation Enthalpies

There is an increase in ionisation enthalpy along each series of the transition elements from left to right due to an increase in nuclear charge. However, many small variations occur.

  • First Ionisation Enthalpy: The trend is irregular. The value for Zn is very high because electrons are removed from a fully filled 4s level of a completely filled 3d10 shell.
  • Second Ionisation Enthalpy: The 2nd IE of Chromium (Cr) and Copper (Cu) are unusually high.
    Reason: After removing one electron (4s1), Cr becomes stable 3d5 and Cu becomes stable 3d10. Removing a second electron disrupts this stability.
  • Third Ionisation Enthalpy: The 3rd IE of Manganese (Mn) is very high.
    Reason: After losing two 4s electrons, Mn2+ has a highly stable half-filled 3d5 configuration. Breaking this stable core requires massive energy.

6. Oxidation States

One of the most notable features of transition elements is their ability to show a wide variety of oxidation states in their compounds. This is because the energies of (n-1)d and ns orbitals are very close, allowing electrons from both to participate in bonding.

Trends in the 3d Series:

  • Scandium (Sc): Exhibits only +3 oxidation state (losing both 4s and the one 3d electron to achieve noble gas configuration).
  • Manganese (Mn): Exhibits the highest number of oxidation states (+2 to +7) because it has the maximum number of unpaired electrons (3d5 4s2) available for bonding.
  • Zinc (Zn): Exhibits only +2 oxidation state (losing only 4s electrons, keeping the 3d10 core intact).
Stability of Oxidation States:
1. Lower Oxidation States: Transition metals in their lower oxidation states (+2, +3) generally form ionic bonds and their oxides are basic.
2. Higher Oxidation States: Transition metals in their highest oxidation states (+5, +6, +7) form covalent bonds. They are stabilized by highly electronegative atoms like Oxygen and Fluorine. Their oxides are acidic (e.g., Mn2O7, CrO3).

Why are higher oxidation states stabilized by Oxygen more than Fluorine?
While fluorine is more electronegative, oxygen is better at stabilizing the highest oxidation states (like +7 in MnO4-) because oxygen can form multiple (double) bonds with the metal atom (pπ-dπ bonding), whereas fluorine can only form single bonds.

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

NCERT Example 8.1: On what ground can you say that scandium (Z = 21) is a transition element but zinc (Z = 30) is not?

Solution:
The electronic configuration of Scandium (Sc) is [Ar] 3d1 4s2. In its ground state, it has an incompletely filled d-orbital (3d1), so it is a transition element.
The electronic configuration of Zinc (Zn) is [Ar] 3d10 4s2. Both in its ground state and in its only common oxidation state (+2), it has a completely filled d-orbital (3d10). Therefore, according to the IUPAC definition, zinc is not a transition element.

NCERT Example 8.3: Why is the highest oxidation state of a metal exhibited in its oxide or fluoride only?

Solution:
Fluorine and oxygen are the most electronegative elements in the periodic table. They have a small size and very high electron affinity. Because of this high electronegativity, they can easily pull electrons away from transition metals, thus bringing out their highest oxidation states. Additionally, oxygen can stabilize high oxidation states by forming multiple (double) bonds with the metal.

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

Part A: Conceptual (1-2 Marks)

[CBSE 2018, 2020]

Q1. Why do transition elements have high enthalpies of atomization?

Answer: Transition metals have high enthalpies of atomization because they have a large number of unpaired electrons in their (n-1)d orbitals. These unpaired electrons participate extensively in interatomic metallic bonding. Stronger metallic bonding results in a higher enthalpy of atomization.
[CBSE 2017, 2019]

Q2. Account for the following: Zirconium (Zr) and Hafnium (Hf) have almost identical atomic radii.

Answer: Zr (4d series) and Hf (5d series) have almost identical atomic radii due to the Lanthanoid Contraction. Before Hf in the periodic table, the 4f subshell is filled (lanthanoids). The 4f electrons offer very poor shielding effect. As a result, the increased nuclear charge strongly pulls the outer electrons inward, perfectly cancelling out the expected increase in size from adding a new principal shell.
[CBSE 2015, 2021]

Q3. Why is the third ionization enthalpy of Manganese (Mn) unexpectedly high?

Answer: The electronic configuration of Mn is [Ar] 3d5 4s2. After losing two electrons to form Mn2+, its configuration becomes [Ar] 3d5. This is an exactly half-filled d-orbital configuration, which is exceptionally stable. Removing a third electron from this highly stable core requires a massive amount of energy, making the third IE unexpectedly high.

Part B: Assertion-Reason Type (1 Mark)

[CBSE Sample Paper 2024]

Q4. Assertion (A): Chromium has the electronic configuration [Ar] 3d5 4s1 instead of [Ar] 3d4 4s2.
Reason (R): Half-filled d-orbitals have extra stability due to greater exchange energy.

Answer: Both Assertion and Reason are correct, and Reason is the correct explanation for Assertion. The shift of one electron from 4s to 3d creates a half-filled 3d subshell. The symmetrical distribution of electrons and the maximum number of parallel spins maximize the exchange energy, conferring extra stability to the Chromium atom.

Part C: Application Based (2-3 Marks)

[CBSE 2016, 2022]

Q5. (a) Name a transition element which does not exhibit variable oxidation states.
(b) Why do transition metals form oxides that are basic in lower oxidation states and acidic in higher oxidation states?

Answer:
(a) Scandium (Sc). It only exhibits a +3 oxidation state, as losing three electrons gives it a stable noble gas (Argon) configuration.
(b) In lower oxidation states, transition metals have low effective nuclear charge, holding their valence electrons relatively loosely, allowing them to form ionic bonds (acting as electron pair donors/bases). In higher oxidation states, the high effective nuclear charge strongly polarizes the electron cloud of oxygen, leading to covalent bonding. Such high oxidation state oxides (like Mn2O7, CrO3) act as electron pair acceptors, making them acidic.

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This module is strictly mapped to the latest rationalised NCERT syllabus for Class 12 Chemistry.
Coming up in Module 2: Standard Electrode Potentials, Magnetic Properties, and Coloured Ions.

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