CBSE Class 12 Chemistry Chapter 8: d and f-Block Elements NCERT Solutions

Reviewed by NCERT Help Editorial Team · Updated on 29 Jul 2026

NCERT Solutions PDF Class 12 PDF

CBSE Class 12 Chemistry, Chapter 8, d and f-Block Elements, introduces students to the fascinating world of transition and inner transition metals. This chapter delves into their electronic configurations, explaining how these determine their variable oxidation states and unique properties. You'll explore why certain ions are more stable than others, like Cu(I) versus Cu(II), and understand the factors that contribute to the high densities of these metals. A key concept covered is the origin of color in transition metal compounds, which arises from the presence of unpaired electrons. The solutions also guide you through calculating magnetic moments using the spin-only formula and identifying common oxidation states within the lanthanoid series. Furthermore, you'll learn to recognize characteristic reactions like disproportionation. This chapter is crucial for grasping the fundamental characteristics and chemical behavior of d and f-block elements, providing a solid foundation for your exam preparation.

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Quick info

Board	CBSE
Class	Class 12
Subject	Chemistry Exemplar
Session	2026
Language	English
Type	NCERT Solutions
Chapter	Chapter 8

Chapter summary

Chapter 8 of the NCERT Class 12 Chemistry textbook deals with d and f-Block Elements. These NCERT Solutions provide detailed answers to MCQs, explaining concepts like electronic configurations, oxidation states, stability of ions, metallic radii and density trends, the origin of color in transition metal compounds, magnetic properties, and the characteristic behavior of lanthanoids. The solutions also cover disproportionation reactions, offering a thorough understanding of the chapter's core topics.

Learning outcomes

Understand the electronic configurations of transition elements and their ions.
Explain the stability of different oxidation states of transition metals.
Relate metallic radii and atomic mass to the density of transition metals.
Identify the cause of color in transition metal compounds.
Calculate magnetic moments based on unpaired electrons.
Recognize common oxidation states of lanthanoids.
Identify and explain disproportionation reactions.

Topics covered

Paper topics

General electronic configuration of d-block elements
Oxidation states of transition metals
Formation of coloured ions
Magnetic properties of transition metals
Metallic radii and density trends
Catalytic properties
Formation of alloys
Lanthanoids
Actinoids
General characteristics of f-block elements
Oxidation states of lanthanoids
Disproportionation reactions

Important topics

Electronic configurations and oxidation states
Color and magnetic properties
Stability of oxidation states
Lanthanoid characteristics
Disproportionation reactions

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Questions and Solutions

Multiple Choice Questions (MCQs) - Q. 1

Q. 1 Electronic configuration of a transition element X in +3 oxidation state is [Ar]3d⁵. What is its atomic number?

25 (b) 26 (c) 27 (d) 24

Solution: The electronic configuration of the transition element X in its +3 oxidation state is given as [Ar]3d⁵. This means that the ion X³⁺ has 18 electrons from the Argon core plus 5 electrons in the 3d subshell, totaling 18 + 5 = 23 electrons. Since this is the configuration after losing 3 electrons, the neutral atom X must have 23 + 3 = 26 electrons. The atomic number of an element is equal to the number of electrons in its neutral atom. Therefore, the atomic number of element X is 26. Option (b) is correct.

Multiple Choice Questions (MCQs) - Q. 2

Q. 2 The electronic configuration of Cu(II) is 3d⁹ whereas that of Cu(I) is 3d¹⁰. Which of the following is correct?

Cu(II) is more stable
Cu(II) is less stable

Cu(l) and Cu(ll) are equally stable

(d) Stability of Cu(I) and Cu(II) depends on nature of copper salts

Solution: While a completely filled d-subshell (3d¹⁰) in Cu(I) suggests high stability, Cu(II) (3d⁹) is often found to be more stable in aqueous solutions. This is attributed to the higher effective nuclear charge in Cu²⁺ compared to Cu⁺. The increased attraction of the nucleus for the electrons in Cu²⁺ leads to a greater release of energy upon hydration, which makes the Cu²⁺ ion more stable overall, despite not having a completely filled d-subshell.

Multiple Choice Questions (MCQs) - Q. 3

Q. 3 Metallic radii of some transition elements are given below. Which of these elements will have highest density?

Element Fe Co Ni Cu

Metallic radii/pm 126 125 125 128

Fe (b) Ni (c) Co (d) Cu

Solution: Density is generally influenced by both atomic mass and atomic volume (related to metallic radius). As we move from left to right across a period in the transition metals, the atomic mass generally increases, while the metallic radius tends to decrease or remain similar, leading to an increase in density. Among the given elements (Fe, Co, Ni, Cu), Cu is located furthest to the right. Although its metallic radius (128 pm) is slightly larger than Ni and Co (125 pm), its atomic mass is significantly higher. The trend of increasing density across the period is dominant, making Copper (Cu) the element with the highest density among these choices.

Multiple Choice Questions (MCQs) - Q. 4

Q. 4 Generally, transition elements form coloured salts due to the presence of unpaired electrons. Which of the following compounds will be coloured in solid state?

<math>Ag_2SO_4</math> (b) <math>CuF_2</math> (c) <math>ZnF_2</math> (d) Cu2Cl2

Solution: The colour of transition metal compounds arises from d-d electronic transitions, which occur when there are unpaired electrons in the d-orbitals. Let's examine the oxidation states and electronic configurations of the metal ions:

Ag₂SO₄: Silver (Ag) is in the +1 oxidation state. Ag⁺ has the electronic configuration [Kr]4d¹⁰. All d-orbitals are filled, so no d-d transitions occur. It is colourless.
CuF₂: Copper (Cu) is in the +2 oxidation state. Cu²⁺ has the electronic configuration [Ar]3d⁹. There is one unpaired electron in the 3d subshell, allowing for d-d transitions. Thus, CuF₂ is coloured.
ZnF₂: Zinc (Zn) is in the +2 oxidation state. Zn²⁺ has the electronic configuration [Ar]3d¹⁰. All d-orbitals are filled, so no d-d transitions occur. It is colourless.
Cu₂Cl₂: In this compound, copper is in the +1 oxidation state. Cu⁺ has the electronic configuration [Ar]3d¹⁰. All d-orbitals are filled, so no d-d transitions occur. It is colourless.

Therefore, CuF₂ is the coloured compound.

Multiple Choice Questions (MCQs) - Q. 5

Q. 5 On addition of small amount of KMnO₄ to concentrated H₂SO₄, a green oily compound is obtained which is highly explosive in nature. Identify the compound from the following.

<math>Mn_2O_7</math> (b) <math>MnO_2</math> (c) <math>MnSO_4</math> (d) Mn2O3

Solution: When potassium permanganate (KMnO₄) is treated with concentrated sulfuric acid (H₂SO₄), a highly reactive and explosive green oily compound is formed. This compound is manganese heptoxide, Mn₂O₇. The reaction can be represented as:

Mn₂O₇ contains manganese in the +7 oxidation state and is a powerful oxidizing agent.

Multiple Choice Questions (MCQs) - Q. 6

Q. 6 The magnetic nature of elements depends on the presence of unpaired electrons. Identify the configuration of transition element, which shows highest magnetic moment.

<math>3d^{7}</math> (b) <math>3d^5</math> (c) 3d8 (d) <math>3d^2</math>

Solution: The magnetic moment (μ) of a transition metal ion is primarily determined by the number of unpaired electrons ('n') and is calculated using the spin-only formula: μ = √{n(n+2)} BM. A higher number of unpaired electrons leads to a higher magnetic moment. Let's determine the number of unpaired electrons for each configuration:

3d⁷: 3 unpaired electrons (t_2g³ e_g² or t_2g⁴ e_g³ depending on crystal field splitting, but always 3 unpaired in high spin).
3d⁵: 5 unpaired electrons (maximum possible for d-orbitals, t_2g³ e_g² in high spin).
3d⁸: 2 unpaired electrons (t_2g⁶ e_g²).
3d²: 2 unpaired electrons (t_2g² e_g⁰ in high spin).

The configuration 3d⁵ has the maximum number of unpaired electrons (5), and therefore, it will exhibit the highest magnetic moment. For 3d⁵, μ = √{5(5+2)} = √{35} ≈ 5.92 BM.

Multiple Choice Questions (MCQs) - Q. 7

Q. 7 Which of the following oxidation state is common for all lanthanoids? <math>(a) +2</math> (b) <math>+3</math> <math>(c) +4</math> <math>(d) +5</math>

Solution: All lanthanoids exhibit a common oxidation state of +3. This is because the removal of three electrons (two from the outermost 6s subshell and one from the inner 4f or 5d subshell) leads to a stable electronic configuration, often a half-filled (4f⁷) or a completely filled (4f¹⁴) f-subshell. For example, Ce⁴⁺ ([Xe]4f⁰), Eu²⁺ ([Xe]4f⁷), and Yb²⁺ ([Xe]4f¹⁴) show +2 or +4 oxidation states, but the +3 state is characteristic of the entire series.

Multiple Choice Questions (MCQs) - Q. 8

Q. 8 Which of the following reactions are disproportionation reactions?

<math>Cu^+ \longrightarrow Cu^{2+} + Cu</math>
<math>3Mn0_4^- + 4H^+ \longrightarrow 2Mn0_4^- + Mn0_2^- + 2H_20</math>
<math>2KMnO_4 \longrightarrow K_2MnO_4 + MnO_2 + O_2</math>
<math>2Mn0_{4}^{-} + 3Mn^{2+} + 2H_{2}0 \longrightarrow 5Mn0_{2} + 4H^{+}</math>

(i) (b) (i), (ii) and (iii) (c) (ii), (iii) and (iv) (d) (i) and (iv)

Solution: A disproportionation reaction is a redox reaction where a single species is simultaneously oxidized and reduced. Let's analyze each reaction:

<math>Cu^+ \longrightarrow Cu^{2+} + Cu</math>: Here, copper in the +1 oxidation state is converted into Cu²⁺ (oxidation state +2, oxidized) and Cu (oxidation state 0, reduced). This is a disproportionation reaction.
<math>3Mn0_4^- + 4H^+ \longrightarrow 2Mn0_4^- + Mn0_2^- + 2H_20</math>: This reaction involves permanganate (MnO₄^-, Mn is +7) reacting to form permanganate (MnO₄^-, Mn is +7) and manganese dioxide (MnO₂, Mn is +4). This is not a disproportionation as the initial MnO₄^- is not reduced. (Note: The provided reaction seems unbalanced or incorrectly written, but the core species involved do not show disproportionation of a single Mn species).
<math>2KMnO_4 \longrightarrow K_2MnO_4 + MnO_2 + O_2</math>: In this reaction, permanganate ion (MnO₄^-, Mn is +7) decomposes. While MnO₂ has Mn in +4 oxidation state (reduced), the formation of O₂ (oxidation state 0) from the oxygen atoms in KMnO₄ is not a disproportionation of manganese itself. However, if we consider the decomposition of MnO₄^- where Mn(+7) goes to Mn(+6) in K₂MnO₄ and Mn(+4) in MnO₂, it can be considered a disproportionation of Mn(+7) into Mn(+6) and Mn(+4) if the oxygen is considered separately. A more common example of disproportionation involving Mn(+7) is in acidic medium. Given the options, this reaction is often cited as a disproportionation in some contexts where Mn(+7) is considered to disproportionate into Mn(+6) and Mn(+4).
<math>2Mn0_{4}^{-} + 3Mn^{2+} + 2H_{2}0 \longrightarrow 5Mn0_{2} + 4H^{+}</math>: This is a redox reaction where MnO₄^- (Mn is +7) reacts with Mn²⁺ (Mn is +2). MnO₄^- is reduced to MnO₂ (Mn is +4), and Mn²⁺ is oxidized to MnO₂ (Mn is +4). This is a comproportionation reaction, not disproportionation, as two different oxidation states of manganese are combining to form a single intermediate oxidation state.

Based on the clear definition of disproportionation, reaction (i) is a definite example. Reaction (iii) is also considered a disproportionation of Mn(+7) into Mn(+6) and Mn(+4) under certain conditions. Therefore, the most appropriate answer considering typical textbook examples is that (i) is a disproportionation. If (iii) is also considered a disproportionation, then option (b) might be intended, but the wording and typical examples lean towards (i) being the primary example.

Revisiting the options and common understanding: Reaction (i) is a classic example of disproportionation. Reaction (iii) involves the decomposition of KMnO₄. While Mn(+7) goes to Mn(+6) and Mn(+4), the oxygen is released as O₂. This is often classified as a decomposition rather than a pure disproportionation of manganese alone. However, some sources may classify it as such. Given the options, and the unambiguous nature of (i), it's the most certain disproportionation. If we strictly follow the definition where one element is both oxidized and reduced, (i) fits perfectly. Let's assume the question intends to include (i) as the primary example. If we consider the possibility of (iii) also being a disproportionation, then option (b) would be selected. However, without further context or clarification on how reaction (iii) is interpreted, (i) is the most straightforward answer. Let's re-evaluate based on common textbook treatments. Reaction (i) is always listed as disproportionation. Reaction (iii) is often listed as decomposition. Therefore, option (a) which includes only (i) is the most likely correct answer if only one is a clear disproportionation.

Final check: Reaction (i) is definitely disproportionation. Reaction (iii) is decomposition. Therefore, only (i) is a disproportionation reaction among the clear examples. Option (a) is (i).

Common mistakes

Confusing electronic configurations of neutral atoms versus ions.
Incorrectly predicting stability based solely on a filled d-orbital.
Miscalculating the number of unpaired electrons.
Difficulty in identifying disproportionation reactions.
Overlooking the influence of effective nuclear charge on stability.

Revision tips

Focus on understanding the relationship between electronic configuration and properties like color and magnetic moment.
Practice calculating magnetic moments for different d-electron configurations.
Review the common and exceptional oxidation states of transition metals and lanthanoids.
Pay close attention to the definitions and examples of disproportionation reactions.
Use the provided solutions to cross-check your understanding of each MCQ.

Practice MCQs

Q1. The electronic configuration of a transition element X in its +3 oxidation state is [Ar]3d5. What is the atomic number of this element X?

Q2. The electronic configuration of Copper(II) is 3d9, whereas that of Copper(I) is 3d10. Which of the following statements is correct regarding their stability?

Q3. Given the metallic radii of Fe (126 pm), Co (125 pm), Ni (125 pm), and Cu (128 pm), which of these elements will have the highest density?

Q4. Generally, transition elements form coloured salts due to the presence of unpaired electrons. Which of the following compounds will be coloured in the solid state?

Q5. Upon addition of a small amount of potassium permanganate (KMnO4) to concentrated sulfuric acid (H2SO4), a green oily compound is obtained which is highly explosive. Identify this compound.

Q6. The magnetic nature of elements depends on the presence of unpaired electrons. Identify the electronic configuration of a transition element that shows the highest magnetic moment.

Q7. Which of the following oxidation states is common for all lanthanoids?

Q8. Which of the following reactions represent disproportionation reactions?

Frequently asked questions

What are the key concepts covered in the NCERT Solutions for Class 12 Chemistry Chapter 8?

These solutions cover the general characteristics of d and f-block elements, including their electronic configurations, variable oxidation states, formation of coloured ions, magnetic properties, catalytic activity, and alloy formation. Specific focus is given to lanthanoids and their properties, as well as concepts like metallic radii, density, and disproportionation reactions.

Why are transition metal compounds often coloured?

Transition metal compounds are often coloured due to the presence of unpaired electrons in their d-orbitals. These unpaired electrons can absorb certain wavelengths of visible light, promoting electrons to higher energy d-orbitals (d-d transitions), and the transmitted or reflected light appears coloured.

How does the number of unpaired electrons affect the magnetic moment?

The magnetic moment of a transition metal ion is directly proportional to the number of unpaired electrons. The formula μ = √{n(n+2)} BM shows that as 'n' (number of unpaired electrons) increases, the magnetic moment (μ) also increases, indicating stronger paramagnetic character.

What is the most common oxidation state of lanthanoids?

The most common and stable oxidation state for all lanthanoids is +3. This is because losing three electrons (two from the 6s subshell and one from the 4f or 5d subshell) often leads to a stable electronic configuration like a half-filled (f7) or fully-filled (f14) f-subshell.

How can these NCERT Solutions help in exam preparation?

These solutions provide clear, step-by-step explanations for MCQs, helping students understand the underlying principles and problem-solving techniques. They reinforce key concepts and offer insights into common mistakes, making them an excellent tool for revision and self-assessment before exams.

What is a disproportionation reaction?

A disproportionation reaction is a type of redox reaction where an element in a specific oxidation state is simultaneously oxidized and reduced to form products with higher and lower oxidation states, respectively. An example is Cu+ disproportionating into Cu2+ and Cu.

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