CBSE Class 12 Chemistry Chapter 16: d and f Block Elements NCERT Solutions

NCERT Solutions PDF Class 12 PDF

This resource provides detailed NCERT Solutions for Class 12 Chemistry, Chapter 16, focusing on the d and f Block Elements. It covers essential concepts such as determining the electronic configurations of various ions, explaining the enhanced stability of certain oxidation states (like Mn^2+ over Fe^2+), and analyzing the trend of increasing stability of the +2 oxidation state across the first half of the first-row transition elements. The solutions break down complex topics into understandable steps, helping students grasp the underlying principles of electronic structure and stability in transition metals. These solutions are designed to aid students in their exam preparation by offering clear explanations and accurate answers to key questions from the NCERT textbook.

Quick info

BoardCBSE
ClassClass 12
SubjectChemistry
Session2026
LanguageEnglish
TypeNCERT Solutions
ChapterChapter 16

Chapter summary

Chapter 16 of the NCERT Class 12 Chemistry textbook delves into the d and f block elements. These NCERT Solutions provide clear explanations for questions related to their electronic configurations, the stability of various oxidation states, and trends within the first-row transition elements. The solutions focus on understanding why certain ions are more stable and how oxidation states evolve with increasing atomic number, offering a solid foundation for students studying these important elements.

Learning outcomes

  • Determine the electronic configuration of transition metal ions.
  • Explain the stability of half-filled and fully-filled electronic configurations.
  • Compare the stability of different oxidation states of transition metals.
  • Analyze the trend in stability of the +2 oxidation state in the first-row transition elements.

Topics covered

Paper topics

  • Electronic configuration of d-block elements
  • Electronic configuration of f-block elements
  • Electronic configuration of transition metal ions
  • Stability of half-filled orbitals
  • Stability of fully-filled orbitals
  • Oxidation states of transition metals
  • Stability of Mn^2+ ion
  • Stability of Fe^2+ ion
  • Trends in oxidation states
  • d and f block elements

Important topics

  • Electronic configurations of ions
  • Stability of half-filled and fully-filled orbitals
  • Comparison of stability of oxidation states (e.g., Mn^2+ vs Fe^2+)
  • Trends in +2 oxidation state stability

PDF preview

Read page by page below. PDF is streamed from the official NCERT website — no download button on this page.

Loading document …
Page of
Loading page …

Questions and Solutions

Question 8.1

Write down the electronic configuration of:
  1. Cr3+
  2. Pm3+
  3. Cu+
  4. Ce4+
  5. Co2+
  6. Lu2+
  7. Mn2+
  8. Th4+
Solution:

To determine the electronic configuration of ions, we first write the configuration of the neutral atom and then remove electrons from the outermost shell(s). For transition metals, electrons are typically removed from the 4s orbital before the 3d orbital.

  1. Cr3+: The electronic configuration of neutral Chromium (Cr) is [Ar] 3d5 4s1. To form Cr3+, we remove one electron from the 4s orbital and two electrons from the 3d orbital. Thus, the configuration is [Ar] 3d3.
  2. Pm3+: Promethium (Pm) is a lanthanide. Its electronic configuration is [Xe] 4f5 6s2. To form Pm3+, we remove two 6s electrons and one 4f electron. Thus, the configuration is [Xe] 4f4.
  3. Cu+: The electronic configuration of neutral Copper (Cu) is [Ar] 3d10 4s1. To form Cu+, we remove the single electron from the 4s orbital. Thus, the configuration is [Ar] 3d10. This fully filled d-orbital configuration makes it very stable.
  4. Ce4+: Cerium (Ce) has the electronic configuration [Xe] 4f1 5d1 6s2. To form Ce4+, all four valence electrons are removed. Thus, the configuration is [Xe].
  5. Co2+: The electronic configuration of neutral Cobalt (Co) is [Ar] 3d7 4s2. To form Co2+, we remove two electrons from the 4s orbital. Thus, the configuration is [Ar] 3d7.
  6. Lu2+: Lutetium (Lu) has the electronic configuration [Xe] 4f14 5d1 6s2. To form Lu2+, we remove two electrons from the 6s orbital. Thus, the configuration is [Xe] 4f14 5d1.
  7. Mn2+: The electronic configuration of neutral Manganese (Mn) is [Ar] 3d5 4s2. To form Mn2+, we remove two electrons from the 4s orbital. Thus, the configuration is [Ar] 3d5. This half-filled d-orbital configuration is very stable.
  8. Th4+: Thorium (Th) is an actinoid. Its electronic configuration is [Rn] 6d2 7s2. To form Th4+, all four valence electrons are removed. Thus, the configuration is [Rn].

Question 8.2

Why are Mn2+ compounds more stable than Fe2+ compounds towards oxidation to their +3 state?
Solution:

The stability of an ion is often related to its electronic configuration, particularly the stability of its d-orbitals. Half-filled and fully-filled orbitals are known to have extra stability due to symmetry and exchange energy. Let's examine the electronic configurations of Mn2+ and Fe2+:

  • The electronic configuration of Mn2+ is [Ar] 3d5. This configuration features a half-filled d-orbital (d5), which is a particularly stable arrangement.
  • The electronic configuration of Fe2+ is [Ar] 3d6. This configuration has one electron more than a half-filled d-orbital.

When Mn2+ is oxidized to Mn3+, its configuration changes from the stable 3d5 to 3d4. This transition involves moving away from a stable state, making it less favorable. Conversely, when Fe2+ (3d6) is oxidized to Fe3+, its configuration changes to 3d5. This transition from a 3d6 configuration to a stable, half-filled 3d5 configuration is energetically favorable. Therefore, Fe2+ is more readily oxidized to Fe3+ than Mn2+ is to Mn3+, indicating that Mn2+ compounds are more stable towards oxidation to the +3 state.

Question 8.3

Explain briefly how the +2 state becomes more and more stable in the first half of the first row transition elements with increasing atomic number?
Solution:

In the first half of the first-row transition elements (from Scandium, Sc, to Manganese, Mn), the +2 oxidation state generally becomes more stable as the atomic number increases. This trend can be understood by looking at the electronic configurations of these elements and their +2 ions.

The +2 oxidation state arises from the loss of the two 4s electrons. As we move across the period from Sc (atomic number 21) to Mn (atomic number 25), electrons are progressively added to the 3d subshell:

  • Sc (+2 state): [Ar] 3d1
  • Ti (+2 state): [Ar] 3d2
  • V (+2 state): [Ar] 3d3
  • Cr (+2 state): [Ar] 3d4
  • Mn (+2 state): [Ar] 3d5

The stability of an ion is influenced by its electronic configuration. A half-filled d-orbital (d5) configuration is particularly stable. As we progress from Sc2+ to Mn2+, the number of d-electrons increases, and the electronic configuration approaches the highly stable half-filled state (d5) in Mn2+. This increasing stability of the d-electron configuration contributes to the increasing stability of the +2 oxidation state for these elements.

While other oxidation states are also possible and often more stable for some elements (like Cr and Mn), the stability of the +2 state specifically shows a clear increasing trend in this segment of the transition series due to the filling of the 3d orbitals towards the stable d5 configuration.

Common mistakes

  • Incorrectly determining the order of electron removal from orbitals when forming ions.
  • Not recognizing the significance of half-filled (d^5) and fully-filled (d^10) configurations for stability.
  • Confusing the stability of different oxidation states without considering electronic configurations.

Revision tips

  • Memorize the general electronic configurations of transition elements.
  • Practice writing electronic configurations for various ions, paying attention to electron removal.
  • Understand the principle of orbital stability (half-filled and fully-filled) and apply it to oxidation states.
  • Review the trends in oxidation states across the first-row transition elements.

Practice MCQs

Q1. Which of the following ions has a stable electronic configuration due to a half-filled d-orbital?

Q2. The electronic configuration of Cr^3+ is:

Q3. Why is Ce^4+ ion colorless?

Q4. Which of the following is the electronic configuration of Cu^+?

Q5. The stability of +2 oxidation state increases in the first row of transition elements from:

Frequently asked questions

What is the electronic configuration of Cr^3+?

The electronic configuration of Cr^3+ is [Ar] 3d^3. This is derived from Chromium's configuration ([Ar] 3d^5 4s^1) by removing one 4s electron and two 3d electrons.

Why are Mn^2+ compounds more stable than Fe^2+ compounds towards oxidation?

Mn^2+ has a stable half-filled d-orbital configuration ([Ar] 3d^5), while Fe^2+ has a [Ar] 3d^6 configuration. Losing one more electron to become Fe^3+ ([Ar] 3d^5) is energetically favorable for iron, making Fe^2+ easier to oxidize than Mn^2+.

How does the stability of the +2 oxidation state change in the first half of the first-row transition elements?

The stability of the +2 oxidation state generally increases from Sc to Mn. This is because as the atomic number increases, more d-electrons are added, and the d^n configuration becomes more stable, especially reaching the highly stable d^5 configuration in Mn^2+.

What is the electronic configuration of Ce^4+?

The electronic configuration of Ce^4+ is [Xe]. This is because Cerium (Ce) has the configuration [Xe] 4f^1 5d^1 6s^2, and Ce^4+ loses all these valence electrons.

What is the significance of a half-filled or fully-filled orbital in terms of stability?

Half-filled (e.g., d^5) and fully-filled (e.g., d^10) orbitals possess extra stability due to symmetry and exchange energy effects, making ions with such configurations more resistant to changes in oxidation state.

Content reviewed by the NCERT Help team. Editorial Team and update policy

NCERT Solutions PDF PDF on NCERT Help. URL unchanged for search indexing.