Class 12 Chemistry Chapter 8 The d and f Block Elements introduces students to the unique properties of transition and inner-transition elements. These elements play a vital role in chemical reactivity, catalysis, and industrial processes. I

Class 12 Chemistry Chapter 8 The d and f Block

n this chapter, you will explore their electronic configurations, variable oxidation states, coloured compounds, complex formation, and magnetic properties. The solved NCERT problems provided here will help you understand key concepts clearly and boost your exam preparation.

Class 12 Chemistry Chapter 8 The d and f Block Elements Textbook Answers

Problem 8.1

What is the lowest oxidation state shown by Mn and why?
The lowest oxidation state of Mn is +2. This is because the Mn²⁺ ion has a stable 3d⁵ half-filled configuration, which is relatively more stable due to exchange energy.

Problem 8.2

How will you justify that second and third rows of transition elements resemble each other more than the first row?
Second and third row transition elements show similar atomic and ionic sizes due to the lanthanoid contraction, leading to similar chemical properties. Their oxidation states, complex formation, and other properties resemble more closely than with the first row.

Problem 8.3

Explain the occurrence of highest oxidation state in oxoanions of transition metals.
In oxoanions, transition metals form multiple bonds with oxygen, stabilizing their highest oxidation states. For example, in MnO₄⁻, Mn is in +7 oxidation state, stabilized by strong M=O π-bonding.

Problem 8.4

Which is a stronger reducing agent, Cr²⁺ or Fe²⁺ and why?
Cr²⁺ is a stronger reducing agent than Fe²⁺ because it can be easily oxidized to the more stable Cr³⁺ (d³), which has a half-filled t₂g configuration in an octahedral field.

Problem 8.5

How would you account for the increasing oxidising power in the series VO²⁺ < Cr₂O₇²⁻ < MnO₄⁻?
The oxidising power increases with increase in the oxidation state of the central metal ion. VO²⁺ (+5), Cr₂O₇²⁻ (+6), and MnO₄⁻ (+7) show increasing tendency to gain electrons (i.e., get reduced), thus stronger oxidising ability

Problem 8.6

Predict which is more stable: Fe³⁺ or Fe²⁺ and why?
Fe³⁺ is more stable in aqueous solution due to higher hydration energy and formation of more stable complexes. However, Fe²⁺ is also quite stable and can exist under reducing conditions.

Problem 8.7

Why is the highest oxidation state of a metal exhibited in its oxide or fluoride only?
Oxygen and fluorine are highly electronegative and can stabilize high oxidation states of metals through strong bonding. Oxides and fluorides can form multiple bonds and thus allow the metal to attain its highest oxidation state.

Problem 8.8

Which is a stronger oxidising agent: Fe³⁺ or Cu²⁺ and why?
Fe³⁺ is a stronger oxidising agent than Cu²⁺. Fe³⁺ tends to gain an electron to form Fe²⁺ (which is more stable), while Cu²⁺ is already stable and less likely to get reduced.

Problem 8.9

Why is Cr²⁺ reducing while Mn³⁺ oxidising when both have d⁴ configuration?
Cr²⁺ (d⁴) prefers to lose an electron and form Cr³⁺ (d³) which has a stable t₂g³ configuration. Mn³⁺ (d⁴) tends to gain an electron to form Mn²⁺ (d⁵), a stable half-filled configuration. Hence, Cr²⁺ is reducing and Mn³⁺ is oxidising.

Problem 8.10

How would you explain the fact that [MnO₄]⁻ is purple, while [Cr₂O₇]²⁻ is orange in colour?
Both ions absorb light in the visible region due to charge transfer transitions. The energy difference between the transitions leads to absorption of different wavelengths, giving [MnO₄]⁻ a purple colour and [Cr₂O₇]²⁻ an orange colour.

Problem 8.11

The E°(M²⁺/M) value for copper is positive (+0.34 V). What does it indicate?
It indicates that Cu²⁺ is more stable than Cu metal in aqueous solution. Copper does not easily get oxidised, and hence it is less reactive.

Problem 8.12

Out of Mn²⁺, Fe²⁺, and Cr²⁺, which ion is most stable and why?
Mn²⁺ is most stable because it has a d⁵ half-filled configuration, which is highly stable due to symmetry and exchange energy.

Problem 8.13

Why is Zn²⁺ not considered a transition element?
Zn²⁺ has a completely filled d¹⁰ configuration, and it does not show variable oxidation states or form coloured ions. Hence, it is not a transition element as per IUPAC definition.

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Understanding the unique behavior of transition and inner-transition elements becomes easier with a clear explanation of their chemical properties. This page has provided a comprehensive Class 12 Chemistry Chapter 8 The d and f Block solution to help students grasp key concepts like variable oxidation states, colour formation, and magnetic properties.

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