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The d- and f-Block Elements Complete NCERT Resource Guide
Mastering inorganic chemistry can feel like a mountain to climb, but The d- and f-Block Elements is one of the most high-scoring chapters in your CBSE Class 12 Chemistry syllabus. This comprehensive guide provides the ultimate Updated NCERT Solutions, detailed concept breakdowns, and vital Board Exam Questions 2026. Whether you are aiming to ace your CBSE board exams or tracking towards competitive exams like NEET and JEE, these student-friendly, step-by-step solutions will help you secure top marks.
Chapter NameThe d- and f-Block Elements
SubjectChemistry
ClassClass 12
BoardCBSE & State Boards
Important TopicsElectronic Configuration, Lanthanoid Contraction, Transition Element Properties, Potassium Permanganate ($KMnO_4$), Potassium Dichromate ($K_2Cr_2O_7$)
Difficulty LevelModerate (Requires conceptual clarity)
Exam WeightageApprox. 7 Marks
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Learning Objectives
After completing this chapter, students will be able to:
- Define transition and inner transition elements based on their electronic configurations.
- Explain the general trends in physical and chemical properties of 3d, 4d, and 5d transition series.
- Understand the causes and consequences of Lanthanoid Contraction and Actinoid Contraction.
- Write balanced chemical equations for the preparation and reactions of $K_2Cr_2O_7$ and $KMnO_4$.
- Appreciate the catalytic, magnetic, and interstitial compound-forming properties of d-block elements.
Key Concepts, Definitions & Formulas
Before diving into the NCERT textbook exercises, let us quickly revise the core rules and concepts that examiners love to test.
- Transition Elements: Elements that have partially filled d-orbitals in their ground state or in any of their common oxidation states. Note that Zinc ($Zn$), Cadmium ($Cd$), and Mercury ($Hg$) are not considered transition elements because they have completely filled $d^{10}$ configurations.
- General Electronic Configuration:
- d-Block: $(n-1)d^{1-10} ns^{1-2}$
- f-Block: $(n-2)f^{1-14} (n-1)d^{0-1} ns^2$
- Lanthanoid Contraction: The steady decrease in atomic and ionic radii of lanthanoid elements with an increase in atomic number. This happens due to the poor shielding effect of 4f electrons, pulling the outer shell closer to the nucleus.
- Magnetic Moment Formula: The spin-only magnetic moment ($\mu$) is calculated using the formula: $$\mu = \sqrt{n(n+2)} \text{ BM}$$ Where $n$ is the number of unpaired electrons and BM stands for Bohr Magneton.
- Interstitial Compounds: Compounds formed when small atoms like $H$, $C$, or $N$ get trapped inside the crystal lattices of transition metals. They are hard, have high melting points, and retain metallic conductivity.
Full NCERT Solutions for Class 12 Chemistry Chapter 8
Here are the complete, step-by-step updated NCERT solutions for all text and exercise questions. These answers are written in an ideal board-exam presentation style.
Question 1: Write down the electronic configuration of: (i) $Cr^{3+}$ (ii) $Pm^{3+}$ (iii) $Cu^+$ (iv) $Ce^{4+}$ (v) $Co^{2+}$ (vi) $Lu^{2+}$ (vii) $Mn^{2+}$ (viii) $Th^{4+}$
Step 1: Foundational Rule. To write the ionic configurations, first write the neutral atom's configuration and then remove the required number of electrons starting from the outermost $s$ shell, followed by the $d$ or $f$ shells.
Step 2: Configuration Derivations.
Step 2: Configuration Derivations.
- (i) $Cr^{3+}$: Neutral $Cr = [Ar] 3d^5 4s^1$. Removing 3 electrons gives $[Ar] 3d^3$.
- (ii) $Pm^{3+}$: Neutral $Pm (Z=61) = [Xe] 4f^5 6s^2$. Removing 3 electrons gives $[Xe] 4f^4$.
- (iii) $Cu^+$: Neutral $Cu = [Ar] 3d^{10} 4s^1$. Removing 1 electron gives $[Ar] 3d^{10}$.
- (iv) $Ce^{4+}$: Neutral $Ce (Z=58) = [Xe] 4f^1 5d^1 6s^2$. Removing 4 electrons gives $[Xe]$ (Stable noble gas core).
- (v) $Co^{2+}$: Neutral $Co = [Ar] 3d^7 4s^2$. Removing 2 electrons gives $[Ar] 3d^7$.
- (vi) $Lu^{2+}$: Neutral $Lu (Z=71) = [Xe] 4f^{14} 5d^1 6s^2$. Removing 2 electrons gives $[Xe] 4f^{14} 5d^1$.
- (vii) $Mn^{2+}$: Neutral $Mn = [Ar] 3d^5 4s^2$. Removing 2 electrons gives $[Ar] 3d^5$ (Highly stable half-filled state).
- (viii) $Th^{4+}$: Neutral $Th (Z=90) = [Rn] 6d^2 7s^2$. Removing 4 electrons gives $[Rn]$.
Question 2: Why are $Mn^{2+}$ compounds more stable than $Fe^{2+}$ towards oxidation to their +3 state?
Step 1: Electronic Configuration Check.
$Mn^{2+} = [Ar] 3d^5$
$Fe^{2+} = [Ar] 3d^6$
Step 2: Stability Principle. The $3d^5$ configuration of $Mn^{2+}$ is exactly half-filled, which gives it extra thermodynamic stability. Removing another electron to make it $Mn^{3+}$ ($3d^4$) disrupts this stability, so $Mn^{2+}$ strongly resists oxidation.
Step 3: Comparative Analysis. On the other hand, $Fe^{2+}$ ($3d^6$) readily loses one electron to achieve a highly stable, half-filled $3d^5$ configuration ($Fe^{3+}$). Hence, $Fe^{2+}$ is easily oxidized compared to $Mn^{2+}$.
$Mn^{2+} = [Ar] 3d^5$
$Fe^{2+} = [Ar] 3d^6$
Step 2: Stability Principle. The $3d^5$ configuration of $Mn^{2+}$ is exactly half-filled, which gives it extra thermodynamic stability. Removing another electron to make it $Mn^{3+}$ ($3d^4$) disrupts this stability, so $Mn^{2+}$ strongly resists oxidation.
Step 3: Comparative Analysis. On the other hand, $Fe^{2+}$ ($3d^6$) readily loses one electron to achieve a highly stable, half-filled $3d^5$ configuration ($Fe^{3+}$). Hence, $Fe^{2+}$ is easily oxidized compared to $Mn^{2+}$.
Question 3: Explain briefly how +2 oxidation state becomes more and more stable in the first half of the first row transition elements with increasing atomic number?
Step 1: Structural Progression. In the first transition series ($Sc$ to $Mn$), as the atomic number increases, the $3d$ orbital gets progressively filled with electrons while the outermost $4s^2$ shell remains constant.
Step 2: Ion Formation Mechanics. When these elements lose their two $4s$ electrons, they form $+2$ ions. As we move from $Sc$ to $Mn$, the effective nuclear charge increases, which holds the $3d$ electrons more tightly.
Step 3: Stabilization Matrix. Furthermore, the electronic configuration approaches a highly symmetrical, stable half-filled $3d^5$ state at $Mn^{2+}$. Consequently, the tendency to lose further $3d$ electrons decreases, making the $+2$ oxidation state increasingly stable across the first half of the series.
Step 2: Ion Formation Mechanics. When these elements lose their two $4s$ electrons, they form $+2$ ions. As we move from $Sc$ to $Mn$, the effective nuclear charge increases, which holds the $3d$ electrons more tightly.
Step 3: Stabilization Matrix. Furthermore, the electronic configuration approaches a highly symmetrical, stable half-filled $3d^5$ state at $Mn^{2+}$. Consequently, the tendency to lose further $3d$ electrons decreases, making the $+2$ oxidation state increasingly stable across the first half of the series.
Question 4: To what extent do the electronic configurations decide the stability of oxidation states in the first series of the transition elements? Illustrate your answer with examples.
Step 1: Core Dependency. Electronic configurations play a decisive role in determining the stability of oxidation states. The most stable states correspond to empty ($d^0$), half-filled ($d^5$), and completely filled ($d^{10}$) d-orbitals.
Step 2: Verification Example 1 (Empty $d^0$). Scandium ($Sc$) has the configuration $[Ar] 3d^1 4s^2$. It easily loses all 3 electrons to form $Sc^{3+}$ ($[Ar]$ or $3d^0$), which is remarkably stable.
Step 3: Verification Example 2 (Half-filled $d^5$). Manganese ($Mn$) shows a highly stable $+2$ state because $Mn^{2+}$ has a stable half-filled $[Ar] 3d^5$ configuration.
Step 4: Verification Example 3 (Fully-filled $d^{10}$). Zinc ($Zn = [Ar] 3d^{10} 4s^2$) exclusively shows the $+2$ oxidation state because $Zn^{2+}$ retains the completely filled stable $3d^{10}$ configuration.
Step 2: Verification Example 1 (Empty $d^0$). Scandium ($Sc$) has the configuration $[Ar] 3d^1 4s^2$. It easily loses all 3 electrons to form $Sc^{3+}$ ($[Ar]$ or $3d^0$), which is remarkably stable.
Step 3: Verification Example 2 (Half-filled $d^5$). Manganese ($Mn$) shows a highly stable $+2$ state because $Mn^{2+}$ has a stable half-filled $[Ar] 3d^5$ configuration.
Step 4: Verification Example 3 (Fully-filled $d^{10}$). Zinc ($Zn = [Ar] 3d^{10} 4s^2$) exclusively shows the $+2$ oxidation state because $Zn^{2+}$ retains the completely filled stable $3d^{10}$ configuration.
Question 5: What may be the stable oxidation state of the transition element with the following d electron configurations in the ground state of their atoms: $3d^3$, $3d^5$, $3d^8$ and $3d^4$?
Step 1: Analysis of $3d^3$. For Vanadium ($V$): Stable oxidation states are $+2, +3, +4,$ and $+5$. The $+5$ state ($3d^0$) and $+3$ state (stable $t_{2g}^3$ configuration in coordination compounds) are highly stable.
Step 2: Analysis of $3d^5$. For Chromium ($Cr$, $3d^5 4s^1$), $+3$ and $+6$ are stable. For Manganese ($Mn$, $3d^5 4s^2$), $+2$ and $+7$ are exceptionally stable.
Step 3: Analysis of $3d^8$. For Nickel ($Ni$): The most common and highly stable oxidation state is $+2$ ($3d^8 4s^0$).
Step 4: Analysis of $3d^4$. Note that ground state atoms typically do not have a $3d^4$ ground state due to shifting (like $Cr$ which becomes $3d^5 4s^1$). However, if an atom has a $3d^4$ valence structure hypothetically, it stabilizes at $+6$ ($d^0$) or $+3$ ($d^1$).
Step 2: Analysis of $3d^5$. For Chromium ($Cr$, $3d^5 4s^1$), $+3$ and $+6$ are stable. For Manganese ($Mn$, $3d^5 4s^2$), $+2$ and $+7$ are exceptionally stable.
Step 3: Analysis of $3d^8$. For Nickel ($Ni$): The most common and highly stable oxidation state is $+2$ ($3d^8 4s^0$).
Step 4: Analysis of $3d^4$. Note that ground state atoms typically do not have a $3d^4$ ground state due to shifting (like $Cr$ which becomes $3d^5 4s^1$). However, if an atom has a $3d^4$ valence structure hypothetically, it stabilizes at $+6$ ($d^0$) or $+3$ ($d^1$).
Question 6: Name the oxometal anions of the first series of the transition metals in which the metal exhibits the oxidation state equal to its group number.
Step 1: Group 6 Evaluation. Chromate ion ($CrO_4^{2-}$): Chromium is in Group 6. Oxidation state of $Cr$ in $CrO_4^{2-}$ is $+6$.
Step 2: Group 7 Evaluation. Permanganate ion ($MnO_4^-$): Manganese is in Group 7. Oxidation state of $Mn$ in $MnO_4^-$ is $+7$.
Step 3: Group 5 Evaluation. Vanadate ion ($VO_4^{3-}$): Vanadium is in Group 5. Oxidation state of $V$ in $VO_4^{3-}$ is $+5$.
Step 2: Group 7 Evaluation. Permanganate ion ($MnO_4^-$): Manganese is in Group 7. Oxidation state of $Mn$ in $MnO_4^-$ is $+7$.
Step 3: Group 5 Evaluation. Vanadate ion ($VO_4^{3-}$): Vanadium is in Group 5. Oxidation state of $V$ in $VO_4^{3-}$ is $+5$.
Question 7: What is lanthanoid contraction? What are the consequences of lanthanoid contraction?
Step 1: Scientific Definition. Lanthanoid contraction refers to the gradual, steady decrease in the atomic and ionic radii of lanthanoids with the increase in atomic number from Lanthanum ($Z=57$) to Lutetium ($Z=71$).
Step 2: Root Cause Mechanics. This happens because of the poor shielding effect of 4f electrons. As the nuclear charge increases by one unit at each step, the inner 4f electrons fail to shield the outer electrons effectively from the growing positive pull of the nucleus.
Step 3: Primary Consequences.
Step 2: Root Cause Mechanics. This happens because of the poor shielding effect of 4f electrons. As the nuclear charge increases by one unit at each step, the inner 4f electrons fail to shield the outer electrons effectively from the growing positive pull of the nucleus.
Step 3: Primary Consequences.
- Similarity of 2nd and 3rd transition series: Due to this contraction, the atomic radii of elements of the 4d series (e.g., Zirconium, $Zr$) and 5d series (e.g., Hafnium, $Hf$) become nearly identical. As a result, they exhibit very similar chemical and physical properties.
- Separation difficulty: Because their chemical properties are nearly identical, separating pure lanthanoids from natural mixtures is extremely difficult.
- Basic strength trend: The basic strength of their hydroxides decreases from $La(OH)_3$ to $Lu(OH)_3$. $La(OH)_3$ is the most basic, while $Lu(OH)_3$ is the least basic.
Question 8: What are the characteristics of the transition elements and why are they called transition elements? Which of the d-block elements may not be regarded as the transition elements?
Step 1: Core Characteristics. They exhibit variable oxidation states, form colored ions and complex compounds, show paramagnetic properties due to unpaired electrons, and are widely used as industrial catalysts.
Step 2: Naming Logic. They are called transition elements because their properties represent a transition (bridge) between the highly electropositive s-block metals and the highly electronegative p-block non-metals. Technically, they are defined as elements with partially filled d-orbitals.
Step 3: Exceptions Matrix. Zinc ($Zn$), Cadmium ($Cd$), and Mercury ($Hg$) are not regarded as transition elements because they have completely filled d-orbitals ($d^{10}$) in their ground states as well as in their common oxidation states (like $Zn^{2+}$).
Step 2: Naming Logic. They are called transition elements because their properties represent a transition (bridge) between the highly electropositive s-block metals and the highly electronegative p-block non-metals. Technically, they are defined as elements with partially filled d-orbitals.
Step 3: Exceptions Matrix. Zinc ($Zn$), Cadmium ($Cd$), and Mercury ($Hg$) are not regarded as transition elements because they have completely filled d-orbitals ($d^{10}$) in their ground states as well as in their common oxidation states (like $Zn^{2+}$).
Question 9: In what way is the electronic configuration of the transition elements different from that of the non-transition elements?
Step 1: Transition Elements Profile. They possess an incompletely filled $(n-1)d$ orbital. Their general valence configuration is $(n-1)d^{1-10} ns^{1-2}$.
Step 2: Non-Transition Elements Profile. In s and p-block elements, their inner d-orbitals are either completely empty or completely filled, and their valence electrons reside purely in the outermost $s$ or $p$ shells. Their general configuration is $ns^{1-2}$ or $ns^2 np^{1-6}$.
Step 2: Non-Transition Elements Profile. In s and p-block elements, their inner d-orbitals are either completely empty or completely filled, and their valence electrons reside purely in the outermost $s$ or $p$ shells. Their general configuration is $ns^{1-2}$ or $ns^2 np^{1-6}$.
Question 10: What are the different oxidation states exhibited by the Lanthanoids?
Step 1: Principal State. The principal and most stable oxidation state exhibited by all lanthanoids is $+3$.
Step 2: Variable States. However, some lanthanoids occasionally exhibit $+2$ and $+4$ oxidation states in solution or solid compounds when such states lead to extra stable empty ($f^0$), half-filled ($f^7$), or fully-filled ($f^{14}$) configurations.
Step 3: Concrete Examples.
Step 2: Variable States. However, some lanthanoids occasionally exhibit $+2$ and $+4$ oxidation states in solution or solid compounds when such states lead to extra stable empty ($f^0$), half-filled ($f^7$), or fully-filled ($f^{14}$) configurations.
Step 3: Concrete Examples.
- $Ce^{4+}$ is stable because it achieves an $f^0$ configuration.
- $Eu^{2+}$ is stable because it achieves an $f^7$ configuration.
Question 11: Explain giving reasons: (i) Transition metals and many of their compounds show paramagnetic behaviour. (ii) The enthalpies of atomisation of the transition metals are high. (iii) The transition metals generally form coloured compounds. (iv) Transition metals and their many compounds act as good catalysts.
Step 1: Paramagnetic Behavior. This is due to the presence of unpaired electrons in their $(n-1)d$ orbitals. These unpaired spins create a net magnetic moment when placed in an external magnetic field.
Step 2: High Enthalpies of Atomisation. Transition metals have a large number of unpaired electrons in their d-orbitals. This allows them to form strong metallic bonds and covalent links between atoms, which requires a huge amount of thermal energy to break apart into individual gaseous atoms.
Step 3: Coloured Compounds. Due to the presence of unpaired d-electrons, when visible light falls on these compounds, the electrons absorb specific wavelengths of light and jump from a lower-energy d-orbital to a higher-energy d-orbital. This is called a d-d transition. The transmitted light appears as the complementary color.
Step 4: Catalytic Properties. Transition metals make great catalysts because they can exhibit variable oxidation states (allowing them to form unstable intermediate active complexes easily) and they offer a large surface area with free valencies for reactant molecules to adsorb and react efficiently.
Step 2: High Enthalpies of Atomisation. Transition metals have a large number of unpaired electrons in their d-orbitals. This allows them to form strong metallic bonds and covalent links between atoms, which requires a huge amount of thermal energy to break apart into individual gaseous atoms.
Step 3: Coloured Compounds. Due to the presence of unpaired d-electrons, when visible light falls on these compounds, the electrons absorb specific wavelengths of light and jump from a lower-energy d-orbital to a higher-energy d-orbital. This is called a d-d transition. The transmitted light appears as the complementary color.
Step 4: Catalytic Properties. Transition metals make great catalysts because they can exhibit variable oxidation states (allowing them to form unstable intermediate active complexes easily) and they offer a large surface area with free valencies for reactant molecules to adsorb and react efficiently.
Question 12: What are interstitial compounds? Why are such compounds well known for transition metals?
Step 1: Definition. Interstitial compounds are non-stoichiometric chemical compounds formed when small non-metal atoms like hydrogen ($H$), carbon ($C$), nitrogen ($N$), or boron ($B$) get trapped inside the vacant interstitial spaces (voids) within the crystal lattice of transition metals.
Step 2: Lattice Compatibility. They are well known for transition metals because transition metals crystallize in close-packed structures (like HCP or FCC) which naturally leave behind distinct interstitial voids. Because transition metal atoms are relatively large and their crystal lattices are structurally stable, they easily accommodate these small non-metal atoms without breaking their basic metallic framework.
Step 2: Lattice Compatibility. They are well known for transition metals because transition metals crystallize in close-packed structures (like HCP or FCC) which naturally leave behind distinct interstitial voids. Because transition metal atoms are relatively large and their crystal lattices are structurally stable, they easily accommodate these small non-metal atoms without breaking their basic metallic framework.
Question 13: How is the variability in oxidation states of transition metals different from that of the non-transition metals? Illustrate with examples.
Step 1: Transition Metals Mechanics. The variability in oxidation states arises because electrons from both the $(n-1)d$ and $ns$ orbitals can participate in bonding since their energy levels are very close. Crucially, these oxidation states differ by unity (1 unit). For example, Manganese shows $+2, +3, +4, +5, +6, +7$.
Step 2: Non-Transition Metals Mechanics. The variable oxidation states in p-block elements occur due to the inert pair effect and typically differ by two units. For example, Lead exhibits $+2$ and $+4$ states, and Tin exhibits $+2$ and $+4$ states.
Step 2: Non-Transition Metals Mechanics. The variable oxidation states in p-block elements occur due to the inert pair effect and typically differ by two units. For example, Lead exhibits $+2$ and $+4$ states, and Tin exhibits $+2$ and $+4$ states.
Question 14: Describe the preparation of potassium dichromate from chromite ore. What is the effect of increasing pH on a solution of potassium dichromate?
Step 1: Ore Fusion Phase. Conversion of Chromite ore to Sodium Chromate: The ore is fused with sodium carbonate ($Na_2CO_3$) in the presence of excess air.
$$4FeCr_2O_4 + 8Na_2CO_3 + 7O_2 \rightarrow 8Na_2CrO_4 + 2Fe_2O_3 + 8CO_2$$
Step 2: Acidification Phase. Conversion of Sodium Chromate to Sodium Dichromate: The yellow solution of sodium chromate is filtered and acidified with concentrated sulfuric acid.
$$2Na_2CrO_4 + H_2SO_4 \rightarrow Na_2Cr_2O_7 + Na_2SO_4 + H_2O$$
Step 3: Precipitation Phase. Conversion of Sodium Dichromate to Potassium Dichromate: The highly soluble sodium dichromate is treated with a calculated amount of Potassium Chloride ($KCl$) to precipitate out the less soluble orange crystals of $K_2Cr_2O_7$.
$$Na_2Cr_2O_7 + 2KCl \rightarrow K_2Cr_2O_7 \downarrow + 2NaCl$$
Step 4: pH Variation Dynamics. In aqueous solution, chromate ($CrO_4^{2-}$) and dichromate ($Cr_2O_7^{2-}$) ions exist in a reversible equilibrium depending on the pH:
$$2CrO_4^{2-} \text{ (Yellow)} + 2H^+ \rightleftharpoons Cr_2O_7^{2-} \text{ (Orange)} + H_2O$$
Increasing pH (Adding Base) drops the $H^+$ concentration, shifting the equilibrium to the left. The orange dichromate converts into yellow chromate ($CrO_4^{2-}$). Decreasing pH (Adding Acid) shifts the equilibrium to the right yielding orange dichromate.
Question 15: Describe the oxidizing action of potassium dichromate and write the ionic equations for its reaction with: (i) Iodide (ii) Iron(II) solution (iii) $H_2S$
Step 1: Core Action Profile. Potassium dichromate ($K_2Cr_2O_7$) acts as a strong oxidizing agent in an acidic medium. The core reduction half-reaction is:
$$Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$$
Step 2: Ionic Equation - Reaction with Iodide ($I^-$ to $I_2$).
$$Cr_2O_7^{2-} + 14H^+ + 6I^- \rightarrow 2Cr^{3+} + 3I_2 + 7H_2O$$
Step 3: Ionic Equation - Reaction with Iron(II) ($Fe^{2+}$ to $Fe^{3+}$).
$$Cr_2O_7^{2-} + 14H^+ + 6Fe^{2+} \rightarrow 2Cr^{3+} + 6Fe^{3+} + 7H_2O$$
Step 4: Ionic Equation - Reaction with $H_2S$ ($H_2S$ to $S$).
$$Cr_2O_7^{2-} + 8H^+ + 3H_2S \rightarrow 2Cr^{3+} + 3S \downarrow + 7H_2O$$
Question 16: Describe the preparation of potassium permanganate. How does acidified permanganate solution react with (i) iron(II) ions (ii) $SO_2$ (iii) oxalic acid? Write the ionic equations for the reactions.
Step 1: Commercial Preparation Phase. Potassium permanganate is prepared commercially from Pyrolusite ore ($MnO_2$).
Step 3: Reaction with Iron(II) ions ($Fe^{2+}$). $$MnO_4^- + 8H^+ + 5Fe^{2+} \rightarrow Mn^{2+} + 5Fe^{3+} + 4H_2O$$ Step 4: Reaction with Sulfur Dioxide ($SO_2$). $$2MnO_4^- + 5SO_2 + 2H_2O \rightarrow 2Mn^{2+} + 5SO_4^{2-} + 4H^+$$ Step 5: Reaction with Oxalic Acid ($C_2O_4^{2-}$). $$2MnO_4^- + 16H^+ + 5C_2O_4^{2-} \rightarrow 2Mn^{2+} + 10CO_2 \uparrow + 8H_2O$$
- Pyrolusite ore ($MnO_2$) is fused with potassium hydroxide ($KOH$) in the presence of an oxidizing agent like $O_2$ or $KNO_3$ to form dark-green Potassium Manganate ($K_2MnO_4$).
$$2MnO_2 + 4KOH + O_2 \rightarrow 2K_2MnO_4 \text{ (Green)} + 2H_2O$$ - The green potassium manganate is then disproportionated in an acidic or neutral solution (or oxidized electrolytically) to yield purple Potassium Permanganate ($KMnO_4$).
$$3MnO_4^{2-} + 4H^+ \rightarrow 2MnO_4^- \text{ (Purple)} + MnO_2 + 2H_2O$$
Step 3: Reaction with Iron(II) ions ($Fe^{2+}$). $$MnO_4^- + 8H^+ + 5Fe^{2+} \rightarrow Mn^{2+} + 5Fe^{3+} + 4H_2O$$ Step 4: Reaction with Sulfur Dioxide ($SO_2$). $$2MnO_4^- + 5SO_2 + 2H_2O \rightarrow 2Mn^{2+} + 5SO_4^{2-} + 4H^+$$ Step 5: Reaction with Oxalic Acid ($C_2O_4^{2-}$). $$2MnO_4^- + 16H^+ + 5C_2O_4^{2-} \rightarrow 2Mn^{2+} + 10CO_2 \uparrow + 8H_2O$$
Question 17: For $M^{2+}/M$ and $M^{3+}/M^{2+}$ systems, the $E^\ominus$ values for some metals are as follows:
$Cr^{2+}/Cr = -0.9V$, $Cr^{3+}/Cr^{2+} = -0.4V$
$Mn^{2+}/Mn = -1.2V$, $Mn^{3+}/Mn^{2+} = +1.5V$
$Fe^{2+}/Fe = -0.4V$, $Fe^{3+}/Fe^{2+} = +0.8V$
Use this data to comment upon: (i) The stability of $Fe^{3+}$ in acid solution as compared to that of $Cr^{3+}$ or $Mn^{3+}$. (ii) The ease with which iron can be oxidized as compared to a similar process for either chromium or manganese metal.
Step 1: Analyzing +3 Ion Stability. The $E^\ominus$ value for $Mn^{3+}/Mn^{2+}$ is highly positive ($+1.5V$), meaning $Mn^{3+}$ is extremely unstable and eagerly changes into stable $Mn^{2+}$ ($3d^5$). The value for $Fe^{3+}/Fe^{2+}$ is $+0.8V$, indicating it is moderately stable. However, the $E^\ominus$ value for $Cr^{3+}/Cr^{2+}$ is negative ($-0.4V$), meaning $Cr^{3+}$ is incredibly stable and does not want to reduce to $Cr^{2+}$. Thus, the order of stability of $+3$ ions toward reduction is: $Cr^{3+} > Fe^{3+} > Mn^{3+}$.
Step 2: Analyzing Oxidation Ease. Looking at the $M^{2+}/M$ reduction potential values:
Step 2: Analyzing Oxidation Ease. Looking at the $M^{2+}/M$ reduction potential values:
- $Mn^{2+}/Mn = -1.2V$
- $Cr^{2+}/Cr = -0.9V$
- $Fe^{2+}/Fe = -0.4V$
Question 18: Predict which of the following will be coloured in aqueous solution? $Ti^{3+}, V^{3+}, Sc^{3+}, Mn^{2+}, Fe^{3+}, Co^{2+}, Cu^+, Zn^{2+}$.
Step 1: Color Mechanics Rule. An ion is colored in aqueous solution if it contains unpaired d-electrons capable of undergoing d-d transitions. Ions with $d^0$ or $d^{10}$ configurations are colorless.
Step 2: Configuration Breakdown.
Step 2: Configuration Breakdown.
| Ion | Configuration | Unpaired Electrons | Color Status |
|---|---|---|---|
| $Ti^{3+}$ | $[Ar] 3d^1$ | 1 | Coloured (Purple) |
| $V^{3+}$ | $[Ar] 3d^2$ | 2 | Coloured (Green) |
| $Sc^{3+}$ | $[Ar] 3d^0$ | 0 | Colourless |
| $Mn^{2+}$ | $[Ar] 3d^5$ | 5 | Coloured (Pink) |
| $Fe^{3+}$ | $[Ar] 3d^5$ | 5 | Coloured (Yellow) |
| $Co^{2+}$ | $[Ar] 3d^7$ | 3 | Coloured (Pink/Red) |
| $Cu^+$ | $[Ar] 3d^{10}$ | 0 | Colourless |
| $Zn^{2+}$ | $[Ar] 3d^{10}$ | 0 | Colourless |
Question 19: Compare the chemistry of actinoids with that of the lanthanoids with special reference to: (i) electronic configuration (ii) atomic and ionic sizes and (iii) oxidation state.
Step 1: Structured Comparison Analysis. Here is a structured comparison outlining the key differences between Lanthanoids and Actinoids:
| Feature | Lanthanoids | Actinoids |
|---|---|---|
| Electronic Configuration | Involves the progressive filling of the 4f orbital. General form: $[Xe] 4f^{1-14} 5d^{0-1} 6s^2$. | Involves the progressive filling of the 5f orbital. General form: $[Rn] 5f^{1-14} 6d^{0-1} 7s^2$. |
| Atomic/Ionic Sizes | Size gradually decreases due to Lanthanoid contraction. Shielding by 4f is poor. | Size decreases at a greater rate due to Actinoid contraction. Shielding by 5f is even poorer. |
| Oxidation States | Principally show $+3$. Rarely show $+2$ and $+4$. | Show a much wider variety of oxidation states ($+3, +4, +5, +6, +7$) due to low energy gaps between 5f, 6d, and 7s shells. |
Question 20: What is meant by ‘disproportionation’? Give two examples of disproportionation reaction in aqueous solution.
Step 1: Chemical Definition. A chemical reaction in which the same substance (element) simultaneously undergoes both oxidation and reduction to form two different products containing the element in different oxidation states is called a disproportionation reaction.
Step 2: Example 1 Matrix. Disproportionation of Manganese(VI) in Acid: Green manganate ion converts into purple permanganate and manganese dioxide. $$3MnO_4^{2-} \text{ (+6 state)} + 4H^+ \rightarrow 2MnO_4^- \text{ (+7 state)} + MnO_2 \text{ (+4 state)} + 2H_2O$$
Step 3: Example 2 Matrix. Disproportionation of Copper(I) Ion: Copper(I) is unstable in aqueous medium and instantly forms Copper(II) and solid copper metal. $$2Cu^+ \text{ (+1 state)} \rightarrow Cu^{2+} \text{ (+2 state)} + Cu \text{ (0 state)}$$
Step 2: Example 1 Matrix. Disproportionation of Manganese(VI) in Acid: Green manganate ion converts into purple permanganate and manganese dioxide. $$3MnO_4^{2-} \text{ (+6 state)} + 4H^+ \rightarrow 2MnO_4^- \text{ (+7 state)} + MnO_2 \text{ (+4 state)} + 2H_2O$$
Step 3: Example 2 Matrix. Disproportionation of Copper(I) Ion: Copper(I) is unstable in aqueous medium and instantly forms Copper(II) and solid copper metal. $$2Cu^+ \text{ (+1 state)} \rightarrow Cu^{2+} \text{ (+2 state)} + Cu \text{ (0 state)}$$
Extra Important Questions (Board Exam Style 2026)
Target verification tasks matching high-yield board testing frameworks engineered to lock down full marks in your 2026 exams.
Section A: Multiple Choice Questions (MCQs)
Q1. Which of the following ions has the maximum magnetic moment?
A) $Mn^{2+}$
B) $Fe^{2+}$
C) $Ti^{2+}$
D) $Cr^{2+}$
Correct Answer Choice: A
Step 1: Configuration Analysis. $Mn^{2+}$ has a $3d^5$ configuration containing 5 unpaired electrons (maximum possible in d-series).
Step 2: Calculation Matrix. Applying the spin-only formula: $\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \text{ BM}$.
Step 1: Configuration Analysis. $Mn^{2+}$ has a $3d^5$ configuration containing 5 unpaired electrons (maximum possible in d-series).
Step 2: Calculation Matrix. Applying the spin-only formula: $\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \text{ BM}$.
Q2. The element which does NOT show variable oxidation states in the d-block is:
A) $Fe$
B) $Mn$
C) $Sc$
D) $Cu$
Correct Answer Choice: C
Step 1: Structural Verification. Scandium ($Sc$) exhibits only a $+3$ oxidation state because it quickly loses all three outer valence electrons to attain a stable noble gas configuration.
Step 1: Structural Verification. Scandium ($Sc$) exhibits only a $+3$ oxidation state because it quickly loses all three outer valence electrons to attain a stable noble gas configuration.
Section B: Assertion-Reason Questions
Directions: Choose option (A) if both (A) and (R) are true and (R) is the correct explanation. Choose (B) if both are true but independent. Choose (C) if (A) is true but (R) is false. Choose (D) if (A) is false.
Q3. Assertion (A): $Zr$ (Zirconium) and $Hf$ (Hafnium) have almost identical atomic radii.
Reason (R): It is due to the lanthanoid contraction.
Correct Answer Choice: A
Step 1: Validation. Hafnium follows the lanthanoid series. Due to the very poor shielding effect of the 14 f-electrons, a significant contraction occurs, neutralizing the expected size increase down the group. Hence, $Zr$ and $Hf$ have nearly identical sizes.
Step 1: Validation. Hafnium follows the lanthanoid series. Due to the very poor shielding effect of the 14 f-electrons, a significant contraction occurs, neutralizing the expected size increase down the group. Hence, $Zr$ and $Hf$ have nearly identical sizes.
Section C: Short & Long Answer Questions
Q4. Explain why $Cu^+$ ion is not stable in aqueous solutions. Is it paramagnetic or diamagnetic?
Step 1: Stability Mechanics. $Cu^+$ undergoes disproportionation in an aqueous solution. The high hydration enthalpy of $Cu^{2+}$ (due to higher charge density) more than compensates for the second ionization enthalpy of Copper, driving the reaction forward.
Step 2: Magnetic State Verification.
Step 2: Magnetic State Verification.
- $Cu^{2+} = [Ar] 3d^9 \rightarrow$ Contains 1 unpaired electron (paramagnetic).
- $Cu^+ = [Ar] 3d^{10} \rightarrow$ All electrons are completely paired. No net magnetic spin exists, making it diamagnetic.
Q5. Why do transition elements form interstitial compounds? State any two physical properties of these compounds.
Step 1: Structural Reason. They form interstitial compounds because their crystal lattices contain open spaces (voids) where small non-metal atoms like $H, C,$ or $N$ easily become trapped.
Step 2: Physical Properties.
Step 2: Physical Properties.
- They have exceptionally high melting points, higher than pure parent metals.
- They are extremely hard (some approach diamond-grade hardness).
Section D: Case-Based Questions
Q6. Read the passage and answer the questions:
The d-block elements form a unique family of metallic elements. They are famous for their vibrant colors in solution and their catalytic abilities. The stability of their oxidation states depends largely on their electronic configuration.
(a) Identify the metal from the 3d series which is an essential component of Vitamin B12.
(b) Calculate the magnetic moment of an element with atomic number 26 in its +2 oxidation state.
Step 1: Resolving Part (a). The central metal atom present in Vitamin B12 is Cobalt ($Co$).
Step 2: Resolving Part (b). Atomic number 26 corresponds to Iron ($Fe$).
Step 3: Calculating Magnetic Moment. $$\mu = \sqrt{n(n+2)} = \sqrt{4(4+2)} = \sqrt{24} \approx 4.9 \text{ BM}$$
Step 2: Resolving Part (b). Atomic number 26 corresponds to Iron ($Fe$).
- Neutral $Fe = [Ar] 3d^6 4s^2$
- $Fe^{2+} = [Ar] 3d^6$
Step 3: Calculating Magnetic Moment. $$\mu = \sqrt{n(n+2)} = \sqrt{4(4+2)} = \sqrt{24} \approx 4.9 \text{ BM}$$
Common Mistakes Students Make
- Treating Zn, Cd, Hg as Transition Metals: Students frequently call them transition metals. Remember, they are d-block elements but NOT transition elements because their d-orbitals are fully saturated ($d^{10}$).
- Incorrect Electronic Configurations of Cr and Cu: Do not write $Cr$ as $3d^4 4s^2$ or $Cu$ as $3d^9 4s^2$. Due to the stability of half-filled and fully-filled sublevels, the correct configurations are $Cr = 3d^5 4s^1$ and $Cu = 3d^{10} 4s^1$.
- Confusing Lanthanoid Contraction: Students often mention "high effective nuclear charge" without specifying the "poor shielding effect of 4f electrons." You must state both to secure full marks.
Frequently Asked Questions (FAQs)
1. Is the d- and f-Block Elements chapter important for CBSE Class 12 Boards?
Yes! This chapter carries a weightage of about 7 marks. It is highly scoring because it consists almost entirely of conceptual reasoning questions and direct chemical reactions.
2. What is the main cause of Lanthanoid Contraction?
The primary cause is the extremely poor shielding effect of 4f electrons, which cannot effectively counter-balance the increasing nuclear charge as you move across the series.
3. Why do transition metals exhibit variable oxidation states?
This is because the energy difference between the inner $(n-1)d$ orbital and outer $ns$ orbital is incredibly small. Electrons from both shells can easily take part in forming chemical bonds.
4. Are there any shortcuts to calculate the spin-only magnetic moment ($\mu$)?
Yes! The value of $\mu$ always starts with the number of unpaired electrons ($n$). For example, if $n=3$, $\mu \approx 3.87\text{ BM}$. If $n=5$, $\mu \approx 5.92\text{ BM}$.
More Class 12 Chemistry Chapters
Ch 1: The Solid State
Ch 2: Solutions
Ch 3: Electrochemistry
Ch 4: Chemical Kinetics
Ch 5: Surface Chemistry
Ch 6: General Principles
Ch 7: The p-Block Elements
Ch 8: The d- & f-Block Elements
Ch 9: Coordination Compounds
Ch 10: Haloalkanes
Ch 11: Alcohols & Phenols
Ch 12: Aldehydes & Ketones
Ch 13: Amines
Ch 14: Biomolecules
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