Chapter Weightage: 7 Marks

The d- and f-Block Elements

Complete Revision Guide for CBSE Board Exams

1. Introduction

The d-block elements are located in the middle section of the periodic table, spanning groups 3 to 12. They are characterized by the filling of inner d-orbitals by electrons. They are commonly referred to as Transition Elements because they exhibit transitional behaviour between the highly reactive s-block metals and the elements of the p-block.

Strict Definition of a Transition Element

A transition element is defined as an element that has an incompletely filled d-orbital in its ground state or in any one of its common oxidation states.

CBSE Favorite: Zinc (Zn), Cadmium (Cd), and Mercury (Hg) of Group 12 have completely filled $d^{10}$ configurations in their ground state as well as in their common oxidation states (e.g., $Zn^{2+}$ is $3d^{10}$). Therefore, they are NOT considered transition elements, although they are part of the d-block.

The f-block elements (Lanthanoids and Actinoids) are placed separately at the bottom of the periodic table. They involve the filling of inner f-orbitals and are known as Inner Transition Elements.

2. Master Cheat Sheet: Formulas & Key Equations

A. Electronic Configurations & Magnetic Properties

General Electronic Configuration (d-block): $(n-1)d^{1-10} ns^{1-2}$
General Electronic Configuration (f-block): $(n-2)f^{1-14} (n-1)d^{0-1} ns^2$
Anomalous Configurations:
Chromium (Cr, Z=24): $[Ar] 3d^5 4s^1$ (Half-filled stability)
Copper (Cu, Z=29): $[Ar] 3d^{10} 4s^1$ (Fully-filled stability)
Spin-Only Magnetic Moment ($\mu$):
$$\mu = \sqrt{n(n+2)} \text{ B.M.}$$
(Where $n$ is the number of unpaired electrons; B.M. is Bohr Magneton).

B. Potassium Dichromate ($K_2Cr_2O_7$) Key Reactions

pH Dependent Equilibrium: Chromate (yellow) and dichromate (orange) interconvert based on pH.
$$2CrO_4^{2-} (\text{Yellow}) + 2H^+ \rightleftharpoons Cr_2O_7^{2-} (\text{Orange}) + H_2O$$

$$Cr_2O_7^{2-} (\text{Orange}) + 2OH^- \rightleftharpoons 2CrO_4^{2-} (\text{Yellow}) + H_2O$$
Powerful Oxidizing Agent (Acidic Medium):
$$Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O \quad (E^\circ = 1.33 \text{ V})$$
It oxidizes $I^-$ to $I_2$, $S^2-$ to $S$, $Sn^{2+}$ to $Sn^{4+}$, and $Fe^{2+}$ to $Fe^{3+}$.

C. Potassium Permanganate ($KMnO_4$) Key Reactions

Acidic Medium (Strongest action, n-factor = 5):
$$MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$$
Oxidizes oxalates ($C_2O_4^{2-}$) to $CO_2$, $Fe^{2+}$ to $Fe^{3+}$.
Neutral / Faintly Alkaline Medium (n-factor = 3):
$$MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2 + 4OH^-$$
Oxidizes $I^-$ directly to iodate ($IO_3^-$), and thiosulphate ($S_2O_3^{2-}$) to sulphate ($SO_4^{2-}$).
Disproportionation of Manganate in Acidic Medium:
$$3MnO_4^{2-} + 4H^+ \rightarrow 2MnO_4^- + MnO_2 + 2H_2O$$

3. General Properties of Transition Elements (d-Block)

This section is entirely about "Give Reason" questions. Pay close attention to the explanations.

3.1 Atomic and Ionic Radii

As we move along a period from left to right in a transition series, the atomic radii decrease initially, become roughly constant in the middle, and then slightly increase towards the end.

Initial Decrease: The new electrons enter the inner $3d$ orbitals. The shielding effect of a $d$-electron is poor. Hence, the effective nuclear charge ($Z_{eff}$) increases, pulling the electron cloud inward.
Constant in Middle: The increased nuclear charge is balanced by the increased inter-electronic repulsion between the added $3d$ electrons.
Increase at the End (e.g., Zn): The $3d$ orbitals become completely filled. Electron-electron repulsion dominates over the effective nuclear charge, causing a slight expansion.

3.2 Enthalpy of Atomization

Transition metals exhibit very high enthalpies of atomization. Why? Because they have a large number of unpaired electrons in their (n-1)d orbitals. These unpaired electrons participate in the formation of strong interatomic metallic bonds.
Trend: Greater the number of unpaired electrons, stronger is the metallic bonding, and higher is the enthalpy of atomization. (Note: Zn, Cd, Hg have 0 unpaired electrons, hence very low enthalpy of atomization and are soft metals).

3.3 Variable Oxidation States

Unlike s-block elements, transition metals exhibit a wide variety of oxidation states. This is because the energy difference between the $(n-1)d$ and $ns$ orbitals is very small. Therefore, after the $ns$ electrons are lost, the $(n-1)d$ electrons can also participate in bonding.

Scandium (Sc): Only shows +3 because losing 3 electrons gives it the stable noble gas configuration of Argon.
Manganese (Mn): Shows the maximum number of oxidation states (+2 to +7) in the 3d series because it has the maximum number of unpaired electrons ($3d^5 4s^2$).
Higher oxidation states (like +6 in $CrO_4^{2-}$ or +7 in $MnO_4^-$) are generally stabilized by highly electronegative elements like Oxygen or Fluorine.

3.4 Colour of Transition Metal Ions

Most transition metal compounds are highly colored in solid state or aqueous solutions. When an electron from a lower energy d-orbital is excited to a higher energy d-orbital, the energy of excitation corresponds to the frequency of light absorbed from the visible region. This is called the d-d transition. The complementary color of the absorbed light is observed.

Exception Alert! Compounds with $d^0$ (e.g., $Sc^{3+}, Ti^{4+}$) or $d^{10}$ (e.g., $Zn^{2+}, Cu^+$) configurations are colorless because there are either no d-electrons to transition, or no empty d-orbitals to jump into.

3.5 Catalytic Properties

Transition metals and their compounds act as excellent catalysts (e.g., $V_2O_5$ in Contact Process, Finely divided Fe in Haber's Process). Reasons:

They can adopt multiple oxidation states, allowing them to form unstable intermediate compounds with reactants, providing a lower activation energy pathway.
They provide a large surface area for the adsorption of reactant molecules, bringing them closer and weakening their bonds.

3.6 Formation of Interstitial Compounds

Transition metals form interstitial compounds when small atoms like H, C, or N get trapped inside the empty spaces (interstices) of their metallic crystal lattices.
Characteristics: They are chemically inert, non-stoichiometric (e.g., $TiC$, $Mn_4N$, $VH_{0.56}$), extremely hard (some approaching diamond), and have melting points higher than the pure metals.

4. Inner Transition Elements (f-Block)

4.1 The Lanthanoids

The series of 14 elements following Lanthanum (La, Z=57) from Cerium (Ce, Z=58) to Lutetium (Lu, Z=71) are called lanthanoids. The 4f subshell is progressively filled.

Oxidation States: The most common and characteristic oxidation state is +3. Some elements show +2 (e.g., $Eu^{2+}$, $Yb^{2+}$) or +4 (e.g., $Ce^{4+}$) to achieve stable empty ($f^0$), half-filled ($f^7$), or fully filled ($f^{14}$) subshells. However, they act as strong reducing or oxidizing agents to revert back to the stable +3 state. (e.g., $Ce^{4+}$ is a good analytical oxidizing agent because it easily reduces to $Ce^{3+}$).

Lanthanoid Contraction (Highly Important)

Definition: The steady and continuous decrease in the atomic and ionic radii of lanthanoids with increasing atomic number is called the lanthanoid contraction.

Cause: As we move along the lanthanoid series, the new electron enters the inner 4f orbital. The shielding effect of one 4f electron by another is extremely poor due to the highly diffused shape of f-orbitals. Consequently, the effective nuclear charge increases steadily, drawing the outer electrons closer and shrinking the radius.

Consequences:

Similarity in 2nd and 3rd transition series: The atomic radii of elements of the 3rd transition series (5d) become almost identical to the corresponding elements of the 2nd transition series (4d). E.g., Zirconium (Zr, 160 pm) and Hafnium (Hf, 159 pm) have identical radii, making their separation extremely difficult.
Decrease in basic character of oxides/hydroxides: As the size of $Ln^{3+}$ ion decreases from $La^{3+}$ to $Lu^{3+}$, the covalent character of the Ln-OH bond increases. Therefore, basic strength decreases. $La(OH)_3$ is the most basic, and $Lu(OH)_3$ is the least basic.

4.2 The Actinoids

The 14 elements following Actinium (Z=89) from Thorium (Th, Z=90) to Lawrencium (Lr, Z=103) are called actinoids. The 5f subshell is progressively filled. All actinoids are radioactive.

Actinoid Contraction: There is a gradual decrease in size across the series. The actinoid contraction is greater from element to element than the lanthanoid contraction. This is because 5f orbitals have an even poorer shielding effect than 4f orbitals. Actinoids exhibit a larger number of oxidation states (up to +7) because the 5f, 6d, and 7s levels have comparable energies.

5. Most Important CBSE Board Questions

Curated based on frequency in past papers and latest CBSE patterns.

Section A: 1 Mark Questions (MCQs & A-R)

Q1. Which of the following is NOT regarded as a transition element? 1 Mark

Iron (Fe)
Silver (Ag)
Zinc (Zn)
Copper (Cu)

Answer: (c) Zinc (Zn)
Explanation: Zinc has a fully filled $3d^{10}$ electronic configuration in its ground state ($[Ar] 3d^{10} 4s^2$) as well as in its common +2 oxidation state ($[Ar] 3d^{10}$). Thus, it does not fit the definition of a transition element.

Q2. The magnetic moment of a transition metal ion is found to be 3.87 B.M. The number of unpaired electrons present in it is: 1 Mark

Answer: (b) 3
Explanation: Use the formula $\mu = \sqrt{n(n+2)}$.
If $n = 3$, $\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87$ B.M. (Shortcut trick: The integer part of the magnetic moment usually matches the number of unpaired electrons).

Assertion-Reasoning

Q3. Assertion (A): $Cu^{2+}$ is colored while $Cu^+$ is colorless in aqueous solution.
Reason (R): $Cu^{2+}$ has a $d^9$ configuration allowing d-d transitions, while $Cu^+$ has a fully filled $d^{10}$ configuration. 1 Mark

Answer: (A) Both A and R are true, and R is the correct explanation of A.
Explanation: Color in transition metal ions arises from the excitation of electrons from lower d-orbitals to higher d-orbitals (d-d transition). $Cu^+$ ($3d^{10}$) has no empty space in d-orbitals for an electron to jump into, hence no light is absorbed, and it is colorless.

Section B: 2 Mark Questions (Very Short Answer)

Q4. Assign reasons for the following:
(i) Transition metals and many of their compounds act as good catalysts.
(ii) Transition metals generally form colored compounds. 2 Marks

Answer: (i) Transition metals have vacant d-orbitals and show variable oxidation states. This allows them to form unstable intermediate compounds with reactants, providing an alternate reaction pathway with lower activation energy. They also provide a large surface area for adsorption.
(ii) They form colored compounds due to the presence of unpaired d-electrons. When visible light falls on them, these electrons absorb specific wavelengths to jump from lower to higher energy d-orbitals (d-d transition). The transmitted/reflected light shows the complementary color.

Q5. Explain why Zirconium (Zr, Z=40) and Hafnium (Hf, Z=72) have almost identical atomic radii (160 pm and 159 pm respectively). 2 Marks

Answer: This is a direct consequence of the Lanthanoid Contraction. Before the filling of the 5d orbitals (where Hf is located), the 14 elements of the lanthanoid series (4f orbitals) are filled. Due to the extremely poor shielding effect of the 4f electrons, the effective nuclear charge increases significantly. This pulls the electron cloud inward, canceling out the normal expected increase in size down the group. Hence, the radius of the 5d element (Hf) shrinks to become nearly equal to that of the 4d element (Zr) just above it.

Section C: 3 Mark Questions (Short Answer)

Q6. Account for the following:
(i) The lowest oxide of a transition metal is basic, the highest is amphoteric/acidic.
(ii) Scandium (Z=21) does not exhibit variable oxidation states.
(iii) The $E^\circ(M^{2+}/M)$ value for copper is positive (+0.34 V), unlike other first series transition metals. 3 Marks

Answer: (i) In lower oxidation states, the metal has low polarizing power, so the Metal-Oxygen bond is highly ionic, making the oxide basic. In higher oxidation states, the highly positive metal ion strongly polarizes the oxygen electron cloud, giving the bond a highly covalent character, which makes the oxide acidic (e.g., $MnO$ is basic, $Mn_2O_7$ is acidic).

(ii) Scandium ($[Ar] 3d^1 4s^2$) has only 3 valence electrons. By losing all three, it achieves the highly stable noble gas configuration of Argon. Removing fewer electrons leaves it unstable, and removing more requires too much energy. Thus, it only shows the +3 oxidation state.

(iii) The $E^\circ(M^{2+}/M)$ value depends on the sum of enthalpy of sublimation, ionization enthalpies (IE1 + IE2), and enthalpy of hydration. For Copper, the high energy required to atomize it and ionize it (high IE1+IE2) is not balanced by its hydration enthalpy. Hence, the overall process is endothermic, resulting in a positive $E^\circ$ value. (This means Cu cannot displace hydrogen from dilute acids).

Q7. Complete the following chemical equations:
(i) $MnO_4^- + C_2O_4^{2-} + H^+ \rightarrow$
(ii) $Cr_2O_7^{2-} + Sn^{2+} + H^+ \rightarrow$
(iii) $2CrO_4^{2-} + 2H^+ \rightarrow$ 3 Marks

Answer: (i) Potassium permanganate in acidic medium oxidizes oxalate ions to carbon dioxide gas.
$$2MnO_4^- + 5C_2O_4^{2-} + 16H^+ \rightarrow 2Mn^{2+} + 10CO_2 + 8H_2O$$
(ii) Dichromate in acidic medium oxidizes Stannous ($Sn^{2+}$) to Stannic ($Sn^{4+}$).
$$Cr_2O_7^{2-} + 3Sn^{2+} + 14H^+ \rightarrow 2Cr^{3+} + 3Sn^{4+} + 7H_2O$$
(iii) In acidic medium, yellow chromate ions convert to orange dichromate ions.
$$2CrO_4^{2-} + 2H^+ \rightleftharpoons Cr_2O_7^{2-} + H_2O$$

Section D: 4 Mark Case-Based Questions

Q8. Read the passage given below and answer the following questions:

The transition elements (d-block) exhibit unique characteristics due to their partially filled d-orbitals. These elements can form a wide variety of complex compounds, act as industrial catalysts, and form alloys easily. One interesting property is the variability of their oxidation states, unlike the s-block elements which show fixed oxidation states. The stability of a particular oxidation state heavily depends on the nature of the solvent and the atoms the metal is bonding with. Transition metals often show their highest oxidation states when combined with highly electronegative elements.

(i) Why do transition metals show variable oxidation states? (1 Mark)

(ii) Name an element of the 3d series which shows the maximum number of oxidation states. (1 Mark)

(iii) Explain why the highest oxidation states of transition metals are observed only in their oxides and fluorides. (2 Marks)

Answers: (i) Transition metals show variable oxidation states because the energy difference between the outer ns orbitals and the inner (n-1)d orbitals is very small. Therefore, electrons from both energy levels can participate in bonding.

(ii) Manganese (Mn). It can exhibit oxidation states from +2 to +7.

(iii) Oxygen and Fluorine are highly electronegative elements and possess very small sizes. They have strong oxidizing power, meaning they can unpair electrons in the d-orbitals of the transition metal and force them to form bonds, effectively stabilizing the metal in its highest possible oxidation state (e.g., $OsF_6$, $Mn_2O_7$).

Section E: 5 Mark Long Answer Questions

Q9. (a) Describe the preparation of potassium dichromate ($K_2Cr_2O_7$) from chromite ore. Write the chemical equations for the reactions involved. (3 Marks)

(b) Give reasons:
(i) Cerium (Ce, Z=58) exhibits a +4 oxidation state.
(ii) Actinoid contraction is greater from element to element than lanthanoid contraction. (2 Marks)

5 Marks

Answer: (a) Preparation of $K_2Cr_2O_7$ from Chromite ore ($FeCr_2O_4$):
This involves three main steps:
Step 1: Conversion of chromite ore to sodium chromate.
The finely powdered ore is fused with sodium carbonate in the presence of air (roasting).
$$4FeCr_2O_4 + 8Na_2CO_3 + 7O_2 \rightarrow 8Na_2CrO_4 (\text{yellow}) + 2Fe_2O_3 + 8CO_2$$
Step 2: Conversion of sodium chromate to sodium dichromate.
The yellow solution of sodium chromate is filtered and acidified with dilute sulfuric acid to give orange sodium dichromate.
$$2Na_2CrO_4 + 2H^+ \rightarrow Na_2Cr_2O_7 (\text{orange}) + 2Na^+ + H_2O$$
Step 3: Conversion of sodium dichromate to potassium dichromate.
Sodium dichromate is highly soluble. It is treated with potassium chloride. Because $K_2Cr_2O_7$ is less soluble than $Na_2Cr_2O_7$, it crystallizes out as orange crystals on cooling.
$$Na_2Cr_2O_7 + 2KCl \rightarrow K_2Cr_2O_7 \downarrow + 2NaCl$$

(b) Reasoning:
(i) Cerium ($[Xe] 4f^1 5d^1 6s^2$) can easily lose four electrons to achieve the highly stable noble gas configuration of Xenon ($f^0$). This extra stability allows it to exhibit the +4 oxidation state ($Ce^{4+}$), although it acts as an oxidizing agent and eventually reverts to the most stable +3 state.
(ii) The 5f electrons in actinoids have a much more diffused shape compared to the 4f electrons in lanthanoids. As a result, the mutual shielding effect of 5f electrons is even poorer than that of 4f electrons. Thus, the effective nuclear charge felt by outer electrons increases more sharply in actinoids, leading to a greater contraction in size from element to element.

Mastering 'd and f Block Elements'

•
The 'Because of unpaired d-electrons' hack: For almost any question asking "Why do transition metals form colored ions?", "Why do they form complexes?", "Why do they show paramagnetism?", or "Why do they act as catalysts?", the core of your answer must include "due to the presence of unpaired (n-1)d electrons and empty d-orbitals."
•
Memorize the Big Two Preparations: The 5-mark question in this chapter is almost always either the preparation of $K_2Cr_2O_7$ from chromite ore OR $KMnO_4$ from pyrolusite ore ($MnO_2$). Memorize these step-by-step equations.
•
Lanthanoid Contraction is Guaranteed: Expect at least a 2 or 3 mark question on what it is and its consequences (Zr/Hf similarity, basicity trends).
•
Balance Ionic Equations Properly: When writing the oxidizing action of $KMnO_4$ or $K_2Cr_2O_7$, always balance the H+ and $H_2O$ molecules. Acidic medium reactions are asked 90% of the time.

Search This Blog

Class 12 Chemistry: Complete Revision of d and f Block Elements