D and F Block Elements Class 12 Notes, IIT JEE & NEET

The d- and f-Block Elements is Unit 4 of 10 in the current NCERT Class 12 Chemistry book (it was Chapter 8 in older books), and it carries 7 marks in the CBSE theory paper. This page is written so that you can learn the whole chapter from it: the definitions, the trends and their reasons, the important compounds with their equations, the f-block, and the questions that exams ask. NCERT calls the f-block series lanthanoids and actinoids, and many books call them lanthanides and actinides. The two names mean the same thing.

D and F Block Elements Class 12 Notes: Chapter Overview and What Each Exam Needs

Check the latest syllabi of CBSE, NTA (JEE Main) and NMC (NEET) for the exact lists. Numericals in this chapter are mostly spin-only magnetic moments, oxidation-state counting and titration stoichiometry of $\mathrm{KMnO_4}$ and $\mathrm{K_2Cr_2O_7}$.

Tip from the eSaral Chemistry team: This chapter is mostly reasons. For every trend, learn the one-line reason (for example “half-filled $d^5$ is extra stable”), because exam answers are marked on the reason.

What Are D Block and F Block Elements? in D and F Block Elements Class 12 Notes

D Block Elements (Transition Metals)

The d-block elements are the elements in which the last electron enters the d orbital of the penultimate shell. They occupy Groups 3 to 12. A transition element is defined as an element that has an incomplete d subshell in the atom or in any of its common ions. The four series are:

Zn, Cd and Hg are d-block elements, but they are not transition elements, because their d subshell is full ($d^{10}$) in the atom and in the common ion ($\mathrm{Zn^{2+}}$, $\mathrm{Cd^{2+}}$, $\mathrm{Hg^{2+}}$). Always check the ion as well as the atom.

F Block Elements (Inner Transition Metals)

In the f-block elements, the last electron enters the f orbital of the pre-penultimate shell, $(n-2)f$. They are the lanthanoids (4f series, Ce to Lu, Z = 58 to 71) and the actinoids (5f series, Th to Lr, Z = 90 to 103). They are placed at the bottom of the periodic table.

D and F Block Elements Class 12 Notes

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Electronic Configuration of Transition Metals in D and F Block Elements Class 12 Notes

The general configuration is:

$$(n-1)d^{1\text{–}10}\,ns^{0\text{–}2}$$

For the first series, it is $[\mathrm{Ar}]\,3d^{1\text{–}10}\,4s^{1\text{–}2}$. Two elements of the first series are exceptions, because half-filled and fully filled d subshells are extra stable:

Ions: remove the $ns$ electrons first, and then the $(n-1)d$ electrons. For example, $\mathrm{Fe^{2+}}$ is $[\mathrm{Ar}]\,3d^{6}$, $\mathrm{Fe^{3+}}$ is $[\mathrm{Ar}]\,3d^{5}$, $\mathrm{Cu^{+}}$ is $[\mathrm{Ar}]\,3d^{10}$, $\mathrm{Cu^{2+}}$ is $[\mathrm{Ar}]\,3d^{9}$ and $\mathrm{Mn^{2+}}$ is $[\mathrm{Ar}]\,3d^{5}$.

General Properties of Transition Metals in D and F Block Elements Class 12 Notes

Physical Properties, Radii and Ionisation Enthalpy

• Melting and boiling points are high, because both the $(n-1)d$ and the $ns$ electrons take part in metallic bonding. The melting point rises to a maximum near the middle of the series ($d^5$) and then falls. Mn and Zn are exceptions with low melting points, and Hg is a liquid.
• Atomic radii decrease from Sc to Cr, remain nearly constant from Mn to Cu, and rise slightly at Zn. Added d electrons screen the nuclear charge only partly, and electron–electron repulsion balances the pull in the middle of the series.
• Ionisation enthalpies rise slowly across the series, because the increasing nuclear charge is partly cancelled by the d-electron screening.

Oxidation States in D and F Block Elements Class 12 Notes

Transition metals show variable oxidation states, because the $(n-1)d$ and $ns$ electrons have similar energies and can all take part in bonding. The highest oxidation state rises from Sc ($+3$) to Mn ($+7$) and then falls. Higher states are stabilised by oxygen and fluorine.

Standard Electrode Potentials and Stability of Oxidation States

• Copper is the only first-row metal with a positive $E^\circ(\mathrm{M^{2+}/M})$, because its sum of sublimation, ionisation and hydration enthalpies is not enough to make it a good reducing agent. Cu does not liberate hydrogen from acids.
• $\mathrm{Cr^{3+}}$ is stable, because of the $t_{2g}^3$ configuration, so $\mathrm{Cr^{2+}}$ is a strong reducing agent.
• $\mathrm{Mn^{3+}}$ and $\mathrm{Co^{3+}}$ are strong oxidising agents, because $\mathrm{Mn^{2+}}$ ($d^5$) and $\mathrm{Co^{2+}}$ are more stable.
• Disproportionation of $\mathrm{Cu^+}$: in water, $\mathrm{Cu^+}$ is unstable, because the hydration enthalpy of $\mathrm{Cu^{2+}}$ is much more negative:

$$2\,\mathrm{Cu^{+}} \longrightarrow \mathrm{Cu^{2+}} + \mathrm{Cu}$$

Magnetic Properties in D and F Block Elements Class 12 Notes

A substance with unpaired electrons is paramagnetic, and a substance with all electrons paired is diamagnetic. The spin-only magnetic moment is:

$$\mu = \sqrt{n(n+2)}\ \text{BM}$$

where $n$ is the number of unpaired electrons.

Among the first-row d-block ions, the high-spin $d^5$ ions have the highest moment (5.92 BM).

Colour of Transition Metal Ions in D and F Block Elements Class 12 Notes

Most transition metal ions are coloured, because an electron absorbs visible light and moves between the split d orbitals (a d–d transition). The colour seen is the complementary colour of the light absorbed.

$d^0$ and $d^{10}$ ions have no d–d transition, so they are colourless. Some $d^0$ ions, such as $\mathrm{CrO_4^{2-}}$ (yellow) and $\mathrm{MnO_4^{-}}$ (purple), are coloured because of charge transfer from oxygen to the metal, not because of d–d transitions.

Catalytic Property, Complex Formation, Interstitial Compounds and Alloys

• Catalysis: variable oxidation states and the ability to form intermediates make transition metals good catalysts. Examples: $\mathrm{V_2O_5}$ in the Contact process, Fe in the Haber process, Ni in hydrogenation, and $\mathrm{TiCl_4}$ with $\mathrm{Al(CH_3)_3}$ in the Ziegler–Natta catalyst.
• Complex formation: small size, high charge and vacant d orbitals favour the formation of complexes (see eSaral’s Coordination Compounds Class 12 notes).
• Interstitial compounds: small atoms (H, C, N) are trapped in the voids of the metal lattice. They are hard, have high melting points, keep metallic conductivity and are chemically inert. They are non-stoichiometric (for example $\mathrm{TiC}$ and steel).
• Alloys: transition metals form alloys easily, because their atomic radii are similar (within about 15%). Examples: brass (Cu and Zn), bronze (Cu and Sn) and steel.

Important Compounds: KMnO₄ and K₂Cr₂O₇ in D and F Block Elements Class 12 Notes

Potassium Permanganate ($\mathrm{KMnO_4}$)

Preparation: pyrolusite ore ($\mathrm{MnO_2}$) is fused with KOH in air to give green potassium manganate. It is then oxidised electrolytically, or it disproportionates in acid:

$$2\,\mathrm{MnO_2} + 4\,\mathrm{KOH} + \mathrm{O_2} \longrightarrow 2\,\mathrm{K_2MnO_4} + 2\,\mathrm{H_2O}$$ $$3\,\mathrm{MnO_4^{2-}} + 4\,\mathrm{H^{+}} \longrightarrow 2\,\mathrm{MnO_4^{-}} + \mathrm{MnO_2} + 2\,\mathrm{H_2O}$$

Properties: dark purple crystals. On heating at about 513 K, it decomposes: $2\,\mathrm{KMnO_4} \longrightarrow \mathrm{K_2MnO_4} + \mathrm{MnO_2} + \mathrm{O_2}$. It is a strong oxidising agent, and its product depends on the medium:

Important oxidations in acidic medium:

$$5\,\mathrm{Fe^{2+}} + \mathrm{MnO_4^{-}} + 8\,\mathrm{H^{+}} \longrightarrow 5\,\mathrm{Fe^{3+}} + \mathrm{Mn^{2+}} + 4\,\mathrm{H_2O}$$ $$2\,\mathrm{MnO_4^{-}} + 5\,\mathrm{C_2O_4^{2-}} + 16\,\mathrm{H^{+}} \longrightarrow 2\,\mathrm{Mn^{2+}} + 10\,\mathrm{CO_2} + 8\,\mathrm{H_2O}$$ $$2\,\mathrm{MnO_4^{-}} + 10\,\mathrm{I^{-}} + 16\,\mathrm{H^{+}} \longrightarrow 2\,\mathrm{Mn^{2+}} + 5\,\mathrm{I_2} + 8\,\mathrm{H_2O}$$

Potassium Dichromate ($\mathrm{K_2Cr_2O_7}$) in D and F Block Elements Class 12 Notes

Preparation: chromite ore ($\mathrm{FeCr_2O_4}$) is fused with sodium carbonate in air, and the sodium chromate is acidified and treated with KCl:

$$4\,\mathrm{FeCr_2O_4} + 8\,\mathrm{Na_2CO_3} + 7\,\mathrm{O_2} \longrightarrow 8\,\mathrm{Na_2CrO_4} + 2\,\mathrm{Fe_2O_3} + 8\,\mathrm{CO_2}$$ $$2\,\mathrm{Na_2CrO_4} + 2\,\mathrm{H^{+}} \longrightarrow \mathrm{Na_2Cr_2O_7} + 2\,\mathrm{Na^{+}} + \mathrm{H_2O}$$ $$\mathrm{Na_2Cr_2O_7} + 2\,\mathrm{KCl} \longrightarrow \mathrm{K_2Cr_2O_7} + 2\,\mathrm{NaCl}$$

Properties: orange crystals. Chromate and dichromate ions are interconverted by the pH of the solution (orange in acid and yellow in alkali):

$$2\,\mathrm{CrO_4^{2-}} + 2\,\mathrm{H^{+}} \rightleftharpoons \mathrm{Cr_2O_7^{2-}} + \mathrm{H_2O}$$

In acidic medium, it is a strong oxidising agent, and chromium goes from $+6$ to $+3$ (orange to green):

$$\mathrm{Cr_2O_7^{2-}} + 14\,\mathrm{H^{+}} + 6e^{-} \longrightarrow 2\,\mathrm{Cr^{3+}} + 7\,\mathrm{H_2O}$$ $$\mathrm{Cr_2O_7^{2-}} + 6\,\mathrm{Fe^{2+}} + 14\,\mathrm{H^{+}} \longrightarrow 2\,\mathrm{Cr^{3+}} + 6\,\mathrm{Fe^{3+}} + 7\,\mathrm{H_2O}$$ $$\mathrm{Cr_2O_7^{2-}} + 14\,\mathrm{H^{+}} + 6\,\mathrm{I^{-}} \longrightarrow 2\,\mathrm{Cr^{3+}} + 7\,\mathrm{H_2O} + 3\,\mathrm{I_2}$$

The dichromate ion has two $\mathrm{CrO_4}$ tetrahedra that share one oxygen atom.

F Block Elements: Lanthanoids and Actinoids in D and F Block Elements Class 12 Notes

Lanthanoids

• Configuration: $[\mathrm{Xe}]\,4f^{1\text{–}14}\,5d^{0\text{–}1}\,6s^{2}$.
• Oxidation states: $+3$ is the common state. Ce ($+4$) and Tb ($+4$) give $+4$, and Eu ($+2$) and Yb ($+2$) give $+2$, because of the stability of the empty, half-filled or full 4f subshell.
• Chemical reactivity: they are reactive, electropositive metals, and the reactivity decreases from La to Lu. They are good reducing agents.
• Uses: mischmetall (about 95% lanthanoids and 5% iron, with traces of S, C, Ca and Al) is used in magnesium alloys and in lighter flints.

Lanthanoid Contraction in D and F Block Elements Class 12 Notes

The lanthanoid contraction is the steady decrease in the atomic and ionic radii from La to Lu (for the $\mathrm{M^{3+}}$ ions, from about 106 pm to 85 pm). Each added 4f electron screens the increased nuclear charge poorly, because the f orbitals have a diffuse shape, so the outer electrons are pulled in.

Consequences:

1. The 4d and 5d metals have almost the same radii (Zr and Hf), so the pairs Zr and Hf, Nb and Ta, and Mo and W have very similar properties and are hard to separate.
2. The 5d metals have a higher density and a higher ionisation enthalpy than the 4d metals.
3. The basic strength of the hydroxides $\mathrm{Ln(OH)_3}$ decreases from La to Lu, because the size of the ion decreases.

Actinoids

• Configuration: $[\mathrm{Rn}]\,5f^{1\text{–}14}\,6d^{0\text{–}1}\,7s^{2}$.
• Oxidation states: $+3$ is common, but the actinoids show a wider range ($+3$ to $+7$), because the 5f, 6d and 7s orbitals have similar energies.
• Radioactivity: all actinoids are radioactive. Most elements after uranium are synthetic.

Key Differences: D Block vs F Block in D and F Block Elements Class 12 Notes

Key Formulas and Facts in D and F Block Elements Class 12 Notes

Solved Examples from D and F Block Elements Class 12 Notes

Example 1 (configuration): Write the configuration of $\mathrm{Cr}$ and $\mathrm{Cu^{+}}$.

$\mathrm{Cr}$: $[\mathrm{Ar}]\,3d^{5}4s^{1}$ (half-filled stability). $\mathrm{Cu^{+}}$: $[\mathrm{Ar}]\,3d^{10}$ (the $4s$ electron is removed first).

Example 2 (magnetic moment): Find the spin-only magnetic moment of $\mathrm{Fe^{2+}}$.

$\mathrm{Fe^{2+}}$ is $3d^{6}$, with $n = 4$ unpaired electrons. $\mu = \sqrt{4 \times 6} = \sqrt{24} \approx 4.90$ BM.

Example 3 (oxidation state): Find the oxidation state of Mn in $\mathrm{KMnO_4}$ and of Cr in $\mathrm{K_2Cr_2O_7}$.

$\mathrm{KMnO_4}$: $+1 + x + 4(-2) = 0$, so $x = +7$. $\mathrm{K_2Cr_2O_7}$: $2(+1) + 2x + 7(-2) = 0$, so $x = +6$.

Example 4 (titration with $\mathrm{KMnO_4}$): How many moles of $\mathrm{Fe^{2+}}$ are oxidised by 25 mL of 0.1 M $\mathrm{KMnO_4}$ in acidic medium?

Moles of $\mathrm{MnO_4^-} = 0.025 \times 0.1 = 2.5 \times 10^{-3}$. The ratio is $1 : 5$, so moles of $\mathrm{Fe^{2+}} = 5 \times 2.5 \times 10^{-3} = 1.25 \times 10^{-2}$.

Example 5 (titration with $\mathrm{K_2Cr_2O_7}$): How many moles of $\mathrm{Fe^{2+}}$ are oxidised by 20 mL of 0.05 M $\mathrm{K_2Cr_2O_7}$?

Moles of $\mathrm{Cr_2O_7^{2-}} = 0.020 \times 0.05 = 1.0 \times 10^{-3}$. The ratio is $1 : 6$, so moles of $\mathrm{Fe^{2+}} = 6.0 \times 10^{-3}$.

Example 6 (colour): Why is $\mathrm{Sc^{3+}}$ colourless, and why is $\mathrm{Cu^{2+}}$ blue?

$\mathrm{Sc^{3+}}$ is $d^{0}$, so no d–d transition is possible. $\mathrm{Cu^{2+}}$ is $d^{9}$, so it has a vacant position in the d subshell and absorbs visible light by a d–d transition.

Example 7 (lanthanoid contraction): State one consequence of the lanthanoid contraction.

Zr and Hf have almost the same ionic radius, so they have very similar properties and are difficult to separate.

Common Mistakes in D and F Block Elements Class 12 Notes Problems

How to Study D and F Block Elements Class 12 Notes

1. Read the NCERT chapter first, and download the textbook from eSaral’s NCERT books for Class 12 page.
2. Learn the reasons for each trend, and write the equations of $\mathrm{KMnO_4}$ and $\mathrm{K_2Cr_2O_7}$ from memory.
3. Practise the titration and magnetic moment numericals, and then the next unit with eSaral’s Coordination Compounds Class 12 notes.
4. Revise with eSaral’s Class 12 Chemistry notes, and test yourself with eSaral’s NEET chapter-wise PYQs and JEE Main chapter-wise PYQs.