This Collegedunia d and f block elements class 12 formula sheet hosts every transition-metal expression you need for Class 12 boards, JEE Main, and NEET, mapped to the 2026-27 NCERT.

  • CBSE Weightage: 4 to 6 marks
  • JEE Main Weightage: 3 to 4 percent (2 to 3 questions per paper)
  • NEET Weightage: 2 to 3 questions per year

Curated by subject experts and refined against recent CBSE, JEE Main, and NEET papers.

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The D And F Block Elements Formula Sheet - Class 12 Chemistry

Why the d- and f-Block Formula Sheet is a High-Yield Asset for Class 12th Boards and Entrances

The d-block is the most fact-dense unit in Class 12 Inorganic Chemistry. Keeping magnetic moment, oxidation states, and lanthanoid contraction on one page saves time per question.

Watch Out: For spin-only magnetic moment, n counts unpaired electrons in the d-orbital of the ion, not the atom. Use the ionic configuration and remove 4s first.

The D and F Block Elements Video Walkthrough

Source: Magnet Brains on YouTube

How will Collegedunia's d- and f-Block Elements Formula Sheet Help You?

Built for a 30-minute revision before any Chemistry paper.

  • 2026-27 NCERT Alignment: matches the current syllabus print.
  • One-Page Printability: master tables fit one A4 landscape sheet.
  • Trend Tagging: every property tagged increases / decreases / irregular.
  • Expert Verification: checked against NCERT 4.1 to 4.8 and recent JEE and NEET papers.

d- and f-Block Symbol Glossary for Class 12 Chemistry

Mixing up Mn+ with the neutral-atom configuration is the most common magnetic-moment slip.

SymbolMeaningTypical Unit / Value
nNumber of unpaired electrons in d-orbital of Mn+dimensionless
μSpin-only magnetic momentBohr Magneton (BM)
μBOne Bohr Magneton9.274 x 10-24 J T-1
(n-1)d, nsOuter electron shells of a d-block element-
4f, 5fFilling orbitals of lanthanoids and actinoids-
Ln, AnGeneric lanthanoid / actinoid-
Standard electrode potentialVolt
ZeffEffective nuclear chargedimensionless
riIonic radiuspm
Spin-only magnetic moment formula - Class 12 Chemistry Chapter 4 Formula Sheet

d- and f-Block Elements All Important Formulae for Class 12 Chemistry

The master table below lists every formula in NCERT Chapter 4 with units, section reference, and exam-use cue. All entries are retained in the 2026-27 syllabus.

ConceptFormula / RelationValue or UnitNCERT RefCommon Use
General d-block configuration (n-1)d1-10 ns1-2 -4.1Identify block
Lanthanoid configuration [Xe] 4f1-14 5d0-1 6s2 -4.5Z = 58 to 71
Actinoid configuration [Rn] 5f1-14 6d0-1 7s2 -4.7Z = 90 to 103
Cr exception [Ar] 3d5 4s1 -4.1Half-filled stability
Cu exception [Ar] 3d10 4s1 -4.1Fully-filled stability
Spin-only magnetic moment μ = n(n+2) BM4.2From n unpaired e-
Magnetic moment for n = 1 μ = 1.73 BM4.2Ti3+, V4+, Cu2+
Magnetic moment for n = 2 μ = 2.83 BM4.2V3+, Ni2+
Magnetic moment for n = 3 μ = 3.87 BM4.2Cr3+, Co2+
Magnetic moment for n = 4 μ = 4.90 BM4.2Fe2+, Co3+
Magnetic moment for n = 5 μ = 5.92 BM4.2Mn2+, Fe3+
Lanthanoid contraction r(La3+) = 103 pm → r(Lu3+) = 86 pm pm4.5Steady decrease across Ln
K2Cr2O7 preparation 4FeCr2O4 + 8Na2CO3 + 7O2 → 8Na2CrO4 + 2Fe2O3 + 8CO2 -4.3From chromite ore
Cr orange to yellow Cr2O72- + 2OH- 2CrO42- + H2O -4.3pH-dependent eq.
KMnO4 oxidation (acidic) MnO4- + 8H+ + 5e- → Mn2+ + 4H2O n-factor = 54.3Acidic titrations
KMnO4 (neutral / faintly alkaline) MnO4- + 2H2O + 3e- → MnO2 + 4OH- n-factor = 34.3Baeyer's reagent
KMnO4 (strongly alkaline) MnO4- + e- → MnO42- n-factor = 14.3Manganate green
Standard reduction potential (acid) E(MnO4-/Mn2+) = +1.51 V V4.3Strong oxidiser

Reverse shortcut: square the BM value and solve n(n+2) for n. For μ = 5.92 BM, n = 5.

d- and f-Block Quick-Fact Cards for MCQ Recall

The five facts below are the ones JEE Main and NEET rotate as 1-mark MCQs.

√n(n+2)
Spin-only magnetic moment in Bohr Magneton, where n = unpaired d-electrons of Mn+
+1.51 V
E° of MnO4-/Mn2+ in acid: KMnO4 is a strong oxidiser
+2, +3
Most stable lanthanoid oxidation state is +3; +2 / +4 exceptions only at Eu, Yb, Ce, Tb
17 pm
Total lanthanoid contraction: La3+ (103 pm) to Lu3+ (86 pm)
+2 to +7
Mn shows the widest range of oxidation states in the first transition series

Class 12th d- and f-Block Property Trend Table

The five trends below are frequent CBSE one-mark VSAQs. Memorise the direction.

PropertyTrend Across 3d SeriesReasonNCERT Ref
Atomic radiusDecreases then nearly constant (Mn to Cu)Zeff rise balanced by d-d repulsion4.2
Ionisation enthalpyIrregular rise overallHalf- / fully-filled stability of Mn, Zn4.2
Melting pointPeaks at Cr (3d54s1), low at ZnNumber of unpaired d electrons4.2
DensityIncreases steadily across the periodMass rise outweighs radius decrease4.2
Oxidation state rangeMaximum at Mn (+2 to +7), drops afterPairing of d-electrons after Mn4.2
Quick Tip: Lanthanoid contraction is the top NEET MCQ here. Always cite the cause (poor 4f shielding) and one consequence (Zr and Hf have near-identical radii).
KMnO4 oxidation of Fe2+ reaction - Class 12 Chemistry Chapter 4 Formula Sheet

When to Use Which Formula in d- and f-Block Elements

  • "Magnetic moment of Mn+": count unpaired d-electrons, apply μ = n(n+2) BM.
  • "Why +3 stable for lanthanoids": ionisation plus hydration enthalpy favours Ln3+.
  • "KMnO4 n-factor": acidic / neutral / alkaline gives n = 5 / 3 / 1.
  • "Why Zr similar to Hf": lanthanoid contraction gives near-identical radii.
  • "Cr and Cu exceptions": half / full d-subshell stability beats Aufbau order.

d- and f-Block Common-Numerical Pattern Templates

The four setups below dominate CBSE, JEE Main, and NEET papers.

PatternWhat the question givesFormula to applyCommon trap
μ from ionIdentity of Mn+Count unpaired d-electrons, μ = n(n+2) Using neutral-atom configuration
n from μμ in BMSolve n(n+2) = μ2Forgetting to take +ve root
n-factor of KMnO4Medium (acidic / neutral / alkaline)5 / 3 / 1 respectivelyDefaulting to 5 in every medium
Identify exceptionElement symbol (Cr, Cu, Mo, Ag, Au)Half / full d-subshell ruleListing Mn or Fe as exception
Watch Out: With KMnO4 in a redox titration, check the medium first. Using n-factor 5 in neutral medium (correct value 3) is the top mark-losing slip.

Student Feedback

In a Collegedunia poll of 900 Class 12 students, 78% said the spin-only magnetic moment and n-factor cards on this d- and f-Block sheet were the fastest to revise the night before the exam.

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Other Resources for The d- and f-Block Elements Class 12 Chemistry

NCERT Formula Sheet for Class 12 Chemistry: All Chapters

Jump to the formula sheet for any other chapter of Class 12 Chemistry below.

d- and f-Block Elements Class 12 Chemistry Formula Sheet FAQs

Ques. Where can I download the d- and f-Block Elements Class 12 Chemistry Formula Sheet PDF?

Ans. You can download the d- and f-Block Elements Class 12 Chemistry Formula Sheet PDF directly from this Collegedunia page. Both the Normal and HD versions are available and free.

Ques. Is this Formula Sheet aligned with the 2026-27 NCERT?

Ans. Yes. This page reflects the current 2026-27 syllabus for Class 12 Chemistry. The d- and f-Block Elements chapter is fully retained in the new edition with no formula cuts; every relation in Sections 4.1 to 4.8 of the NCERT remains examinable.

Ques. How many pages is the Class 12th Chemistry d- and f-Block Elements Formula Sheet PDF?

Ans. The Formula Sheet PDF runs approximately 8 pages and covers the master formula table, symbol glossary, magnetic-moment quick cards, trend table, when-to-use decision tree, and four common numerical pattern templates.

Ques. What is the formula for spin-only magnetic moment in d- and f-Block Elements?

Ans. The spin-only magnetic moment is given by μ = √n(n+2) Bohr Magneton, where n is the number of unpaired electrons in the d-orbital of the metal ion Mn+. For example, Fe3+ (3d5) has n = 5, so μ = √35 = 5.92 BM.

Ques. What is lanthanoid contraction and why does it matter?

Ans. Lanthanoid contraction is the steady decrease in atomic and ionic radii of the 4f-block elements from La to Lu (a total drop of about 17 pm). It arises because 4f electrons shield the nuclear charge poorly. The consequence is that 4d and 5d transition elements of the same group (e.g. Zr and Hf, Nb and Ta) have nearly identical sizes and similar chemistry.

Ques. Why are Cr and Cu electronic configurations exceptions in the 3d series?

Ans. Cr is [Ar]3d54s1 and Cu is [Ar]3d104s1. A half-filled (d5) or fully filled (d10) d-subshell is extra-stable due to exchange energy and symmetric distribution of electrons. The 4s electron promotes to 3d to achieve this stable configuration.

Ques. What is the n-factor of KMnO4 in different media?

Ans. KMnO4 changes its n-factor with the medium: 5 in acidic solution (Mn7+ to Mn2+), 3 in neutral or faintly alkaline solution (Mn7+ to Mn4+, forming MnO2), and 1 in strongly alkaline solution (Mn7+ to Mn6+, forming MnO42-). Always check the medium before solving a redox titration.

Ques. Why do transition metals show variable oxidation states?

Ans. Transition metals have (n-1)d and ns electrons of comparable energy, so both shells can lose electrons. Mn shows the widest range (+2 to +7) because its 3d54s2 configuration provides seven outer electrons that can all participate in bonding.

Ques. What are the Mn oxidation states and which one is most stable?

Ans. Mn shows oxidation states +2, +3, +4, +6 and +7 in standard compounds. Mn(+2) is the most stable in aqueous solution because losing the two 4s electrons gives 3d5 high-spin half-filled configuration with maximum exchange energy. Mn(+7) is the highest oxidation state in the 3d series and is realised in KMnO4.

Ques. What are the consequences of lanthanide contraction?

Ans. The lanthanoid contraction consequences are: (i) nearly identical sizes of 4d and 5d transition metals of the same group, so Zr/Hf and Nb/Ta are chemically inseparable; (ii) decreasing basicity of trivalent hydroxides from La(OH)3 to Lu(OH)3; (iii) closely spaced ionisation enthalpies that make individual lanthanoid separation industrially difficult.

Ques. How does the chromate-dichromate equilibrium shift with pH?

Ans. The equilibrium 2CrO42- + 2H+ Cr2O72- + H2O is pH-controlled. Adding acid shifts it right, giving the orange Cr2O72- ion in dichromate solutions. Adding base shifts it left, giving the yellow CrO42- ion in chromate solutions.

Ques. What is the color of transition metal compounds and which ions are colourless?

Ans. Most transition metal compounds are coloured because of d-d electronic transitions in the crystal field, where an electron absorbs a part of the visible spectrum and jumps from one d-orbital to another of higher energy. Sc3+, Ti4+ (d0) and Cu+, Zn2+ (d10) lack a possible d-d transition and are colourless.

Ques. What is the catalytic activity of transition metals?

Ans. Transition metals are good catalysts because they have variable oxidation states (allowing electron exchange in a reaction cycle) and partially filled d-orbitals (providing adsorption sites). V2O5 catalyses the Contact process, Fe catalyses the Haber process, Ni catalyses hydrogenation, and Pt catalyses catalytic converters.

Ques. What are interstitial compounds and alloys of transition metals?

Ans. Interstitial compounds are non-stoichiometric solids formed when small atoms (H, C, N, B) occupy lattice voids of transition metals, giving hard, high-melting, chemically inert materials like TiC and Mn4N. Alloy formation is favoured by the similar atomic radii of transition metals. Brass (Cu-Zn), bronze (Cu-Sn), stainless steel (Fe-Cr-Ni) and misch metal (95% Ln + 5% Fe) are textbook alloys.

Ques. What are the actinoids vs lanthanoids differences for formula recall?

Ans. Lanthanoids have the configuration [Xe] 4f1-14 5d0-1 6s2 and are predominantly +3 with a few +2/+4 exceptions. Actinoids have [Rn] 5f1-14 6d0-1 7s2 and show oxidation states up to +7 (Np, Pu). All actinoids are radioactive; actinoid contraction is larger per element than lanthanoid contraction because 5f electrons shield even less effectively than 4f.