A
The five-day first pass
Inorganic chapters reward a different method from Physics or Organic. Almost every question here is an NCERT statement or a reason attached to an exception — so the goal is recall under time pressure, not conceptual derivation.
Day 1
NCERT read, nothing else
Read the chapter line by line — small print, in-text tables, solved examples. Take no notes yet. Mark every sentence that states an exception or supplies a reason.
≈ 2 hours
Day 2
Build one master table
One sheet: 3d series across the top, rows for configuration, oxidation states, both E° values, unpaired electrons, μ, ion colour. Building it yourself is the study — a printed one is worth far less.
≈ 2 hours
Day 3
Equations, by hand
K₂Cr₂O₇ and KMnO₄ preparation, the three medium-dependent half reactions, the standard oxidations. Write each three times from memory, then check against the book.
≈ 2 hours
Day 4
The f-block
Lanthanoid contraction with all consequences, the exceptional oxidation states, the lanthanoid–actinoid table. Short and pure memory — treat as flashcards, not reading.
≈ 1.5 hours
Day 5
Questions only
Work the bank in §D, then a full previous-year compilation. The repetition rate in this chapter is unusually high — the same four or five ideas rotate.
≈ 2 hours
Why the order matters
Reading before tabulating means the table is a recall exercise rather than a copying exercise. Tabulating before the equations means the oxidation-state logic behind the KMnO₄ and K₂Cr₂O₇ reactions is already in place. Leaving the f-block until day 4 keeps it fresh — it is the part most often forgotten by exam day.
B
Maintenance loop and what goes wrong
The first pass is the cheap part. Retention is where this chapter is won or lost.
The revisit schedule
- Revisit on roughly day 10, day 25, and day 50, then once more in the final month.
- Every revisit is recall first, book second: close the NCERT, reproduce the master table and the equation set on blank paper, then correct in red.
- Re-reading feels productive and is not. The correction step is where the retention comes from — the red marks are the actual study material for the next revisit.
- Each revisit should take less time than the last. If it doesn't, the first pass was passive.
Three ways students lose marks here
- Learning trends and skipping exceptions. Roughly a third of the questions live in the exceptions. For every trend, immediately ask "who breaks this, and why" — Cr and Cu break configuration, Mn and Zn break melting point, Cu breaks E°, MnO₄⁻ breaks the colour explanation, lanthanoids break the magnetic formula.
- Treating KMnO₄ and K₂Cr₂O₇ as optional detail. That block alone outweighs the entire trends section in question count.
- Reading the stem too fast. A large share of this chapter's questions are phrased with "not", "except" or "incorrect". Underline the negative word before looking at the options.
Timing in the exam
- Target 35–40 seconds per question from this chapter. These are recall questions; if the answer isn't surfacing in 40 seconds, it is not going to, so mark and move.
- Take the direct factual ones on the first sweep and leave assertion–reason items for the second.
Check before you start
Syllabus rationalisation has trimmed parts of this chapter in recent editions, and some coaching sources report that the chemical reactivity of the lanthanoids has been dropped. Confirm the current scope against the official NTA syllabus PDF before spending time on anything that looks marginal.
C
Topic-wise weightage map
Share of this chapter's questions, estimated from the recurring pattern across recent papers. Use it to allocate revision time, not as a guarantee.
| Band | Topics | Share of questions | Share of revision time |
|---|---|---|---|
| Non-negotiable | KMnO₄ / K₂Cr₂O₇, oxidation states, lanthanoids, magnetism | ≈ 61% | About two-thirds |
| Solid coverage | Configuration, colour, electrode potentials | ≈ 26% | About a quarter |
| One clean read | Physical trends, catalysis, interstitials, alloys, actinoids, applications | ≈ 13% | The remainder |
D
Question bank, sorted by topic
Thirty-six questions in the exact shapes this chapter is asked in, distributed to match the weightage map above.
Read this first
These are written in the recurring NEET patterns for this chapter — they are not reproductions of actual question papers. Work them as a diagnostic, then go through an official previous-year compilation alongside for the real wording and difficulty calibration.
1 · KMnO₄ and K₂Cr₂O₇
- The oxidation state of chromium in Cr2O72− is
a+3b+6c+7d+12
Answer
(b) +6. 2x + 7(-2) = -2 ⇒ x = +6. The same +6 is present in CrO42− — which is why their interconversion is not a redox change. - Acidified KMnO4 oxidising Fe2+ gains how many electrons per permanganate ion?
a1b3c5d6
Answer
(c) 5. MnO4− + 8H+ + 5e− → Mn2+ + 4H2O. Manganese falls from +7 to +2. - When KMnO4 acts as an oxidant in a neutral or faintly alkaline medium, the manganese product is
aMn2+bMnO2cMnO42−dMn2O3
Answer
(b) MnO2. Three electrons are gained: MnO4− + 2H2O + 3e− → MnO2 + 4OH−. Strongly alkaline conditions instead give green MnO42− by a one-electron change. - On adding KOH to an orange solution of K2Cr2O7, the solution turns yellow. The oxidation state of Cr
afalls to +3brises to +7cremains +6dfalls to +2
Answer
(c) remains +6. Cr2O72− + 2OH− ⇌ 2CrO42− + H2O — an acid–base equilibrium, not a redox reaction. A classic distractor. - The ore used industrially to prepare K2Cr2O7 is
apyrolusitebchromitechaematitedsiderite
Answer
(b) chromite, FeCr2O4. Pyrolusite, MnO2, is the starting material for KMnO4 — the two are routinely swapped in options. - Which statement about the dichromate ion is correct?
aCr is octahedrally coordinatedbTwo tetrahedra share one oxygencIt is paramagneticdIt contains a Cr–Cr bond
Answer
(b). Two CrO4 tetrahedra joined through a Cr–O–Cr bridge, the bridging bond being longer. Cr(VI) is d⁰, so the ion is diamagnetic and there is no metal–metal bond. - The equivalent mass of KMnO4 acting as an oxidant in acidic medium is (M = 158)
a158b79c52.7d31.6
Answer
(d) 31.6. E = M/5 = 158/5 = 31.6. In neutral medium the change is 3 electrons, giving 158/3 = 52.7 — option (c) is the trap for a misread medium.
2 · Oxidation states
- The element of the first transition series showing the maximum number of oxidation states is
aCrbMncFedSc
Answer
(b) Mn. Configuration 3d⁵4s²; all seven electrons can participate, giving +2 through +7. - The highest oxidation state attained by any d-block element is
a+6b+7c+8d+9
Answer
(c) +8, shown by Ru and Os in RuO4 and OsO4. The +7 of Mn is only the 3d-series maximum. - Which of the following compounds does not exist?
aCuIbCuI2cCuCl2dCuF2
Answer
(b) CuI2. Iodide is a reducing anion and cannot stabilise the higher oxidation state; it reduces Cu(II) to Cu(I). By the same logic FeI3 does not exist. - Among the manganese oxides, the most acidic is
aMnObMn2O3cMnO2dMn2O7
Answer
(d) Mn2O7. Acidity rises with oxidation state as bonding becomes more covalent: MnO basic → MnO₂ amphoteric → Mn₂O₇ acidic. - The highest oxidation state of a transition metal is generally found in its
aiodidebbromidecfluoride or oxidedsulphide
Answer
(c). Only the most electronegative and least polarisable partners — F and O — can stabilise a high oxidation state: VF5, CrO3, Mn2O7, OsF6.
3 · Lanthanoids and lanthanoid contraction
- Lanthanoid contraction is caused by
aperfect shielding by 4f electronsbimperfect shielding by 4f electronscdecreasing nuclear chargedincreasing 5d occupancy
Answer
(b). The diffuse, buried 4f orbitals screen the outer electrons poorly, so effective nuclear charge rises steadily and the radius shrinks from La³⁺ to Lu³⁺. - Zr and Hf have almost identical chemical properties because
athey belong to the same group and periodbof lanthanoid contractioncboth are radioactivedboth have a d¹⁰ configuration
Answer
(b). Lanthanoid contraction cancels the expected size increase from period 5 to 6, giving radii of 160 and 159 pm — which is why the two are so hard to separate. Nb/Ta and Mo/W behave the same way. - Ce4+ is a strong oxidising agent because
ait has a 4f⁷ configurationbit readily reverts to the common +3 statecit is radioactivedit is the largest lanthanoid ion
Answer
(b). E°(Ce4+/Ce3+) = +1.74 V; the +4 state is unusual for a lanthanoid, so it accepts an electron readily to reach the characteristic +3. - Which lanthanoid ion is both colourless and diamagnetic?
aCe3+bEu2+cLu3+dGd3+
Answer
(c) Lu3+, which is 4f¹⁴. La3+ (4f⁰) is the other one. Everything between has f electrons, hence colour and paramagnetism. - Misch metal consists principally of
a95% lanthanoid metal with 5% Febequal parts Ce and FecCu and Zndpure cerium
Answer
(a). Roughly 95% lanthanoid metal and 5% iron, with traces of S, C, Ca and Al; used in Mg-based alloys for tracer bullets and shells, and in lighter flints.
4 · Magnetic properties
- The spin-only magnetic moment of Fe3+ is
a1.73 BMb3.87 BMc4.90 BMd5.92 BM
Answer
(d) 5.92 BM. Fe3+ is 3d⁵, so n = 5 and μ = \sqrt{5×7} = \sqrt{35} = 5.92. - Which of the following ions is diamagnetic?
aCu2+bTi3+cZn2+dFe2+
Answer
(c) Zn2+, 3d¹⁰ with no unpaired electrons. Cu+, Sc3+ and Ti4+ are the other stock answers to this question. - An aqueous 3d ion has μ = 2.84 BM. The ion is
aCr3+bNi2+cMn2+dCu2+
Answer
(b) Ni2+. μ = 2.84 ⇒ n = 2, and Ni2+ is 3d⁸ with two unpaired electrons. V3+ (3d²) fits equally and is the usual alternative. - The spin-only formula cannot be applied to lanthanoid ions because
athey are diamagneticbthe orbital contribution is not quenchedcthey have no unpaired electronsdthey are radioactive
Answer
(b). The 4f orbitals are shielded by 5s and 5p, so orbital angular momentum contributes and μ = g\sqrt{J(J+1)} must be used instead.
5 · Electronic configuration and position
- Which of the following is not regarded as a transition element?
aScbCucZndCr
Answer
(c) Zn. Both the atom (3d¹⁰4s²) and the only ion (3d¹⁰) have a full d subshell. Sc qualifies through the atom, and Cu through the d⁹ Cu²⁺ ion. - The ground-state configuration of chromium is
a[Ar] 3d⁴4s²b[Ar] 3d⁵4s¹c[Ar] 3d⁶4s⁰d[Ar] 3d³4s²
Answer
(b) [Ar] 3d⁵4s¹, from the extra stability of the exactly half-filled d subshell. Copper is the parallel case, [Ar] 3d¹⁰4s¹. - Which element has a completely vacant outermost s orbital in its ground state?
aPtbPdcAgdNi
Answer
(b) Pd, which is [Kr] 4d¹⁰5s⁰ — unique in the block. Pt is 5d⁹6s¹ and Ag is 4d¹⁰5s¹. - The number of unpaired electrons in Cu2+ is
a0b1c2d3
Answer
(b) 1. Cu2+ is 3d⁹. Note that Cu+ is 3d¹⁰ with zero — the two are commonly confused.
6 · Colour of ions
- Which aqueous ion is colourless?
aSc3+bTi3+cCr3+dNi2+
Answer
(a) Sc3+, 3d⁰ — no d–d transition is possible. The other stock colourless ions are Ti4+, Zn2+, Cu+ and Ag+. - The intense purple colour of MnO4− arises from
aa d–d transitionbcharge transfercunpaired d electronsdlattice defects
Answer
(b) charge transfer. Mn(VII) is d⁰, so no d–d transition is available; light promotes an electron from oxygen to manganese. The same holds for orange Cr2O72− and yellow CrO42−. - Hydrated copper(II) sulphate is blue while anhydrous copper(II) sulphate is white because
aCu is reduced on hydrationbwater acts as a ligand and splits the d orbitalscthe anhydrous salt is d¹⁰dthe anhydrous salt is ionic
Answer
(b). Colour needs a ligand field. With no coordinated water there is no splitting, so no d–d transition and no colour, even though Cu is still d⁹.
7 · Standard electrode potentials
- Which 3d metal does not liberate hydrogen from dilute acids?
aFebZncCudNi
Answer
(c) Cu. It is the only 3d metal with a positive E°(M2+/M) = +0.34 V, because its high sublimation and ionisation enthalpies are not offset by its hydration enthalpy. - The strongest reducing agent among the following aqueous ions is
aCr2+bMn2+cFe2+dCo2+
Answer
(a) Cr2+. E°(Cr3+/Cr2+) = -0.41 V; losing one electron gives the stable half-filled t₂g d³ configuration. - E°(M3+/M2+) is most positive for
aCrbFecMndTi
Answer
(c) Mn, at +1.57 V — Mn3+ is strongly oxidising because reduction gives the stable d⁵ Mn2+. If Co appears among the options it wins instead, at +1.97 V.
8 · Physical and atomic trends
- The element with the highest melting point in the 3d series is
aMnbCrcFedZn
Answer
(b) Cr. Manganese is the trap: despite sitting at d⁵ it has an anomalously low melting point, because the stable half-filled shell holds its electrons back from metallic bonding. - The lowest enthalpy of atomisation in the 3d series belongs to
aScbMncCudZn
Answer
(d) Zn. With 3d¹⁰ full, no d electrons take part in metallic bonding — the same reason Hg is a liquid.
9 · Catalysis, interstitials, alloys
- The Ziegler–Natta catalyst is
aV2O5bTiCl4 with Al(C2H5)3cfinely divided NidMnO2
Answer
(b). Used for the stereospecific polymerisation of alkenes to polythene. V2O5 belongs to the contact process, Ni to hydrogenation, and MnO2 to the decomposition of KClO3. - Which is not a property of interstitial compounds?
aNon-stoichiometric compositionbHigher melting point than the parent metalcLoss of metallic conductivitydGreater hardness than the parent metal
Answer
(c). They retain metallic conductivity. Note the "not" in the stem — this phrasing is very common in this chapter.
10 · Actinoids and applications
- Compared with the lanthanoids, the actinoids
ashow a smaller contractionbshow a greater contraction and a wider range of oxidation statescare all non-radioactivedhave higher ionisation enthalpies
Answer
(b). The 5f orbitals shield even more poorly than 4f, so the contraction is greater; oxidation states run beyond +3 up to +7 (Np). All actinoids are radioactive, and their ionisation enthalpies are lower than those of the lanthanoids.
How to score this
Mark each topic block out of its question count. Any block below 70% goes back into the day-10 revisit. A wrong answer in blocks 1 to 4 matters roughly three times as much as one in blocks 8 to 10 — weight the correction effort accordingly.