Every chemistry question flagged on this paper, rebuilt in full. Each one carries what was given, what was asked, the concept behind it, the formula, the steps written out, a table justifying the right option and ruling out each of the others, the fastest route through, and an animated figure wherever seeing the thing settles the answer.
9Questions lost
3Attempted, wrong
6Left blank
39Marks at stake
The shape of this paper
Nine questions, and the balance is heavily one-sided: six were left blank against three
attempted. On NEET marking the six blanks were worth 24 marks, and the three wrong answers
cost 12 more plus 3 in negatives — 39 marks in total.
A blank rate of two in three is the dominant fact here, and it is worth separating from
accuracy. Of the six blanks, four are short: Q139 is a configuration you already know, Q149 is
a trend with no arithmetic at all, Q164 is pure ordering, and Q168 is a single division once
the electron count is written down. Only Q170 genuinely takes reading time, and even that
collapses to one filter.
The three wrong answers, by contrast, were all close. Each had one half right.
Consider the ground state of chromium atom (Z = 24). How many electrons are with Azimuthal quantum number l = 1 and l = 2 respectively?
12 and 4
16 and 4
Correct12 and 5
16 and 5
Given
Chromium, Z = 24, in its ground state.
Azimuthal quantum number l = 1 and l = 2.
Asked
The number of electrons with l = 1, and with l = 2.
Concept to use
The azimuthal quantum number l labels the type of subshell, not the shell: l = 0 is s, l = 1 is p, l = 2 is d, l = 3 is f. So “how many electrons have l = 1” means “how many p electrons are there in the whole atom” — across every shell, not just the outermost. The second thing the question is testing is whether you remember that chromium is one of the two configuration exceptions.
DiagramAnimatedChromium's orbital diagram, with the two subshell types counted.
Formula to usel = 0 → s l = 1 → p l = 2 → d l = 3 → f
Baby steps
Write the ground-state configuration. Chromium is an exception: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s¹, not 3d⁴4s². A half-filled d subshell is extra stable, so one 4s electron moves across.
Check the total: 2 + 2 + 6 + 2 + 6 + 5 + 1 = 24. Correct for Z = 24.
Electrons with l = 1 are all the p electrons: 2p⁶ and 3p⁶, giving 6 + 6 = 12.
Electrons with l = 2 are all the d electrons: 3d⁵ only, giving 5.
So the answer is 12 and 5.
Answer
12 and 5
Why this option and not the others
Option
Verdict
Reason
12 and 4
rule out
The p count is right, but 4 d electrons means the configuration was written as 3d⁴4s² — the chromium exception was missed.
16 and 4
rule out
Wrong on both counts. 16 would require a third full p subshell, and there is no 4p electron in chromium.
12 and 5
keep
2p⁶ + 3p⁶ = 12 for l = 1, and 3d⁵ = 5 for l = 2, using the correct 3d⁵4s¹ ground state.
16 and 5
rule out
The d count is right but the p count is not. There is no 4p occupancy in a ground-state chromium atom.
Shortcut
Half of this question is free and always will be. Every element from Sc to Zn has the same l = 1 count: 12, because 2p and 3p are both full for all of them and 4p is empty. So for any 3d-series atom you only ever need to work out the d count. That instantly eliminates the two options offering 16, and the remaining choice is just 3d⁴ against 3d⁵ — which is the chromium exception you already know.
The d- and f-Block Elements
3 wrong · 5 blank
Eight questions, six of them left blank. Three separate questions turn on the same ladder: what manganese becomes in acidic, neutral and alkaline media. That one ladder is the highest-value thing on this paper.
Q 149Left blankTrends · density
Which of the following pairs of d-block elements have higher density?
Sc and Ti
Cr and Mn
Fe and Co
CorrectNi and Cu
Given
Four pairs drawn from the first transition series.
Asked
Which pair has the higher density.
Concept to use
Density is mass packed into volume, and across the 3d series both factors push the same way for a while. Atomic mass climbs steadily from Sc to Zn, while the metallic radius shrinks as the growing nuclear charge pulls the electron cloud in tighter. Heavier atoms in a smaller volume means density rises — reaching a maximum around copper before falling away at zinc, whose filled d shell makes for weaker metallic bonding and a looser packing.
DiagramAnimatedDensity across the 3d series, with the peak picked out.
Formula to usedensity ∝ atomic mass / atomic volume mass ↑ , radius ↓ → density ↑
Baby steps
Sc 2.99 and Ti 4.51 — the lightest pair by a wide margin, at the start of the series.
Cr 7.19 and Mn 7.21 — middling.
Fe 7.87 and Co 8.90 — high, but iron pulls the pair down.
Ni 8.90 and Cu 8.95 — both near the top of the series.
Ni and Cu is the densest pair.
Answer
Ni and Cu
Why this option and not the others
Option
Verdict
Reason
Sc and Ti
rule out
The two least dense metals of the whole series — largest radii, smallest masses.
Cr and Mn
rule out
Around 7.2 g cm⁻³ each, comfortably below the late-series metals.
Fe and Co
rule out
Cobalt is genuinely dense at 8.90, but iron at 7.87 drags the pair below Ni–Cu. This is the closest distractor.
Ni and Cu
keep
8.90 and 8.95 — the two highest values among the pairs offered, sitting at the density maximum of the series.
Shortcut
You do not need the numbers, only the shape of the trend: density rises across the series and peaks near copper. So the answer is whichever pair sits furthest to the right without reaching zinc. Reading the four options left to right, Ni–Cu is the rightmost pair on offer, and that is enough.
Q 153Attempted · wrongMagnetic moment · matching species
Worth knowing the key is arguableWorking every moment out gives two matched sets, not one. The keyed answer of 2 is right if the question means “how many matching groups”. If it means “how many individual species share a value with another”, the answer would be 4. What is certain is that 3 is not obtainable on either reading — see the full table in the steps below.
In the following species, how many species have same magnetic moment? (i) Cr²⁺ (ii) Mn³⁺ (iii) Ni³⁺ (iv) Sc²⁺ (v) Zn²⁺ (vi) V³⁺ (vii) Ti⁺
1
Her answer3
Correct2
4
Given
Seven transition-metal ions.
Magnetic moment is to be taken as the spin-only value.
Asked
How many of them share the same magnetic moment.
Concept to use
The spin-only magnetic moment depends on one number and one number only: n, the count of unpaired electrons. So two ions match if and only if their unpaired counts match — which means the whole question reduces to writing out seven d-configurations and counting. Note that d-configurations either side of d⁵ give the same n: d² and d⁸ both give 2, d³ and d⁷ both give 3.
DiagramAnimatedThe spin-only curve. Equal n means equal moment — always.
Formula to useμ = √[n(n + 2)] BM, n = number of unpaired electrons
Baby steps
Cr²⁺ (Z = 24): d⁴, n = 4, μ = 4.90 BM.
Mn³⁺ (Z = 25): d⁴, n = 4, μ = 4.90 BM. Matches Cr²⁺.
Ni³⁺ (Z = 28): d⁷, n = 3, μ = 3.87 BM.
Sc²⁺ (Z = 21): d¹, n = 1, μ = 1.73 BM.
Zn²⁺ (Z = 30): d¹⁰, n = 0, μ = 0 — diamagnetic.
V³⁺ (Z = 23): d², n = 2, μ = 2.83 BM.
Ti⁺ (Z = 22): d³, n = 3, μ = 3.87 BM. Matches Ni³⁺.
So there are two matching sets: {Cr²⁺, Mn³⁺} at 4.90 BM and {Ni³⁺, Ti⁺} at 3.87 BM. The keyed answer, 2, counts the sets.
Answer
2 (as keyed — two matching sets)
Why this option and not the others
Option
Verdict
Reason
1
rule out
There is more than one match. {Cr²⁺, Mn³⁺} alone would give this, but Ni³⁺ and Ti⁺ also agree.
3
rule out
No reading of the question produces 3. There are 2 matching sets, 4 species involved in a match, and 5 distinct moment values among the seven ions — but never 3.
2
keep
Two sets of ions share a moment: the d⁴ pair at 4.90 BM and the d³/d⁷ pair at 3.87 BM.
4
rule out
This is the count of individual species caught up in a match rather than the count of matching groups. Defensible as a reading of the wording, but not the keyed answer.
Shortcut
Do not compute seven square roots. Just write the seven n values in a row — 4, 4, 3, 1, 0, 2, 3 — and look for repeats. Equal n means equal μ, always, so the moments themselves never need to be evaluated. And remember the mirror rule: d¹ and d⁹ both give n = 1, d² and d⁸ both give 2, d³ and d⁷ both give 3, d⁴ and d⁶ both give 4. Half the configurations you meet have a twin.
Where it went wrong
3 is not reachable by any counting of this list. The most likely route to it is stopping the tally early — spotting the Cr²⁺/Mn³⁺ pair, noticing one more coincidence, and counting species and sets in the same breath. The fix is mechanical: write all seven n values down in a row before looking at the options at all. It takes about forty seconds and it makes the tally visible instead of held in the head.
Q 154Left blankKMnO₄ · reactions in acidic medium
Which of the following reactions of KMnO₄ occurs in acidic medium?
Oxidation of thiosulphate to sulphate
CorrectPrecipitation of sulphur from H₂S
Oxidation of iodide to iodate
Oxidation of manganous salt to MnO₂
Given
Potassium permanganate acting as an oxidising agent.
Four candidate reactions, one of which occurs in acidic medium.
Asked
Which reaction takes place in acidic medium.
Concept to use
There is a single test that sorts all four, and it does not require remembering the reactions individually. In acidic medium manganese always ends up as Mn²⁺. In neutral or faintly alkaline medium it stops at MnO₂. So look at what happens to the manganese in each option: if the product named is MnO₂ or MnO₄²⁻, the reaction is not the acidic one.
DiagramAnimatedFollow the manganese: the medium decides where it stops.
Thiosulphate → sulphate. The standard reaction is 8MnO₄⁻ + 3S₂O₃²⁻ + H₂O → 8MnO₂ + 6SO₄²⁻ + 2OH⁻. Manganese lands at MnO₂, and hydroxide is produced — this is the neutral/faintly alkaline reaction.
Iodide → iodate. 2MnO₄⁻ + H₂O + I⁻ → 2MnO₂ + 2OH⁻ + IO₃⁻. Again MnO₂ and hydroxide — neutral/alkaline. (In acidic medium iodide goes only as far as I₂, not iodate.)
Mn²⁺ → MnO₂. This is manganous salt being oxidised up to MnO₂, which again happens in neutral or alkaline conditions.
H₂S → S. 2MnO₄⁻ + 5H₂S + 6H⁺ → 2Mn²⁺ + 5S + 8H₂O. Manganese lands at Mn²⁺ and H⁺ is consumed — this is the acidic reaction.
So the answer is the precipitation of sulphur from H₂S.
Answer
Precipitation of sulphur from H₂S
Why this option and not the others
Option
Verdict
Reason
thiosulphate → sulphate
rule out
Manganese ends as MnO₂ and OH⁻ is released. That is the neutral/faintly alkaline reaction.
sulphur from H₂S
keep
H⁺ appears on the left and Mn²⁺ on the right — the signature of the acidic-medium half reaction.
iodide → iodate
rule out
Iodate forms only in neutral/alkaline conditions. In acid, permanganate takes iodide only to I₂.
manganous salt → MnO₂
rule out
Manganese finishes at +4, not +2, so the medium cannot be acidic.
Shortcut
Ignore the sulphur and iodine entirely and follow only the manganese. Acidic finishes at Mn²⁺; neutral finishes at MnO₂; alkaline finishes at MnO₄²⁻. Three of the four options here name MnO₂ as the manganese product either explicitly or through the standard equation, so they eliminate themselves at a glance.
Q 164Left blankKMnO₄ · products in three media
Potassium permanganate acts as an oxidant in neutral, alkaline as well as acidic media. The final product obtained in the three conditions respectively are
CorrectMnO₂, MnO₄⁻² and Mn⁺²
Mn⁺², MnO₂ and MnO₄⁻²
MnO₄⁻², Mn⁺² and MnO₂
MnO₂, Mn⁺² and MnO₄⁻²
Given
KMnO₄ as an oxidant, where Mn starts at +7.
Three media, named in the order neutral, alkaline, acidic.
Asked
The final manganese product in each of the three media, in that order.
Concept to use
Manganese begins at +7 and falls by a different amount in each medium. The more acidic the conditions, the further it falls. In acid it goes all the way to +2 (Mn²⁺); in neutral conditions it stops halfway at +4 (MnO₂); in alkali it barely moves, dropping just one step to +6 (MnO₄²⁻, manganate). Every option offers the same three products — the whole question is whether you can put them in the order the stem asks for.
DiagramAnimatedThe same ladder, now read in the order the question asks.
Formula to useacidic → Mn²⁺ (+2) neutral → MnO₂ (+4) alkaline → MnO₄²⁻ (+6)
Baby steps
Learn the ladder once, in oxidation-state order: +7 falls to +6 in alkali, to +4 in neutral, to +2 in acid.
Now read the order the question actually asks for: neutral, alkaline, acidic — not the order you memorised it in.
Neutral → MnO₂.
Alkaline → MnO₄²⁻.
Acidic → Mn²⁺.
So the sequence is MnO₂, MnO₄²⁻, Mn²⁺.
Answer
MnO₂, MnO₄²⁻ and Mn²⁺
Why this option and not the others
Option
Verdict
Reason
MnO₂, MnO₄²⁻, Mn²⁺
keep
Matches the stem's order of neutral, alkaline, acidic exactly.
Mn²⁺, MnO₂, MnO₄²⁻
rule out
This is the right set of products in the order acidic, neutral, alkaline — the sequence most people memorise, which is precisely why it is offered.
MnO₄²⁻, Mn²⁺, MnO₂
rule out
Alkaline, acidic, neutral — a third permutation, matching no sensible reading.
MnO₂, Mn²⁺, MnO₄²⁻
rule out
Neutral is right but alkaline and acidic are swapped. Assigning Mn²⁺ to alkali is the single commonest slip here.
Shortcut
Every option contains the same three species, so this is purely an ordering question — and that means you should underline the order in the stem before looking at the options at all. Then use one anchor to test each option instead of checking all three slots: acidic is always Mn²⁺, so whichever option puts Mn²⁺ in the third position is the answer. One check, not three.
Q 166Attempted · wrongMatch the following · d-configurations
Match the following: A. Zn²⁺ I. d⁸ configuration B. Cu²⁺ II. Colourless C. Ni²⁺ III. μ = 1.73 BM
A-I, B-II, C-III
CorrectA-II, B-III, C-I
Her answerA-II, B-I, C-III
A-I, B-III, C-II
Given
Zn²⁺ (Z = 30), Cu²⁺ (Z = 29), Ni²⁺ (Z = 28).
Three properties: a d⁸ configuration, colourlessness, and μ = 1.73 BM.
Asked
The correct matching.
Concept to use
Write the three d-configurations first and every property falls out of them. Zn²⁺ is d¹⁰ — completely full, so there are no d-d transitions and it is colourless. Cu²⁺ is d⁹ — exactly one unpaired electron, so μ = √[1(3)] = 1.73 BM. Ni²⁺ is d⁸, which is the configuration named in the list.
DiagramAnimated1.73 BM sits at n = 1, which is d⁹ — copper.
Formula to used-electrons = (Z − 18) − charge μ = √[n(n+2)]
Baby steps
Zn²⁺: 30 − 18 − 2 = d¹⁰. All paired, no vacant d orbital to jump into → colourless → A–II.
Cu²⁺: 29 − 18 − 2 = d⁹. One unpaired electron, so μ = √3 = 1.73 BM → B–III.
Ni²⁺: 28 − 18 − 2 = d⁸ → C–I.
The matching is A–II, B–III, C–I.
Answer
A-II, B-III, C-I
Why this option and not the others
Option
Verdict
Reason
A-I, B-II, C-III
rule out
Gives d⁸ to Zn²⁺ (it is d¹⁰) and calls Cu²⁺ colourless (copper sulphate solutions are famously blue).
A-II, B-III, C-I
keep
d¹⁰ → colourless, d⁹ → 1.73 BM, d⁸ → Ni²⁺. All three consistent.
A-II, B-I, C-III
rule out
Zinc is placed correctly, but copper and nickel are swapped: Cu²⁺ is d⁹ not d⁸, and Ni²⁺ has two unpaired electrons giving 2.83 BM, not 1.73.
A-I, B-III, C-II
rule out
The copper match is right, but Zn²⁺ is not d⁸ and Ni²⁺ is not colourless — nickel salts are green.
Shortcut
1.73 BM is the fingerprint of exactly one unpaired electron. Among the common divalent 3d ions only Cu²⁺ (d⁹) fits, so lock that pairing in first. Then colourless can only be the full shell, Zn²⁺. Two matches made, and the third is forced — you never have to think about nickel at all.
Where it went wrong
Zinc was placed correctly, so the colourless-means-full-shell idea was secure. The error is a straight swap of copper and nickel: Cu²⁺ was given the d⁸ configuration and Ni²⁺ the 1.73 BM. Both halves of that swap are wrong for the same underlying reason — the d-electron count was off by one. Cu²⁺ is d⁹ (29 − 18 − 2) and Ni²⁺ is d⁸ (28 − 18 − 2). Writing the arithmetic out for all three ions before looking at the properties would have caught it; matching questions reward doing all the spadework first and only then reading the options.
Q 167Attempted · wrongRedox · SO₂ with acidified KMnO₄
In the reaction between moist SO₂ and acidified permanganate solution:
CorrectSO₂ is oxidised to SO₄²⁻ ; MnO₄⁻ is reduced to Mn²⁺
SO₂ is reduced to S ; MnO₄⁻ is oxidised to Mn²⁺
Her answerSO₂ is oxidised to SO₃²⁻ ; MnO₄⁻ is reduced to MnO₄⁻²
SO₂ is reduced to H₂S ; MnO₄⁻ is oxidised to MnO₄
Given
Moist SO₂, in which sulphur is at +4.
Acidified permanganate, in which manganese is at +7.
Asked
What is oxidised, what is reduced, and to what.
Concept to use
Permanganate is a powerful oxidising agent, so it must be the thing that gets reduced, and SO₂ the thing that gets oxidised. Then the medium fixes the destinations: acidified means manganese falls all the way to Mn²⁺, and sulphur, already at +4, rises to its maximum of +6 as sulphate.
DiagramAnimatedElectrons leave sulphur and arrive at manganese.
Formula to use2MnO₄⁻ + 5SO₂ + 2H₂O → 2Mn²⁺ + 5SO₄²⁻ + 4H⁺
Baby steps
Assign the starting oxidation numbers: S in SO₂ is +4; Mn in MnO₄⁻ is +7.
Permanganate is the oxidising agent, so it is reduced; SO₂ is therefore the reducing agent and is oxidised.
Sulphur oxidised means it goes up from +4. The available step is to +6, which is SO₄²⁻ (sulphate).
Manganese reduced in acidic medium goes to Mn²⁺ (+2).
Each sulphur gives up 2 electrons and each manganese takes 5, so the balancing numbers are 5 and 2.
Answer
SO₂ is oxidised to SO₄²⁻ and MnO₄⁻ is reduced to Mn²⁺
Why this option and not the others
Option
Verdict
Reason
SO₄²⁻ / Mn²⁺
keep
S rises +4 → +6 (a genuine oxidation) and Mn falls +7 → +2 (the acidic-medium product). Electrons balance at 5 : 2.
S / Mn²⁺
rule out
Backwards. SO₂ going to S would be a reduction of sulphur, and permanganate cannot be oxidised — +7 is already the maximum for manganese.
SO₃²⁻ / MnO₄²⁻
rule out
Self-contradictory. Sulphur is +4 in SO₂ and still +4 in SO₃²⁻, so nothing has been oxidised. And MnO₄²⁻ is the alkaline product, not the acidic one.
H₂S / MnO₄
rule out
Two errors again: sulphur falling to −2 is a reduction, and manganese cannot be oxidised above +7.
Shortcut
You can destroy two options here without knowing any chemistry beyond one fact: +7 is the highest oxidation state manganese can reach, so any option saying MnO₄⁻ is oxidised is dead on arrival. That kills two of the four. For the remaining pair, check the oxidation number of sulphur in the named product — if it has not changed, the option has contradicted itself.
Where it went wrong
The chosen option claims SO₂ is oxidised to SO₃²⁻, but sulphur is +4 in both. Nothing has been oxidised, so the option refutes itself the moment oxidation numbers are written on it. The second half compounds it: MnO₄²⁻ is manganate, the product in alkaline medium, while the stem says acidified. Both halves would have been caught by the same thirty-second habit — write the oxidation number above every species in the option before deciding. Note the link to Q164 and Q154 on this paper: three questions turned on the same acidic-versus-alkaline manganese product, which makes that one ladder the highest-value thing to fix in this chapter.
Q 168Left blankRedox stoichiometry · dichromate and iodide
How many moles of acidified K₂Cr₂O₇ is required to liberate 6 moles of I₂ from an aqueous solution of I⁻?
Correct2
1
0.25
0.5
Given
Acidified K₂Cr₂O₇ reacting with iodide.
6 moles of I₂ are to be liberated.
Asked
The number of moles of dichromate required.
Concept to use
This is electron bookkeeping. Dichromate takes chromium from +6 down to +3, and there are two chromium atoms per ion, so one dichromate accepts 6 electrons. Each iodide gives up 1 electron, and it takes two iodides to make one I₂ molecule, so each I₂ released costs 2 electrons. Match the electrons and the ratio appears.
DiagramAnimatedSix electrons in, two per iodine molecule out.
Formula to useCr₂O₇²⁻ + 6I⁻ + 14H⁺ → 2Cr³⁺ + 3I₂ + 7H₂O
Baby steps
Chromium: +6 → +3 is a drop of 3 per atom, and there are 2 atoms, so 1 mole of dichromate accepts 6 moles of electrons.
Iodine: 2I⁻ → I₂ + 2e⁻, so each mole of I₂ releases 2 moles of electrons.
6 electrons in, 2 electrons per I₂ → 1 mole dichromate liberates 3 moles of I₂.
For 6 moles of I₂: 6 ÷ 3 = 2 moles of K₂Cr₂O₇.
Answer
2 moles
Why this option and not the others
Option
Verdict
Reason
2
keep
1 mole of dichromate gives 3 moles of I₂, so 6 moles of I₂ needs 2 moles of dichromate.
1
rule out
1 mole would give only 3 moles of I₂, half of what is asked for.
0.25
rule out
This inverts the ratio and then halves it. Since one dichromate produces more I₂ than itself, the answer must still be well below 6 but above zero — 0.25 would mean one dichromate produces 24 I₂.
0.5
rule out
The 1 : 3 ratio has been read as 1 : 12. A sensible sanity check: 0.5 mole could only give 1.5 moles of I₂.
Shortcut
Skip the balanced equation and count electrons instead: dichromate is a 6-electron oxidant (two Cr, three each) and I₂ is a 2-electron product. 6 ÷ 2 = 3, so one dichromate always liberates three I₂. That single ratio — 1 : 3 — answers every version of this question, whatever number of moles they ask for. Compare with permanganate, which is a 5-electron oxidant in acid and so gives 2.5 I₂ per mole.
Q 170Left blankMatch the following · ores and oxidation states
Match the following: I. K₂MnO₄ P. Transition element in +6 oxidation state II. KMnO₄ Q. Oxidising agent in acid medium III. K₂Cr₂O₇ R. Manufactured from pyrolusite ore IV. K₂CrO₄ S. Manufactured from chromite ore
I-PQR; II-QR; III-PQ; IV-PR
CorrectI-PQR; II-QR; III-PQS; IV-PQS
I-PQR; II-QR; III-PQS; IV-PQR
I-PQR; II-QR; III-PS; IV-PR
Given
Four salts: potassium manganate, permanganate, dichromate and chromate.
Four properties: +6 oxidation state, oxidising in acid, from pyrolusite, from chromite.
Asked
The correct set of matches.
Concept to use
Two independent questions are being asked about each salt: what oxidation state is the metal in, and which ore does it come from? The ore half is the easier one and it settles the question on its own. Pyrolusite (MnO₂) is a manganese ore; chromite (FeCr₂O₄) is a chromium ore. A chromium salt can never be matched to R, and a manganese salt can never be matched to S.
DiagramAnimatedTwo ores, two metals — and they never cross over.
Formula to usepyrolusite MnO₂ → Mn salts chromite FeCr₂O₄ → Cr salts
Baby steps
I. K₂MnO₄ — manganate. Mn is +6 (P ✓), from pyrolusite (R ✓), and acts as an oxidant in acid (Q ✓). So I–PQR.
II. KMnO₄ — permanganate. Mn is +7, so P fails. From pyrolusite (R ✓) and a classic acidic oxidant (Q ✓). So II–QR.
III. K₂Cr₂O₇ — dichromate. Cr is +6 (P ✓), an acidic oxidant (Q ✓), from chromite (S ✓). So III–PQS.
IV. K₂CrO₄ — chromate. Cr is +6 (P ✓), from chromite (S ✓), and in acid it converts to dichromate and oxidises (Q ✓). So IV–PQS.
The answer is I-PQR; II-QR; III-PQS; IV-PQS.
Answer
I-PQR; II-QR; III-PQS; IV-PQS
Why this option and not the others
Option
Verdict
Reason
III-PQ; IV-PR
rule out
IV is a chromium salt matched to R, the manganese ore. Impossible. III also drops S, which it needs.
III-PQS; IV-PQS
keep
Both chromium salts correctly take S and neither takes R; only KMnO₄ is denied P, because Mn sits at +7 there.
III-PQS; IV-PQR
rule out
III is right, but IV pairs the chromium salt K₂CrO₄ with pyrolusite. Chromate does not come from a manganese ore.
III-PS; IV-PR
rule out
IV again takes R wrongly, and III drops Q even though dichromate is the standard acidic oxidising agent.
Shortcut
One rule kills three of the four options in a single sweep: never pair a chromium salt with pyrolusite. Scan only the III and IV entries and reject any option containing R there. Options 1, 3 and 4 all fail that test, leaving option 2 without checking a single oxidation state. In a long matching question, look for the property that appears in only one column and use it as a filter before doing anything else.
What the nine have in common
Reading the paper as a whole
One ladder is worth three questions
Q154, Q164 and Q167 all turn on the same fact, and between them they carry 12 marks:
Medium
Manganese ends at
As
Acidic
+2
Mn²⁺ — pale pink to colourless
Neutral
+4
MnO₂ — brown precipitate
Alkaline
+6
MnO₄²⁻ — green solution
The more acidic the medium, the further manganese falls from its starting +7. Learn the
ladder in that order and then read carefully which order the question wants it in — Q164
asks for neutral, alkaline, acidic, which is nobody's memorised order, and that is the whole
difficulty of the question.
The three wrong answers were each half right
Q
What was secure
What slipped
153
That the moment depends on unpaired electrons
The tally itself. Writing the seven n values in a row would have made the count visible
instead of held in the head.
166
Zinc: full d shell means colourless
The d-count for copper and nickel, off by one each, which swapped the two.
167
That permanganate is the oxidising agent
The destinations. SO₃²⁻ leaves sulphur at +4, so the option contradicted
itself; and MnO₄²⁻ is the alkaline product, not the acidic one.
All three are caught by the same physical act: write the oxidation number, or the
d-electron count, above every species in the option before choosing. None of the three
needed extra knowledge.
Three filters that turn long questions into short ones
l = 1 is always 12 for a 3d-series atom. 2p and 3p are full for every element from
Sc to Zn and 4p is empty, so only the d count ever varies. That halves Q139 before you start.
Never pair a chromium salt with pyrolusite. Pyrolusite is manganese ore, chromite is
chromium ore. That single rule kills three of the four options in Q170 without checking a
single oxidation state.
+7 is the ceiling for manganese. Any option claiming MnO₄⁻ is oxidised is
dead on arrival — two of the four options in Q167 go straight away.
In each case the filter works on the options rather than on the chemistry, which is
exactly what makes it fast enough to be worth using under time pressure.
A note on Q153
Worth knowing that the key to this one is arguable. Computing every moment gives
two matched sets — Cr²⁺ with Mn³⁺ at 4.90 BM, and
Ni³⁺ with Ti⁺ at 3.87 BM. The keyed answer of 2 counts the sets; counting the
species involved would give 4. Both are defensible readings of “how many species have
same magnetic moment”.
If a question like this appears again, work out all the values first and then pick whichever
option matches one of the two honest readings. Arguing with the paper costs time you do not
have.