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Chapter test · Chemistry · Class 12 · Ch 9 + Ch 8

Coordination Compounds + d- and f-Block Elements — 80-Question Test

Aamirah Fathima · NEET 2027 · every question lost on this paper, rebuilt in full

257marks / 320
67correct
11attempted, wrong
2left blank
63marks lost (incl. −11)
Target was 265+, danger line 240: this lands between them, 8 marks short. Carbonyls and EAN (4/4), the d-block data tables (8/8) and the lanthanoids (5/5) were clean. Everything lost sits in four places — the permanganate/dichromate block, assertion–reason, statement counts, and two formula-writing steps.

Weak areas from this paper

KMnO₄ and K₂Cr₂O₇ — the weakest block at 6 / 10 new

Q64, Q65, Q66, Q71 — both blanks and two of the numericals are here, 20 marks in all. All four turn on the same skill: writing the oxidation numbers and counting electrons.

Fix to drill. One card: MnO₄⁻ takes 5 e⁻ in acid (→ Mn²⁺), 3 e⁻ in neutral (→ MnO₂); Cr₂O₇²⁻ takes 6 e⁻ (→ 2Cr³⁺). FeC₂O₄ gives 3 e⁻; each I₂ gives 2 e⁻. Then every mole-ratio question is one division.

Assertion–reason, still unsettled outside Biology recurring

Q23, Q33, Q59 — three met, two wrong, and in opposite directions: verdicts exchanged twice and a real link denied once. Her Biology score on this format is now 8/8 plus 10/10 on the drill, so the method exists; it is not being used in Chemistry.

Fix to drill. Write the same three lines in the Chemistry margin, and test R against one fact: the spectrochemical series, the standard potential, or the d-count.

Statement counts — one statement too many, twice watch

Q5 (three instead of one) and Q41 (all four instead of three) — 8 marks. Both times a statement with one altered word was accepted.

Fix to drill. Mark T/F with a two-word reason beside each line before counting. Words that almost always signal a plant: always, ionic, low spin next to tetrahedral.

Two formula-writing steps 3rd miss

Q1 (coordination sphere) and Q9 (cross-multiplying charges). Q9 has now been missed or skipped on three successive coordination papers.

Fix to drill. For AgCl problems: chlorides outside = moles of AgCl, then fill the sphere to the coordination number. For name → formula: cation charge, anion charge, swap as subscripts.

The thirteen at a glance

QStatusMarkedKeyTopicError family
Q1wrongBAWernerCoordination sphere counted from the AgCl figure alone
Q5wrongCAWernerStatement count — one statement too many accepted
Q9wrongBDIUPACCharges not cross-multiplied
Q23wrongCDIsomer countAssertion–Reason: geometrical vs stereo count
Q33wrongDCVBTAssertion–Reason: field strength of CN⁻
Q34wrongCDMagnetismField strength ignored in a spin-only calculation
Q59wrongBAOxidation statesAssertion–Reason: a real link denied
Q41wrongDCCFTStatement count — 'all four' accepted
Q64blank—BKMnO₄Blank — oxidation numbers not written down
Q65wrongBDK₂Cr₂O₇Electron ratio guessed instead of counted
Q66wrongBAKMnO₄Only one of the two oxidisable ions counted
Q71blank—AK₂Cr₂O₇Blank — four dichromate statements
Q80wrongDBActinoidsSeries maximum overshot

Question by question

Werner's Theory, Ligands and Nomenclature

Q1Q5Q9
Q1Attempted · wrong Coordination sphere counted from the AgCl figure alone
One mole of PtCl₄·yNH₃ gives 3 moles of ions in solution and 2 moles of AgCl with excess AgNO₃. Taking Pt(IV) to have coordination number 6, the value of y is
  1. 4 ✓ key
  2. 2 her answer
  3. 6
  4. 5
Given
  • 1 mol PtCl₄·yNH₃ gives 3 mol of ions in solution.
  • It gives 2 mol AgCl with excess AgNO₃.
  • Pt(IV) has coordination number 6.
Asked
The value of y.
Concept
Silver nitrate precipitates only the chlorides outside the bracket. Two moles of AgCl mean two ionisable Cl⁻, and with one complex ion that makes the three ions counted in solution. The remaining two chlorides are still bonded inside the sphere, so they occupy two of the six coordination positions and the ammonia molecules fill the other four.
Diagram
INSIDE the bracket(coordination sphere)4 NH₃ + 2 Cl → CN = 6not precipitated by AgNO₃OUTSIDE the bracket(ionisable)2 Cl⁻ → 2 mol AgCl1 complex ion + 2 Cl⁻ = 3 ions[Pt(NH₃)₄Cl₂]Cl₂ : y = 4AgCl counts only the chlorides OUTSIDE; the sphere still needs six donors
Animated · the still frame carries the full message.
Baby steps
  1. 2 mol AgCl → 2 Cl⁻ outside the bracket; 1 complex ion + 2 Cl⁻ = 3 ions ✓
  2. So 2 Cl remain inside the sphere
  3. CN 6 = y NH₃ + 2 Cl → y = 4; formula [Pt(NH₃)₄Cl₂]Cl₂
Answer
(A) 4
Why not the others
A · keepCN 6 = 4 NH₃ + 2 Cl inside → y = 4.
B · rule outy = 2 leaves the sphere with only four donors.
C · rule outy = 6 ignores the two chlorides that are inside.
D · rule outy = 5 counts only one inner chloride.
Shortcut
Two lines in the margin: chlorides outside = moles of AgCl; donors inside = coordination number minus inner chlorides.
Trap (as built)
y = 5 counts only one inner chloride — bound chlorides still fill the sphere.
from: Coord B1 ILTS-09 Q46, Q47
Where it went wrong
Marked (B): 2.
Marked y = 2, which fills only four of the six coordination positions. The AgCl figure was used for the whole formula instead of for the outside of the bracket only.
Q5Attempted · wrong Statement count — one statement too many accepted
How many of the following statements are correct?
A. In Werner's theory, ligands are bound to the metal by ionic bonds.
B. Primary valencies are directional.
C. Secondary valencies are non-ionisable.
D. A coordination entity is always cationic.
  1. Only one ✓ key
  2. Only two
  3. Only three her answer
  4. All four
Given
  • Four statements about Werner's theory.
Asked
How many are correct.
Concept
Only the statement that secondary valencies are non-ionisable survives. Ligands are held by coordinate bonds, not ionic bonds; it is the secondary valencies that are directional, not the primary ones; and a coordination entity may be cationic, anionic or neutral, as [Ni(CO)₄] shows.
Baby steps
  1. A false: ligands are held by coordinate bonds
  2. B false: SECONDARY valencies are directional
  3. C true
  4. D false: entities can be cationic, anionic or neutral ([Ni(CO)₄])
Answer
(A) Only one
Why not the others
A · keepOnly C is correct.
B · rule outTwo would need one more true line.
C · rule outThree accepts two of the planted statements.
D · rule outAll four cannot be right — the first says 'ionic'.
Shortcut
Circle the planted words first: ionic, primary … directional, always. Each one kills its statement.
Trap (as built)
Only C survives. 'Always' and 'ionic' are the planted words.
from: Coord B1 ILTS-08 Q69; B1 ILTS-09 Q90
Where it went wrong
Marked (C): Only three.
Counted three. The same slip as Q41 on this paper and as HHD Q78: statements that read fluently are accepted without checking the one altered word.
Q9Attempted · wrong Charges not cross-multiplied
The formula of tris(ethane-1,2-diamine)cobalt(III) sulphate is
  1. [Co(en)₃]₃(SO₄)₂
  2. [Co(en)₃](SO₄)₃ her answer
  3. [Co(en)₃]SO₄
  4. [Co(en)₃]₂(SO₄)₃ ✓ key
Given
  • Name: tris(ethane-1,2-diamine)cobalt(III) sulphate.
  • en is a neutral bidentate ligand.
Asked
The formula of the compound.
Concept
Three neutral en ligands leave the cobalt(III) charge untouched, so the cation is [Co(en)₃]³⁺. Sulphate is 2−. To balance 3+ against 2−, two cations pair with three anions.
Diagram
cation[Co(en)₃]³⁺anionSO₄²⁻takes subscript 2takes subscript 3[Co(en)₃]₂(SO₄)₃ — 2 × 3+ = 6+ = 3 × 2−
Animated · the still frame carries the full message.
Baby steps
  1. en is neutral → cation is [Co(en)₃]³⁺
  2. Sulphate is 2−
  3. Cross-multiply 3+ and 2− → 2 cations : 3 anions
Answer
(D) [Co(en)₃]₂(SO₄)₃
Why not the others
A · rule out[Co(en)₃]₂(SO₄)₃ balances 6+ against 6−.
B · rule out[Co(en)₃](SO₄)₃ gives 3+ against 6−.
C · rule out[Co(en)₃]SO₄ gives 3+ against 2−.
D · keep[Co(en)₃]₃(SO₄)₂ reverses the two subscripts.
Shortcut
Write the cation charge, write the anion charge, then swap them as subscripts.
Trap (as built)
Writing the cation charge first prevents [Co(en)₃](SO₄)₃. This question was left blank twice on one paper.
from: Coord B1 ILTS-08 Q46, Q73 (both blank)
Where it went wrong
Marked (B): [Co(en)₃](SO₄)₃.
Marked [Co(en)₃](SO₄)₃, which leaves the formula with a net 3− charge. This question was left blank twice on the earlier coordination papers, so it is now a third miss — the cross-multiplication step is the one to drill.

Isomerism

Q23
Q23Attempted · wrong Assertion–Reason: geometrical vs stereo count
Assertion (A): The total number of geometrical isomers shown by [Co(en)₂Cl₂]⁺ is three.
Reason (R): [Co(en)₂Cl₂]⁺ has an octahedral geometry.
  1. Both A and R are true and R is the correct explanation of A
  2. Both A and R are true but R is NOT the correct explanation of A
  3. A is true but R is false her answer
  4. A is false but R is true ✓ key
Given
  • A: the total number of geometrical isomers of [Co(en)₂Cl₂]⁺ is three.
  • R: [Co(en)₂Cl₂]⁺ is octahedral.
Asked
Truth of A and R, and the link.
Concept
Geometrical isomerism here gives only cis and trans, that is two. The third isomer appears because the cis form is chiral and exists as a d and an l form — that is optical isomerism. So the stereoisomer count is three but the geometrical count is two, and the assertion is false. The reason is a plain fact about the complex and is true.
Diagram
GEOMETRICAL isomers = 2cis and transSTEREOisomers = 3trans + cis-d + cis-lOnly the cis form is chiral, so the extra isomer is OPTICAL, not geometrical.The assertion says 'geometrical … is three' → false.Count what the sentence names: geometrical (2) or stereo (3)CoClClCoClClPQ
Animated · the still frame carries the full message.
Three-step defence
StepVerdictOne-line reason
1 · Is A true?FALSEGeometrical isomers are cis and trans only → 2, not 3.
2 · Is R true?TRUECo(III) with CN 6 is octahedral.
3 · 'A because R'?NOT ASKEDOnly asked when both are true.
Baby steps
  1. A false: geometrical isomers are only cis and trans (2); 3 is the stereoisomer count
  2. R true: Co(III), CN 6
Answer
(D) A is false but R is true
Why not the others
A · rule outR is true, so this cannot be right.
B · rule outA is false.
C · rule outA is false, not true — this is the exact reversal.
D · keepA false, R true.
Shortcut
Underline the word geometrical or stereo in the assertion before counting anything.
Trap (as built)
She marked the exact reverse ('A true, R false') on this pair. Judge A alone, then R alone.
from: Coord B1 ILTS-09 Q54
Where it went wrong
Marked (C): A is true but R is false.
Marked 'A true, R false' — both verdicts reversed. This is the third appearance of this same pair (ILTS-09 Q54 and the Evolution error notes), and each time the two verdicts were exchanged rather than judged separately.

Bonding and Magnetism

Q33Q34Q59
Q33Attempted · wrong Assertion–Reason: field strength of CN⁻
Assertion (A): [Fe(CN)₆]³⁻ is an inner orbital complex.
Reason (R): CN⁻ is a weak field ligand.
  1. Both A and R are true and R is the correct explanation of A
  2. Both A and R are true but R is NOT the correct explanation of A
  3. A is true but R is false ✓ key
  4. A is false but R is true her answer
Given
  • A: [Fe(CN)₆]³⁻ is an inner orbital complex.
  • R: CN⁻ is a weak field ligand.
Asked
Truth of A and R, and the link.
Concept
Cyanide is among the strongest field ligands in the spectrochemical series. It forces the d electrons of Fe(III) to pair in the t₂g set, leaving two 3d orbitals free for d²sp³ hybridisation — an inner orbital complex. The assertion is therefore true and the reason false.
Three-step defence
StepVerdictOne-line reason
1 · Is A true?TRUEd²sp³ hybridisation → inner orbital complex.
2 · Is R true?FALSECN⁻ is a strong field ligand, near the top of the series.
3 · 'A because R'?NOT ASKEDOnly asked when both are true.
Baby steps
  1. A true: d²sp³
  2. R false: CN⁻ is among the strongest field ligands
Answer
(C) A is true but R is false
Why not the others
A · rule outR is false.
B · rule outR is false.
C · keepA true, R false.
D · rule outA is true — d²sp³ is an inner orbital complex.
Shortcut
Check R against the series: I⁻ < Cl⁻ < F⁻ < H₂O < NH₃ < en < CN⁻ < CO. Cyanide sits at the strong end.
Trap (as built)
Judge R on its own — it is simply wrong.
from: Coord B1 ILTS-09 Q72 (verdicts flipped)
Where it went wrong
Marked (D): A is false but R is true.
Marked 'A false, R true' — again the two verdicts exchanged, in the opposite direction to Q23 on the same paper. Judging A on its own, then R on its own, would have caught both.
Q34Attempted · wrong Field strength ignored in a spin-only calculation
The spin-only magnetic moment of [Mn(CN)₆]³⁻ is about
  1. 0 BM
  2. 5.92 BM
  3. 4.90 BM her answer
  4. 2.83 BM ✓ key
Given
  • Complex [Mn(CN)₆]³⁻.
  • Spin-only formula μ = √n(n+2) BM.
Asked
The spin-only magnetic moment.
Concept
Mn(III) is 3d⁴. With the strong-field CN⁻ ligand the electrons pair up inside the t₂g set, giving t₂g⁴: one pair and two unpaired electrons. Hence n = 2 and μ = √8 = 2.83 BM. Only with a weak-field ligand would the d⁴ ion keep four unpaired electrons.
Diagram
[Mn(CN)₆]³⁻ : CN⁻ strong fielde_g↑↓↑↑t₂gt₂g⁴ → 2 unpaired → 2.83 BMweak-field d⁴ (for comparison)↑e_g↑↑↑t₂g4 unpaired → μ = 4.90 BM
Animated · the still frame carries the full message.
Baby steps
  1. Mn(III) = 3d⁴; CN⁻ strong → t₂g⁴ (low spin)
  2. t₂g⁴: one pair + 2 unpaired
  3. μ = √(2 × 4) = 2.83 BM
Answer
(D) 2.83 BM
Why not the others
A · rule out2.83 BM — low spin, n = 2.
B · rule out4.90 BM is the weak-field d⁴ value.
C · rule out0 BM cannot arise from an odd arrangement of four electrons in three orbitals.
D · keep5.92 BM belongs to d⁵ with five unpaired electrons.
Shortcut
Four steps, one line each: oxidation state → d-count → strong or weak ligand → unpaired electrons.
Trap (as built)
4.90 BM is the high-spin d⁴ value; 0 BM would need an even pairing that d⁴ cannot achieve.
from: Coord B1 ILTS-09 Q88
Where it went wrong
Marked (C): 4.90 BM.
Marked 4.90 BM, the high-spin value. The oxidation state and d-count were right; the ligand was not consulted before pairing.
Q59Attempted · wrong Assertion–Reason: a real link denied
Assertion (A): Mn³⁺ is a strong oxidising agent.
Reason (R): Mn²⁺ has the extra-stable half-filled 3d⁵ configuration.
  1. Both A and R are true and R is the correct explanation of A ✓ key
  2. Both A and R are true but R is NOT the correct explanation of A her answer
  3. A is true but R is false
  4. A is false but R is true
Given
  • A: Mn³⁺ is a strong oxidising agent.
  • R: Mn²⁺ has the extra-stable half-filled 3d⁵ configuration.
Asked
Truth of A and R, and the link.
Concept
Mn³⁺ is d⁴. Gaining one electron takes it to the half-filled d⁵ arrangement of Mn²⁺, which is extra stable — and that is exactly why the reduction happens so readily, giving E°(Mn³⁺/Mn²⁺) = +1.57 V. The reason is the mechanism behind the assertion.
Three-step defence
StepVerdictOne-line reason
1 · Is A true?TRUEE°(Mn³⁺/Mn²⁺) = +1.57 V.
2 · Is R true?TRUEHalf-filled d⁵ is an extra-stable configuration.
3 · 'A because R'?READS RIGHTR is the reason behind A.
Baby steps
  1. A true: E°(Mn³⁺/Mn²⁺) = +1.57 V
  2. R true
  3. Mn³⁺ (d⁴) readily gains an electron BECAUSE the product d⁵ is extra stable → R explains A
Answer
(A) Both A and R are true and R is the correct explanation of A
Why not the others
A · keepBoth true, and R is the reason.
B · rule outDenies a link that is real.
C · rule outR is a standard NCERT statement, so it is true.
D · rule outA is true — the potential is strongly positive.
Shortcut
Say it aloud: 'Mn³⁺ oxidises because the product d⁵ is stable.' If the sentence reads as a cause, choose (A).
Trap (as built)
The 'because' sentence reads correctly — accept the link.
from: AR link denied
Where it went wrong
Marked (B): Both A and R are true but R is NOT the correct explanation of A.
Marked (B). The same denial pattern as the Evolution Q63 pair and the HHD placenta pair: when the reason is the mechanism, the link must be accepted.

Crystal Field Theory

Q41
Q41Attempted · wrong Statement count — 'all four' accepted
How many of the following statements are correct?
A. Δo depends on the nature of the ligand and the charge on the metal ion.
B. A low-spin octahedral complex forms when Δo > P (pairing energy).
C. Tetrahedral complexes are usually low spin.
D. Anhydrous CuSO₄ is white, but CuSO₄·5H₂O is blue.
  1. Only one
  2. Only two
  3. Only three ✓ key
  4. All four her answer
Given
  • Four statements about crystal field theory.
Asked
How many are correct.
Concept
Δo does depend on the ligand and on the charge of the metal ion; a low-spin complex does form when Δo exceeds the pairing energy; and anhydrous CuSO₄ is white while the hydrate is blue. The false statement is the one about tetrahedral complexes: Δt is only about 4/9 of Δo, so it is almost always smaller than the pairing energy and tetrahedral complexes are high spin.
Baby steps
  1. A true
  2. B true
  3. C false: Δt is small, so they are usually high spin
  4. D true: no ligand field without water → no d–d band in the visible
Answer
(C) Only three
Why not the others
A · rule outOnly one leaves three NCERT statements unaccounted for.
B · rule outTwo is one short.
C · keepThree — A, B and D are correct.
D · rule outAll four accepts the tetrahedral claim.
Shortcut
Any statement that puts 'tetrahedral' and 'low spin' in the same sentence is false.
Trap (as built)
Tetrahedral → small splitting → high spin.
from: CFT
Where it went wrong
Marked (D): All four.
Marked 'all four'. Together with Q5, this is 8 marks lost to accepting one extra statement — the same direction of error both times.

KMnO₄ and K₂Cr₂O₇

Q64Q65Q66Q71
Q64Left blank Blank — oxidation numbers not written down
When moist SO₂ is passed through acidified KMnO₄,
  1. SO₂ is reduced to S and MnO₄⁻ is oxidised to Mn²⁺
  2. SO₂ is oxidised to SO₄²⁻ and MnO₄⁻ is reduced to Mn²⁺ ✓ key
  3. SO₂ is oxidised to SO₃²⁻ and MnO₄⁻ is reduced to MnO₄²⁻
  4. SO₂ is oxidised to SO₄²⁻ and MnO₄⁻ is reduced to MnO₂
Given
  • Moist SO₂ passed through acidified KMnO₄.
Asked
The products of the reaction.
Concept
Sulphur in SO₂ is +4 and is oxidised to +6 as sulphate. Manganese in MnO₄⁻ is +7 and, in acid medium, is reduced all the way to Mn²⁺: 2MnO₄⁻ + 5SO₂ + 2H₂O → 2Mn²⁺ + 5SO₄²⁻ + 4H⁺.
Diagram
SO₂S is +4oxidisedSO₄²⁻S is +6MnO₄⁻Mn is +7reduced (acid)Mn²⁺Mn is +2SO₃²⁻ is also +4→ no oxidation at allMnO₂ is the NEUTRALmedium productWrite the oxidation number on every species before choosing
Animated · the still frame carries the full message.
Baby steps
  1. S: +4 → +6 (sulphate)
  2. Mn: +7 → +2 in acid
  3. 2MnO₄⁻ + 5SO₂ + 2H₂O → 2Mn²⁺ + 5SO₄²⁻ + 4H⁺
Answer
(B) SO₂ is oxidised to SO₄²⁻ and MnO₄⁻ is reduced to Mn²⁺
Why not the others
A · rule outSO₄²⁻ and Mn²⁺ — the acid-medium pair.
B · keepSO₃²⁻ still has sulphur at +4, so nothing has been oxidised.
C · rule outSulphur is not reduced here; SO₂ is the reducing agent.
D · rule outMnO₂ is the neutral or faintly alkaline product, not the acidic one.
Shortcut
Two ladders in the margin: S +4 → +6, Mn +7 → +2 in acid. Then read the options.
Trap (as built)
S is +4 in SO₃²⁻ too — that option oxidises nothing. MnO₂ is the neutral product.
from: d-block ILTS Q167 (marked SO₃²⁻)
Q65Attempted · wrong Electron ratio guessed instead of counted
How many moles of acidified K₂Cr₂O₇ are needed to liberate 9 moles of I₂ from iodide?
  1. 1.5
  2. 4.5 her answer
  3. 18
  4. 3 ✓ key
Given
  • Acidified K₂Cr₂O₇ liberating I₂ from iodide.
  • 9 moles of I₂ are to be produced.
Asked
The moles of K₂Cr₂O₇ needed.
Concept
Dichromate is a six-electron oxidant: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O. Each I₂ formed releases two electrons, so one mole of dichromate can produce three moles of I₂. Nine moles of I₂ therefore need three moles of dichromate.
Diagram
Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂Oone dichromate accepts 6 electrons2I⁻ → I₂ + 2e⁻each I₂ releases 2 electrons6 ÷ 2 = 3 mol I₂ per mol dichromate9 mol I₂ → 3 mol K₂Cr₂O₇FeC₂O₄ : Fe²⁺ → Fe³⁺ (1e⁻)C₂O₄²⁻ → 2CO₂ (2e⁻) → 3e⁻ in allMnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂Oeach permanganate takes 5 electrons3 ÷ 5 = 0.6 mol KMnO₄ per mol FeC₂O₄Count electrons on both sides, then divide — never guess the ratio
Animated · the still frame carries the full message.
Baby steps
  1. Cr₂O₇²⁻ + 6I⁻ + 14H⁺ → 2Cr³⁺ + 3I₂ + 7H₂O
  2. 1 mol dichromate (6 e⁻) → 3 mol I₂ (2 e⁻ each)
  3. 9 ÷ 3 = 3 mol
Answer
(D) 3
Why not the others
A · rule out3 mol — from the 1 : 3 ratio.
B · rule out1.5 uses a 1 : 6 ratio, counting iodide ions instead of I₂ molecules.
C · rule out18 multiplies by 2 instead of dividing by 3.
D · keep4.5 uses a 1 : 2 ratio.
Shortcut
Write both half-reactions, divide 6 by 2, then scale.
Trap (as built)
1.5 uses a 1 : 6 ratio (per I⁻, not per I₂); 4.5 uses 1 : 2.
from: d-block ILTS Q168 (blank)
Where it went wrong
Marked (B): 4.5.
Marked 4.5, which is the 1 : 2 ratio — the electrons released per I₂ were counted but the six electrons accepted by dichromate were halved somewhere. This question was left blank on the earlier paper, so this is the first attempt at it.
Q66Attempted · wrong Only one of the two oxidisable ions counted
The number of moles of KMnO₄ (acidic medium) needed to oxidise one mole of ferrous oxalate (FeC₂O₄) completely is
  1. 0.6 ✓ key
  2. 0.2 her answer
  3. 1.5
  4. 0.4
Given
  • 1 mole of ferrous oxalate, FeC₂O₄.
  • Acidic KMnO₄ as the oxidant.
Asked
The moles of KMnO₄ required.
Concept
Ferrous oxalate has two oxidisable parts. Fe²⁺ → Fe³⁺ releases one electron and C₂O₄²⁻ → 2CO₂ releases two, so each formula unit gives three electrons. In acid medium MnO₄⁻ accepts five electrons, so the requirement is 3/5 = 0.6 mol.
Diagram
Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂Oone dichromate accepts 6 electrons2I⁻ → I₂ + 2e⁻each I₂ releases 2 electrons6 ÷ 2 = 3 mol I₂ per mol dichromate9 mol I₂ → 3 mol K₂Cr₂O₇FeC₂O₄ : Fe²⁺ → Fe³⁺ (1e⁻)C₂O₄²⁻ → 2CO₂ (2e⁻) → 3e⁻ in allMnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂Oeach permanganate takes 5 electrons3 ÷ 5 = 0.6 mol KMnO₄ per mol FeC₂O₄Count electrons on both sides, then divide — never guess the ratio
Animated · the still frame carries the full message.
Baby steps
  1. Fe²⁺ → Fe³⁺ gives 1 e⁻; C₂O₄²⁻ → 2CO₂ gives 2 e⁻ → 3 e⁻ per FeC₂O₄
  2. MnO₄⁻ → Mn²⁺ takes 5 e⁻
  3. moles KMnO₄ = 3/5 = 0.6
Answer
(A) 0.6
Why not the others
A · keep0.6 mol — three electrons given, five taken.
B · rule out0.4 counts only the oxalate (2 e⁻).
C · rule out0.2 counts only the iron (1 e⁻).
D · rule out1.5 inverts the ratio.
Shortcut
Underline every oxidisable species in the formula before counting electrons; FeC₂O₄ contains two.
Trap (as built)
0.4 oxidises only the oxalate; 0.2 only the iron — both ions are oxidised.
from: Electron bookkeeping
Where it went wrong
Marked (B): 0.2.
Marked 0.2, which counts only the Fe²⁺. Oxalate is the part most often forgotten because it looks like a spectator.
Q71Left blank Blank — four dichromate statements
How many of the following statements are correct?
A. Acidified dichromate oxidises iodide to iodate.
B. Acidified dichromate oxidises Sn²⁺ to Sn⁴⁺.
C. In acidic medium, chromium in dichromate is reduced from +6 to +2.
D. Dichromate acts as a five-electron oxidant.
  1. Only one ✓ key
  2. Only two
  3. Only three
  4. All four
Given
  • Four statements about acidified dichromate.
Asked
How many are correct.
Concept
Only the tin statement is right: acidified dichromate oxidises Sn²⁺ to Sn⁴⁺. Iodide is oxidised to I₂, not to iodate; chromium goes from +6 to +3, not to +2; and dichromate is a six-electron oxidant, not a five-electron one — that figure belongs to permanganate.
Baby steps
  1. A false: iodide → I₂
  2. B true
  3. C false: +6 → +3
  4. D false: six electrons per Cr₂O₇²⁻
Answer
(A) Only one
Why not the others
A · keepOnly one — statement B.
B · rule outTwo would need a second true line.
C · rule outThree accepts two false statements.
D · rule outAll four cannot be right.
Shortcut
Check the electron count first: 6 for dichromate, 5 for permanganate. That single number settles the last statement.
Trap (as built)
Only B survives; 'five-electron' belongs to permanganate.
from: d-block ILTS Q168

Actinoids

Q80
Q80Attempted · wrong Series maximum overshot
The highest oxidation state shown by actinoids is
  1. +5
  2. +7 ✓ key
  3. +6
  4. +8 her answer
Given
  • Actinoid series.
Asked
The highest oxidation state shown.
Concept
Actinoids show a wider range of oxidation states than lanthanoids because the 5f, 6d and 7s levels have comparable energies. Uranium reaches +6, and neptunium and plutonium reach +7, which is the maximum for the series.
Diagram
+3+4+5+6+7most commonU reaches +6Np, Pu reach +7Range of oxidation states of the actinoidsThe series maximum is +7 — there is no +8
Animated · the still frame carries the full message.
Baby steps
  1. Np and Pu show +7
  2. U reaches +6
Answer
(B) +7
Why not the others
A · rule out+7 — shown by Np and Pu.
B · keep+6 is uranium's maximum.
C · rule out+8 does not occur in the series.
D · rule out+5 is well within the range.
Shortcut
Remember the top of the ladder: U +6, Np and Pu +7, nothing beyond.
Trap (as built)
+6 is uranium's maximum, not the series maximum.
from: Recall
Where it went wrong
Marked (D): +8.
Marked +8. The reasoning 'actinoids show a wide range, so the maximum must be high' was carried one step past the actual limit.

What to carry into the next paper

HabitWould have saved
Electron card for the two oxidants (5, 3 and 6 electrons) and for FeC₂O₄ and I₂.Q64, Q65, Q66, Q71 — 20 marks
Three margin lines on every assertion–reason item, with R tested against one fact.Q23, Q33, Q59 — 14 marks
T/F with a two-word reason beside each statement before counting.Q5, Q41 — 10 marks
Oxidation state → d-count → ligand strength → unpaired electrons, one line each.Q34 — 5 marks
AgCl counts only outside the bracket; cation and anion charges swap as subscripts.Q1, Q9 — 10 marks