Error Solutions Workbook
Chemistry

• The lanthanoid series La → Lu; each successive element adds one proton and one 4f electron.
• Observation: ionic and atomic radii shrink steadily across the series.

Shielding ability follows the order s > p > d > f. The 4f orbitals are diffuse and oddly shaped, so a 4f electron screens the outer electrons from the nucleus badly. Each step across the series adds +1 to the nuclear charge but the extra 4f electron cancels only part of it — so the effective nuclear charge Z_eff felt by the outer shell keeps creeping up, pulling the electron cloud inward.

1. Moving one step right: nuclear charge rises by +1.
2. The added electron goes into the inner 4f subshell, not the valence shell.
3. 4f shielding is poor, so it cancels well under one unit of the added charge.
4. Net effect: Z_eff on the outer electrons increases.
5. Stronger pull ⇒ smaller radius. Repeat 14 times ⇒ a cumulative contraction.

• Logic check the distractors: if 4f shielding were effective (option 1), Z_eff would stay flat and there would be no contraction at all. The option contradicts the phenomenon it claims to explain — an instant elimination.
• Option 4 is a category error: mass has nothing to do with size.
• Carry-forward consequence: lanthanoid contraction is why Zr and Hf (and Nb/Ta, Mo/W) have almost identical radii. Questions often test the consequence rather than the cause — learn both together.
• Mnemonic: “f shields feebly.”

Noble gases have closed-shell configurations (ns²np⁶, or 1s² for He). Two consequences follow directly:

• No bonding partner needed ⇒ they exist as single atoms (monoatomic), unlike H₂, N₂, O₂.
• The only force between atoms is weak London dispersion ⇒ very low melting and boiling points and poor solubility.
• Full shells also mean no low-energy electronic transitions ⇒ colourless, and being chemically inert ⇒ odourless and tasteless.

1. Closed shell ⇒ no covalent bonding to itself ⇒ monoatomic. Rules out “diatomic” and “polyatomic”.
2. Only dispersion forces ⇒ low boiling points. Rules out every “high boiling/melting” option.
3. Non-polar and inert ⇒ only sparingly soluble in water. Rules out “strongly soluble”.
4. No accessible transitions ⇒ colourless, odourless, tasteless.
5. Only one option survives all four filters.

• Serial elimination on one word each: diatomic ✗, polyatomic ✗, strongly soluble ✗. Three options die on a single word apiece — you never need to assess the whole statement.
• Anchor fact: helium boils at 4.2 K, the lowest of any substance. Any option claiming high boiling points for noble gases is dead on arrival.
• One-line summary for the notes: “Mono, colourless, odourless, low-boiling, sparingly soluble.”

Almost all transition metals adopt one of the three close-packed metallic lattices (bcc, fcc, hcp), giving them typical metallic hardness, lustre and conductivity. NCERT explicitly lists the exceptions: Mn, Zn, Cd and Hg. Manganese has a complex, distorted structure; Zn, Cd and Hg have distorted lattices tied to their full d¹⁰ configurations (and Hg is a liquid at room temperature).

1. Recall the NCERT sentence on metallic character in the d-block chapter.
2. It names the exceptions explicitly: Mn, Zn, Cd, Hg.
3. Scan the options: Fe (bcc), Co (hcp), Ni (fcc) are all textbook-normal metals.
4. Only Mn appears on the exception list.

• Mnemonic: “MZCH” — Mn, Zn, Cd, Hg. The last three are the d¹⁰ group; Mn is the odd one out and therefore the one most often examined.
• Fe, Co and Ni are the classic ferromagnetic trio — they're famous for having normal metallic structure, so they are almost never the “exception” answer.
• Watch the phrasing “specifically mentioned” — this signals a direct recall question from the NCERT line, not a reasoning question.

• X and Y are the energies released when a mole of gaseous halogen molecules is converted to halide ions.
• Each process has two energy stages.

This is not a plain electron-affinity comparison. The overall process is a two-step cycle:

Chlorine has the more negative electron gain enthalpy (F⁻ is small and crowded, so adding an electron to F is less favourable than expected). But the F–F bond is unusually weak (lone-pair repulsion in a very short bond), so step 1 costs much less for fluorine — and that difference dominates.

1. Fluorine. BDE(F–F) ≈ +158 kJ/mol; Δ_egH(F) ≈ −328 kJ/mol.
2. Net: 158 + 2(−328) = 158 − 656 = −498 kJ/mol ⇒ X ≈ 498 kJ released.
3. Chlorine. BDE(Cl–Cl) ≈ +242 kJ/mol; Δ_egH(Cl) ≈ −349 kJ/mol.
4. Net: 242 + 2(−349) = 242 − 698 = −456 kJ/mol ⇒ Y ≈ 456 kJ released.
5. Compare: 498 > 456 ⇒ X > Y.

• Qualitative route, no numbers needed: Cl wins on electron affinity by a small margin (~21 kJ per atom), but F wins on bond weakness by a large margin (~84 kJ per mole of bonds). The bigger effect wins ⇒ X > Y.
• This is exactly why F₂ is the strongest oxidising halogen despite not having the most negative electron gain enthalpy. If you remember that single fact, the answer follows instantly.
• Read what X and Y are. They are energies released (they appear on the product side), so larger released energy = larger symbol value. Treating them as enthalpy changes flips the inequality — which is exactly the error made here.

• Cl, Br, I exhibit −1, +1, +3, +5, +7.
• Fluorine exhibits only −1 (and 0 in F₂).

Fluorine is stuck at −1 for two independent reasons: it is the most electronegative element (nothing can pull electrons away from it, so it can't be positive), and it has no accessible d-orbitals in the n = 2 shell (so it cannot expand its octet to reach +3, +5 or +7). Cl, Br and I have vacant d-orbitals and lower electronegativity, so they access the full positive range.

1. The states available to Cl/Br/I: −1, +1, +3, +5, +7.
2. The state available to F: −1 only.
3. Required set = states in the first list but not the second = +1, +3, +5, +7.
4. Now test each option: any set containing −1 fails immediately, because fluorine does show −1.
5. That eliminates options 1, 3 and 4 at a stroke.
6. Remaining: +1, +3, +5 — all shown by Cl/Br/I, none by F. ✓

• The −1 filter is the entire question. Fluorine's one and only state is −1, so any option containing −1 is automatically wrong. Scan for “−1” first and you're done in five seconds.
• Why fluorine can't go positive: “most electronegative” + “no d-orbitals”. Quote both in any subjective answer.
• Note that +7 would also be a valid member of the answer set — the option simply doesn't list it. Don't reject a correct set for being incomplete; check only that every listed state satisfies both conditions.

Extra stability in the f-block comes from empty (4f⁰), half-filled (4f⁷) or completely filled (4f¹⁴) subshells — exchange energy is maximised in these arrangements. For Eu, losing the two 6s electrons leaves the 4f⁷ core untouched, landing exactly on the half-filled configuration.

1. Neutral Eu: [Xe]4f⁷6s².
2. Forming Eu²⁺ removes the two 6s electrons (always the outermost s first).
3. Result: Eu²⁺ = [Xe]4f⁷.
4. 4f⁷ is half-filled — a maximum-exchange-energy configuration.
5. Reaching +3 would require breaking into that stable half-filled set, which is why Eu resists and the +2 state is unusually stable.

• Partner fact: Yb²⁺ is also unusually stable, because it is 4f¹⁴ (completely filled). Learn Eu²⁺ = 4f⁷ and Yb²⁺ = 4f¹⁴ as a pair — that pairing is exactly what the 4f¹⁴ distractor exploits.
• Sanity count: Z(Eu) = 63; Xe accounts for 54; that leaves 9 electrons. Eu²⁺ has 61 electrons, so 61 − 54 = 7 ⇒ 4f⁷. Pure arithmetic, no memory required.
• Companion trio on the +4 side: Ce⁴⁺ (4f⁰), Tb⁴⁺ (4f⁷). Same stability logic, opposite direction.

• Group 18 element.
• Diffuses through rubber, glass, plastics.
• Boiling point 4.2 K.

Helium is the smallest atom of all (radius ~31 pm), so it slips through the microscopic pores of rubber, glass and plastic — a real nuisance in vacuum work. Its tiny, tightly-held 1s² cloud is also barely polarisable, so its dispersion forces are the weakest of any substance, giving it the lowest boiling point known: 4.2 K.

1. Boiling point 4.2 K is the lowest of any substance ⇒ it must be the least polarisable atom.
2. Polarisability rises with size, so the least polarisable Group 18 atom is the smallest: He.
3. Cross-check with diffusion: smallest atom ⇒ passes most easily through solid barriers ⇒ He again.
4. Both clues converge on helium.

• Memorise the number: He boils at 4.2 K. It appears verbatim in NCERT and recurs across papers; it also underpins helium's use as a cryogenic coolant for MRI superconducting magnets.
• Both clues point the same way — smallest atom explains the diffusion and the boiling point. When two independent clues agree, you can answer with full confidence.
• Bonus facts worth carrying: He has the highest ionisation enthalpy of all elements, and is obtained commercially from natural gas, not from air.

Down any group, two things happen: the atomic radius increases and the inner shells shield more. The outermost electron therefore sits further from the nucleus and feels less of its pull, so it is easier to remove. Hence ionisation enthalpy decreases down the group, with no exceptions in Group 18.

1. Going down the group, n increases ⇒ the valence electron is further out.
2. More filled inner shells ⇒ more shielding ⇒ lower Z_eff on the outer electron.
3. Weaker attraction ⇒ less energy needed to remove it.
4. So IE falls steadily: He > Ne > Ar > Kr > Xe > Rn.
5. Helium has the highest ionisation enthalpy of any element (2372 kJ/mol) — two electrons, tiny radius, no shielding at all.

• Two-second elimination: the correct answer must start with He (highest IE of all elements) and end with Rn (biggest atom). Only one option does both.
• Option 3 breaks the sequence in the middle (Ar before Ne) and option 4 starts with Ne — both fail the “strictly down the group” test.
• Trend pairing: IE decreases down the group while atomic radius, polarisability and boiling point all increase. Learn them as one linked package and any one of them generates the rest.

Key occurrence facts:

• Noble gases make up about 1% by volume of dry air, and argon is overwhelmingly the largest share (~0.93% of air — the third most abundant atmospheric gas after N₂ and O₂).
• Helium is extracted commercially from natural gas deposits, where it accumulates from the radioactive decay of heavy elements underground.
• Radon is radioactive, short-lived, and present only in traces — it is the decay product of ²²⁶Ra, not xenon.

1. Option 1: Radon is radioactive and vanishingly rare — nitrogen is the major constituent of air. False.
2. Option 2: Pitchblende is a uranium ore; commercial helium comes from natural gas. False.
3. Option 3: Argon is ~0.93% of dry air, far more than all the other noble gases combined. TRUE.
4. Option 4: The decay product of ²²⁶Ra is radon, not xenon. False.

• Radioactive ⇒ rare. Any statement calling radon abundant is automatically false. One filter, one option gone.
• Source pairings worth memorising: He ← natural gas; Ar ← air (fractional distillation of liquid air); Rn ← radium decay; Ne/Kr/Xe ← air, in tiny amounts.
• Abundance ladder of dry air: N₂ (78%) > O₂ (21%) > Ar (0.93%) > CO₂ (0.04%). Argon being third overall is the fact this question hangs on.

Noble gas reactivity requires a low ionisation enthalpy plus a very strong oxidising partner. Only the heavier members qualify:

• Xe — rich chemistry: XeF₂, XeF₄, XeF₆, XeO₃, XeOF₄…
• Kr — essentially just KrF₂, the one krypton compound studied in detail.
• Rn — RnF₂ is believed to form but is known only through radiotracer methods; it cannot be isolated in bulk because radon's half-life is ~3.8 days.
• He, Ne, Ar — no true compounds known. Their ionisation enthalpies are simply too high.

1. Option 1: No true compounds of He, Ne or Ar are known. False.
2. Option 2: Kr forms essentially only the fluoride KrF₂ — not “numerous” chlorides and oxides. Only fluorine (and oxygen, with Xe) is oxidising enough. False.
3. Option 3: Matches the facts exactly. TRUE.
4. Option 4: RnF₂ is identified only by radiotracer studies, never isolated as a stable bulk solid. False.

• The F-and-O rule: noble gases bond only to fluorine and oxygen — the two most electronegative elements. Any option mentioning noble-gas chlorides is wrong on sight.
• Reactivity ladder: Xe > Kr > (Rn, trace only) >> Ar, Ne, He (nothing). Runs opposite to ionisation enthalpy, exactly as expected.
• Anchor compound: the first noble gas compound ever made was Xe⁺[PtF₆]⁻ (Bartlett, 1962), reasoned from the fact that O₂ and Xe have almost identical ionisation enthalpies.

Oxidising power of the halogens falls down the group:

A halogen can displace only the halogens below it. The reaction proceeds only if X₂ is the stronger oxidising agent, i.e. X sits above Y in the group.

1. I₂ + Cl⁻: iodine is below chlorine ⇒ weaker oxidant ⇒ no reaction.
2. Br₂ + Cl⁻: bromine is below chlorine ⇒ no reaction.
3. Cl₂ + Br⁻: chlorine is above bromine ⇒ reaction proceeds ⇒ Cl₂ + 2Br⁻ → 2Cl⁻ + Br₂. ✓
4. Br₂ + F⁻: fluorine is the strongest oxidant of all, so F⁻ can never be oxidised by a halogen ⇒ no reaction.

• “Higher displaces lower” — the same logic as the metal reactivity series, running down the halogen group.
• F⁻ is untouchable. Nothing in the halogen family can oxidise fluoride. Any option pairing a halogen with F⁻ is wrong instantly.
• Lab observation to quote: adding chlorine water to KBr solution turns it orange-brown (Br₂ released); adding it to KI gives a brown/violet colour (I₂). Colour change is the standard identification.

Three of the four behave perfectly regularly down the group:

• Ionisation energy — decreases steadily (bigger atom, more shielding).
• Size — increases steadily (new shell each period).
• Electronegativity — decreases steadily.

Electron gain enthalpy is the anomaly: the order is Cl > F > Br > I (magnitude of energy released), not F > Cl > Br > I. Fluorine's 2p shell is so small that the incoming electron suffers strong inter-electronic repulsion, making the process less exothermic than for chlorine.

1. Check (i) IE: F > Cl > Br > I — regular.
2. Check (iii) Size: F < Cl < Br < I — regular.
3. Check (iv) Electronegativity: F > Cl > Br > I — regular.
4. Check (ii) Electron gain enthalpy: Cl > F > Br > I — the first two are swapped ⇒ irregular.
5. Only (ii) breaks the pattern ⇒ answer: Only II.

• One-line rule: in the whole halogen group, only electron gain enthalpy misbehaves, and only because of fluorine's small size. If a question asks for the anomaly, the answer is essentially always this.
• Same anomaly, second period generally: O also has a less negative Δ_egH than S, for exactly the same compactness reason. Learn F/Cl and O/S together.
• Read “does not” carefully — the question asks for the irregular property. Answering with the regular ones is the standard trap in negatively-worded questions.

• Aqueous solution containing Cr₂O₇²⁻.
• pH is being increased ⇒ the solution is becoming more basic (fewer H⁺).

The equilibrium is

H⁺ appears on the left. By Le Chatelier's principle, removing H⁺ (raising pH) pulls the equilibrium to the left, toward chromate. No electrons are exchanged — chromium stays at +6 on both sides, so this is not a redox change.

1. Raising pH means lowering [H⁺].
2. H⁺ is a reactant on the left side of the equilibrium as written.
3. Le Chatelier: removing a reactant shifts the system toward that side ⇒ to the left.
4. The left-hand species is CrO₄²⁻ ⇒ chromate is favoured.
5. Visible confirmation: the solution changes from orange to yellow.

• Colour mnemonic: “Orange in Acid, Yellow in alkali.” Dichromate (orange) needs H⁺; chromate (yellow) needs OH⁻. One phrase covers every version of this question.
• Watch the pH direction. “Increasing pH” = more basic. Reading it as “more acidic” produces exactly the answer that was marked here. Convert to “more basic” in words before applying Le Chatelier.
• Not redox: Cr is +6 in both ions. That rules out Cr³⁺, which would require a reducing agent — so two options die on the oxidation-state check alone.

Judge each statement independently, then match to an option.

1. (a) Electron affinity of F < Cl. TRUE — F's compact 2p shell causes extra electron–electron repulsion (this is the Q79 anomaly again).
2. (b) ClF₃: central Cl has 7 valence electrons; 3 are used in bonds, leaving 4 as 2 lone pairs. Steric number = 3 + 2 = 5 ⇒ trigonal bipyramidal geometry, T-shaped molecule. TRUE.
3. (c) Fluorine is pale yellow; it is chlorine that is greenish-yellow. FALSE.
4. (d) F₂ is the strongest oxidising agent among the halogens and displaces every other halogen from its halides. TRUE.
5. Correct set = a, b, d.

• Lone-pair count in one line: lone pairs on the central atom = (valence electrons − number of bonds)/2. For ClF₃: (7 − 3)/2 = 2. Works for BrF₃, ICl₃, XeF₂ and the rest of the family.
• Halogen colour ladder: F₂ pale yellow → Cl₂ greenish-yellow → Br₂ red-brown → I₂ violet. Colour darkens down the group. Statement (c) shifts everything by one place — the classic swap.
• Elimination shortcut: once (c) is known to be false, any option containing c dies — that removes two options immediately.

Two NCERT statements combine here:

• Element order: I > Cl > Br. Iodine oxides are the most stable; bromine oxides are the least stable — note that this is not a simple down-the-group trend, which is exactly why it is examined.
• Within one element: higher oxides are more stable than lower oxides. The halogen in a higher oxidation state is bonded to more oxygens, and the greater delocalisation stabilises the structure.

1. Recall the element order: I > Cl > Br — bromine oxides are the least stable of the three.
2. Screen the options: only options 2 and 4 begin with I. Option 4 has Br > Cl, which is the wrong way round ⇒ eliminated.
3. Recall the second fact: higher oxides are more stable.
4. Option 2 states exactly both facts ⇒ correct.

• Two-filter method for any compound-statement option: apply filter 1 (element order) to cut the list, then filter 2 (higher vs lower). Never try to evaluate a whole sentence at once.
• Memory hook: “I am Cleaner than Br” → I > Cl > Br. The irregularity (Br at the bottom) is the whole point of the question.
• Supporting fact: I₂O₅ is a stable, well-characterised solid used to estimate CO, whereas the bromine oxides decompose readily — a concrete anchor for “I most stable, Br least”.

• Configuration [Kr]4d¹⁰5s²5p⁶.
• Kr has Z = 36.

Count the electrons: 36 + 10 + 2 + 6 = 54 ⇒ Z = 54 = xenon. The outermost electrons are in n = 5, far from the nucleus and screened by all the inner shells, so Z_eff on them is small and they leave easily. Helium's two electrons sit in n = 1 with no shielding whatsoever — hence the highest ionisation enthalpy of any element.

1. Add up: Z(Kr) = 36; +10 (4d) +2 (5s) +6 (5p) = 18.
2. 36 + 18 = 54 ⇒ the element is Xe.
3. Its valence shell is n = 5, versus n = 1 for helium.
4. Larger radius + far more shielding ⇒ much lower Z_eff on the valence electrons.
5. Lower attraction ⇒ lower ionisation enthalpy than He.
6. Both halves of option 2 are correct.

• Noble-gas atomic numbers are worth memorising: He 2, Ne 10, Ar 18, Kr 36, Xe 54, Rn 86. Recognising 54 instantly identifies xenon with no counting at all.
• Shell-number shortcut: the highest n in the configuration gives the period. Here n = 5 ⇒ period 5 ⇒ the period-5 noble gas is Xe. No addition needed.
• Check both halves of the option. Option 3 names Kr and gives a backwards reason (“smaller size” would raise IE). A compound option is wrong if either half fails.

Halogen colour arises from absorption of visible light, and the absorption shifts to longer wavelengths as the atom gets bigger and more polarisable. So the colours deepen down the group:

None of them is colourless.

1. F₂ = yellow — acceptable (pale yellow). Correct.
2. Cl₂ = colourless — chlorine is distinctly greenish-yellow (the name comes from the Greek chloros, pale green). INCORRECT ⇒ this is the answer.
3. Br₂ = red — acceptable (red-brown). Correct.
4. I₂ = violet — correct; iodine vapour is famously violet. Correct.

• No halogen is colourless. That single fact answers the question without evaluating the other three options.
• Etymology as memory aid: chloros = pale green. The element is named for its colour, so “colourless chlorine” is a contradiction in terms.
• Negative-wording discipline: underline the word “incorrectly” and write “find the FALSE one” in the margin. Selecting a true statement here is the commonest error, and it is what happened.

Neutral lanthanoids are [Xe]4fⁿ5d^0–16s². Forming Ln³⁺ removes the two 6s electrons and one more (from 5d if present, otherwise from 4f). What survives is a clean [Xe]4fⁿ core with n running from 1 (Ce³⁺) to 14 (Lu³⁺). That is why +3 is the characteristic oxidation state of the whole series and why their chemistry is so uniform.

1. Neutral: [Xe]4fⁿ5d^0–16s².
2. Remove the two 6s electrons ⇒ Ln²⁺.
3. Remove one more (5d if occupied, otherwise 4f) ⇒ Ln³⁺ = [Xe]4fⁿ.
4. Check the endpoints: Ce³⁺ = 4f¹; Lu³⁺ = 4f¹⁴.
5. So the general form is 4fⁿ with n = 1 to 14.

• Watch the words “common to”. The question wants the general pattern, not one specific ion. Options giving a fixed number (4f⁰, 4f¹⁴) describe individual ions (La³⁺ and Lu³⁺), not the series.
• Option 4 is a neutral-atom fragment, not an ion configuration — eliminate it on sight.
• Anchor endpoints: La³⁺ = 4f⁰, Ce³⁺ = 4f¹, Lu³⁺ = 4f¹⁴. Between them, every value of n appears exactly once.

Lanthanoid colour comes from f–f electronic transitions. Such transitions need partially filled 4f orbitals. If the 4f subshell is empty (4f⁰) or completely full (4f¹⁴), there is no vacancy to promote an electron into, so no visible absorption occurs ⇒ the ion is colourless.

1. La³⁺: Z = 57, so 57 − 3 − 54 = 0 ⇒ 4f⁰ ⇒ colourless.
2. Lu³⁺: Z = 71, so 71 − 3 − 54 = 14 ⇒ 4f¹⁴ ⇒ colourless.
3. Both members of pair 1 qualify. ✓
4. The other pairs all contain partially filled ions — Ce³⁺ (4f¹), Pr³⁺ (4f²), Eu³⁺ (4f⁶), Nd³⁺ (4f³), Sm³⁺ (4f⁵) — all coloured.
5. (Gd³⁺ is 4f⁷, half-filled — still partially filled, so still capable of f–f transitions and not counted as colourless here.)

• Only the two ends of the series are colourless. La³⁺ (the first) and Lu³⁺ (the last). Every ion between them is coloured. One sentence covers the whole topic.
• Instant count: n = Z − 3 − 54 for any Ln³⁺. Colourless only when n = 0 or 14.
• Careful with half-filled: 4f⁷ (Gd³⁺) is extra stable but not colourless — stability and colour are different questions. Option 3 is built on exactly that confusion.

1. Underline incorrect / not / except in the stem and write the word “FALSE” in the margin before reading a single option.
2. When a statement is nearly right, name the element it actually describes. That is how Q81(c) and Q86 both come apart in one second.