NCERT Exemplar Solutions Class 12 Chemistry Ch 4 - Download PDF

Correct option: (ii) 26.

Concept used. A neutral transition atom X has electronic configuration [Ar] 3d^x4s^y. When it forms Xⁿ⁺, electrons are removed first from the outer 4s orbital, then from the inner 3d orbital. So electrons in X³⁺ = Z - 3, where Z is the atomic number of X. Argon (Ar) has 18 electrons, so the [Ar] 3d⁵ ion holds 18+5 = 23 electrons.

1. Count the electrons in X³⁺: n(X³⁺) = 18 (from Ar core) + 5 (from 3d⁵) = 23.
2. Restore the three lost electrons to obtain the neutral atom: Z = n(X) = n(X³⁺) + 3 = 23 + 3 = 26.
3. Identify the element: Z = 26 is iron (Fe). The neutral atom is [Ar] 3d⁶4s² (lose 2× 4s and 1× 3d to land at [Ar] 3d⁵, the half-filled stable Fe³⁺). Check: Fe³⁺ is indeed [Ar] 3d⁵.

Atomic number of X is 26, i.e. iron (Fe); option (ii).

Correct option: (i) Cu(II) is more stable.

Concept used. The stability of a metal ion in aqueous solution is decided not only by its gas-phase electron configuration but also by the hydration enthalpy _hydH. A more highly charged, smaller cation is hydrated far more exothermically. For copper, _hydH(Cu²⁺) ≈ -2100 kJ/mol, _hydH(Cu⁺) ≈ -582 kJ/mol. The difference of ∼ 1500 kJ/mol swamps the second ionisation energy needed to make Cu²⁺ from Cu⁺.

1. Write the disproportionation reaction in water: 2 Cu⁺(aq) -> Cu²⁺(aq) + Cu(s). This reaction is spontaneous: E^∘_cell = E^∘(Cu⁺/Cu) - E^∘(Cu²⁺/Cu⁺) = 0.52 - 0.16 = +0.36 V > 0.
2. Apply the thermodynamic identity: _rG^∘ = -nFE^∘ = -(1)(96485)(0.36) = -34.7 kJ/mol < 0. So Cu⁺ disproportionates in water, leaving Cu²⁺ as the stable species.
3. Compare to the gas-phase argument: 3d¹⁰ (closed shell) looks more stable than 3d⁹, which would predict Cu⁺ favoured. But the high _hydH of Cu²⁺ flips the order in solution. Hence option (i) is correct.

In aqueous medium Cu(II) is more stable than Cu(I); option (i).

Correct option: (iv) Cu.

Concept used. For a metal that crystallises in a close-packed lattice, density scales as ρ = Z MN_A V_cell, V_cell ∝ r³, where Z is atoms per unit cell, M is atomic mass, r is metallic radius and V_cell is unit-cell volume. With the four elements packing in essentially the same lattice and Z, N_A the same, density is driven by the ratio M/r³. A larger atomic mass and a smaller radius both push density up.

1. Tabulate atomic masses (g/mol) and radii (pm): Fe=55.85, r=126; Co=58.93, r=125; Ni=58.69, r=125; Cu=63.55, r=128.
2. Compute the proxy M/r³ (in g mol^-1/pm³, absolute scale irrelevant since we only compare): Fe: 55.85/126³=2.79× 10^-5, Co: 58.93/125³=3.02× 10^-5, Ni: 58.69/125³=3.01× 10^-5, Cu: 63.55/128³=3.03× 10^-5.
3. The largest proxy is for Cu, consistent with the observed densities (g/cm³): Fe=7.87, Co=8.90, Ni=8.91, Cu=8.95. Cu wins because its ∼ 8% heavier mass outweighs its ∼ 2.4% larger radius.

Copper has the highest density; option (iv).

Correct option: (ii) CuF2.

Concept used. The colour of a transition-metal compound arises from d–d electronic transitions: visible-light photons promote an electron from a lower-energy d orbital (t_2g or e_g depending on geometry) to a higher one. This is only possible when the metal ion has partly filled d orbitals (one or more unpaired electrons, or at least a vacant d orbital to receive an electron). Closed-shell ions (d⁰ or d¹⁰) cannot show d–d transitions; they are usually colourless unless charge-transfer bands fall in the visible.

1. Identify the metal-ion configuration in each option:
   • Ag2SO4: Ag⁺ is [Kr] 4d¹⁰ – closed shell. Colourless.
   • CuF2: Cu²⁺ is [Ar] 3d⁹ – one unpaired electron. Coloured (blue).
   • ZnF2: Zn²⁺ is [Ar] 3d¹⁰ – closed shell. Colourless.
   • Cu2Cl2: Cu⁺ is [Ar] 3d¹⁰ – closed shell. Colourless (white).
2. Only CuF2 has a partly-filled d shell.
3. Hence (ii) is the only coloured solid in the list.

CuF2 is coloured (blue); option (ii).

Correct option: (i) Mn2O7.

Concept used. Concentrated H2SO4 is a strong dehydrating agent. When it acts on KMnO4 it strips water from the permanganate to give the anhydride of permanganic acid, which is Mn2O7 – a dark-green oily liquid that is thermodynamically very unstable and detonates above 55^∘C.

1. Write the dehydration: 2 KMnO4 + H2SO4 (conc.) -> Mn2O7 + K2SO4 + H2O. Manganese stays in the +7 oxidation state on both sides – this is acid–base chemistry, not redox.
2. Check the colour and properties: Mn2O7 is described in textbooks as ``dark green/red oil''; in thin films it appears deep green. It decomposes explosively: 2 Mn2O7 -> 4 MnO2 + 3 O2 .
3. Eliminate distractors: MnO2 is brown/black solid (not oily); MnSO4 is pale pink crystalline salt; Mn2O3 is black-brown solid. Only Mn2O7 matches ``green oily explosive''.

The green oily explosive is Mn2O7; option (i).

Correct option: (ii) 3d⁵.

Concept used. For a first-row transition ion the spin-only magnetic moment is _s.o. = √n(n+2) B.M., where n is the number of unpaired d electrons (B.M. = Bohr magneton). μ is maximised by maximising n; for a d^x ion the maximum value of n is min(x, 10-x)× 2 – but more simply, the highest possible unpaired count in a d subshell is 5 (one in each of the five d orbitals, in line with Hund's rule).

1. Count maximum unpaired electrons (high-spin, weak ligand field) for each option:
   • 3d⁷: orbitals fill as ↑↓, ↑↓, ↑, ↑, ↑; n = 3.
   • 3d⁵: ↑, ↑, ↑, ↑, ↑; n = 5 (half-filled).
   • 3d⁸: ↑↓, ↑↓, ↑↓, ↑, ↑; n = 2.
   • 3d²: ↑, ↑, three empty; n = 2.
2. Compute _s.o. for each: aligned d⁷: μ &= √3· 5 = √15 = 3.87 B.M.
   d⁵: μ &= √5· 7 = √35 = 5.92 B.M.
   d⁸: μ &= √2· 4 = √8 = 2.83 B.M.
   d²: μ &= √2· 4 = √8 = 2.83 B.M. aligned
3. Largest is d⁵ at 5.92 B.M.

3d⁵ has the highest magnetic moment, 5.92 B.M.; option (ii).

Correct option: (ii) +3.

Concept used. Lanthanoids (Ce to Lu, Z=58–71) have the general atomic configuration [Xe] 4f^1--14 5d^0--1 6s². The 4f electrons are well shielded, the 6s electrons easy to remove and the 5d¹ (where present) likewise. The energetically cheapest ionisation removes the two 6s electrons and (when present) the 5d¹, leaving the 4f core intact and producing the +3 ion.

1. For a typical lanthanoid, sum of the first three ionisation enthalpies ₁³ IE ≈ 3500 kJ/mol – comparable to that of group 13 elements and easily provided by lattice/hydration energy of Ln³⁺ compounds.
2. Going beyond +3 requires ionising a 4f electron, which is much harder (IE₄ 4000 kJ/mol), so +4 is seen only when it produces a particularly stable configuration (e.g. Ce⁴⁺ becomes 4f⁰). Similarly +2 exists only when a stable 4f⁷ or 4f¹⁴ shell is reached (Eu²⁺, Yb²⁺).
3. The single oxidation state shared by every lanthanoid from La/Ce through Lu is therefore +3.

+3 is the characteristic oxidation state of every lanthanoid; option (ii).

Correct option: (i) a, b.

Concept used. A disproportionation reaction is one in which the same element is simultaneously oxidised and reduced: i.e. a single species in an intermediate oxidation state splits into one higher and one lower oxidation state of itself.

1. Reaction (a) Cu⁺ -> Cu²⁺ + Cu: Cu starts at +1, ends as +2 (oxidised) and 0 (reduced). Same element, two new states. Disproportionation. (Stoichiometry must be balanced as 2 Cu⁺ -> Cu²⁺ + Cu.)
2. Reaction (b) 3 MnO4^2- + 4 H+ -> 2 MnO4^- + MnO2 + 2 H2O (note: the question's left side should read MnO4^2-): Mn starts at +6 on the left, ends as +7 (in MnO4^-, oxidised) and +4 (in MnO2, reduced). Same element, two new states. Disproportionation.
3. Reaction (c) 2 KMnO4 -> K2MnO4 + MnO2 + O2: Mn goes from +7 to +6 (K2MnO4) and +4 (MnO2), which is reduction in both cases. The oxidation here happens on oxygen (from -2 to 0), not on Mn. So this is a coupled redox, not a disproportionation. 55
4. Reaction (d) 2 MnO4^- + 3 Mn²⁺ + 2 H2O -> 5 MnO2 + 4 H+: Mn goes from +7 (in MnO4^-) and +2 (in Mn²⁺) both to +4 (in MnO2). This is a comproportionation (the reverse of disproportionation): two different starting states converge to one common state. 55

Only (a) and (b) are disproportionations; option (i).

Correct option: (iv) Mn²⁺ acts as autocatalyst.

Concept used. An autocatalyst is a product of the reaction that itself catalyses that very reaction. The hallmark of autocatalysis is a kinetic curve that starts slow (no catalyst yet) and accelerates as more product accumulates, until reactants begin to deplete.

1. Write the titration reaction: 2 MnO4^- + 5 C2O4^2- + 16 H+ -> 2 Mn²⁺ + 10 CO2 + 8 H2O. At t=0 the only oxidising agent is MnO4^-, which is slow because the reaction is a many-electron process (5e^-/MnO4^-).
2. As Mn²⁺ accumulates, it acts as a redox shuttle:
   • [] Step (i): Mn²⁺ is oxidised by MnO4^- to Mn³⁺. [] Step (ii): Mn³⁺ oxidises C2O4^2-, returning to Mn²⁺.
   The single-electron Mn³⁺/Mn²⁺ couple is kinetically much faster than the direct MnO4^- + C2O4^2- encounter, so the overall rate jumps once [Mn²⁺] builds up.
3. Eliminate distractors: CO2 leaves as gas and does not participate in further chemistry; the reaction is mildly exothermic but the change of T in a dilute aqueous titration is small; MnO4^- is a reactant, not a catalyst.

The catalyst is Mn²⁺, formed as the reaction proceeds; option (iv).

Correct option: (iii) Tm.

Concept used. The actinoid series consists of the 14 elements that follow actinium (Ac, Z=89) in which the 5f orbitals are progressively filled. These are Th (Z=90) through Lr (Z=103). Thulium (Tm, Z=69) is a lanthanoid (filling 4f), not an actinoid.

1. Check each option's atomic number:
   • U (uranium): Z=92, 5f³6d¹7s² – actinoid.
   • Np (neptunium): Z=93, 5f⁴6d¹7s² – actinoid.
   • Tm (thulium): Z=69, 4f¹³6s² – lanthanoid.
   • Fm (fermium): Z=100, 5f¹²7s² – actinoid.
2. Only Tm sits in the 4f block, not the 5f block.

Thulium (Tm) is not an actinoid; option (iii).

Correct option: (i) 25.

Concept used. Balance the redox by matching electrons. In acid, MnO4^- is reduced by 5 electrons per ion: MnO4^- + 8 H+ + 5 e^- -> Mn²⁺ + 4 H2O. Sulphide S^2- is oxidised to elemental sulphur with loss of 2 electrons: S^2- -> S + 2 e^-.

1. Equalise electrons: multiply the MnO4^- half-equation by 2 (⇒ 10 e^-) and the S^2- half-equation by 5 (⇒ 10 e^-).
2. Add: 2 MnO4^- + 16 H+ + 5 S^2- -> 2 Mn²⁺ + 8 H2O + 5 S.
3. Read off the stoichiometric ratio: 2 mol KMnO4 reacts with 5 mol S^2-. Per mol of S^2-, moles of KMnO4 = 2/5.

25 mol KMnO4 per mol S^2-; option (i).

Correct option: (i) V2O5 and Cr2O3.

Concept used. An amphoteric oxide reacts with both acids and bases. For transition-metal oxides, amphoteric character is seen at intermediate oxidation states: very low states are basic (give cations in acid), very high states are acidic (give oxoanions in base), and intermediate states straddle both.

1. Tag each oxide by Mn or Cr/V oxidation state and acid–base nature:
   • Mn2O7 – Mn⁺⁷, strongly acidic (anhydride of HMnO4).
   • CrO3 – Cr⁺⁶, strongly acidic (anhydride of chromic/dichromic acid).
   • Cr2O3 – Cr⁺³, amphoteric (reacts with HCl giving CrCl3, with NaOH giving Na[Cr(OH)4]).
   • CrO – Cr⁺², basic.
   • V2O5 – V⁺⁵, amphoteric (gives VO2+ in acid, vanadate VO4^3- in base).
   • V2O4 – V⁺⁴, basic/weakly amphoteric.
2. The two clearly amphoteric oxides are Cr2O3 and V2O5.
3. Verify with sample reactions: Cr2O3 + 6 HCl -> 2 CrCl3 + 3 H2O, Cr2O3 + 2 NaOH + 3 H2O -> 2 Na[Cr(OH)4]. V2O5 similarly reacts with H+ (gives VO2+) and with OH- (gives vanadate).

V2O5 and Cr2O3 are amphoteric; option (i).

Correct option: (i) [Xe] 4f⁷ 5d¹ 6s².

Concept used. For lanthanoids the filling order is typically 6s² first, then 4f, with 5d¹ inserted whenever that produces a half-filled or fully-filled 4f shell (extra exchange-energy stability, analogous to Cr and Cu in the 3d row). At Z=64 exactly, putting one electron in 5d lets 4f stop at 4f⁷ – the half-filled stable arrangement.

1. Electron count: Z=64 gadolinium. Xe accounts for 54 electrons, leaving 64-54 = 10 to distribute over 4f, 5d, 6s.
2. Naive Aufbau: 6s² 4f⁸ uses 10 electrons. But that gives 4f⁸ (8 electrons among 7 f orbitals ⇒ pairing in one orbital). Promoting one 4f electron to 5d costs little energy and breaks no pair because the resulting 4f⁷ is half-filled (every f orbital singly occupied, maximum exchange energy).
3. Hence Gd is anomalous: [Xe] 4f⁷ 5d¹ 6s², with 4f⁷ providing the half-filled stability.

Gadolinium: [Xe] 4f⁷ 5d¹ 6s²; option (i).

Correct option: (iv) They are chemically very reactive.

Concept used. Interstitial compounds of transition metals (e.g. TiC, ZrN, Fe3H) form when small atoms like H, B, C, or N occupy the octahedral or tetrahedral holes in the metal lattice. The small atoms wedge the lattice, locking metal atoms in place: this raises the melting point, increases hardness, preserves the metallic sea of electrons (so conductivity stays), and – crucially – makes the compound chemically inert, not reactive.

1. Check (i) – True: e.g. TiC melts at 3160^∘C, far above pure Ti at 1668^∘C.
2. Check (ii) – True: most interstitial carbides and nitrides have hardness near or above diamond, used as cutting tool coatings.
3. Check (iii) – True: the conduction-band electrons of the metal lattice are largely untouched, so metallic conductivity survives.
4. Check (iv) – False: interstitial compounds are notoriously unreactive. WC, TiC, Fe3C etc. resist acid attack and oxidation up to high temperatures. Option (iv) is the wrong characterisation.

Option (iv) is the false claim; interstitial compounds are chemically inert.

Correct option: (ii) 3.87 B.M.

Concept used. Cr has Z=24, neutral atom configuration [Ar] 3d⁵ 4s¹. The Cr³⁺ ion loses the 4s¹ and two 3d electrons, leaving [Ar] 3d³. The spin-only magnetic moment formula is _s.o. = √n(n+2) B.M., where n is the number of unpaired electrons.

1. Place 3 electrons in five d orbitals using Hund's rule: ↑,↑,↑,_,. Three unpaired electrons, so n=3.
2. Substitute in the formula: μ = √3 (3+2) = √15.
3. Compute the square root: √15 = 3.873. To 2 d.p., μ = 3.87 B.M.

_s.o.(Cr³⁺) = 3.87 B.M.; option (ii).

Correct option: (iii) IO3^-.

Concept used. In alkaline (or neutral) medium, KMnO4 is a strong oxidising agent that oxidises iodide I^- (-1 state) all the way to iodate IO3^- (+5 state). The Mn side goes from +7 to +4 (MnO2).

1. Half-reactions in alkaline medium: aligned MnO4^- + 2 H2O + 3 e^- &-> MnO2 + 4 OH^- (reduction),
   I^- + 6 OH^- &-> IO3^- + 3 H2O + 6 e^- (oxidation). aligned
2. Balance electrons (multiply Mn half by 2, I half by 1; total 6 e^-) and add: 2 MnO4^- + I^- + H2O -> 2 MnO2 + IO3^- + 2 OH^-.
3. Iodine goes -1 → +5, a 6-electron oxidation that lands at iodate. Distractor analysis: IO^- (hypoiodite, +1) would require only 2e^- – insufficient against a strong oxidant; IO4^- (periodate, +7) would need an even stronger oxidant such as ozone or F2, more than MnO4^- can deliver in alkali.

Iodide is oxidised to iodate IO3^-; option (iii).

Correct option: (i) Copper liberates hydrogen from acids – this is incorrect.

Concept used. A metal liberates H2 from a non-oxidising acid only if its standard reduction potential E^∘(Mⁿ⁺/M) is negative relative to H+/H2 (taken as 0 V). Copper has E^∘(Cu²⁺/Cu) = +0.34 V, which is positive, so Cu cannot reduce H+ to H2. The other three statements are textbook-correct.

1. Tag each statement:
   • (i) Wrong: E^∘(Cu²⁺/Cu) > 0, so Cu does not displace H2 from dilute HCl or H2SO4.
   • (ii) Correct: Mn⁺⁷ (Mn2O7, MnO4^-) and Mn⁺⁴ (MnO2, MnF4) etc. are stable; oxide and fluoride ligands stabilise high oxidation states.
   • (iii) Correct: E^∘(Mn³⁺/Mn²⁺) = +1.51 V, E^∘(Co³⁺/Co²⁺) = +1.81 V; both are strong oxidants.
   • (iv) Correct: E^∘(Ti³⁺/Ti²⁺) = -0.37 V, E^∘(Cr³⁺/Cr²⁺) = -0.41 V; both are reductants in water.
2. Only (i) misrepresents the chemistry.

Statement (i) is incorrect: Cu does not liberate H2 from acids.

Correct option: (iii) Sn⁴⁺.

Concept used. Acidified dichromate is a strong oxidising agent: E^∘(Cr2O7^2-/Cr³⁺) = +1.33 V. It oxidises Sn²⁺ (a known reductant, E^∘(Sn⁴⁺/Sn²⁺) = +0.15 V) all the way to Sn⁴⁺.

1. Half-reactions: aligned Cr2O7^2- + 14 H+ + 6 e^- &-> 2 Cr³⁺ + 7 H2O (reduction),
   Sn²⁺ &-> Sn⁴⁺ + 2 e^- (oxidation). aligned
2. Balance electrons (multiply Sn half by 3, total 6 e^-) and add: Cr2O7^2- + 14 H+ + 3 Sn²⁺ -> 2 Cr³⁺ + 3 Sn⁴⁺ + 7 H2O.
3. Cell potential: E^∘_cell = 1.33 - 0.15 = +1.18 V, strongly spontaneous. Sn²⁺ is driven to Sn⁴⁺. Distractors Sn³⁺ and Sn⁺ are not stable oxidation states for tin.

Sn²⁺ is oxidised to Sn⁴⁺; option (iii).

Correct option: (iv) in covalent compounds fluorine can form single bond only while oxygen forms double bond.

Concept used. The maximum oxidation state of a metal in its fluoride is set by how many F atoms can be packed around it (each F contributes only one bond, so σ-only). In an oxide, oxygen can form a double bond (one σ + one π), so one oxygen ``carries two oxidation units''. To reach +7, manganese needs to form 7 M–X bonds: with fluorine that means MnF7 (sterically impossible, ∼ 7 F around one Mn won't fit), but with oxygen it means Mn2O7 where each oxygen ties up via Mn=O (π-bond).

1. Count bonds in Mn2O7: there are several terminal Mn=O double bonds plus a bridging Mn-O-Mn. Each Mn=O counts 2 toward Mn's oxidation state (Mn donates 2 electrons to that oxygen). With three terminal Mn=O and one bridging Mn-O-Mn per Mn, oxidation state = 3(2) + 1 = 7.
2. For fluorine: each Mn-F contributes only 1 to Mn's oxidation state, and 7 F atoms cannot fit around one Mn atom (coordination too crowded). The known highest manganese fluoride is MnF4 at oxidation state +4.
3. Eliminate distractors:
   • (i) ``F is more electronegative than O'' – true but not the operative reason; higher electronegativity pulls more, but oxygen still wins via π.
   • (ii) F has no d-orbitals – true, but neither does O in normal bonding; not the reason.
   • (iii) F stabilises lower oxidation states – a vague slogan; the real reason is bonding multiplicity.

Reason (iv): oxygen's ability to form M=O double bonds lets it stabilise higher oxidation states than fluorine.

Correct option: (iii) both have similar atomic radius.

Concept used. Lanthanoid contraction: the steady decrease in Ln³⁺ ionic radius across the 4f series. The contraction cancels the increase in shell number that would normally make a 5d atom larger than its 4d counterpart. As a result, the two members of group 4 (Zr, Hf) end up with almost identical atomic radii (160 pm vs 159 pm).

1. Tabulate metallic radii: r(Zr) = 160 pm, r(Hf) = 159 pm.
2. Almost identical radius ⇒ almost identical lattice energies, hydration energies, M–X bond lengths, crystal packings.
3. Hence ionisation energies, melting points, densities, even chemical reactivity track each other very closely.

Same radius (option iii) is the operative reason.

Correct option: (ii) KMnO4 oxidises HCl into Cl2 which is also an oxidising agent.

Concept used. KMnO4 in acid is reduced to Mn²⁺ with E^∘ = +1.51 V. HCl provides Cl^-, which has E^∘(Cl2/Cl^-) = +1.36 V – low enough for MnO4^- to oxidise it to Cl2. This wastes some of the permanganate's oxidising power and adds Cl2 as a competing oxidant, ruining quantitative titrations. Sulphuric acid, in contrast, has SO4^2- which is not oxidisable, so it stays as a spectator acid.

1. Side-reaction balance in acidic medium: 2 MnO4^- + 10 Cl^- + 16 H+ -> 2 Mn²⁺ + 5 Cl2 + 8 H2O. Cell potential: 1.51 - 1.36 = +0.15 V, spontaneous.
2. Consequence in titration: some KMnO4 is consumed by HCl before it can act on the intended reductant (Fe²⁺, C2O4^2-, ...). The titre comes out higher than the true value, and the liberated Cl2 can oxidise the analyte too – a double error.
3. Eliminate distractors: (i) is partially true but doesn't describe the cause; (iii) is false (E^∘(KMnO4) > E^∘(HCl/Cl2)); (iv) is false. Only (ii) explains the practical avoidance of HCl.

Cause: KMnO4 oxidises HCl to Cl2 (also an oxidant); option (ii).

Correct options: (i) and (ii).

Concept used. A transition-metal or lanthanoid compound is coloured if (a) the metal ion has partly-filled d or f orbitals (so d–d / f–f transitions are possible), or (b) the compound exhibits a strong charge-transfer band in the visible (LMCT or MLCT). Closed-shell ions (d⁰, d¹⁰, f⁰, f¹⁴) are usually colourless from d–d alone, but can still be intensely coloured by charge-transfer.

1. KMnO4: Mn⁺⁷ is formally d⁰. No d–d possible. Yet KMnO4 is intensely purple because of strong ligand-to-metal charge transfer (LMCT) from the four O^2- to Mn⁺⁷ in the visible region. Coloured.
2. Ce(SO4)2: Ce⁴⁺ is 4f⁰. No f–f possible. Yet the compound is orange-yellow due to O^2- → Ce⁴⁺ LMCT. Coloured.
3. TiCl4: Ti⁺⁴ is 3d⁰. Cl^- to Ti LMCT falls in the UV, not the visible. The pure liquid is colourless.
4. Cu2Cl2: Cu⁺ is 3d¹⁰. No d–d, no charge-transfer in visible. Colourless (white).

The coloured compounds are KMnO4 and Ce(SO4)2; options (i) and (ii).

Correct options: (i) and (iv).

Concept used. For first-row transition ions in a weak ligand field, the spin-only magnetic moment is μ = √n(n+2) B.M. Two ions have ``almost the same'' moment if they have the same number of unpaired electrons n (modulo small orbital corrections).

1. Count n for each ion (high-spin, neutral electron count):
   • Co²⁺: 3d⁷, n = 3, μ = √15 = 3.87 B.M.
   • Cr²⁺: 3d⁴, n = 4, μ = √24 = 4.90 B.M.
   • Mn²⁺: 3d⁵, n = 5, μ = √35 = 5.92 B.M.
   • Cr³⁺: 3d³, n = 3, μ = √15 = 3.87 B.M.
2. Match the same-n pair: Co²⁺ and Cr³⁺ both have n=3, both μ ≈ 3.87 B.M.
3. The other two (Cr²⁺ and Mn²⁺) have different n values, so different moments.

Co²⁺ and Cr³⁺ share μ ≈ 3.87 B.M.; options (i) and (iv).

Correct options: (ii) and (iii).

Concept used. Within a d-block group, the higher oxidation states become more stable as we move down (i.e., from 3d to 4d to 5d). The reason is that the heavier members have larger, more diffuse d orbitals that can engage in stronger π-bonding with O ligands, stabilising the high state.

1. Apply to group 6: Cr, Mo, W. Cr⁺⁶ in Cr2O7^2- oxidises easily back to Cr⁺³ (E^∘(Cr2O7^2-/Cr³⁺) = +1.33 V) – a strong oxidant. Mo⁺⁶ in MoO3 has no driving force to drop to a lower state; it just sits stable. Same for W⁺⁶ in WO3.
2. Statement (ii) – Mo(VI) and W(VI) are more stable than Cr(VI) – is true. Statement (iii) – higher oxidation states of heavier members of group 6 are more stable – is the general principle behind (ii). Both correct.
3. Statements (i) and (iv) state the opposite trend; they are false.

Both (ii) and (iii) describe the same fact: heavier d-block members are more stable in higher oxidation states.

Correct options: (ii) and (iv).

Concept used. Early actinoids exhibit a much wider range of oxidation states than lanthanoids because the 5f orbitals are more diffuse and energetically closer to 6d and 7s. This lets electrons from several shells be removed before encountering a prohibitive ionisation barrier.

1. Tabulate maximum oxidation state per actinoid up to Am:
   • U (Z=92, 5f³6d¹7s²): max +6 (e.g. UO2²⁺, UF6).
   • Np (Z=93, 5f⁴6d¹7s²): max +7 (e.g. NpO4^- in alkali).
   • Pu (Z=94, 5f⁶7s²): max +7 (e.g. PuO4^-).
   • Am (Z=95, 5f⁷7s²): max +6 commonly (occasional reports of +7 in highly oxidising conditions but it is not the standard high state).
2. NCERT lists Np and Pu as the actinoids that routinely reach +7.
3. Options (ii) Pu and (iv) Np are correct.

Np and Pu reach +7; options (ii) and (iv).

Correct options: (i) and (ii).

Concept used. Actinoid ground-state configurations are empirically determined and follow no clean rule – a 6d¹ electron appears in some actinoids (Th, Pa, U, Np, Cm, Lr) and is absent in others (Pu, Am, Bk–No). Memorise the few that have 6d¹.

1. Stated configurations (NCERT and IUPAC):
   • U: [Rn] 5f³ 6d¹ 7s² – has 6d¹.
   • Np: [Rn] 5f⁴ 6d¹ 7s² – has 6d¹.
   • Pu: [Rn] 5f⁶ 7s² – no 6d. 55
   • Am: [Rn] 5f⁷ 7s² – no 6d; instead 5f⁷ half-filled stable. 55
2. Note the parallel with the lanthanoid pattern: 4f⁶ at Sm corresponds to 5f⁶ at Pu (both prefer fⁿ s² without d¹); 4f⁷ at Eu and Gd compares to 5f⁷ at Am/Cm.
3. Hence U and Np host 6d¹; Pu and Am do not.

U and Np host one 6d electron; options (i) and (ii).

Correct options: (ii) and (iii).

Concept used. The +2 state appears in a lanthanoid when the resulting Ln²⁺ has an especially stable 4f configuration – specifically 4f⁷ (half-filled) for Eu²⁺ or 4f¹⁴ (fully filled) for Yb²⁺. Ce and Ho do not have such a stable 4f shell at +2.

1. Eu neutral: [Xe] 4f⁷ 6s². Lose two 6s electrons ⇒ Eu²⁺ as [Xe] 4f⁷, a half-filled stable configuration. Eu²⁺ is well-known (silvery-white reductant).
2. Yb neutral: [Xe] 4f¹⁴ 6s². Lose two 6s electrons ⇒ Yb²⁺ as [Xe] 4f¹⁴, a fully-filled stable configuration. Also well-known.
3. Ce: Ce²⁺ would be 4f² – not a stable configuration; instead Ce shows +3 and +4 (with Ce⁴⁺ stable at 4f⁰). Ho²⁺ would be 4f¹¹ – not specially stable, so Ho stays at +3.

Eu and Yb show +2; options (ii) and (iii).

Correct options: (ii) and (iii).

Concept used. Spin-only magnetic moment μ = √n(n+2) B.M. is monotonically increasing in n. ``Higher value'' is interpreted relative to the others in the list, so we look for the ions with the highest n.

1. Tabulate n for each (high-spin, neutral electron count):
   • Ti³⁺: 3d¹, n=1, μ=1.73 B.M.
   • Mn²⁺: 3d⁵, n=5, μ=5.92 B.M.
   • Fe²⁺: 3d⁶ high-spin, n=4, μ=4.90 B.M.
   • Co³⁺: 3d⁶ high-spin, n=4, μ=4.90 B.M. (low-spin in strong field → n=0, but in aqueous Co³⁺ is typically high-spin in [Co(H2O)6]³⁺ where its weak-field approximation gives 4 unpaired electrons).
2. Sort: Mn²⁺ (5.92) > Fe²⁺/Co³⁺ (4.90) > Ti³⁺ (1.73).
3. The two ``higher'' values among the four are Mn²⁺ and Fe²⁺ (with Co³⁺ tied with Fe²⁺, but the NCERT answer key specifies (ii) and (iii) because Fe²⁺ is reliably high-spin in water whereas Co³⁺ commonly low-spins).

Mn²⁺ (μ = 5.92) and Fe²⁺ (μ = 4.90); options (ii) and (iii).

Correct options: (i) and (ii).

Concept used. A metal forms an MF3 binary fluoride only if the +3 oxidation state is stable for that metal and is compatible with the small, strongly oxidising fluoride ligand. Fluorine pushes its partner to a moderate (not necessarily highest) oxidation state.

1. CrF3: Cr⁺³ is the most stable Cr oxidation state (configuration 3d³, half-t_2g stability in octahedral field). CrF3 is a well-known violet/red solid.
2. CoF3: Co⁺³ is accessible with the strongly oxidising F^- (Co prefers +2 normally, but F^- stabilises +3). CoF3 is a known brown fluorinating agent.
3. CuF3: would require Cu⁺³ (3d⁸), which is unstable; Cu commonly stops at +2 (CuF2, blue). CuF3 does not exist as a stable binary. 55
4. NiF3: Ni⁺³ is rare; Ni typically gives NiF2 and (very rarely) NiF4. NiF3 is not a standard binary fluoride. 55

Cr and Co form MF3; options (i) and (ii).

Correct options: (ii) and (iii).

Concept used. An oxide MO_x acts as an oxidant only if the metal in that oxide can be reduced to a lower oxidation state under standard conditions. For group 6, Cr(VI) readily reduces to Cr(III) (strongly oxidising); Mo(VI) and W(VI) are thermodynamically the most stable oxidation states for these metals – they are essentially unreduced and not oxidising under ordinary conditions.

1. Recall from Q24 that within group 6, the higher oxidation state stabilises down the group.
2. CrO3 is a powerful oxidant (anhydride of chromic acid, E^∘(Cr2O7^2-/Cr³⁺) = +1.33 V).
3. MoO3, WO3: Mo⁺⁶ and W⁺⁶ are stable, so reduction is hard. They do not oxidise typical substrates under standard conditions.
4. CrO4^2-: chromate. In neutral/alkaline solution it is a weaker oxidant than Cr2O7^2- but can oxidise (e.g. alcohols, in slow reactions). Hence CrO4^2- is still an oxidising agent (textbook). Options (ii) and (iii) are the only non-oxidisers.

MoO3 and WO3 are not oxidising agents; options (ii) and (iii).

Correct options: (ii) and (iii).

Concept used. Ce neutral: [Xe] 4f¹ 5d¹ 6s². Losing four electrons gives Ce⁴⁺ = [Xe], which is both a noble-gas configuration (Xe core) and a 4f⁰ closed subshell. These two phrasings describe the same underlying stable arrangement.

1. Configuration analysis: Ce (4f¹5d¹6s²) → Ce⁴⁺ (4f⁰). The 4f shell, which is normally populated across the lanthanoid series, is empty here – a special stable empty-subshell configuration.
2. The same configuration is identical to the Xe noble-gas core [Xe]. So the ion has attained both ``noble-gas'' and ``f⁰'' stability simultaneously.
3. Statements (ii) and (iii) capture two faces of the same truth. (i) is vague and (iv) is irrelevant (Ce is not Pb).

Ce⁴⁺ attains the stable 4f⁰ / noble-gas Xe configuration; options (ii) and (iii).

Concept used. A metal M displaces H2 from a non-oxidising acid (dilute HCl, H2SO4) only when its standard reduction potential E^∘(Mⁿ⁺/M) is more negative than E^∘(H+/H2) = 0 V. The displacement reaction is M + nH+ → Mⁿ⁺ + (n/2)H2 with cell potential E^∘_cell = -E^∘(Mⁿ⁺/M).

1. Look up Cu's reduction potential: E^∘(Cu²⁺/Cu) = +0.34 V. This is positive, meaning Cu²⁺ is more easily reduced than H+.
2. Cell potential for the displacement: E^∘_cell = E^∘(H+/H2) - E^∘(Cu²⁺/Cu) = 0 - 0.34 = -0.34 V < 0. Negative E^∘_cell means the reaction is non-spontaneous.
3. Conclusion: Cu cannot reduce H+ to H2. With oxidising acids (HNO3, hot conc. H2SO4), Cu reacts – but the oxidising agent is then the NO3^- or SO4^2-, not H+, so the gas evolved is NO/NO2/SO2, not H2.

Cu has positive E^∘(Cu²⁺/Cu) = +0.34 V, so it cannot reduce H+ to H2.

Concept used. The standard reduction potential E^∘(M²⁺/M) for first-row transition metals depends on three thermochemical steps:

• [] (a) M(s) → M(g) with enthalpy of atomisation +_aH; [] (b) M(g) → M²⁺(g) with +∑ IE (first two ionisation enthalpies); [] (c) M²⁺(g) → M²⁺(aq) with hydration enthalpy +_hydH,

where _aH is enthalpy of atomisation, ∑ IE is sum of first two ionisation enthalpies, and _hydH is hydration enthalpy of the dipositive ion. A more negative E^∘ means the dipositive ion is harder to form (stabilising the metal).

1. For Mn²⁺ (3d⁵) and Zn²⁺ (3d¹⁰): the special stability of the half-filled and fully-filled d shells means a lot of extra energy is released by producing Mn²⁺ and Zn²⁺. In thermochemical terms, the IE step toward these stable ions costs less, and once formed they resist further oxidation. Equivalently, the ground-state Mn²⁺/Zn²⁺ are thermodynamically favoured as ions, making E^∘ for their M²⁺/M couple more negative.
2. For Ni²⁺: the reason is different – nickel's Ni²⁺ has the most exothermic hydration enthalpy among the late-3d divalent ions (_hydH(Ni²⁺) ≈ -2105 kJ/mol). That very negative hydration term drags E^∘ more negative.
3. In each case the unusually negative E^∘ has a different microscopic origin (electron-configuration stability for Mn and Zn; large hydration energy for Ni).

For Mn and Zn: half-filled (3d⁵) and fully-filled (3d¹⁰) stability of M²⁺; for Ni: highly exothermic hydration enthalpy of Ni²⁺.

Concept used. The first ionisation enthalpy (IE₁) depends on (a) which orbital the electron leaves and (b) how tightly that orbital is bound in the specific atom. Cr neutral is [Ar] 3d⁵ 4s¹; the 4s¹ is loosely held (single, no pairing penalty). Zn neutral is [Ar] 3d¹⁰ 4s²; removing the first 4s electron breaks the stable 4s² pair and takes one electron from a fuller, more tightly bound shell.

1. Cr: electron removed is the lone 4s¹. Low energy cost (single, unpaired, exposed). IE₁(Cr) = 653 kJ/mol.
2. Zn: electron removed is one of the 4s² pair. Because the 3d¹⁰ subshell shields the 4s pair less effectively than a single 4s¹ is shielded by 3d⁵, 4s in Zn feels a stronger effective nuclear charge and sits at a lower energy. IE₁(Zn) = 906 kJ/mol.
3. Hence IE₁(Cr) < IE₁(Zn) by ∼ 250 kJ/mol.

Cr's lone 4s¹ is loosely held (and 3d⁵ stability is intact after ionisation); Zn's 4s pair experiences a higher Z_eff owing to poor 3d¹⁰ shielding, so IE₁(Zn) > IE₁(Cr).

Concept used. The melting point of a metal is set by the strength of the metallic bond, which itself depends on how many electrons the atoms contribute to the delocalised ``sea'' of conduction electrons. Transition metals contribute not only their ns electrons but also their inner (n-1)d electrons, giving more electrons per atom in the metallic bond.

1. In alkali metals (Na, K) only one s-electron is available per atom for metallic bonding; m.p. are low (∼ 98 ^∘C for Na).
2. In transition metals, both ns and (n-1)d electrons contribute. For example, V (3d³4s²) contributes 5 electrons per atom – m.p. 1910 ^∘C. Cr (5 unpaired 3d + 1 4s) and W (5 5d + 1 6s) push m.p. to 1907 ^∘C and 3422 ^∘C respectively, among the highest of all elements.
3. More electrons in the metallic bond ⇒ stronger cohesion ⇒ higher melting point.

High m.p. arises because both ns and (n-1)d electrons participate in metallic bonding, giving many cohesive electrons per atom.

Concept used. Cu²⁺ is a mild oxidant (E^∘(Cu²⁺/Cu⁺) = +0.15 V) and I^- is a moderate reductant (E^∘(I2/I^-) = +0.54 V). At first glance the cell potential is negative (0.15 - 0.54 = -0.39 V) and the reaction should not happen. But the formation of the very insoluble Cu2I2 (white, K_sp ∼ 10^-12) drags the equilibrium forward by removing Cu⁺ from solution.

1. Half-reactions: 2 Cu²⁺ + 2 e^- -> 2 Cu⁺, 2 I^- -> I2 + 2 e^-.
2. In solution, Cu⁺ is unstable; here it precipitates with the second I^- as Cu2I2 (cuprous iodide, white): 2 Cu²⁺ + 4 I^- -> Cu2I2(s) + I2. The Cu2I2 is the ``white precipitate''; the I2 is yellow-brown (in solution it turns starch blue-black).
3. Net: Cu²⁺ is reduced to Cu⁺ (stabilised as solid Cu2I2), and I^- is oxidised to I2. This is the basis of the iodometric titration for Cu(II).

2 Cu²⁺ + 4 I^- -> Cu2I2(s)↓ + I2. The white precipitate is Cu2I2.

Concept used. In aqueous solution, the stability comparison of Cu⁺ vs Cu²⁺ is controlled by hydration enthalpy (from Q2). Cu²⁺ has _hydH ≈ -2100 kJ/mol vs Cu⁺'s -582 kJ/mol. The 1500 kJ/mol hydration advantage outweighs the second ionisation energy needed to make Cu²⁺, so Cu²⁺ is the stable aqueous species.

1. Disproportionation: 2 Cu⁺(aq) -> Cu²⁺(aq) + Cu(s) has E^∘_cell = +0.36 V ⇒ _rG^∘ = -34.7 kJ/mol < 0. So Cu⁺ is unstable in water.
2. Translated to chlorides: Cu2Cl2 (containing Cu⁺) is stable as the solid but in water it disproportionates. CuCl2 (containing Cu²⁺) is fully water-stable.
3. Hence in aqueous medium, CuCl2 is more stable than Cu2Cl2. Note: in the solid state at room temperature, Cu2Cl2 is perfectly stable (white solid); the instability is solvolytic.

CuCl2 is more stable in aqueous medium because Cu²⁺ has a much more exothermic hydration enthalpy than Cu⁺, more than compensating for the higher IE₂.

Concept used. (A) brown manganese compound that reacts with HCl to give a gas is MnO2 (manganese dioxide, brown-black). (B) the gas is chlorine Cl2. (C) excess Cl2 with NH3 gives NCl3 (nitrogen trichloride), a yellow oil that is notoriously explosive.

1. Reaction of (A) with HCl: MnO2 + 4 HCl -> MnCl2 + Cl2 + 2 H2O. Mn goes from +4 to +2; Cl^- is oxidised to Cl2. This is the classic lab method to make chlorine.
2. (B) is the chlorine gas. With excess Cl2 and NH3: NH3 + 3 Cl2 -> NCl3 + 3 HCl. With excess NH3 instead (the opposite stoichiometry) the product would be N2; the explosive route is favoured only with excess chlorine.
3. Identifications: A = MnO2, B = Cl2, C = NCl3.

A = MnO2, B = Cl2, C = NCl3.

Concept used. Electronegativity tells you how strongly an atom pulls bonding electrons, but stabilisation of high oxidation state needs multiple bonds (high bond order) to one ligand atom. Oxygen can form σ + π (i.e. double bond, bond order 2) to a transition metal; fluorine can only form σ (single bond, bond order 1). So per atom, oxygen ``buys'' twice the oxidation-state increment that fluorine buys.

1. Compare maximum oxidation states with O vs F:
   • Manganese: Mn2O7 at +7 vs MnF4 at +4.
   • Osmium: OsO4 at +8 vs OsF8 unknown, OsF7 at +7.
   • Iron: FeO4^2- at +6 vs FeF3 at +3.
2. The pattern: oxygen reaches higher because each M=O double bond counts 2 toward the oxidation state, and steric requirements are lower (fewer atoms around M).
3. Hence high electronegativity of F is irrelevant for stabilising high states; bonding multiplicity is the key.

Oxygen forms multiple (σ+π) bonds to metals, so each O atom contributes 2 to oxidation state while each F contributes only 1. This is why oxides reach higher oxidation states than fluorides.

Concept used. The pure spin-only formula μ = √n(n+2) accounts only for the spin contribution. For ions where the ground state has unquenched orbital angular momentum, the experimentally observed μ exceeds the spin-only prediction. The orbital contribution is significant when the ground-state t_2g subshell has asymmetric occupancy (i.e. not half-filled or fully-filled).

1. Configurations:
   • Cr³⁺: 3d³. In octahedral [Cr(H2O)6]³⁺ the t_2g³ is symmetric (each of the three t_2g orbitals has one electron, S=3/2, L=0). Orbital contribution quenched; μ ≈ _s.o. = √15 = 3.87 B.M. (matches experiment).
   • Co²⁺: 3d⁷. In octahedral [Co(H2O)6]²⁺, t_2g⁵ e_g². The t_2g⁵ subshell is asymmetric (one orbital pair, two singly-filled), so L ≠ 0. Orbital contribution not quenched; observed μ ≈ 4.87 B.M., higher than the spin-only √15 = 3.87 B.M.
2. The extra 1 B.M. in Co²⁺ is the orbital contribution.

Cr³⁺'s t_2g³ is symmetric (L=0, only spin contributes); Co²⁺'s t_2g⁵ is asymmetric (L ≠ 0, spin + orbital contribute), giving the extra ∼ 1 B.M.

Concept used. For lanthanoids the 4f electrons sit much closer to the nucleus (poorly shielded by the inner core), while for actinoids the 5f electrons extend further out (more shielded from the core and from each other). Consequently the outer 6s/5d of lanthanoids feel a higher effective nuclear charge than the 7s/6d of the corresponding actinoids – so lanthanoid IEs are higher.

1. Compare 5f vs 4f shielding: 5f orbitals have one extra radial node, so on average their electron density penetrates less to the nucleus. They shield the outer 6d/7s electrons more effectively.
2. Consequence: the outer electrons of actinoids are less tightly held than the outer electrons of the corresponding lanthanoids. Hence IE of Th, Pa, U < IE of Ce, Pr, Nd.
3. Numerical check (sum of first three IE in kJ/mol): Ce ≈ 3528, Pr ≈ 3531, Nd ≈ 3578; Th ≈ 2780, Pa ≈ 2790, U ≈ 2826. Lanthanoid trio is ∼ 700 kJ/mol higher.

The 5f electrons in actinoids penetrate less and shield the outer electrons better, so the outer electrons of actinoids are less firmly held than those of the lanthanoids; hence IE_1-3(Ce, Pr, Nd) > IE_1-3(Th, Pa, U).

Concept used. Chemical separation methods (precipitation, crystallisation, extraction) all rely on differences in physical and chemical properties between two ions. When two ions have nearly identical sizes, charges, and bonding preferences, no separation method works easily. Zr and Hf are exactly such a pair, thanks to lanthanoid contraction.

1. Compare metallic radii: r(Zr) = 160 pm, r(Hf) = 159 pm – essentially identical. Ionic radii of Zr⁴⁺ and Hf⁴⁺ are also nearly equal (∼ 84 pm vs ∼ 83 pm).
2. Consequence: Zr and Hf form indistinguishable oxides, halides, oxohalides, with almost identical solubilities, lattice energies, and reaction rates.
3. Hence ordinary fractional crystallisation, precipitation, etc. fail. Industrial separation uses solvent extraction with Bu_nP=O/methyl isobutyl ketone or ion-exchange chromatography – methods that exploit tiny differences in extraction equilibria.

Lanthanoid contraction makes r(Hf) ≈ r(Zr) (∼ 159–160 pm); their compounds are nearly indistinguishable, so separation is difficult.

Concept used. Lanthanoid +3 is the universal state; +4 or +2 shows up only when the resulting Lnⁿ⁺ has a stable closed-shell or half-filled 4f configuration. Ce neutral is [Xe] 4f¹ 5d¹ 6s². Removing four electrons gives Ce⁴⁺ = [Xe] – the noble-gas Xe core, which is simultaneously 4f⁰ (an empty stable subshell).

1. Ce → Ce⁴⁺: lose all four valence electrons (4f¹, 5d¹, 6s²). Resulting ion is [Xe], the closed-shell xenon-like configuration.
2. The extra stability of 4f⁰ explains why Ce⁴⁺ is observable in solution (orange/yellow) and is used as a common laboratory oxidant (E^∘(Ce⁴⁺/Ce³⁺) = +1.74 V in HClO4).
3. Other lanthanoids would need to ionise into the 4f shell to reach +4, costing too much energy without similar special-stability rewards (except Tb, where Tb⁴⁺ → 4f⁷ half-filled stable).

Ce⁴⁺ is stabilised because the configuration [Xe] (i.e. 4f⁰, the noble-gas/empty-subshell stability) is reached.

Concept used. KMnO4 is intensely purple because of LMCT involving Mn⁺⁷ (no d-electrons but strongly oxidising O ligands). When oxalic acid reduces Mn⁺⁷ to Mn⁺², the LMCT band shifts well out of the visible region. The product Mn²⁺ has d⁵ but its d–d transitions are Laporte- and spin-forbidden, giving only a very pale pink absorbance. Net visual effect: purple → colourless.

1. Write the redox equation in acidic medium: 2 MnO4^- + 5 C2O4^2- + 16 H+ -> 2 Mn²⁺ + 10 CO2 + 8 H2O. Mn drops from +7 to +2; C goes from +3 in oxalate to +4 in CO2.
2. Colour change: MnO4^- (deep purple, LMCT) → Mn²⁺ (very pale pink, d–d but spin-forbidden because all five electrons are unpaired in d⁵). At normal concentrations the solution appears colourless.
3. Net observable: titration end-point is reached when the last drop of KMnO4 is reduced (purple ceases to persist), the basis of redox titration of oxalic acid with permanganate.

KMnO4 (purple, Mn⁺⁷) is reduced by oxalic acid to Mn²⁺ (pale-pink/colourless); the colour fades because the strongly absorbing LMCT band of MnO4^- is gone.

Concept used. Dichromate Cr2O7^2- (orange) and chromate CrO4^2- (yellow) are in pH-dependent equilibrium: Cr2O7^2-(orange) + 2 OH^- <=> 2 CrO4^2-(yellow) + H2O. Adding OH^- shifts the equilibrium to chromate (yellow); adding H+ removes OH^- and shifts the equilibrium back to dichromate (orange). Cr stays at +6 throughout – this is acid–base chemistry, not redox.

1. Forward: Cr2O7^2- + 2 OH^- -> 2 CrO4^2- + H2O (orange to yellow on adding alkali).
2. Reverse: 2 CrO4^2- + 2 H+ -> Cr2O7^2- + H2O (yellow to orange on adding acid).
3. Le Chatelier shifts: alkali consumes H+ on the right, pushing equilibrium to chromate; acid consumes OH^- on the left, pushing equilibrium to dichromate.

The pair Cr2O7^2-CrO4^2- is a pH-driven equilibrium: alkali → yellow chromate, acid → orange dichromate.

Concept used. KMnO4's reduction pathway depends on the medium because the half-reaction (and the standard reduction potential) changes with pH. The three observable products correspond to Mn⁺² (acidic, colourless), Mn⁺⁴ (neutral, brown MnO2), and Mn⁺⁶ (alkaline, green MnO4^2-).

1. In acidic medium (pH < 7): 5e^- reduction, MnO4^- + 8 H+ + 5 e^- -> Mn²⁺ + 4 H2O. E^∘ = +1.51 V. Mn²⁺ (d⁵, μ high but d–d spin-forbidden) is essentially colourless in dilute solution.
2. In neutral medium (pH ≈ 7): 3e^- reduction, MnO4^- + 2 H2O + 3 e^- -> MnO2(s, brown) + 4 OH^-. Manganese settles as the brown precipitate MnO2.
3. In alkaline medium (pH > 7): 1e^- reduction, MnO4^- + e^- -> MnO4^2-(green). Mn drops only to +6, giving the green manganate ion.

Three stages: acidic → Mn²⁺ (colourless); neutral → MnO2 (brown ppt); alkaline → MnO4^2- (green).

Concept used. The second and third d-block rows (4d and 5d) span the lanthanoid block in between, which contracts the 5d atomic radii so much that they end up nearly equal to the corresponding 4d radii. The first (3d) row has no such contraction acting on it, so its radii are visibly smaller. Result: 4d and 5d rows have nearly identical sizes (and therefore nearly identical chemistry), whereas 3d stands apart.

1. Compare ``Zr–Hf'' (group 4) radii: 160 pm vs 159 pm. Compare with Ti (group 4, 3d): 147 pm – a ∼ 13 pm gap.
2. Similarly Nb–Ta (group 5): 146–146 pm; V (3d): 134 pm. Mo–W: 139–139 pm; Cr (3d): 128 pm. In every group, 4d and 5d members have nearly equal radii (lanthanoid-contraction-driven) while 3d is smaller.
3. Almost-equal radii imply almost-equal lattice and hydration energies, bond lengths, electronegativities, ionisation energies – hence very similar chemistry between 4d and 5d congeners.

Lanthanoid contraction makes r(5d) ≈ r(4d) > r(3d), so 4d and 5d rows behave as near-twins while the 3d row stays distinct.

Concept used. The reduction potential E^∘(M²⁺/M) decomposes into (i) atomisation enthalpy _aH, (ii) sum of ionisation enthalpies ∑ IE_1,2 and (iii) hydration enthalpy of M²⁺. For Zn the high hydration enthalpy and the stability of the Zn²⁺ state (losing the 4s² pair to reach d¹⁰ closed shell) make E^∘ strongly negative. For Cu the analogous transformation requires opening the d¹⁰ shell of Cu⁺ (high IE₂), which is not adequately compensated by hydration enthalpy, leaving E^∘ positive.

1. Zn neutral 4s² 3d¹⁰ → Zn²⁺ 3d¹⁰: lose only the 4s pair, reach a stable closed shell. Easy ionisation, strong hydration. Strongly negative E^∘ = -0.76 V (Zn is a good reductant).
2. Cu neutral 4s¹ 3d¹⁰ → Cu²⁺ 3d⁹: lose the lone 4s and a d electron, breaking the closed d¹⁰ shell. High second IE (IE₂(Cu) = 1958 kJ/mol). Hydration of Cu²⁺ is exothermic but not enough. Net: E^∘ = +0.34 V (Cu is a poor reductant, non-displacer of H2).
3. Two factors: _aH(Cu) = 339 kJ/mol is also higher than _aH(Zn) = 130 kJ/mol, hurting Cu further.

Zn loses two 4s electrons (closed d¹⁰ retained) with low IE and high hydration ⇒ negative E^∘. Cu has to break 3d¹⁰ to reach Cu²⁺ (high IE₂, high _aH) which hydration cannot compensate ⇒ positive E^∘.

Concept used. Fajans' rules say that a bond becomes more covalent if (i) the cation is small, (ii) the cation has a high charge, or (iii) the anion is large and polarisable. As a transition metal's oxidation state increases, the metal ion's size shrinks and charge rises – both push it from ionic toward covalent bonding with the same halide.

1. Track ionic radii of Mn across oxidation states (pm): Mn⁺² 83, Mn⁺³ 65, Mn⁺⁴ 53, Mn⁺⁷ 25. Charge density z/r grows steeply with n.
2. Apply Fajans: at high z/r the cation polarises the halide electron cloud heavily, distorting it toward the cation and sharing electron density ⇒ covalent character.
3. Example series: MnCl2 (ionic, pink solid), MnCl3 (more covalent, unstable), Mn2O7 (covalent, oily liquid). Each step up in oxidation state increases covalent character.

Higher oxidation state ⇒ smaller, more charged metal cation ⇒ greater polarisation of the halide (Fajans) ⇒ more covalent bonding.

Concept used. The order in which orbitals fill a neutral atom is fixed by the n+ rule (Madelung/Klechkowski): the orbital with smaller n+ is filled first; ties are broken by smaller n. For ionisation we instead look at the actual orbital energies of the cation: 4s is higher in energy than 3d in the ion because adding d electrons also raises 4s via electron-electron repulsion.

1. Apply the n+ rule to the neutral atom: 4s: n+= 4+0 = 4; 3d: n+= 3+2 = 5. Filling order: 4s before 3d. So Sc neutral is [Ar] 3d¹ 4s².
2. After the 3d orbital is populated, screening between 3d and 4s rearranges; 4s becomes higher in energy in the cation. The first electrons to leave are therefore the 4s ones. E.g. Fe²⁺ is [Ar] 3d⁶ (lost two 4s electrons), not [Ar] 3d⁴ 4s².
3. This is sometimes phrased as ``electrons fill 4s before 3d but ionise from 4s first''. The underlying physics is that the 4s/3d energy levels swap order between the neutral atom (without all 3d electrons) and the cation (with most 3d electrons).

Filling order (n+ rule): 4s < 3d. Ionisation order (cation energies): 4s is destabilised by 3d repulsion and leaves first.

Concept used. ``Reactivity'' of a metal at this level means its tendency to undergo oxidation – to be the reductant. That tendency is controlled by ionisation enthalpy: lower IE means easier oxidation means greater reactivity. Across the 3d row, IE rises roughly regularly (because the effective nuclear charge on the valence shell grows). Hence reactivity falls from Sc to Cu.

1. Sum of first two IE (kJ/mol) across the row: Sc 1866, Ti 1969, V 2059, Cr 2241, Mn 2226, Fe 2320, Co 2407, Ni 2490, Cu 2703, Zn 2639. The general trend is upward (with small kinks at Mn and Zn).
2. Reactivity (e.g. ability to liberate H2 from dilute HCl) tracks -E^∘(M²⁺/M). Empirically: Sc reacts vigorously with dilute acid; Ti, V, Cr react more slowly; Mn, Fe, Co, Ni need warming; Cu does not react at all (positive E^∘).
3. Hence the regular decrease in reactivity from Sc to Cu.

IE rises across Sc → Cu (regularly with small kinks), making oxidation progressively harder; reactivity falls accordingly.

Concept used. Each catalyst is matched to the canonical industrial / textbook process it accelerates.

1. (i) Ni / H2 → (c) Vegetable oil to ghee. Heterogeneous hydrogenation of C=C in vegetable oils to give saturated fats (margarine, vanaspati).
2. (ii) Cu2Cl2 → (d) Sandmeyer reaction. Diazonium salts ArN2+ converted to ArCl, ArBr, ArCN via Cu(I) halide catalysis.
3. (iii) V2O5 → (b) Contact process. 2 SO2 + O2 V2O5 2 SO3, the key step in industrial H2SO4 manufacture.
4. (iv) Finely divided iron → (e) Haber's process. N2 + 3 H2 Fe 2 NH3 at high T, P.
5. (v) TiCl4 + Al(CH3)3 → (a) Ziegler–Natta catalyst. Stereospecific polymerisation of ethene and propene.

(i)→(c), (ii)→(d), (iii)→(b), (iv)→(e), (v)→(a).

Concept used. Lanthanoid metals and their oxides have a few canonical applications: phosphors in colour-TV and X-ray screens, catalysts for FCC (petroleum cracking), and pyrophoric alloys (misch metal) in lighter flints and tracer bullets.

1. (i) Lanthanoid oxide (e.g. Y2O3 doped with Eu) → (b) Television (CRT) screen phosphors.
2. (ii) Lanthanoid metal → (a) Production of iron alloys.
3. (iii) Misch metal (Ce ∼ 50%, La, Nd, Pr + ∼ 5% Fe) → (d) ``Lanthanoid metal + iron'' composition.
4. (iv) Magnesium-based alloy with misch metal → (e) Bullets (tracer ammunition).
5. (v) Mixed oxides of lanthanoids → (c) Petroleum cracking.

(i)→(b), (ii)→(a), (iii)→(d), (iv)→(e), (v)→(c).

Concept used. Three textbook records of the d-block – the only element to reach +8 (Os, in OsO4), the 3d element reaching +7 (Mn, in KMnO4/Mn2O7), and the 3d element with the highest melting point (Cr).

1. (i) +8 oxidation state: only Os and Ru reach +8 (OsO4 and RuO4). Among the options, Os. Match (c).
2. (ii) +7 in 3d block: Mn⁺⁷ in MnO4^- and Mn2O7. Match (a).
3. (iii) Highest melting point in 3d: Cr at 1907 ^∘C. Match (b).

(i)→(c), (ii)→(a), (iii)→(b).

Concept used. Compute or recall the oxidation states.

1. (i) MnO2: Mn + 2(-2) = 0 ⇒ Mn = +4. Match (c).
2. (ii) Most stable Mn oxidation state: +2 (special d⁵ stability). Match (a).
3. (iii) Most stable Mn oxidation state in oxides: +7 (Mn2O7). Match (e).
4. (iv) Characteristic oxidation state of lanthanoids: +3. Match (b).

(i)→(c), (ii)→(a), (iii)→(e), (iv)→(b).

Concept used. Aqueous-solution colours of first-row transition-metal hexaaqua complexes are textbook facts.

1. (i) FeSO4.7H2O: [Fe(H2O)6]²⁺, d⁶ high-spin, pale green. Match (d).
2. (ii) NiCl2.4H2O: [Ni(H2O)6]²⁺, d⁸, intense green. Match (a).
3. (iii) MnCl2.4H2O: [Mn(H2O)6]²⁺, d⁵ high-spin, faint pink (spin-forbidden). Match (b).
4. (iv) CoCl2.6H2O: [Co(H2O)6]²⁺, d⁷, rich pink. Match (e).
5. (v) Cu2Cl2: Cu⁺, d¹⁰, colourless. Match (f).

(i)→(d), (ii)→(a), (iii)→(b), (iv)→(e), (v)→(f).

Concept used. Specific lanthanoid identities by configuration.

1. (i) +4: Ce⁴⁺ (4f⁰). Match (b).
2. (ii) +2: Eu²⁺ (4f⁷). Match (d).
3. (iii) Radioactive: Pm (no stable isotope). Match (a).
4. (iv) 4f⁷ in +3: Gd³⁺ is [Xe] 4f⁷. Match (e).
5. (v) 4f¹⁴ in +3: Lu³⁺ is [Xe] 4f¹⁴. Match (c).

(i)→(b), (ii)→(d), (iii)→(a), (iv)→(e), (v)→(c).

Concept used. Four metallic-property records.

1. (i) Highest IE₂: Cu (1958 kJ/mol); breaking Cu⁺'s 3d¹⁰ shell. Match (c).
2. (ii) Highest IE₃: Zn (3833 kJ/mol); breaking Zn²⁺'s 3d¹⁰ shell. Match (d).
3. (iii) M(CO)6: 18-electron rule ⇒ M needs 18 - 12 = 6 valence electrons ⇒ group 6 = Cr. Match (b).
4. (iv) Highest heat of atomisation: among options (Co, Cr, Cu, Zn, Ni), Co has _aH ≈ 425 kJ/mol – the highest in the listed set (NCERT key). Match (a).

(i)→(c), (ii)→(d), (iii)→(b), (iv)→(a).

Correct option: (i) Both A and R true; R correctly explains A.

Concept used. CuI2 is unstable – mixing Cu²⁺ with I^- leads to internal redox (Q36): Cu⁺ precipitates as Cu2I2 while iodide oxidises to I2.

1. Net reaction: 2 Cu²⁺ + 4 I^- -> Cu2I2 + I2. CuI2 never crystallises.
2. A and R both true; R is the cause of A. Hence option (i).

Option (i): both true, reason correctly explains assertion.

Correct option: (ii) Both true; R is not the correct explanation.

Concept used. A: true (lanthanoid contraction, Q42). R: true (Zr, Hf are both group-4). But ``same group'' alone doesn't imply inseparability – Ti–Zr (also group 4) are easily separable. The real reason is lanthanoid contraction making r(Hf) ≈ r(Zr).

1. Check A: true. Check R: true.
2. Test causation: Ti–Zr (same group) are separable, so ``same group'' isn't sufficient for inseparability.
3. Hence option (ii).

Option (ii).

Correct option: (iii) Assertion is not true but reason is true.

Concept used. Actinoids actually form more stable complexes than lanthanoids – precisely because 5f and 6d participate in bonding (the reason given).

1. Assertion A: false. Actinoid complexes are typically more stable than lanthanoid analogues; they show wider variety of complexes too.
2. Reason R: true. 5f orbitals are spatially extended (unlike the buried 4f), and they overlap with 6d, both contributing to M–L bonding.
3. A false + R true ⇒ option (iii).

Option (iii).

Correct option: (i) Both true; reason correctly explains assertion.

Concept used. E^∘(Cu²⁺/Cu) = +0.34 V is positive; Cu cannot reduce H+.

1. A: ``Cu cannot liberate H2.'' True.
2. R: ``Cu has positive E^∘.'' True.
3. R causes A directly via E^∘_cell = -0.34 V < 0.

Option (i).

Correct option: (ii) Both true; R is not the complete explanation of A.

Concept used. A: true (OsO4 shows Os⁺⁸). R: true (Os is 5d). But not every 5d element reaches +8. The full reason combines 5d-row with Os's group-8 position (8 valence electrons).

1. Check A: OsO4 exists. True.
2. Check R: Os is 5d⁶ 6s². True.
3. Does R explain A? Only partially – group-8 valence count is the missing piece. Hence option (ii).

Option (ii).

Concept used. Trace each arrow of the flowchart and identify what reaction it represents. The flowchart shows two branches off CuCO3: a left branch generating CuO then Cu metal (A), then Cu(NO₃)₂ (B), then the tetraamminecopper(II) complex (C, blue); and a right branch producing CO2 (D), which gives milky CaCO3 (E) with Ca(OH)2 and then a clear Ca(HCO3)2 solution on adding excess CO2.

1. Thermal decomposition of CuCO3: CuCO3 Δ CuO + CO2 ^. CO2 is the right-hand product (D).
2. Left branch: reduction of CuO with CuS to give A = Cu metal: 2 CuO + CuS -> 3 Cu + SO2 ^. This is the self-reduction step used industrially in copper smelting.
3. B = Cu(NO3)2: oxidation of A by conc. HNO3: Cu + 4 HNO3(conc.) -> Cu(NO3)2 + 2 NO2 + 2 H2O. Cu²⁺ is the blue cupric ion.
4. C = [Cu(NH3)4]²⁺: complexation of Cu²⁺ with aqueous NH3 gives the deep-blue tetraamminecopper(II) ion: Cu(NO3)2 + 4 NH3 -> [Cu(NH3)4](NO3)2. The intense blue colour matches the ``blue solution'' label.
5. Right branch: D = CO2, then with limewater: Ca(OH)2 + CO2 -> CaCO3 v + H2O, giving the white precipitate E = CaCO3 (``milky'').
6. Further reaction with excess CO2: CaCO3 + CO2 + H2O -> Ca(HCO3)2. Ca(HCO3)2 is soluble, so the ``clear solution''.

A = Cu; B = Cu(NO3)2; C = [Cu(NH3)4](NO3)2 (deep-blue); D = CO2; E = CaCO3.