NCERT Solutions · Class 12 Chemistry The d-and f-Block Elements

Paramagnetism, high atomisation enthalpy, colour, and catalytic power in transition metals all trace back to the same feature: a partly filled d subshell.(i) Transition metals and many of their compounds show paramagnetic behaviourA substance is paramagnetic when its atoms or ions carry unpaired electrons — each unpaired electron acts as a tiny magnet (from its spin), and an external magnetic field aligns these to give a net attraction into the field.Transition metal atoms and ions have partly filled \(\displaystyle (n-1)d \) orbitals. Because the five d orbitals are degenerate, electrons occupy them singly before pairing (Hund's rule), so most transition-metal species end up with unpaired electrons. The number of unpaired electrons, \(\displaystyle n \), sets the magnetic moment through the spin-only formula \[\mu = \sqrt{n(n+2)} \ \text{BM} \] where \(\displaystyle \mu \) is the magnetic moment in Bohr magnetons (BM) and \(\displaystyle n \) is the number of unpaired electrons. For example \(\displaystyle \mathrm{Ti^{3+}} \) (\(\displaystyle 3d^1\)) has \(\displaystyle n=1 \) and \(\displaystyle \mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73\ \text{BM} \), while \(\displaystyle \mathrm{Mn^{2+}} \) (\(\displaystyle 3d^5\), all five d electrons unpaired) has \(\displaystyle n=5 \) and \(\displaystyle \mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92\ \text{BM} \). Only ions with a completely filled (\(\displaystyle d^{10}\), e.g. \(\displaystyle \mathrm{Zn^{2+}} \)) or completely empty (\(\displaystyle d^0\), e.g. \(\displaystyle \mathrm{Sc^{3+}} \)) d subshell have no unpaired electrons and are diamagnetic — this is the exception people forget, not the rule.(ii) The enthalpies of atomisation of the transition metals are highEnthalpy of atomisation is the energy needed to convert one mole of metal in the solid state into free gaseous atoms — it is a direct measure of the strength of metallic bonding holding the lattice together.In transition metals, both the \(\displaystyle (n-1)d \) electrons and the \(\displaystyle ns \) electrons are available for metallic bonding, not just the outer \(\displaystyle s \) electrons as in the s-block. A larger number of unpaired electrons means a larger number of electrons can take part in the delocalised interatomic bonding across the lattice, so the metallic bond is stronger. This is why atomisation enthalpies rise from Sc towards the middle of a series (peaking around Cr/Mo/W, where the number of unpaired d electrons is at a maximum) and fall again toward the end of the series as pairing sets in — the trend tracks the count of unpaired d electrons, not atomic mass or size.(iii) The transition metals generally form coloured compoundsColour arises when a compound absorbs specific wavelengths of visible light and the eye perceives the complementary, transmitted colour.In an isolated transition-metal ion the five d orbitals are degenerate, but when the ion is surrounded by ligands (in a salt or complex), the electric field of the ligands splits the d orbitals into two sets of different energy — this is crystal field splitting, with the energy gap denoted \(\displaystyle \Delta_o \) (octahedral) or \(\displaystyle \Delta_t \) (tetrahedral). If the d subshell is partly filled, an electron can absorb a photon of visible light and jump from the lower-energy set of d orbitals to the higher-energy set — a d–d transition. The wavelength absorbed corresponds to \(\displaystyle \Delta E = h\nu \), and the colour seen is the complement of the absorbed wavelength. A \(\displaystyle d^0\) ion (e.g. \(\displaystyle \mathrm{Sc^{3+}}, \mathrm{Ti^{4+}} \)) or a \(\displaystyle d^{10}\) ion (e.g. \(\displaystyle \mathrm{Zn^{2+}}, \mathrm{Cu^{+}} \)) has no such transition available and is colourless — again the partly-filled d subshell, not the metal identity, is the deciding factor.(iv) Transition metals and their many compounds act as good catalystsTwo features of transition metals together make them effective catalysts: Answer: (i) unpaired electrons in the partly filled \(\displaystyle (n-1)d\) subshell give a net magnetic moment \(\displaystyle \mu=\sqrt{n(n+2)} \) BM, causing paramagnetism; (ii) both \(\displaystyle ns\) and \(\displaystyle (n-1)d\) electrons take part in metallic bonding, and a larger number of unpaired d electrons strengthens this bonding, raising the atomisation enthalpy; (iii) partly filled d orbitals allow ligand-field-split d–d electronic transitions that absorb visible light, producing colour; (iv) variable oxidation states (providing low-energy alternate reaction pathways) and the ability of partly filled d orbitals to adsorb and activate reactants at the metal surface make transition metals good catalysts.

Interstitial compounds form when small atoms occupy the empty spaces (voids) inside a metal's crystal lattice, without destroying that lattice.What interstitial compounds areWhen small atoms such as hydrogen (H), carbon (C), nitrogen (N), or boron (B) are trapped inside the crystal lattice of a metal, the resulting solids are called interstitial compounds. The word "interstitial" refers to the empty spaces — voids — that lie between the metal atoms packed in a crystal lattice. Instead of replacing a metal atom, the small atom slips into one of these gaps.Examples: \(\displaystyle \text{TiC} \), \(\displaystyle \text{ZrH}_{1.92} \), \(\displaystyle \text{Mn}_4\text{N} \), \(\displaystyle \text{Fe}_3\text{H} \), tungsten carbide (\(\displaystyle \text{WC} \)).These are not ordinary ionic or covalent compounds in the usual sense — they don't obey simple stoichiometric ratios (hence formulas like \(\displaystyle \text{ZrH}_{1.92} \) with a non-integer subscript), because the number of small atoms that can squeeze into the voids need not correspond to a fixed whole-number ratio with the metal atoms.Why transition metals are especially known for thisTransition metal atoms pack together in close-packed crystal structures (hcp or ccp) or in body-centred cubic arrangements. Close packing of spheres necessarily leaves small gaps between them — the octahedral and tetrahedral voids. In transition metals these voids are large enough to hold small atoms like H, C, N, or B comfortably, yet the transition metal atoms themselves are large enough that fitting a small atom into a void does not force the lattice to distort or break apart.This is the point to be careful about: the small atoms are not chemically bonded to specific metal atoms in a normal valence sense — they are physically accommodated in the geometric gaps of an existing metallic lattice. That is exactly why the compositions come out non-stoichiometric.Properties that follow from this structureBecause the underlying metallic lattice survives largely intact, interstitial compounds keep metallic characteristics while gaining new ones from the trapped atoms: These properties — extreme hardness plus retained metallic conductivity — are the practical signature that identifies a compound as interstitial, and they follow directly from the small atom sitting in a void of an otherwise-normal transition-metal lattice rather than substituting into it or bonding into a totally new structure.**Answer: Interstitial compounds are formed when small atoms (H, C, N, B) occupy the interstitial voids in a metal's crystal lattice; they are non-stoichiometric (e.g., \(\displaystyle \text{TiC} \), \(\displaystyle \text{ZrH}_{1.92} \), \(\displaystyle \text{Mn}_4\text{N} \)) and typically show high melting points, extreme hardness, retained metallic conductivity, and chemical inertness. Transition metals are well known for forming them because their crystal lattices (close-packed or bcc) contain octahedral/tetrahedral voids large enough to trap these small atoms without disrupting the metallic structure.

Transition metals lose electrons from two sub-shells of nearly equal energy, one at a time, so their oxidation states climb in steps of $\displaystyle 1$; non-transition (representative) metals lose or keep a whole \(\displaystyle ns^2\) pair at once, so their oxidation states differ in steps of $\displaystyle 2$ — the inert-pair effect.Why transition metals show many, closely-spaced statesIn a transition atom the \(\displaystyle (n-1)d\) and \(\displaystyle ns\) orbitals lie so close in energy that electrons from both are available for bond formation. Ionisation removes the \(\displaystyle ns\) electrons first, and then the \(\displaystyle (n-1)d\) electrons come off one by one — so every removal adds exactly $\displaystyle 1$ to the oxidation number, and a single element can display almost every state in a run.Manganese, \(\displaystyle 3d^{5}4s^{2}\), is the clearest case: it shows +$\displaystyle 2$, +$\displaystyle 3$, +$\displaystyle 4$, +$\displaystyle 6$ and +$\displaystyle 7$ —\[\text{Mn}^{2+}\ (3d^{5}) \to \text{Mn}_2\text{O}_3\ (+3) \to \text{MnO}_2\ (+4) \to \text{MnO}_4^{2-}\ (+6) \to \text{KMnO}_4\ (+7) \]each step simply one more \(\displaystyle d\)-electron given up. Iron (\(\displaystyle 3d^{6}4s^{2}\)) shows +$\displaystyle 2$ and +$\displaystyle 3$; chromium shows +$\displaystyle 2$, +$\displaystyle 3$, +$\displaystyle 6$; copper (\(\displaystyle 3d^{10}4s^{1}\)) shows +$\displaystyle 1$ and +$\displaystyle 2$ — always unit-by-unit. This is also why the number of oxidation states rises across the first half of a series and peaks near the middle (Mn), where the most \(\displaystyle d\)- and \(\displaystyle s\)-electrons together are loosely enough held to be removable, then falls off toward Cu and Zn, where the stable \(\displaystyle d^{10}\) configuration resists further ionisation.Why non-transition metals jump by $\displaystyle 2$A representative \(\displaystyle p\)-block metal's valence shell is only \(\displaystyle ns^{2}np^{x}\); there is no partly-filled \(\displaystyle d\) or \(\displaystyle f\) shell in between to buffer the energy gap. Its highest oxidation state — the group oxidation number — comes from losing all the \(\displaystyle ns^{2}np^{x}\) electrons. The only other state available comes from losing just the \(\displaystyle np\) electrons and leaving the \(\displaystyle ns^{2}\) pair untouched, because poor shielding by inner \(\displaystyle d\)/\(\displaystyle f\) electrons pulls that pair in tightly and makes it reluctant to ionise or bond (the inert-pair effect). Since a full pair is either given up or kept, the two available states are always the group number and (group number \(\displaystyle -2\)) — a gap of $\displaystyle 2$, never 1.Group $\displaystyle 14$ shows this directly:\[\text{Sn: } +2\ (\text{SnCl}_2)\ \text{and}\ +4\ (\text{SnCl}_4), \qquad \text{Pb: } +2\ (\text{PbO})\ \text{and}\ +4\ (\text{PbO}_2) \]and Group $\displaystyle 13$:\[\text{Tl: } +1\ (\text{Tl}_2\text{O})\ \text{and}\ +3\ (\text{Tl}_2\text{O}_3) \]In every one of these pairs the two oxidation numbers are exactly $\displaystyle 2$ apart, and there is no intermediate state, unlike Mn's unbroken run of +$\displaystyle 2$ through +7.The stability trend also differs. Going down a transition series, no single simple rule governs which oxidation state is favoured — the highest states (Mn\(\displaystyle ^{VII}\), Cr\(\displaystyle ^{VI}\)) are strongest around the middle of a period and weaken toward either end. But going down a p-block group, the inert-pair effect strengthens steadily, so the lower of the two states becomes progressively more stable: \(\displaystyle \mathrm{Ge^{IV}}\) is more stable than \(\displaystyle \mathrm{Ge^{II}}\), while by the time you reach Pb, \(\displaystyle \mathrm{Pb^{II}}\) is more stable than \(\displaystyle \mathrm{Pb^{IV}}\) — the opposite ranking, purely because the inert \(\displaystyle 6s^{2}\) pair is held ever more tightly by the poorly-shielded nuclear charge of the heavier atom.Answer: Transition metals show a larger number of oxidation states that differ from each other by $\displaystyle 1$ (e.g., Mn: +$\displaystyle 2$, +$\displaystyle 3$, +$\displaystyle 4$, +$\displaystyle 6$, +$\displaystyle 7$), because both the \(\displaystyle (n-1)d\) and \(\displaystyle ns\) electrons are removed one at a time from comparable-energy orbitals. Non-transition metals show at most two oxidation states, differing by $\displaystyle 2$ (e.g., Sn: +$\displaystyle 2$, +$\displaystyle 4$; Pb: +$\displaystyle 2$, +$\displaystyle 4$; Tl: +$\displaystyle 1$, +$\displaystyle 3$), because the inert \(\displaystyle ns^2\) pair is either fully retained or fully lost (inert-pair effect), with no intermediate step.

Potassium dichromate is not pulled straight out of the ore — it is built in three separate stages, and the last stage works only because K₂Cr₂O₇ is LESS soluble than the sodium salts sitting next to it in solution.Step $\displaystyle 1$ — Fuse the ore to convert chromium(III) to chromate. The ore is iron chromite, \(\displaystyle \mathrm{FeCr_2O_4} \) (also written \(\displaystyle \mathrm{FeO\cdot Cr_2O_3} \)). It is fused with sodium carbonate in air. The oxygen oxidises chromium from the +$\displaystyle 3$ state in the ore to the +$\displaystyle 6$ state in chromate, while iron ends up as \(\displaystyle \mathrm{Fe_2O_3} \): \[4\,\mathrm{FeCr_2O_4} + 8\,\mathrm{Na_2CO_3} + 7\,\mathrm{O_2} \;\xrightarrow{\ \text{fuse}\ }\; 8\,\mathrm{Na_2CrO_4} + 2\,\mathrm{Fe_2O_3} + 8\,\mathrm{CO_2} \] This gives a yellow melt. It is leached with water, and the insoluble \(\displaystyle \mathrm{Fe_2O_3} \) is filtered off, leaving a yellow solution of sodium chromate.Step $\displaystyle 2$ — Acidify to convert chromate to dichromate. The filtrate (sodium chromate solution) is acidified with sulphuric acid. Acidifying pushes the chromate–dichromate equilibrium (see below) toward dichromate: \[2\,\mathrm{Na_2CrO_4} + \mathrm{H_2SO_4} \;\longrightarrow\; \mathrm{Na_2Cr_2O_7} + \mathrm{Na_2SO_4} + \mathrm{H_2O} \] The solution is now orange, containing sodium dichromate.Step $\displaystyle 3$ — Swap sodium for potassium using a solubility difference, not a reaction with a driving force of its own. This is the step people get wrong: it looks like a normal double-displacement reaction, but the real reason it works is that sodium dichromate is considerably more soluble in water than potassium dichromate. When the sodium dichromate solution is treated with potassium chloride, \[\mathrm{Na_2Cr_2O_7} + 2\,\mathrm{KCl} \;\longrightarrow\; \mathrm{K_2Cr_2O_7}\!\downarrow \;+\; 2\,\mathrm{NaCl} \] potassium dichromate, being the less soluble of the two salts, crystallises out of the concentrated solution as orange crystals, while sodium chloride stays dissolved. This is a fractional crystallisation, driven purely by the solubility gap — cool the solution and the less-soluble salt comes out first.Effect of increasing pH on a dichromate solution. In water, chromate ion (yellow, \(\displaystyle \mathrm{CrO_4^{2-}} \)) and dichromate ion (orange, \(\displaystyle \mathrm{Cr_2O_7^{2-}} \)) are not two different substances sitting side by side — they are the same chromium(VI) species in a pH-dependent equilibrium: \[2\,\mathrm{CrO_4^{2-}}\,(\text{yellow}) \;+\; 2\,\mathrm{H^+} \;\rightleftharpoons\; \mathrm{Cr_2O_7^{2-}}\,(\text{orange}) \;+\; \mathrm{H_2O} \] Reading this by Le Chatelier's principle: \(\displaystyle \mathrm{H^+} \) is a reactant on the chromate side. Raising the pH means lowering \(\displaystyle [\mathrm{H^+}] \), so the equilibrium is pulled to the left — dichromate is converted back into chromate, and the solution turns from orange to yellow. (Conversely, lowering the pH, i.e. adding acid, pushes the equilibrium to the right and turns a yellow chromate solution orange — this is exactly what Step $\displaystyle 2$ above uses.)Answer: K₂Cr₂O₇ is made by (i) fusing chromite ore, \(\displaystyle \mathrm{FeCr_2O_4} \), with \(\displaystyle \mathrm{Na_2CO_3} \) in air to give sodium chromate, (ii) acidifying with \(\displaystyle \mathrm{H_2SO_4} \) to convert it to sodium dichromate, and (iii) treating the sodium dichromate solution with KCl, from which the less soluble K₂Cr₂O₇ crystallises out (fractional crystallisation) while NaCl remains in solution. Increasing the pH of a dichromate solution shifts the equilibrium \(\displaystyle 2\,\mathrm{CrO_4^{2-}} + 2\,\mathrm{H^+} \rightleftharpoons \mathrm{Cr_2O_7^{2-}} + \mathrm{H_2O} \) to the left, converting the orange dichromate ion into the yellow chromate ion.

Potassium permanganate is built in two stages: first oxidise manganese from +$\displaystyle 4$ (in ore) to +$\displaystyle 6$, then oxidise +$\displaystyle 6$ to +$\displaystyle 7$ — the second step is a disproportionation, not a simple oxidation.PreparationStage $\displaystyle 1$ — the ore pyrolusite, \(\displaystyle MnO_2 \), is fused with solid KOH and an oxidising agent (KNO₃, or simply air) to give potassium manganate, \(\displaystyle K_2MnO_4 \), which is dark green: \[2MnO_2 + 4KOH + O_2 \xrightarrow{\text{fuse}} 2K_2MnO_4 + 2H_2O \] Here manganese goes from \(\displaystyle +4 \) (in \(\displaystyle MnO_2 \)) to \(\displaystyle +6 \) (in \(\displaystyle MnO_4^{2-} \)).Stage $\displaystyle 2$ — the green manganate solution is oxidised to purple permanganate. Industrially this is done electrolytically: manganate ion is oxidised at the anode, \[MnO_4^{2-} \longrightarrow MnO_4^{-} + e^{-} \quad (\text{anode}) \] with \(\displaystyle H_2 \) liberated at the cathode. Alternatively, passing chlorine or ozone through the alkaline manganate solution does the same job chemically: \[2K_2MnO_4 + Cl_2 \longrightarrow 2KMnO_4 + 2KCl \] In neutral or acidic solution manganate also disproportionates on its own — \(\displaystyle Mn(+6) \) splits into \(\displaystyle Mn(+7) \) and \(\displaystyle Mn(+4) \): \[3MnO_4^{2-} + 4H^{+} \longrightarrow 2MnO_4^{-} + MnO_2 + 2H_2O \] The commercial salt is crystallised out and purified by recrystallisation.Reactions of acidified \(\displaystyle KMnO_4 \)Acidified permanganate is a strong oxidising agent because \(\displaystyle Mn \) is at its highest oxidation state, \(\displaystyle +7 \); in acid solution it is reduced all the way to \(\displaystyle Mn^{2+} \) (colourless), a $\displaystyle 5$-electron change per \(\displaystyle MnO_4^{-} \): \[MnO_4^{-} + 8H^{+} + 5e^{-} \longrightarrow Mn^{2+} + 4H_2O \] Do not confuse this with the reduction in neutral/faintly alkaline solution, which stops at \(\displaystyle MnO_2 \) (a $\displaystyle 3$-electron change) — the acidified pathway used here always goes to \(\displaystyle Mn^{2+} \).To get the full ionic equation for each reactant, pair this half-reaction with the reactant's own oxidation half-reaction, scaling both so electrons lost equal electrons gained.(i) With iron(II) ionsIron(II) is oxidised to iron(III), a $\displaystyle 1$-electron change: \[Fe^{2+} \longrightarrow Fe^{3+} + e^{-} \] Multiply this by $\displaystyle 5$ to match the $\displaystyle 5$ electrons the permanganate half-reaction needs, then add: \[MnO_4^{-} + 5Fe^{2+} + 8H^{+} \longrightarrow Mn^{2+} + 5Fe^{3+} + 4H_2O \](ii) With sulfur dioxide\(\displaystyle SO_2 \) dissolved in water is oxidised to sulfate, a $\displaystyle 2$-electron change: \[SO_2 + 2H_2O \longrightarrow SO_4^{2-} + 4H^{+} + 2e^{-} \] To balance $\displaystyle 2$ electrons here against $\displaystyle 5$ in the manganese half-reaction, multiply the \(\displaystyle SO_2 \) equation by $\displaystyle 5$ and the \(\displaystyle MnO_4^- \) equation by $\displaystyle 2$ (giving a common $\displaystyle 10$ electrons), then add and cancel the \(\displaystyle H^+ \) and \(\displaystyle H_2O \) that appear on both sides: \[2MnO_4^{-} + 5SO_2 + 2H_2O \longrightarrow 2Mn^{2+} + 5SO_4^{2-} + 4H^{+} \](iii) With oxalic acidOxalate ion is oxidised to carbon dioxide, a $\displaystyle 2$-electron change per oxalate (each carbon goes from \(\displaystyle +3 \) to \(\displaystyle +4 \)): \[C_2O_4^{2-} \longrightarrow 2CO_2 + 2e^{-} \] As in part (ii), $\displaystyle 2$ electrons per oxalate against $\displaystyle 5$ per permanganate means multiplying by $\displaystyle 5$ and $\displaystyle 2$ respectively ($\displaystyle 10$ electrons common), then adding: \[2MnO_4^{-} + 5C_2O_4^{2-} + 16H^{+} \longrightarrow 2Mn^{2+} + 10CO_2 + 8H_2O \] (This is the titration used to standardise \(\displaystyle KMnO_4 \) against oxalic acid — the disappearance of the purple colour marks the end point, since \(\displaystyle Mn^{2+} \) is colourless.)Answer: \(\displaystyle KMnO_4 \) is made by fusing \(\displaystyle MnO_2 \) with KOH and air/KNO₃ to \(\displaystyle K_2MnO_4 \), then oxidising this (electrolytically, or with \(\displaystyle Cl_2 \)/ \(\displaystyle O_3 \), or by disproportionation) to \(\displaystyle KMnO_4 \). Acidified \(\displaystyle KMnO_4 \) reacts as: (i) \(\displaystyle MnO_4^{-} + 5Fe^{2+} + 8H^{+} \rightarrow Mn^{2+} + 5Fe^{3+} + 4H_2O \); (ii) \(\displaystyle 2MnO_4^{-} + 5SO_2 + 2H_2O \rightarrow 2Mn^{2+} + 5SO_4^{2-} + 4H^{+} \); (iii) \(\displaystyle 2MnO_4^{-} + 5C_2O_4^{2-} + 16H^{+} \rightarrow 2Mn^{2+} + 10CO_2 + 8H_2O \).

A more positive \(\displaystyle E^\circ\) for the \(\displaystyle M^{3+}/M^{2+}\) couple means \(\displaystyle M^{3+}\) is easier to reduce — so stability of the +$\displaystyle 3$ state runs OPPOSITE to the size of \(\displaystyle E^\circ\), not the same way.(i) Stability of \(\displaystyle \mathrm{Fe^{3+}}\) versus \(\displaystyle \mathrm{Cr^{3+}}\) and \(\displaystyle \mathrm{Mn^{3+}}\)The relevant data is the set of \(\displaystyle M^{3+}/M^{2+}\) reduction potentials given:\[Cr^{3+} + e^- \rightarrow Cr^{2+} \qquad E^\circ = -0.4\ \text{V} \] \[Mn^{3+} + e^- \rightarrow Mn^{2+} \qquad E^\circ = +1.5\ \text{V} \] \[Fe^{3+} + e^- \rightarrow Fe^{2+} \qquad E^\circ = +0.8\ \text{V} \]A large positive \(\displaystyle E^\circ\) means the forward reaction (reduction of \(\displaystyle M^{3+}\) to \(\displaystyle M^{2+}\)) is strongly favoured — the +$\displaystyle 3$ ion accepts an electron easily and drops to +2. So the ion with the highest \(\displaystyle E^\circ\) here is the one that is hardest to keep as \(\displaystyle M^{3+}\), i.e. the least stable in that state; the ion whose \(\displaystyle E^\circ\) is smallest (or negative) resists being reduced and is the most stable \(\displaystyle M^{3+}\).Comparing the three values, \(\displaystyle +1.5\ \text{V} > +0.8\ \text{V} > -0.4\ \text{V}\), so reading the stability in the opposite order gives\[\text{stability of } M^{3+}: \quad Cr^{3+} > Fe^{3+} > Mn^{3+} \] So compared with \(\displaystyle \mathrm{Cr^{3+}}\), \(\displaystyle \mathrm{Fe^{3+}}\) is distinctly less stable (\(\displaystyle Cr^{3+}\) essentially will not be reduced, while \(\displaystyle \mathrm{Fe^{3+}}\) will, given the chance); compared with \(\displaystyle \mathrm{Mn^{3+}}\), \(\displaystyle \mathrm{Fe^{3+}}\) is considerably more stable, since \(\displaystyle \mathrm{Mn^{3+}}\) is almost completely reduced to \(\displaystyle \mathrm{Mn^{2+}}\) in acid solution while \(\displaystyle \mathrm{Fe^{3+}}\) survives to a much greater extent.(ii) Ease of oxidation of the metal (M → M\(\displaystyle ^{2+}\)) for Fe versus Cr and MnThis time the relevant data is the \(\displaystyle M^{2+}/M\) reduction potentials:\[Cr^{2+} + 2e^- \rightarrow Cr \qquad E^\circ = -0.9\ \text{V} \] \[Mn^{2+} + 2e^- \rightarrow Mn \qquad E^\circ = -1.2\ \text{V} \] \[Fe^{2+} + 2e^- \rightarrow Fe \qquad E^\circ = -0.4\ \text{V} \]Oxidation of the metal is the reverse of this, \(\displaystyle M \rightarrow M^{2+} + 2e^-\). The reverse of a reaction is favoured exactly when the forward (reduction) potential is unfavourable — that is, a strongly negative \(\displaystyle E^\circ(M^{2+}/M)\) means the metal readily gives up electrons and is easily oxidised. This is the step that is easy to get backwards: a more negative \(\displaystyle E^\circ\) here means the metal is a stronger reducing agent (easier to oxidise), not a weaker one.Ranking the three: \(\displaystyle -1.2\ \text{V} \;(\text{Mn}) < -0.9\ \text{V}\;(\text{Cr}) < -0.4\ \text{V}\;(\text{Fe})\), so\[\text{ease of oxidation of the metal:} \quad Mn > Cr > Fe \]Iron is therefore the hardest of the three metals to oxidise to the +$\displaystyle 2$ state. Its \(\displaystyle E^\circ(Fe^{2+}/Fe) = -0.4\) V is far less negative than chromium's \(\displaystyle -0.9\) V or manganese's \(\displaystyle -1.2\) V, so iron metal holds on to its valence electrons more strongly than either chromium or manganese does — oxidising iron to \(\displaystyle \mathrm{Fe^{2+}}\) is thermodynamically less favourable than the same process for Cr or Mn.**Answer: (i) Stability of the +$\displaystyle 3$ ion in acid solution follows \(\displaystyle Cr^{3+} > Fe^{3+} > Mn^{3+}\): \(\displaystyle \mathrm{Cr^{3+}}\) (\(\displaystyle E^\circ = -0.4\) V, resists reduction, stable \(\displaystyle t_{2g}^3\)) is the most stable, \(\displaystyle \mathrm{Fe^{3+}}\) (\(\displaystyle E^\circ = +0.8\) V, stable \(\displaystyle 3d^5\)) is moderately stable, and \(\displaystyle \mathrm{Mn^{3+}}\) (\(\displaystyle E^\circ = +1.5\) V) is the least stable, being readily reduced to \(\displaystyle \mathrm{Mn^{2+}}\) (\(\displaystyle 3d^5\)). (ii) Ease of oxidation of the metal follows \(\displaystyle Mn > Cr > Fe\): since \(\displaystyle E^\circ(Fe^{2+}/Fe) = -0.4\) V is much less negative than \(\displaystyle E^\circ(Cr^{2+}/Cr) = -0.9\) V or \(\displaystyle E^\circ(Mn^{2+}/Mn) = -1.2\) V, iron is the hardest of the three to oxidise to \(\displaystyle M^{2+}\).

Colour in these ions comes from d–d transitions, so the test is simply: does the ion have a partially filled 3d subshell (\(\displaystyle d^1\)–\(\displaystyle d^9\))? An empty \(\displaystyle d^0\) shell has no electron to promote; a full \(\displaystyle d^{10}\) shell has no vacant d level to promote it into. Both extremes are colourless.The rule, stated properly. In an aqueous solution the ion is really the hydrated complex \(\displaystyle [\mathrm{M}(\mathrm{H_2O})_6]^{n+}\). The six water ligands split the five degenerate 3d orbitals into a lower \(\displaystyle t_{2g}\) set and an upper \(\displaystyle e_g\) set, separated by the crystal-field splitting energy \(\displaystyle \Delta_o\). A d electron absorbs a photon of visible light and jumps \(\displaystyle t_{2g} \rightarrow e_g\):\[\Delta_o = h\nu = \frac{hc}{\lambda} \]where \(\displaystyle h\) is Planck's constant, \(\displaystyle c\) the speed of light, \(\displaystyle \nu\) the frequency and \(\displaystyle \lambda\) the wavelength absorbed. The solution shows the complementary colour of the light it removes. This whole mechanism needs one thing: a d electron and an empty d level for it to move to.The step people get wrong: count the d electrons of the ION, not of the atom. For a transition metal the 4s electrons are removed first, before any 3d electron. So Ti (\(\displaystyle [\mathrm{Ar}]3d^2 4s^2\)) does not become \(\displaystyle 3d^0 4s^2\) — it becomes \(\displaystyle 3d^1\) as \(\displaystyle \mathrm{Ti^{3+}}\): strip \(\displaystyle 4s^2\) first, then one 3d.Ion by ion (Z in brackets, configuration of the ion, verdict): Summary of the counts: \(\displaystyle d^0\) \(\displaystyle \mathrm{Sc^{3+}}\) and \(\displaystyle d^{10}\) \(\displaystyle \mathrm{Cu^{+}}\) sit at the two ends where no d–d transition can occur; the other five, \(\displaystyle d^1\) to \(\displaystyle d^7\), all lie in the \(\displaystyle d^1\)–\(\displaystyle d^9\) window and are coloured.A note on the printed NCERT key. The textbook answer says "Except \(\displaystyle \mathrm{Sc^{3+}}\), all others will be coloured … because of incompletely filled 3d-orbitals," which sweeps \(\displaystyle \mathrm{Cu^{+}}\) into the coloured set. That contradicts the reason it gives in the same sentence: \(\displaystyle \mathrm{Cu^{+}}\) is \(\displaystyle 3d^{10}\), a completely filled subshell, so "incompletely filled 3d-orbitals" does not describe it at all. NCERT's own text elsewhere states that \(\displaystyle d^0\) and \(\displaystyle d^{10}\) ions are colourless, and \(\displaystyle \mathrm{Cu^{+}}\) species in solution (e.g. \(\displaystyle [\mathrm{Cu}(\mathrm{CH_3CN})_4]^+\)) are indeed colourless — the familiar blue of copper solutions is \(\displaystyle \mathrm{Cu^{2+}}\) (\(\displaystyle 3d^9\)), not Cu\(\displaystyle ^{+}\). Treat the "except \(\displaystyle \mathrm{Sc^{3+}}\)" phrasing as a slip in the answer key: two of the seven ions are colourless.Answer: Coloured — \(\displaystyle \mathrm{Ti^{3+}}\) (\(\displaystyle 3d^1\)), \(\displaystyle \mathrm{V^{3+}}\) (\(\displaystyle 3d^2\)), \(\displaystyle \mathrm{Mn^{2+}}\) (\(\displaystyle 3d^5\)), \(\displaystyle \mathrm{Fe^{3+}}\) (\(\displaystyle 3d^5\)) and \(\displaystyle \mathrm{Co^{2+}}\) (\(\displaystyle 3d^7\)), all of which have partially filled 3d subshells and so undergo d–d transitions. Colourless — \(\displaystyle \mathrm{Sc^{3+}}\) (\(\displaystyle 3d^0\), no d electron to excite) and \(\displaystyle \mathrm{Cu^{+}}\) (\(\displaystyle 3d^{10}\), no vacant d orbital to excite into).

The "stability" of an \(\displaystyle \mathrm{M^{2+}}\) ion is really a question of how hard it is to strip off one more electron to reach \(\displaystyle M^{3+}\) and that difficulty is set by two things: the third ionisation enthalpy of the metal, and whether the electron count left behind is one of the specially stable configurations \(\displaystyle d^{0}\), \(\displaystyle d^{5}\) (half-filled) or \(\displaystyle d^{10}\) (fully filled).The 3d series runs \(\displaystyle \mathrm{Sc}\ (Z=21)\) to \(\displaystyle \mathrm{Zn}\ (Z=30)\). Writing the outer configuration of each \(\displaystyle M\) and the resulting \(\displaystyle M^{2+}\) (formed by losing the two 4s electrons) and \(\displaystyle M^{3+}\) (one 3d electron lost as well) shows exactly where the extra-stable counts fall:\[\begin{array}{l} \mathrm{Sc}:[\mathrm{Ar}]3d^{1}4s^{2}\ \to\ \mathrm{Sc}^{2+}:3d^{1}\ \to\ \mathrm{Sc}^{3+}:3d^{0}\\ \mathrm{Ti}:[\mathrm{Ar}]3d^{2}4s^{2}\ \to\ \mathrm{Ti}^{2+}:3d^{2}\ \to\ \mathrm{Ti}^{3+}:3d^{1}\\ \mathrm{V}:[\mathrm{Ar}]3d^{3}4s^{2}\ \to\ \mathrm{V}^{2+}:3d^{3}\ \to\ \mathrm{V}^{3+}:3d^{2}\\ \mathrm{Cr}:[\mathrm{Ar}]3d^{5}4s^{1}\ \to\ \mathrm{Cr}^{2+}:3d^{4}\ \to\ \mathrm{Cr}^{3+}:3d^{3}\\ \mathrm{Mn}:[\mathrm{Ar}]3d^{5}4s^{2}\ \to\ \mathrm{Mn}^{2+}:3d^{5}\ \to\ \mathrm{Mn}^{3+}:3d^{4}\\ \mathrm{Fe}:[\mathrm{Ar}]3d^{6}4s^{2}\ \to\ \mathrm{Fe}^{2+}:3d^{6}\ \to\ \mathrm{Fe}^{3+}:3d^{5}\\ \mathrm{Co}:[\mathrm{Ar}]3d^{7}4s^{2}\ \to\ \mathrm{Co}^{2+}:3d^{7}\ \to\ \mathrm{Co}^{3+}:3d^{6}\\ \mathrm{Ni}:[\mathrm{Ar}]3d^{8}4s^{2}\ \to\ \mathrm{Ni}^{2+}:3d^{8}\ \to\ \mathrm{Ni}^{3+}:3d^{7}\\ \mathrm{Cu}:[\mathrm{Ar}]3d^{10}4s^{1}\ \to\ \mathrm{Cu}^{2+}:3d^{9}\ \;(\mathrm{Cu}^{+}:3d^{10})\\ \mathrm{Zn}:[\mathrm{Ar}]3d^{10}4s^{2}\ \to\ \mathrm{Zn}^{2+}:3d^{10} \end{array} \]The quantity that actually measures the stability of \(\displaystyle +2\) relative to \(\displaystyle +3\) is the standard electrode potential \(\displaystyle E^{\circ}(M^{3+}/M^{2+})\) — this is the potential for \(\displaystyle M^{3+}+e^{-}\to M^{2+}\). A large positive value means \(\displaystyle M^{3+}\) is a strong oxidiser that is readily reduced back to \(\displaystyle M^{2+}\), so \(\displaystyle +2\) is the stable state. A negative value means \(\displaystyle M^{2+}\) is a strong reducing agent that is readily oxidised up to \(\displaystyle M^{3+}\), so \(\displaystyle +2\) is unstable relative to \(\displaystyle +3\). The measured values for the first series are:\[E^{\circ}(\mathrm{Ti}^{3+}/\mathrm{Ti}^{2+})=-0.37\ \text{V},\quad E^{\circ}(\mathrm{V}^{3+}/\mathrm{V}^{2+})=-0.26\ \text{V},\quad E^{\circ}(\mathrm{Cr}^{3+}/\mathrm{Cr}^{2+})=-0.41\ \text{V} \] \[E^{\circ}(\mathrm{Mn}^{3+}/\mathrm{Mn}^{2+})=+1.57\ \text{V},\quad E^{\circ}(\mathrm{Fe}^{3+}/\mathrm{Fe}^{2+})=+0.77\ \text{V},\quad E^{\circ}(\mathrm{Co}^{3+}/\mathrm{Co}^{2+})=+1.97\ \text{V} \]Reading this data element by element:Sc does not really have a stable \(\displaystyle +2\) ion at all. \(\displaystyle \mathrm{Sc}^{3+}\) is \(\displaystyle 3d^{0}\) — an inert-gas-like configuration — so the driving force to lose that third electron is so strong that any \(\displaystyle \mathrm{Sc}^{2+}\) formed is an extremely powerful reducing agent, instantly oxidised. The mistake here is assuming every element in the series shows a genuine, isolable \(\displaystyle +2\) state — Sc effectively does not.Ti, V, Cr ( \(\displaystyle E^{\circ}\) all negative) behave the same way, just less extremely: \(\displaystyle \mathrm{Ti}^{2+}\) and \(\displaystyle \mathrm{V}^{2+}\) are strong reducing agents that oxidise readily to \(\displaystyle \mathrm{Ti}^{3+}(d^{1})\) and \(\displaystyle \mathrm{V}^{3+}(d^{2})\), because so little energy is needed to remove the third electron at this end of the series. \(\displaystyle \mathrm{Cr}^{2+}(d^{4})\) is oxidised most easily of the three, because \(\displaystyle \mathrm{Cr}^{3+}(d^{3})\) has an exactly half-filled \(\displaystyle t_{2g}\) level in an octahedral field and is unusually stable — so for chromium the \(\displaystyle +3\) state, not \(\displaystyle +2\), is the one that dominates its chemistry.Mn is the turning point: \(\displaystyle E^{\circ}(\mathrm{Mn}^{3+}/\mathrm{Mn}^{2+})=+1.57\ \text{V}\) is sharply positive because \(\displaystyle \mathrm{Mn}^{2+}\) is \(\displaystyle 3d^{5}\), an exactly half-filled and therefore exchange-stabilised configuration. Oxidising it to \(\displaystyle \mathrm{Mn}^{3+}(d^{4})\) destroys that stability, so it costs a lot of energy — \(\displaystyle +2\) is one of manganese's most stable oxidation states.Fe still has a positive \(\displaystyle E^{\circ}\) \(\displaystyle (+0.77\ \text{V})\), but a smaller one than Mn's, because here the extra-stable \(\displaystyle d^{5}\) configuration belongs to the product rather than the reactant: \(\displaystyle \mathrm{Fe}^{2+}(d^{6})\) is only moderately resistant to oxidation, since going to \(\displaystyle \mathrm{Fe}^{3+}(d^{5}, \text{half-filled})\) is itself favourable. That is why iron shows both \(\displaystyle +2\) and \(\displaystyle +3\) as common, comparably stable states, with \(\displaystyle \mathrm{Fe}^{2+}\) slowly air-oxidising to \(\displaystyle \mathrm{Fe}^{3+}\).Co ( \(\displaystyle +1.97\ \text{V}\), the largest value in the series) shows the opposite extreme from Mn: here it is \(\displaystyle \mathrm{Co}^{3+}(d^{6})\) that is destabilised (it is such a strong oxidiser that in aqueous solution it even oxidises water), so it is reduced very readily to \(\displaystyle \mathrm{Co}^{2+}(d^{7})\) — making \(\displaystyle +2\) overwhelmingly the stable, dominant state for cobalt.Ni continues the trend past cobalt: \(\displaystyle \mathrm{Ni}^{3+}(d^{7})\) is even more strongly oxidising and correspondingly rarer than \(\displaystyle \mathrm{Co}^{3+}\), so \(\displaystyle \mathrm{Ni}^{2+}(d^{8})\) is highly stable and is essentially the only common oxidation state of nickel.Cu looks like an exception on paper: \(\displaystyle \mathrm{Cu}^{+}\) is \(\displaystyle 3d^{10}\) (a "nicer" configuration than \(\displaystyle \mathrm{Cu}^{2+}\)'s \(\displaystyle 3d^{9}\)), yet \(\displaystyle \mathrm{Cu}^{2+}\) is the ion actually found in aqueous solution and in most of copper's compounds. The subtlety students miss: configurational stability isn't the only factor — the much higher hydration and lattice enthalpy of the smaller, higher-charge-density \(\displaystyle \mathrm{Cu}^{2+}\) ion more than compensates for giving up the filled \(\displaystyle d^{10}\) shell, which is why \(\displaystyle 2\mathrm{Cu}^{+}\rightleftharpoons \mathrm{Cu}^{2+}+\mathrm{Cu}\) (disproportionation) is favourable in water. \(\displaystyle +3\) is essentially unknown for copper, so relative to \(\displaystyle +3\), \(\displaystyle +2\) is copper's stable state.Zn ends the series at the other extreme from Sc: \(\displaystyle \mathrm{Zn}^{2+}\) is \(\displaystyle 3d^{10}\), a fully filled and therefore maximally stable d-subshell. There is no accessible \(\displaystyle +3\) state (it would have to break into the filled shell) and no stable \(\displaystyle +1\) state either, so \(\displaystyle +2\) is the sole oxidation state zinc shows.Putting it together, stability of \(\displaystyle +2\) is not a smooth trend across the row — it is low at the start (Sc, Ti, V, Cr, where the metals prefer to go on to \(\displaystyle +3\)), rises to a sharp local peak at Mn (\(\displaystyle d^{5}\)), dips a little at Fe, and then climbs to the highest values of the whole series at Ni, Cu and Zn, peaking at Zn's fully-filled \(\displaystyle d^{10}\).Answer: Stability of the \(\displaystyle +2\) state is lowest for Sc (practically absent) and low for Ti, V, Cr, since these readily lose a third electron to reach the more stable \(\displaystyle \mathrm{M}^{3+}\); it rises sharply at Mn (\(\displaystyle \mathrm{Mn}^{2+}=3d^{5}\), half-filled) and, after a dip at Fe, rises again through Co and Ni to a maximum at Zn (\(\displaystyle \mathrm{Zn}^{2+}=3d^{10}\), fully filled), where \(\displaystyle +2\) is the only oxidation state shown. Overall order of increasing \(\displaystyle +2\)-state stability: \(\displaystyle \mathrm{Sc}<\mathrm{Cr}<\mathrm{Ti}<\mathrm{V}<\mathrm{Fe}<\mathrm{Mn}<\mathrm{Co}<\mathrm{Cu}<\mathrm{Ni}<\mathrm{Zn}\).

Lanthanoids and actinoids both build up an inner f-subshell while the outer shell stays almost frozen — every similarity between the two series comes from that, and every difference comes from whether the inner subshell being filled is 4f or 5f.(i) Electronic configurationThe general configurations are\[\text{Lanthanoids (Ce to Lu, } Z=58\text{–}71\text{):} \quad [Xe]\,4f^{1-14}5d^{0-1}6s^{2} \] \[\text{Actinoids (Th to Lr, } Z=90\text{–}103\text{):} \quad [Rn]\,5f^{1-14}6d^{0-1}7s^{2} \]In the lanthanoids the \(\displaystyle 4f\) orbitals lie deep inside the atom, screened by the filled \(\displaystyle 5s\) and \(\displaystyle 5p\) shells, so the \(\displaystyle 4f\) subshell fills in an almost regular, textbook order with only a couple of exceptions (La, Gd, Lu prefer \(\displaystyle 5d^{1}\) over an extra \(\displaystyle 4f\) electron).In the actinoids the \(\displaystyle 5f\), \(\displaystyle 6d\), and \(\displaystyle 7s\) orbitals are much closer in energy to one another than \(\displaystyle 4f\), \(\displaystyle 5d\), \(\displaystyle 6s\) are in the lanthanoids. This is the point people gloss over: because three sets of orbitals compete for the same electrons, actinoid configurations are far more irregular, with electrons dropping into \(\displaystyle 6d\) instead of \(\displaystyle 5f\) more often (e.g. Th is \(\displaystyle [Rn]6d^{2}7s^{2}\) with no \(\displaystyle 5f\) electron at all, and Pa, U, Np all carry a \(\displaystyle 6d^{1}\) alongside partially filled \(\displaystyle 5f\)).(ii) Atomic and ionic sizesAcross both series the atomic radius and the \(\displaystyle M^{3+}\) ionic radius fall steadily with increasing atomic number — the lanthanoid contraction and, analogously, the actinoid contraction. The cause in both cases is imperfect shielding: an electron entering the \(\displaystyle f\) subshell does not shield another \(\displaystyle f\) electron from the increasing nuclear charge as effectively as an inner-shell electron would, so the effective nuclear charge felt by the outer electrons rises steadily and the ion shrinks at every step.The actinoid contraction per element is somewhat greater than the lanthanoid contraction. The \(\displaystyle 5f\) orbitals are more diffuse and extend further from the nucleus than the \(\displaystyle 4f\) orbitals, so they overlap with each other and shield each other even more poorly — the nuclear charge "leaks through" more, and the size drop per unit increase in \(\displaystyle Z\) is larger.(iii) Oxidation statesLanthanoids show one dominant oxidation state, \(\displaystyle +3\), for the whole series. A few ions depart from this only when doing so gives a specially stable \(\displaystyle f^{0}\), \(\displaystyle f^{7}\), or \(\displaystyle f^{14}\) configuration: \(\displaystyle \mathrm{Ce}^{4+}\) (\(\displaystyle f^{0}\)), \(\displaystyle \mathrm{Eu}^{2+}\) (\(\displaystyle f^{7}\)), \(\displaystyle \mathrm{Tb}^{4+}\) (\(\displaystyle f^{7}\)), \(\displaystyle \mathrm{Yb}^{2+}\) (\(\displaystyle f^{14}\)).Actinoids show a much wider spread of oxidation states, from \(\displaystyle +3\) up to \(\displaystyle +7\), because the near-degenerate \(\displaystyle 5f\), \(\displaystyle 6d\), \(\displaystyle 7s\) orbitals all let their electrons take part in bonding. Uranium, for instance, forms \(\displaystyle +3, +4, +5,\) and \(\displaystyle +6\) compounds (\(\displaystyle \mathrm{UO_2^{2+}}\) is the familiar \(\displaystyle +6\) uranyl ion), and neptunium, plutonium, and americium reach \(\displaystyle +7\). This is the aside worth remembering: the variety of actinoid oxidation states is a direct consequence of orbital energies being close, not of the atoms being "more chemically flexible" in some vague sense. Only in the later actinoids, once the \(\displaystyle 5f\) subshell has sunk in energy and become inert like the \(\displaystyle 4f\) subshell, does \(\displaystyle +3\) (with some \(\displaystyle +4\)) become the standard state, mirroring the lanthanoids.(iv) Chemical reactivityThe early lanthanoids (La, Ce, Pr) are as reactive as calcium; reactivity falls off somewhat across the series as the lanthanoid contraction raises the ionization enthalpy. They tarnish slowly in air, react with the halogens and with dilute acids liberating \(\displaystyle \mathrm{H_2}\), and react only slowly with cold water.The actinoids are markedly more reactive metals, especially the early members (Th, Pa, U, Np): they are highly electropositive, react with boiling water, and combine with most non-metals — hydrogen, carbon, nitrogen, halogens, oxygen — even at fairly moderate temperatures. Finely divided actinoid metal is pyrophoric enough to attack air and water readily, so in practice actinoids are handled with far more care than lanthanoids of similar electropositive character.Answer: (i) Both fill an inner f-subshell ( \(\displaystyle 4f\) for lanthanoids, \(\displaystyle 5f\) for actinoids) with an outer \(\displaystyle [Xe/Rn]\,(n{-}1)d^{0-1}ns^{2}\) shell, but actinoid configurations are more irregular because \(\displaystyle 5f\), \(\displaystyle 6d\), \(\displaystyle 7s\) are closer in energy than \(\displaystyle 4f\), \(\displaystyle 5d\), \(\displaystyle 6s\). (ii) Both show a steady contraction in atomic/ionic size from poor f-electron shielding; the actinoid contraction is larger per element because \(\displaystyle 5f\) shields even more poorly than \(\displaystyle 4f\). (iii) Lanthanoids are almost exclusively \(\displaystyle +3\); actinoids range from \(\displaystyle +3\) to \(\displaystyle +7\) because their three valence-type orbitals are near-degenerate. (iv) Actinoids, particularly the early ones, are more reactive than lanthanoids, reacting with boiling water and most non-metals under mild conditions, whereas lanthanoids react only slowly with cold water and tarnish gradually in air.