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.

A d-electron count of \(\displaystyle d^{0}\), \(\displaystyle d^{5}\), or \(\displaystyle d^{10}\) — empty, exactly half-filled, or completely filled — is unusually stable, because these are the only arrangements with no unpaired-electron asymmetry within the set of five d-orbitals. Every ion below is chasing one of these three counts.(i) \(\displaystyle \mathrm{Cr^{2+}}\) is a strong reducing agent; \(\displaystyle \mathrm{Mn^{3+}}\) is a strong oxidising agent — both because of what they turn into.\(\displaystyle Cr\) (\(\displaystyle Z=24\)) has the anomalous ground configuration \(\displaystyle [Ar]3d^{5}4s^{1}\). Removing the two outer electrons to form \(\displaystyle \mathrm{Cr^{2+}}\) leaves \[Cr^{2+}:[Ar]3d^{4} \] If \(\displaystyle \mathrm{Cr^{2+}}\) loses one more electron, \[Cr^{2+} \rightarrow Cr^{3+} + e^{-}, \qquad Cr^{3+}:[Ar]3d^{3} \] \(\displaystyle \mathrm{Cr^{3+}}\) is \(\displaystyle t_{2g}^{3}e_{g}^{0}\) — the \(\displaystyle t_{2g}\) set exactly half-filled, one electron in each of the three lower orbitals with parallel spin. That is a configuration of unusually low energy, so the electron loss \(\displaystyle Cr^{2+}\rightarrow Cr^{3+}\) is thermodynamically favourable. A species that loses an electron easily is, by definition, a reducing agent — this is why \(\displaystyle E^{\circ}(Cr^{3+}/Cr^{2+}) = -0.41\ \text{V}\), a negative value that confirms \(\displaystyle \mathrm{Cr^{2+}}\) is readily oxidised.\(\displaystyle Mn\) (\(\displaystyle Z=25\)) has ground configuration \(\displaystyle [Ar]3d^{5}4s^{2}\). Removing three electrons gives \[Mn^{3+}:[Ar]3d^{4} \] which is the same \(\displaystyle d^{4}\) count as \(\displaystyle \mathrm{Cr^{2+}}\), but now the favourable move is the opposite one — gaining an electron: \[Mn^{3+} + e^{-} \rightarrow Mn^{2+}, \qquad Mn^{2+}:[Ar]3d^{5} \] \(\displaystyle \mathrm{Mn^{2+}}\) is \(\displaystyle t_{2g}^{3}e_{g}^{2}\), the fully half-filled d-subshell — every one of the five d-orbitals singly occupied with parallel spin, the single most stable d-arrangement there is. Because \(\displaystyle \mathrm{Mn^{3+}}\) is so strongly driven to grab an electron and become \(\displaystyle \mathrm{Mn^{2+}}\), it is a powerful oxidising agent, matching its large positive potential \(\displaystyle E^{\circ}(Mn^{3+}/Mn^{2+}) = +1.57\ \text{V}\).So the same \(\displaystyle d^{4}\) starting point gives opposite behaviour purely because \(\displaystyle \mathrm{Cr^{2+}}\) is one electron away from a stable half-filled \(\displaystyle t_{2g}^{3}\), while \(\displaystyle \mathrm{Mn^{3+}}\) is one electron away from a stable half-filled \(\displaystyle d^{5}\) — one is stabilised by losing, the other by gaining.(ii) \(\displaystyle \mathrm{Co^{2+}}\) survives in water; complexing agents flip the preference to \(\displaystyle Co^{3+}\).\(\displaystyle Co\) (\(\displaystyle Z=27\)): \(\displaystyle \mathrm{Co^{2+}}\) is \(\displaystyle [Ar]3d^{7}\), \(\displaystyle \mathrm{Co^{3+}}\) is \(\displaystyle [Ar]3d^{6}\). In water (weak-field ligand \(\displaystyle H_2O\), high-spin), converting \(\displaystyle \mathrm{Co^{2+}}\) to \(\displaystyle \mathrm{Co^{3+}}\) costs the third ionisation energy of cobalt, which is very large, and the hydration enthalpy released by the small, more highly charged \(\displaystyle \mathrm{Co^{3+}}\) ion does not make up the difference. The net result is a very large positive potential, \[E^{\circ}\big(Co^{3+}/Co^{2+}\big) = +1.97\ \text{V} \] A large positive \(\displaystyle E^{\circ}\) for this couple means \(\displaystyle \mathrm{Co^{3+}(aq)}\) is a strong oxidiser that is itself readily reduced back to \(\displaystyle \mathrm{Co^{2+}(aq)}\) — even by water. So in plain aqueous solution, \(\displaystyle \mathrm{Co^{2+}}\) is the stable form.Strong-field complexing ligands such as \(\displaystyle NH_3\) or \(\displaystyle \mathrm{CN^{-}}\) change the balance. With a strong field, \(\displaystyle \mathrm{Co^{3+}}\) (\(\displaystyle d^{6}\)) becomes low-spin, \(\displaystyle t_{2g}^{6}e_{g}^{0}\) — every electron in the lower, non-antibonding set — which carries a large crystal field stabilisation energy (CFSE) and is also kinetically inert. \(\displaystyle \mathrm{Co^{2+}}\) (\(\displaystyle d^{7}\)) cannot match this: even low-spin it must place one electron in the higher \(\displaystyle e_g\) level (\(\displaystyle t_{2g}^{6}e_{g}^{1}\)), so it gains far less CFSE. This extra stabilisation of the complexed \(\displaystyle \mathrm{Co^{3+}}\) far outweighs the third ionisation energy that was the obstacle in water, so the effective potential collapses — for example \(\displaystyle E^{\circ}\big([Co(NH_3)_6]^{3+}/[Co(NH_3)_6]^{2+}\big) \approx +0.1\ \text{V}\) — small enough that even atmospheric oxygen can oxidise the \(\displaystyle \mathrm{Co^{2+}}\) complex to the \(\displaystyle \mathrm{Co^{3+}}\) complex. The metal hasn't changed; what changed is which oxidation state the ligand field rewards.(iii) A \(\displaystyle d^{1}\) ion is one electron short of the stable empty shell, and it takes the shortcut.Take \(\displaystyle \mathrm{Ti^{3+}}\) as the example: \(\displaystyle Ti\) (\(\displaystyle Z=22\)) is \(\displaystyle [Ar]3d^{2}4s^{2}\), so \[Ti^{3+}:[Ar]3d^{1} \] This single d-electron gives almost no crystal-field stabilisation and no special symmetry to protect it. Losing that one electron takes the ion straight to \[Ti^{3+} \rightarrow Ti^{4+} + e^{-}, \qquad Ti^{4+}:[Ar]3d^{0} \] an empty, perfectly symmetric d-subshell with the stability of a full noble-gas core — the same kind of stability that makes \(\displaystyle d^{5}\) and \(\displaystyle d^{10}\) favourable, just at the other end of the shell. Because this \(\displaystyle d^{0}\) state is so much lower in energy, any \(\displaystyle d^{1}\) ion is under strong thermodynamic pressure to give up its lone d-electron and be oxidised. That is why \(\displaystyle d^{1}\) species such as \(\displaystyle \mathrm{Ti^{3+}}\) or \(\displaystyle \mathrm{V^{4+}}\) are short-lived, strongly reducing, and rarely the stable, isolable form of the metal.Answer: (i) \(\displaystyle Cr^{2+}(d^{4})\) is reducing because losing an electron gives \(\displaystyle Cr^{3+}(t_{2g}^{3})\), a stable half-filled \(\displaystyle t_{2g}\) set; \(\displaystyle Mn^{3+}(d^{4})\) is oxidising because gaining an electron gives \(\displaystyle Mn^{2+}(d^{5})\), the stable half-filled d-subshell. (ii) \(\displaystyle Co^{2+}(d^{7})\) is the stable form in water because \(\displaystyle E^{\circ}(Co^{3+}/Co^{2+}) = +1.97\ \text{V}\) is too high for oxidation to occur, but strong-field ligands (\(\displaystyle NH_3\), \(\displaystyle CN^{-}\)) give low-spin \(\displaystyle Co^{3+}(t_{2g}^{6})\) a much larger CFSE than \(\displaystyle \mathrm{Co^{2+}}\) can match, dropping the effective potential (\(\displaystyle \approx +0.1\ \text{V}\) for the ammine complex) so that \(\displaystyle \mathrm{Co^{2+}}\) is easily oxidised once complexed. (iii) A \(\displaystyle d^{1}\) ion (e.g. \(\displaystyle Ti^{3+}\)) has almost no stabilisation and is one electron from the very stable empty \(\displaystyle d^{0}\) shell, so it readily loses that electron and is oxidised, making \(\displaystyle d^{1}\) configurations inherently unstable in solution.

Disproportionation is a redox reaction in which one element, sitting in a single intermediate oxidation state among the reactants, splits into two different oxidation states of itself among the products — one higher, one lower. The same atom is oxidised and reduced at once, so no external oxidising or reducing agent is needed.Why this happens. For a species \(\displaystyle \mathrm{X}^{n} \) to disproportionate, \[2\,\mathrm{X}^{n} \longrightarrow \mathrm{X}^{n+a} + \mathrm{X}^{n-b} \] this must be thermodynamically favourable — the two extreme oxidation states must together be more stable (lower in energy) than twice the intermediate one. Among the 3d elements this shows up whenever an ion's intermediate oxidation state is not specially stabilised (e.g. by a half-filled or filled d subshell), while its neighbouring oxidation states are.Example $\displaystyle 1$ — manganate disproportionating to permanganate, in acidic aqueous solution. Manganese in manganate, \(\displaystyle \mathrm{MnO_4^{2-}} \), is present in a single oxidation state, \(\displaystyle +6 \). In acid solution this ion is unstable and disproportionates: \[3\,\mathrm{MnO_4^{2-}}(aq) + 4\mathrm{H^+}(aq) \longrightarrow 2\,\mathrm{MnO_4^{-}}(aq) + \mathrm{MnO_2}(s) + 2\mathrm{H_2O}(l) \] Track the oxidation number of Mn on each side: it starts at \(\displaystyle +6 \) in every \(\displaystyle \mathrm{MnO_4^{2-}} \), and ends up at \(\displaystyle +7 \) in \(\displaystyle \mathrm{MnO_4^{-}} \) (permanganate, the oxidised product) and at \(\displaystyle +4 \) in \(\displaystyle \mathrm{MnO_2} \) (the reduced product). One reactant oxidation state has become two product oxidation states of the same element — that split is the signature of disproportionation, and it is easy to miss if you only check that the equation balances without following where each Mn atom's charge actually goes.Example $\displaystyle 2$ — copper(I) disproportionating in aqueous solution. \[2\,\mathrm{Cu^{+}}(aq) \longrightarrow \mathrm{Cu^{2+}}(aq) + \mathrm{Cu}(s) \] Here copper starts at \(\displaystyle +1 \) in every \(\displaystyle \mathrm{Cu^+} \) ion and ends at \(\displaystyle +2 \) (oxidised, in solution) and \(\displaystyle 0 \) (reduced, as metal). \(\displaystyle \mathrm{Cu^{+}} \) is unstable in aqueous solution precisely because \(\displaystyle \mathrm{Cu^{2+}} \) is markedly stabilised by its higher hydration/lattice energy, which is why copper(I) salts are rare in water while copper(II) salts are common — the disproportionation of \(\displaystyle \mathrm{Cu^+} \) is what removes it.Answer: Disproportionation is a reaction in which one element in a single oxidation state is simultaneously oxidised and reduced, ending up in two different oxidation states in the products (e.g., \(\displaystyle 3\mathrm{MnO_4^{2-}} + 4\mathrm{H^+} \rightarrow 2\mathrm{MnO_4^{-}} + \mathrm{MnO_2} + 2\mathrm{H_2O} \), and \(\displaystyle 2\mathrm{Cu^{+}}(aq) \rightarrow \mathrm{Cu^{2+}}(aq) + \mathrm{Cu}(s) \)).

Losing one electron from copper's odd 4s\(\displaystyle ^1\) electron leaves behind a completely filled 3d\(\displaystyle ^{10}\) shell — and a fully filled d-subshell is an unusually stable arrangement, which is why copper is the metal that shows the +$\displaystyle 1$ state most often among the first-series transition metals.Start from copper's ground-state electron configuration (atomic number $\displaystyle 29$): \[\mathrm{Cu}:\ [Ar]\,3d^{10}4s^{1} \]Copper is already one of the exceptions to the normal aufbau order — it adopts \(\displaystyle 3d^{10}4s^{1}\) rather than \(\displaystyle 3d^{9}4s^{2}\) because a completely filled d-subshell lowers the atom's energy more than a filled s-subshell would. That preference for a full \(\displaystyle 3d^{10}\) shell carries straight over into copper's ion chemistry.To form the +$\displaystyle 1$ ion, copper only has to give up its single, loosely-held 4s electron: \[\mathrm{Cu} \longrightarrow \mathrm{Cu}^{+} + e^{-} \] \[[Ar]\,3d^{10}4s^{1} \longrightarrow [Ar]\,3d^{10} \]Here \(\displaystyle [Ar]\) is the argon core (configuration up to \(\displaystyle 3p^{6}\)), and the \(\displaystyle 3d^{10}\) left behind is a completely filled d-subshell — the same kind of extra-stable arrangement (filled or half-filled subshells) that governs anomalous configurations throughout the d-block. Because this ion needs only the first ionisation energy to form (not a second, harder one), and lands on an especially stable electron arrangement, \(\displaystyle \mathrm{Cu}^{+}\) is a genuinely common, isolable species. It turns up across a range of copper(I) compounds: \(\displaystyle \mathrm{Cu_2O}\), \(\displaystyle \mathrm{CuCl}\), \(\displaystyle \mathrm{CuBr}\), \(\displaystyle \mathrm{CuI}\), \(\displaystyle \mathrm{Cu_2S}\), and complexes such as \(\displaystyle [\mathrm{CuCl_2}]^{-}\).Contrast this with the rest of the first transition series. For metals such as Fe, Co, Ni, or Mn, neither the +$\displaystyle 1$ nor the +$\displaystyle 2$ ion corresponds to a filled or half-filled d-subshell, so there is no comparable "reward" for stopping at +$\displaystyle 1$ — these metals instead favour +$\displaystyle 2$ or +$\displaystyle 3$, governed by crystal-field stabilisation and exchange-energy effects, and a +$\displaystyle 1$ state is rare or unknown for them. Copper is the one metal in the series where losing exactly one electron lands on a specially stabilised configuration, which is why +$\displaystyle 1$ recurs for copper and essentially nowhere else in the row.A common mix-up: in aqueous solution, blue \(\displaystyle \mathrm{Cu}^{2+}\) (3d\(\displaystyle ^9\)) salts are actually the more stable, everyday form of copper, not \(\displaystyle \mathrm{Cu}^{+}\). That is not a contradiction — it is a separate energy balance. Forming \(\displaystyle \mathrm{Cu}^{2+}\) costs a second ionisation energy that \(\displaystyle \mathrm{Cu}^{+}\) formation does not, but the smaller, more highly charged \(\displaystyle \mathrm{Cu}^{2+}\) ion is hydrated far more strongly in water, and that hydration enthalpy more than pays back the extra ionisation cost. So in water, \(\displaystyle \mathrm{Cu}^{2+}\) wins. But that hydration bonus only exists in solution. In the solid state and in most non-aqueous/complex environments, there is no large hydration term to tip the balance, and the intrinsic electronic stability of the filled \(\displaystyle 3d^{10}\) shell is what decides the outcome — which is exactly why \(\displaystyle \mathrm{Cu}^{+}\) shows up so frequently across copper(I) compounds, more than any +$\displaystyle 1$ ion does for any other first-series transition metal.Answer: Copper (Cu). Losing its single 4s electron gives Cu\(\displaystyle ^+\) the completely filled, extra-stable \(\displaystyle 3d^{10}\) configuration, so the +$\displaystyle 1$ state recurs often for copper — in compounds like Cu\(\displaystyle _2\)O, CuCl, CuBr, and CuI — far more than for any other first-series transition metal.

When a transition-metal atom loses electrons to form an ion, the 4s electrons go first — always strip 4s completely, then take from 3d.Start from the ground-state electron configuration of each neutral atom and remove three electrons to get the M³⁺ ion.Ti (Z = $\displaystyle 22$): \(\displaystyle [Ar]3d^{2}4s^{2}\). Removing $\displaystyle 2$ electrons from 4s and $\displaystyle 1$ from 3d gives Ti³⁺: \(\displaystyle [Ar]3d^{1}\).V (Z = $\displaystyle 23$): \(\displaystyle [Ar]3d^{3}4s^{2}\). Removing $\displaystyle 2$ from 4s and $\displaystyle 1$ from 3d gives V³⁺: \(\displaystyle [Ar]3d^{2}\).Cr (Z = $\displaystyle 24$): \(\displaystyle [Ar]3d^{5}4s^{1}\). Removing the $\displaystyle 1$ electron in 4s and $\displaystyle 2$ from 3d gives Cr³⁺: \(\displaystyle [Ar]3d^{3}\).Mn (Z = $\displaystyle 25$): \(\displaystyle [Ar]3d^{5}4s^{2}\). Removing $\displaystyle 2$ from 4s and $\displaystyle 1$ from 3d gives Mn³⁺: \(\displaystyle [Ar]3d^{4}\).Counting unpaired electrons uses Hund's rule: each of the $\displaystyle 5$ degenerate d-orbitals takes one electron before any orbital takes a second. Since none of these ions has more than $\displaystyle 5$ d-electrons, no pairing occurs at all — the number of unpaired electrons simply equals the number of d-electrons. Stability of an M³⁺ ion in water is decided by how hard it is to reduce to M²⁺ — read that off the standard reduction potential \(\displaystyle E^{\ominus}(M^{3+}/M^{2+})\), not off the electron count. A large positive \(\displaystyle E^{\ominus}\) means \(\displaystyle M^{3+}\) is a strong oxidising agent that grabs an electron easily, so it collapses to \(\displaystyle M^{2+}\) — it is not stable in the +$\displaystyle 3$ state. A negative \(\displaystyle E^{\ominus}\) means the reverse: \(\displaystyle M^{2+}\) is the species that gets oxidised back up, so \(\displaystyle M^{3+}\) is the one that persists in solution.The standard values for these four couples are:\[E^{\ominus}(Ti^{3+}/Ti^{2+}) = -0.37\ \text{V}, \quad E^{\ominus}(V^{3+}/V^{2+}) = -0.26\ \text{V}, \] \[E^{\ominus}(Cr^{3+}/Cr^{2+}) = -0.41\ \text{V}, \quad E^{\ominus}(Mn^{3+}/Mn^{2+}) = +1.57\ \text{V} \]Mn³⁺/Mn²⁺ at \(\displaystyle +1.57\ \text{V}\) stands far apart from the other three — this is the step that people misjudge, expecting the ion with the most unpaired electrons (Mn³⁺, \(\displaystyle 3d^4\)) to be "extra stable" because half-filled-ish shells are usually praised. In reality this huge positive potential means Mn³⁺ is reduced to Mn²⁺ (which reaches the genuinely stable, fully symmetric \(\displaystyle 3d^5\) half-filled configuration) almost as soon as it is formed, making Mn³⁺ the least stable of the four in solution.Cr³⁺/Cr²⁺ at \(\displaystyle -0.41\ \text{V}\) is the most negative of the set — more negative than Ti's \(\displaystyle -0.37\ \text{V}\) and V's \(\displaystyle -0.26\ \text{V}\). A negative potential here means Cr²⁺ is readily oxidised back to Cr³⁺, so Cr³⁺ is the species that survives in aqueous solution. This is reinforced structurally: Cr³⁺ is \(\displaystyle 3d^{3}\), giving a symmetric, half-filled \(\displaystyle t_{2g}^{3}e_{g}^{0}\) arrangement in its octahedral aqua-complex, which carries a large crystal-field stabilisation energy and makes the ion kinetically inert (its ligands exchange very slowly) — an independent reason \(\displaystyle [Cr(H_2O)_6]^{3+}\) persists once formed.Answer: Ti³⁺ (\(\displaystyle 3d^1\)) has $\displaystyle 1$ unpaired electron, V³⁺ (\(\displaystyle 3d^2\)) has $\displaystyle 2$, Cr³⁺ (\(\displaystyle 3d^3\)) has $\displaystyle 3$, and Mn³⁺ (\(\displaystyle 3d^4\)) has 4. Cr³⁺ is the most stable of the four in aqueous solution, since \(\displaystyle E^{\ominus}(Cr^{3+}/Cr^{2+}) = -0.41\ \text{V}\) is the most negative of the set, while Mn³⁺ (\(\displaystyle E^{\ominus} = +1.57\ \text{V}\)) is the least stable, collapsing readily to Mn²⁺.

In transition-metal chemistry, the oxidation state of the metal decides how ionic or covalent the metal–oxygen (or metal–fluorine) bond is, and that bond character is what fixes whether an oxide is basic or acidic, and where the metal's highest oxidation state actually shows up.(i) Lowest oxide basic, highest oxide amphoteric/acidicIn a low oxidation state the metal ion carries only a small positive charge, so it cannot pull electron density strongly away from oxygen — the M–O bond stays largely ionic. An ionic metal oxide dissolves in water or reacts with acids to give a salt, which is exactly what "basic oxide" means: \[\text{MnO (Mn in +2) + 2HCl} \rightarrow \text{MnCl}_2 + \text{H}_2\text{O} \] As the oxidation state rises, the effective charge on the metal ion increases sharply while its radius shrinks (higher charge packed into a smaller ion). That combination pulls the shared electron pair strongly toward the metal, and the M–O bond becomes increasingly covalent — which is the signature of a non-metallic, acidic oxide. \(\displaystyle \text{Mn}_2\text{O}_7\) (Mn in +$\displaystyle 7$) is a covalent, low-melting liquid that is the acid anhydride of \(\displaystyle \text{HMnO}_4\): \[\text{Mn}_2\text{O}_7 + \text{H}_2\text{O} \rightarrow 2\text{HMnO}_4 \] The same trend appears for chromium: \(\displaystyle \text{CrO}\) (Cr in +$\displaystyle 2$) is basic, while \(\displaystyle \text{CrO}_3\) (Cr in +$\displaystyle 6$) is acidic. In between, an oxide such as \(\displaystyle \text{V}_2\text{O}_5\) (V in +$\displaystyle 5$) is amphoteric — it reacts with both acids and alkalis — because that oxidation state sits at the crossover from ionic to covalent bonding.The trap here is treating "oxide of a metal" as automatically basic — that is only true at the metal's lowest oxidation states; the same element's highest oxide behaves like a non-metal oxide.(ii) Highest oxidation state shown in oxides and fluoridesOxygen and fluorine are the two elements most able to drag a metal up to its highest possible oxidation state, and for two different reasons. Fluorine has the highest electronegativity of any element and a small atomic size, so it can oxidize the metal fully and still form a stable, small anion (\(\displaystyle F^-\)) that packs closely around a highly charged, small metal cation. Oxygen is also small and highly electronegative, but in addition its \(\displaystyle 2p\) orbitals can overlap with empty metal \(\displaystyle d\) orbitals to form \(\displaystyle \pi\)-bonds (\(\displaystyle p\pi\)–\(\displaystyle d\pi\) back bonding) on top of the \(\displaystyle \sigma\) bond — this extra multiple-bond character removes even more electron density from the metal and stabilizes very high formal charges that a single \(\displaystyle \sigma\) bond to fluorine could not.This is why the very highest oxidation states of transition metals are reached in their oxides rather than their fluorides: manganese reaches +$\displaystyle 7$ in \(\displaystyle \text{Mn}_2\text{O}_7\), but the highest known manganese fluoride is only \(\displaystyle \text{MnF}_4\) (Mn in +$\displaystyle 4$); osmium reaches +$\displaystyle 8$ in \(\displaystyle \text{OsO}_4\), but only +$\displaystyle 6$ in \(\displaystyle \text{OsF}_6\). Where a metal's highest state is more moderate, both routes reach it: vanadium reaches +$\displaystyle 5$ in both \(\displaystyle \text{V}_2\text{O}_5\) and \(\displaystyle \text{VF}_5\).The step people skip: it is not "any electronegative element" that does this — it is specifically the small size (close approach, strong electrostatic pull) combined, for oxygen, with the ability to π-bond, that lets these two elements alone push metals to their ceiling oxidation state.(iii) Highest oxidation state in oxoanionsAn oxoanion carries an overall negative charge because several oxide ions surround the central metal, and each of those \(\displaystyle \text{O}^{2-}\) ligands can donate into empty metal \(\displaystyle d\) orbitals through \(\displaystyle p\pi\)–\(\displaystyle d\pi\) bonding, exactly as in a neutral oxide. Having several oxygens do this at once removes a large amount of electron density from the metal, so an oxoanion structure can support a formal oxidation state on the metal that would be unstable in a simple binary compound. This is why the highest oxidation states of transition metals so often turn up written as an oxoanion rather than a simple salt: \[\text{MnO}_4^{-}\ (\text{Mn in } +7), \qquad \text{Cr}_2\text{O}_7^{2-} \text{ and } \text{CrO}_4^{2-}\ (\text{Cr in } +6) \] The negative charge of the anion is not incidental — it is the electrostatic and π-donation environment that makes such a high positive oxidation state on the metal viable at all.Answer: (i) Because bond character shifts from ionic to covalent as oxidation state rises — \(\displaystyle \text{MnO}\) (Mn²⁺) is basic while \(\displaystyle \text{Mn}_2\text{O}_7\) (Mn⁷⁺) is acidic; \(\displaystyle \text{CrO}\) (Cr²⁺) is basic while \(\displaystyle \text{CrO}_3\) (Cr⁶⁺) is acidic; \(\displaystyle \text{V}_2\text{O}_5\) (V⁵⁺) is amphoteric. (ii) O and F are small and highly electronegative, and O additionally forms \(\displaystyle p\pi\)–\(\displaystyle d\pi\) multiple bonds with the metal, so both can push a metal to its highest oxidation state — e.g., Mn reaches +$\displaystyle 7$ in \(\displaystyle \text{Mn}_2\text{O}_7\) but only +$\displaystyle 4$ in \(\displaystyle \text{MnF}_4\); V reaches +$\displaystyle 5$ in both \(\displaystyle \text{V}_2\text{O}_5\) and \(\displaystyle \text{VF}_5\). (iii) In oxoanions such as \(\displaystyle \text{MnO}_4^{-}\) and \(\displaystyle \text{Cr}_2\text{O}_7^{2-}\), multiple \(\displaystyle \text{O}^{2-}\) ligands π-bond to the metal and the anionic charge electrostatically stabilizes the metal's highest oxidation state, so the extreme oxidation states of transition metals are most often seen in these oxoanions.

Both preparations follow the same two-stage logic: first fuse the ore with an alkali under oxidizing conditions to pull the metal out as an alkali-metal oxo-salt, then oxidize/convert that salt to the final compound.(i) \(\displaystyle K_2Cr_2O_7\) from chromite oreChromite ore is \(\displaystyle FeCr_2O_4\) (iron(II) chromite) — chromium here is trapped inside a mixed iron–chromium oxide, so the very first job is to break it open and pull the chromium out as a soluble chromate.Step $\displaystyle 1$ — Fusion with sodium carbonate in air. The powdered ore is fused with \(\displaystyle Na_2CO_3\) with a free flow of air. The oxygen oxidizes chromium(III) inside the ore all the way to chromium(VI), while the carbonate ties up the iron as insoluble \(\displaystyle Fe_2O_3\): \[4FeCr_2O_4 + 8Na_2CO_3 + 7O_2 \xrightarrow{\Delta} 8Na_2CrO_4 + 2Fe_2O_3 + 8CO_2 \] This gives a yellow melt of sodium chromate, \(\displaystyle Na_2CrO_4\), which is leached out with water and filtered — the iron oxide stays behind as an insoluble residue.Step $\displaystyle 2$ — Acidification to the dichromate. The yellow chromate solution is acidified with sulphuric acid. Chromate and dichromate are just the same chromium(VI) in different protonation states — acidifying pushes two tetrahedral chromate ions to condense (sharing one oxygen) into the orange dichromate ion: \[2Na_2CrO_4 + 2H^+ \longrightarrow Na_2Cr_2O_7 + 2Na^+ + H_2O \] The aside people miss here: chromium's oxidation state is unchanged (+$\displaystyle 6$ throughout) — acidification is a condensation reaction, not a redox step.Step $\displaystyle 3$ — Swap the cation to get the low-solubility salt. Sodium dichromate is highly soluble and will not crystallize out cleanly, so the solution is treated with potassium chloride. Potassium dichromate is far less soluble than sodium chromide, so it crystallizes out on cooling while \(\displaystyle NaCl\) stays in solution: \[Na_2Cr_2O_7 + 2KCl \longrightarrow K_2Cr_2O_7\downarrow + 2NaCl \](ii) \(\displaystyle KMnO_4\) from pyrolusite orePyrolusite is manganese dioxide, \(\displaystyle MnO_2\), with manganese already at +4. To reach permanganate (+$\displaystyle 7$) it must be oxidized twice, so this preparation happens in two oxidizing stages instead of one.Step $\displaystyle 1$ — Oxidative fusion with \(\displaystyle KOH\). Powdered pyrolusite is fused with potassium hydroxide in the presence of air (or an oxidizing agent such as \(\displaystyle KNO_3\)). Atmospheric oxygen lifts manganese from +$\displaystyle 4$ to +$\displaystyle 6$, giving dark-green potassium manganate: \[2MnO_2 + 4KOH + O_2 \xrightarrow{\Delta} 2K_2MnO_4 + 2H_2O \]Step $\displaystyle 2$ — Oxidize manganate(VI) to permanganate(VII). The manganate ion, \(\displaystyle MnO_4^{2-}\), is unstable outside strongly alkaline solution — this is the step that decides which route is used industrially: \[2K_2MnO_4 + 2H_2O \xrightarrow{\text{electrolysis}} 2KMnO_4 + 2KOH + H_2 \] \[3MnO_4^{2-} + 4H^+ \longrightarrow 2MnO_4^- + MnO_2 + 2H_2O \] The aside worth flagging: this disproportionation step is why the electrolytic route is preferred industrially — it converts manganate to permanganate quantitatively instead of losing a third of the manganese back to \(\displaystyle MnO_2\).The purple potassium permanganate is then crystallized out of the alkaline solution.Answer: \(\displaystyle K_2Cr_2O_7\) is obtained by fusing chromite ore (\(\displaystyle FeCr_2O_4\)) with \(\displaystyle Na_2CO_3\) in air to sodium chromate, acidifying to sodium dichromate, then treating with \(\displaystyle KCl\) to precipitate potassium dichromate. \(\displaystyle KMnO_4\) is obtained by fusing pyrolusite (\(\displaystyle MnO_2\)) with \(\displaystyle KOH\) and air to green potassium manganate (\(\displaystyle K_2MnO_4\)), which is then oxidized — electrolytically in alkaline solution, or by disproportionation on acidification — to purple potassium permanganate.

An alloy is a homogeneous mixture of two or more metals (or a metal and a non-metal), made by melting the components together and letting the mixture solidify — it is a solid solution, not a chemical compound with a fixed formula.Because the atoms of the different metals simply pack together in one crystal lattice rather than forming true chemical bonds in a fixed ratio, an alloy's composition can vary somewhat, and its properties (hardness, melting point, resistance to corrosion) differ from those of the pure metals that went into it — usually alloys are harder and more resistant to corrosion than the parent metals. This is the point people mix up: an alloy is a physical blend of metals, not a compound held together by chemical bonding in fixed stoichiometric proportions.The important lanthanoid-containing alloy is mischmetal.Its composition is: \[\text{Mischmetal} \approx 95\% \text{ lanthanoid metal} + 5\% \text{ iron}, \text{ with traces of S, C, Ca, and Al} \]Here "lanthanoid metal" means a mixture of the naturally-occurring lanthanoid elements (mainly cerium, along with lanthanum, neodymium, and praseodymium) in roughly the proportions they occur together in the ore — they are not separated before making the alloy, which is what keeps mischmetal cheap compared to the pure separated lanthanoids.Use. Mischmetal is alloyed with magnesium to produce a magnesium-based alloy that is used to make bullets, shells, and lighter flints (the sparking flint in cigarette lighters and gas lighters) — the cerium in the mix is pyrophoric, meaning fine shavings of it ignite spontaneously in air, which is exactly the spark-producing behaviour a flint needs.Answer: An alloy is a homogeneous solid solution formed by melting two or more metals together and cooling the melt, giving properties different from (usually superior to) the constituent metals. The important lanthanoid-containing alloy is mischmetal (≈$\displaystyle 95$% lanthanoid metal + $\displaystyle 5$% iron, with traces of S, C, Ca, Al), used chiefly in Mg-based alloys to make bullets, shells, and lighter flints.

Inner transition elements are the ones where the last electron enters an f orbital, not a d orbital — so the label depends on electron configuration, not on which block of the periodic table you'd guess by eye.Inner transition elements are the lanthanides (\(\displaystyle Z = 58 \) to \(\displaystyle 71 \), filling the \(\displaystyle 4f \) subshell) and the actinides (\(\displaystyle Z = 90 \) to \(\displaystyle 103 \), filling the \(\displaystyle 5f \) subshell). Their general configuration is \(\displaystyle (n-2)f^{1-14}(n-1)d^{0-1}ns^2 \) — the defining feature is a partly filled f subshell in the ground state or in a common oxidation state. This is different from the ordinary transition elements (the d-block), where the last electron enters an \(\displaystyle (n-1)d \) orbital instead.The aside people trip on: an element sitting near the lanthanides/actinides on the table is not automatically "inner transition" — you have to check whether the differentiating electron actually goes into the f subshell or spills into the d subshell instead, which happens right at the boundaries of these series.Checking each atomic number against its ground-state configuration:\(\displaystyle Z = 29 \) (Cu): \(\displaystyle [Ar]\,3d^{10}4s^1 \) — the last electron fills \(\displaystyle 3d \). This is an ordinary transition (\(\displaystyle d \)-block) element, not inner transition.\(\displaystyle Z = 59 \) (Pr): \(\displaystyle [Xe]\,4f^3 6s^2 \) — the last electron fills \(\displaystyle 4f \). This is a lanthanide, so it is an inner transition element.\(\displaystyle Z = 74 \) (W): \(\displaystyle [Xe]\,4f^{14}5d^4 6s^2 \) — the \(\displaystyle 4f \) subshell is already complete; the last electron fills \(\displaystyle 5d \). This is a \(\displaystyle d \)-block (transition) element, not inner transition.\(\displaystyle Z = 95 \) (Am): \(\displaystyle [Rn]\,5f^7 7s^2 \) — the last electron fills \(\displaystyle 5f \). This is an actinide, so it is an inner transition element.\(\displaystyle Z = 102 \) (No): \(\displaystyle [Rn]\,5f^{14}7s^2 \) — the last electron fills \(\displaystyle 5f \). This is an actinide, so it is an inner transition element.\(\displaystyle Z = 104 \) (Rf): \(\displaystyle [Rn]\,5f^{14}6d^2 7s^2 \) — the \(\displaystyle 5f \) subshell is already complete (filled at No, \(\displaystyle Z=102 \)); the last electron fills \(\displaystyle 6d \). This is the first member of a new \(\displaystyle d \)-block series (a transactinide), not inner transition, even though it comes right after the actinides.So among $\displaystyle 29$, $\displaystyle 59$, $\displaystyle 74$, $\displaystyle 95$, $\displaystyle 102$, $\displaystyle 104$, the atomic numbers whose last electron enters an f orbital are $\displaystyle 59$, $\displaystyle 95$, and 102.**Answer: Inner transition elements are the lanthanides and actinides, whose last electron enters an \(\displaystyle (n-2)f \) orbital. Of the given atomic numbers, $\displaystyle 59$, $\displaystyle 95$, and $\displaystyle 102$ are inner transition elements ($\displaystyle 29$ and $\displaystyle 74$ are ordinary \(\displaystyle d \)-block transition elements, and $\displaystyle 104$ is the first \(\displaystyle d \)-block element of a new series, not an inner transition element).

The lanthanoids tell almost one story (+$\displaystyle 3$, again and again); the actinoids tell several different stories per element, because their 5f, 6d and 7s orbitals sit at nearly the same energy and all three can supply electrons for bonding.Why lanthanoid chemistry is smoothAcross La to Lu, the electrons added go into the deeply buried \(\displaystyle 4f \) subshell, which lies well below the \(\displaystyle 5d \) and \(\displaystyle 6s \) levels and is shielded strongly enough that it rarely takes part in bond formation. As a result almost every lanthanoid shows one dominant oxidation state, \(\displaystyle +3 \), and the whole series behaves in a chemically uniform way. Only two elements deviate, and both deviations have a clean explanation: These are the only common departures from \(\displaystyle +3 \); the aside is that they arise from special stability of empty/half-filled/full \(\displaystyle f \) subshells, not from any general willingness of \(\displaystyle 4f \) electrons to bond.Why actinoid chemistry is not smoothIn the actinoids, the \(\displaystyle 5f \), \(\displaystyle 6d \) and \(\displaystyle 7s \) orbitals are comparable in energy (a much smaller gap than the \(\displaystyle 4f\)–\(\displaystyle 5d\) gap in the lanthanoids), and the \(\displaystyle 5f \) orbitals themselves are more diffuse and shield the nucleus less effectively than \(\displaystyle 4f \) orbitals do. This lets electrons from all three subshells participate in bonding, so most actinoids show three, four, or more oxidation states instead of one. The actual pattern, element by element, is where the "not smooth" shows up directly: So while a lanthanoid keeps essentially the same oxidation state whatever compound it forms, an actinoid such as uranium or plutonium can be pushed between \(\displaystyle +3 \) and \(\displaystyle +6 \) (or \(\displaystyle +7 \)) by changing the oxidising or reducing conditions — the oxidation state is not fixed to the element the way it is for the lanthanoids. This variability, plus the more irregular ("actinoid") contraction caused by the poorer shielding of the \(\displaystyle 5f \) electrons, is exactly what makes actinoid chemistry harder to systematize than lanthanoid chemistry.Answer: Lanthanoids overwhelmingly show a single oxidation state, \(\displaystyle +3 \) (with \(\displaystyle \text{Ce}^{4+} \), \(\displaystyle \text{Eu}^{2+} \), \(\displaystyle \text{Yb}^{2+} \) as the only stability-driven exceptions), because the \(\displaystyle 4f \) electrons are too shielded and too low in energy to bond. Actinoids show a wide spread of oxidation states — Th (\(\displaystyle +4 \)), Pa (\(\displaystyle +5 \)), U (\(\displaystyle +3 \) to \(\displaystyle +6 \)), Np and Pu (up to \(\displaystyle +7 \)), Am (up to \(\displaystyle +6 \)), settling back to \(\displaystyle +3 \) only from curium onward — because the \(\displaystyle 5f \), \(\displaystyle 6d \) and \(\displaystyle 7s \) orbitals lie at comparable energies and all can take part in bonding. This far greater and more irregular range of accessible oxidation states is what makes actinoid chemistry less smooth than lanthanoid chemistry.

The actinoid series is defined by the progressive filling of the 5f subshell — it runs for exactly fourteen elements after actinium, and the last one is lawrencium.The actinoids are the Period $\displaystyle 7$ elements in which electrons are added into the 5f orbitals, just as the lanthanoids fill 4f in Period 6. The series is taken to start at thorium, \(\displaystyle \mathrm{Th} \) (\(\displaystyle Z=90\)), where 5f-filling begins, and runs through fourteen elements. Counting fourteen places from \(\displaystyle Z=90\) lands on\[Z = 90 + 13 = 103 \]so the last member of the actinoid series is lawrencium, \(\displaystyle \mathrm{Lr} \), atomic number 103.Electronic configurationBuild it the same way you build any configuration beyond a noble-gas core: start from the preceding noble gas, then add the remaining electrons in order of filling. For lawrencium this gives\[\mathrm{Lr}\;(Z=103):\quad [\mathrm{Rn}]\,5f^{14}\,6d^{1}\,7s^{2} \]Check the electron count: \(\displaystyle 86 + 14 + 1 + 2 = 103\), which matches \(\displaystyle Z\) for Lr.Possible oxidation statesAcross most of the actinoid series, the 5f, 6d and 7s orbitals lie close enough in energy that several of those electrons can be lost, which is why actinoids in the middle of the row show a wide spread of oxidation states — for example uranium shows +$\displaystyle 3$, +$\displaystyle 4$, +$\displaystyle 5$, +$\displaystyle 6$, and neptunium, plutonium and americium even reach +7.Lawrencium is the exception, and the reason is exactly the configuration just written. Its 5f subshell is already completely filled (\(\displaystyle 5f^{14}\)), so that shell is no longer available to lose electrons from — a filled f subshell is a stable, low-energy arrangement the atom "protects." The only electrons left outside that stable core are the single \(\displaystyle 6d^{1}\) and the pair \(\displaystyle 7s^{2}\). Losing all three of those,\[\mathrm{Lr} \;-\; 3e^{-} \;\longrightarrow\; \mathrm{Lr}^{3+}\quad\big([\mathrm{Rn}]\,5f^{14}\big) \]leaves the ion with the extra-stable, fully-filled \(\displaystyle 5f^{14}\) core — the same driving force that makes \(\displaystyle +3\) the dominant state for actinoids generally, but here it is essentially the only state observed, because there is no comparable-energy 5f electron left to add extra flexibility. This mirrors lutetium, \(\displaystyle \mathrm{Lu} \) (\(\displaystyle [\mathrm{Xe}]\,4f^{14}5d^{1}6s^{2}\)), the last lanthanoid, which likewise shows only \(\displaystyle +3\) because its 4f shell is already full.A short aside on the trap here: it is tempting to assume every actinoid behaves like uranium or plutonium and shows many oxidation states — but that variability comes specifically from 5f electrons still being "in play" energetically. Once 5f is completely filled, as at the end of the row, that variability disappears and the element falls back to the single, filled-subshell-protected \(\displaystyle +3\) state.Answer: The last actinoid is lawrencium, \(\displaystyle \mathrm{Lr} \) (\(\displaystyle Z=103\)), with configuration \(\displaystyle [\mathrm{Rn}]\,5f^{14}6d^{1}7s^{2}\); it shows only the \(\displaystyle +3\) oxidation state, since losing the \(\displaystyle 6d^{1}\) and \(\displaystyle 7s^{2}\) electrons leaves the highly stable, fully-filled \(\displaystyle 5f^{14}\) core.

Cerium loses its 6s and 5d electrons first, not its 4f electron — so Ce³⁺ keeps exactly one f electron, and that single electron decides everything about its magnetism.Step $\displaystyle 1$ — write the electronic configuration of neutral Ce (Z = $\displaystyle 58$).Cerium is one of the lanthanides with an irregular configuration: instead of filling 4f before 5d, it puts one electron in each,\[\text{Ce (Z = 58)} : [\text{Xe}]\,4f^{1}\,5d^{1}\,6s^{2} \]where \(\displaystyle [\text{Xe}]\) stands for the full xenon core (\(\displaystyle 1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,3d^{10}\,4s^2\,4p^6\,4d^{10}\,5s^2\,5p^6\), $\displaystyle 54$ electrons), and the three extra electrons sit in \(\displaystyle 4f^1\,5d^1\,6s^2\).Step $\displaystyle 2$ — remove three electrons to form Ce³⁺.When a lanthanide atom ionises, electrons are always lost from the outermost occupied shells first — \(\displaystyle 6s\) before \(\displaystyle 5d\) before \(\displaystyle 4f\) — because once the atom is stripped of its outer electrons, the 4f orbitals contract and drop below 5d and 6s in energy. This is the step that trips people up: it is tempting to remove the 4f electron since "f" looks more buried, but 4f is not the outermost shell here — 6s and 5d are, and Ce³⁺ removes exactly those three:\[\text{Ce}^{3+} = \text{Ce} - 3e^{-} : [\text{Xe}]\,4f^{1}\,5d^{0}\,6s^{0} = [\text{Xe}]\,4f^{1} \]So Ce³⁺ has a single electron sitting in the 4f subshell and nothing outside the xenon core besides it.Step $\displaystyle 3$ — apply Hund's rule to the \(\displaystyle 4f^1\) electron.The 4f subshell has \(\displaystyle l = 3\), so it has \(\displaystyle 2l+1 = 7\) degenerate orbitals (\(\displaystyle m_l = -3, -2, -1, 0, +1, +2, +3\)), each able to hold $\displaystyle 2$ electrons of opposite spin. Hund's rule says electrons fill degenerate orbitals singly, with parallel spin, before any pairing occurs. With only one electron to place, it simply drops into one of the seven empty 4f orbitals with its spin unpaired:\[4f:\ \underset{\uparrow}{\Box}\ \ \Box\ \ \Box\ \ \Box\ \ \Box\ \ \Box\ \ \Box \]There is no partner electron available to pair with, so this electron stays unpaired — giving\[n(\text{unpaired electrons}) = 1 \]Step $\displaystyle 4$ — calculate the magnetic moment from the spin-only formula.The spin-only formula for the magnetic moment of an ion is\[\mu = \sqrt{n(n+2)}\ \text{BM} \]where \(\displaystyle n\) is the number of unpaired electrons and BM (Bohr Magneton) is the unit of magnetic moment. This formula ignores orbital contribution — it counts only the spin of unpaired electrons, which is the level expected here.Substituting \(\displaystyle n = 1\):\[\mu = \sqrt{1(1+2)}\ \text{BM} = \sqrt{3}\ \text{BM} \]\[\mu = 1.732\ \text{BM} \]Rounding to three significant figures (the precision the spin-only model itself supports):\[\mu \approx 1.73\ \text{BM} \]Answer: Ce³⁺ has the configuration \(\displaystyle [\text{Xe}]\,4f^{1}\) with $\displaystyle 1$ unpaired electron, giving a spin-only magnetic moment of \(\displaystyle \mu = \sqrt{3} \approx 1.73\) BM.

The characteristic oxidation state of every lanthanoid is +$\displaystyle 3$ — but a few members swing to +$\displaystyle 2$ or +$\displaystyle 4$ whenever that gets them to an empty \(\displaystyle 4f^0\), half‑filled \(\displaystyle 4f^7\), or completely filled \(\displaystyle 4f^{14}\) configuration. These three arrangements have extra stability (empty and completely filled subshells minimise electron–electron repulsion; half‑filled subshells maximise exchange energy), and that stability can outweigh the energy cost of removing one extra electron or of not losing one that would normally be lost.Setting up the bookkeeping. A neutral lanthanoid atom is \(\displaystyle \text{[Xe]}\,4f^{1-14}5d^{0-1}6s^2\). To form \(\displaystyle Ln^{3+}\), an atom always loses its \(\displaystyle 6s^2\) electrons plus one \(\displaystyle 4f\) or \(\displaystyle 5d\) electron, so every \(\displaystyle Ln^{3+}\) ends up as a pure \(\displaystyle \text{[Xe]}\,4f^n\) ion with \(\displaystyle n = Z - 57\). It is the value of this \(\displaystyle n\) — how close it sits to $\displaystyle 0$, $\displaystyle 7$, or $\displaystyle 14$ — that decides whether an element can also be pushed to +$\displaystyle 4$ (one electron poorer than \(\displaystyle Ln^{3+}\)) or pulled back to +$\displaystyle 2$ (one electron richer than \(\displaystyle Ln^{3+}\)).Members showing +$\displaystyle 4$ (they want to lose one more electron than usual): Members showing +$\displaystyle 2$ (they resist losing their normal third electron): The pattern to notice: Ce and Tb are the two elements whose \(\displaystyle Ln^{3+}\) configuration sits one electron above an extra‑stable count (\(\displaystyle f^1\) above \(\displaystyle f^0\); \(\displaystyle f^8\) above \(\displaystyle f^7\)), so they are pulled up to +4. Eu and Yb are the two elements whose neutral atom sits exactly at an extra‑stable count once the \(\displaystyle 6s^2\) pair is removed (\(\displaystyle f^7\), \(\displaystyle f^{14}\)), so they are held back at +2. Pr, Nd, and Sm are one electron further away from the nearest magic number in each direction, so the same effect operates for them but more weakly, giving oxidation states that are known but less common and less stable than Ce(IV)/Tb(IV) or Eu(II)/Yb(II).Answer: +$\displaystyle 4$ oxidation state — Ce, Pr, Nd, Tb (Ce⁴⁺ = \(\displaystyle 4f^0\) and Tb⁴⁺ = \(\displaystyle 4f^7\) are the most stable, since they are one electron short of the parent \(\displaystyle Ln^{3+}\) configuration). +$\displaystyle 2$ oxidation state — Sm, Eu, Yb (Eu²⁺ = \(\displaystyle 4f^7\) and Yb²⁺ = \(\displaystyle 4f^{14}\) are the most stable, since they retain one electron more than the parent \(\displaystyle Ln^{3+}\) configuration). In every case the driving force is the extra stability of an empty, half‑filled, or completely filled 4f subshell.

The row of orbitals being filled — 4f for one series, 5f for the other — is the single fact that drives every difference below: how well it shields the nucleus decides how many oxidation states show up and how reactive the metal is.(i) Electronic configurationLanthanoids (Ce to Lu, following La) have the general configuration \[[\text{Xe}]\,4f^{1-14}\,5d^{0-1}\,6s^{2} \] Actinoids (Th to Lr, following Ac) have the general configuration \[[\text{Rn}]\,5f^{1-14}\,6d^{0-1}\,7s^{2} \] So structurally the two series run in parallel — one is the 4f row, the other the 5f row, each sitting one period below the last d-block element (La and Ac respectively), and each showing the same kind of "extra stability" exceptions: an electron shifts from f into d whenever that empties, half-fills, or fully fills the f subshell. That is why La is \(\displaystyle [\text{Xe}]5d^{1}6s^{2}\) (no f electron at all) and Gd is \(\displaystyle [\text{Xe}]4f^{7}5d^{1}6s^{2}\) (f half-filled); the same pattern gives Ac as \(\displaystyle [\text{Rn}]6d^{1}7s^{2}\) and Cm as \(\displaystyle [\text{Rn}]5f^{7}6d^{1}7s^{2}\).The difference is in how messy this filling gets. In the lanthanoids, \(\displaystyle 4f\), \(\displaystyle 5d\) and \(\displaystyle 6s\) are close in energy, so exceptions are limited. In the actinoids, \(\displaystyle 5f\), \(\displaystyle 6d\) and \(\displaystyle 7s\) are even closer in energy over a much wider stretch of the series — so actinoid configurations are far less regular, with more exceptions and even some configurations still disputed experimentally.The knock-on effect is size: because the \(\displaystyle 5f\) orbitals are more diffuse than \(\displaystyle 4f\) and shield the growing nuclear charge poorly, the actinoid contraction (the steady fall in atomic/ionic radius along Ac→Lr) is greater, element for element, than the lanthanoid contraction.(ii) Oxidation statesLanthanoids are overwhelmingly monotonous: +$\displaystyle 3$ is the characteristic and dominant oxidation state for every member, because losing the \(\displaystyle 6s^{2}\) and one \(\displaystyle 5d\)/\(\displaystyle 4f\) electron reaches this state easily and it is the one from which the fourth ionisation energy rises sharply. A few members also show +$\displaystyle 2$ or +$\displaystyle 4$, but only where it buys a special stability — an empty, half-filled, or fully filled \(\displaystyle 4f\) subshell: \(\displaystyle \text{Ce}^{4+}\) (\(\displaystyle f^{0}\)), \(\displaystyle \text{Tb}^{4+}\) (\(\displaystyle f^{7}\)), \(\displaystyle \text{Eu}^{2+}\) (\(\displaystyle f^{7}\)), \(\displaystyle \text{Yb}^{2+}\) (\(\displaystyle f^{14}\)).Actinoids, by contrast, show a genuinely wide spread of oxidation states — commonly +$\displaystyle 3$ to +$\displaystyle 6$, and up to +$\displaystyle 7$ for some (e.g. uranium shows +$\displaystyle 3$, +$\displaystyle 4$, +$\displaystyle 5$, +$\displaystyle 6$; neptunium and plutonium reach +$\displaystyle 7$ under strong oxidising conditions). This is a direct consequence of point (i): because \(\displaystyle 5f\), \(\displaystyle 6d\) and \(\displaystyle 7s\) are so close in energy, more electrons from all three sets can take part in bonding, not just the outer three. So while +$\displaystyle 3$ is common to both families, it is only one option among many for actinoids, whereas for lanthanoids it is essentially the only option.(iii) Chemical reactivityLanthanoids are electropositive metals whose reactivity falls off along the series: the early members (La, Ce, Pr) react with water almost as vigorously as calcium, while the later, smaller members (a consequence of the lanthanoid contraction) behave more like aluminium. They tarnish in air, burn in oxygen to the oxide \(\displaystyle M_2O_3\), combine with halogens to give \(\displaystyle MX_3\), and react with dilute acids liberating \(\displaystyle H_2\) gas; their oxides and hydroxides, \(\displaystyle M_2O_3\) and \(\displaystyle M(OH)_3\), are distinctly basic.Actinoids are markedly more reactive metals, particularly when finely divided. They react with boiling water to give a mixture of the oxide and the hydride, combine directly with most non-metals at only moderate temperature, and dissolve readily in mineral acids such as hydrochloric acid, though alkalis have little effect on them. This heightened reactivity traces back to the same cause as their oxidation-state variety: the diffuse, poorly-shielding \(\displaystyle 5f\) electrons are more easily drawn into bonding, and the early actinoids in particular are larger and more electropositive than their lanthanoid counterparts. Unlike the lanthanoids — where only promethium is radioactive — every actinoid is radioactive, which is itself an added complication in studying their chemistry (self-heating and radiolytic decomposition of compounds).Answer: Lanthanoids fill \(\displaystyle 4f\) (general configuration \(\displaystyle [Xe]4f^{1-14}5d^{0-1}6s^2\)), show almost exclusively the +$\displaystyle 3$ state (with rare +$\displaystyle 2$/+$\displaystyle 4$ only when an empty, half-filled, or filled \(\displaystyle 4f\) results), and are moderately reactive, electropositive metals of gradually falling reactivity along the series. Actinoids fill \(\displaystyle 5f\) (general configuration \(\displaystyle [Rn]5f^{1-14}6d^{0-1}7s^2\)) with far more irregular configurations and a greater actinoid contraction (poorer \(\displaystyle 5f\) shielding); they show a much wider range of oxidation states (+$\displaystyle 3$ up to +$\displaystyle 7$, since \(\displaystyle 5f\), \(\displaystyle 6d\), \(\displaystyle 7s\) are comparable in energy); and they are more reactive than the lanthanoids, besides being uniformly radioactive.

A lanthanide or actinide's configuration is the previous noble gas core plus the leftover electrons split between the s-orbital and the f (or d) orbital that is filling — and for the early actinides you cannot assume the f-orbital fills before the d one.For any atom, write its configuration as [core]\(\displaystyle +\) remaining electrons, where the core is the noble gas with the atomic number closest below it. Beyond period $\displaystyle 4$, the filling order runs \(\displaystyle 6s < 4f < 5d\) for period $\displaystyle 6$, and \(\displaystyle 7s < 5f < 6d\) for period 7. Each s-subshell holds at most $\displaystyle 2$ electrons, each f-subshell at most $\displaystyle 14$, each d-subshell at most $\displaystyle 10$ — so once you know how many electrons are left after the core and the s-electrons, you know how many are left to place in f (and, for actinides, d).Z = $\displaystyle 61$ — Promethium (Pm), a lanthanideThe last noble gas below $\displaystyle 61$ is xenon, \(\displaystyle Z = 54 \). Electrons remaining: \[61 - 54 = 7 \] Of these, $\displaystyle 2$ fill \(\displaystyle 6s\) (its maximum), leaving \(\displaystyle 7 - 2 = 5\) for \(\displaystyle 4f\): \[\text{Pm: } [\text{Xe}]\,4f^{5}\,6s^{2} \]Z = $\displaystyle 91$ — Protactinium (Pa), an actinideThe last noble gas below $\displaystyle 91$ is radon, \(\displaystyle Z = 86 \). Electrons remaining: \[91 - 86 = 5 \] Two fill \(\displaystyle 7s\), leaving $\displaystyle 3$ electrons to be split between \(\displaystyle 5f\) and \(\displaystyle 6d\). This is the step where the actinides differ from the lanthanides: early in the actinide series \(\displaystyle 5f\) and \(\displaystyle 6d\) sit so close in energy that electrons occupy both before \(\displaystyle 5f\) is anywhere near full — filling all $\displaystyle 3$ straight into \(\displaystyle 5f\) is the mistake to watch for. For protactinium the accepted ground-state split is $\displaystyle 2$ electrons in \(\displaystyle 5f\) and $\displaystyle 1$ in \(\displaystyle 6d\): \[\text{Pa: } [\text{Rn}]\,5f^{2}\,6d^{1}\,7s^{2} \]Z = $\displaystyle 101$ — Mendelevium (Md), an actinideElectrons remaining beyond radon: \[101 - 86 = 15 \] By this point in the actinide series the stray \(\displaystyle 6d\) electron seen in elements like Pa has been reabsorbed, and filling runs regularly through \(\displaystyle 5f\). Two electrons fill \(\displaystyle 7s\), leaving \(\displaystyle 15 - 2 = 13\) for \(\displaystyle 5f\): \[\text{Md: } [\text{Rn}]\,5f^{13}\,7s^{2} \]Z = $\displaystyle 109$ — Meitnerium (Mt), a d-block elementElectrons remaining beyond radon: \[109 - 86 = 23 \] Two fill \(\displaystyle 7s\), leaving $\displaystyle 21$ electrons for \(\displaystyle 5f\) and \(\displaystyle 6d\) together. \(\displaystyle 5f\) fills completely first (its maximum is $\displaystyle 14$), leaving \(\displaystyle 21 - 14 = 7\) for \(\displaystyle 6d\): \[\text{Mt: } [\text{Rn}]\,5f^{14}\,6d^{7}\,7s^{2} \] This also matches where Mt sits on the periodic table — directly below cobalt, rhodium, and iridium in group $\displaystyle 9$ — so a filled \(\displaystyle 5f^{14}\) core plus \(\displaystyle 6d^{7}\) is exactly the configuration its position predicts.Answer: Z = $\displaystyle 61$ (Pm): \(\displaystyle [\text{Xe}]\,4f^{5}\,6s^{2}\); Z = $\displaystyle 91$ (Pa): \(\displaystyle [\text{Rn}]\,5f^{2}\,6d^{1}\,7s^{2}\); Z = $\displaystyle 101$ (Md): \(\displaystyle [\text{Rn}]\,5f^{13}\,7s^{2}\); Z = $\displaystyle 109$ (Mt): \(\displaystyle [\text{Rn}]\,5f^{14}\,6d^{7}\,7s^{2}\).

The first (3d) series looks like a preview of the second (4d) and third (5d) series, but the third series breaks the pattern everywhere — because of the lanthanoid contraction sitting just before it.(i) Electronic configurations. All three series share the general outer configuration \(\displaystyle (n-1)d^{1-10}\,ns^{0-2} \): In a vertical column the outer configuration usually matches — but the 4d and 5d elements deviate from the "expected" pattern far more often than the 3d elements, because in the heavier series the \(\displaystyle (n-1)d \) and \(\displaystyle ns \) orbitals lie closer in energy, so an electron slips from \(\displaystyle ns \) into \(\displaystyle (n-1)d \) whenever that gives a filled or half-filled d subshell. This is the step where students expect a clean, unbroken \(\displaystyle d^{n}s^{2} \) pattern down a group and get caught out: compare \(\displaystyle \mathrm{Ni}\,3d^{8}4s^{2} \) with \(\displaystyle \mathrm{Pd}\,4d^{10}5s^{0} \) and \(\displaystyle \mathrm{Pt}\,5d^{9}6s^{1} \) — three very different splits of the same ten valence electrons in one group.(ii) Oxidation states. Going down a column, transition metals show more oxidation states, and the higher ones become more stable — the opposite of the inert-pair trend in the p-block. In Group $\displaystyle 7$, \(\displaystyle \mathrm{Mn} \) is most stable as \(\displaystyle \mathrm{Mn}^{2+} \), and its \(\displaystyle +7 \) state (\(\displaystyle \mathrm{MnO_4^-} \)) is a strong, easily-reduced oxidiser; but \(\displaystyle \mathrm{Tc} \) and \(\displaystyle \mathrm{Re} \) form stable \(\displaystyle +7 \) oxides (\(\displaystyle \mathrm{Tc_2O_7}, \mathrm{Re_2O_7} \)) and the perrhenate ion is a far weaker oxidant than permanganate. In Group $\displaystyle 8$, iron rarely gets past \(\displaystyle +3 \) in ordinary chemistry (its \(\displaystyle +6 \) ferrate is a strong oxidiser and hard to isolate), while \(\displaystyle \mathrm{Ru} \) and \(\displaystyle \mathrm{Os} \) readily form the stable, isolable \(\displaystyle +8 \) oxides \(\displaystyle \mathrm{RuO_4} \) and \(\displaystyle \mathrm{OsO_4} \). The reason is that in the 4d/5d metals, the valence d and s orbitals are close enough in energy that more electrons can be pulled into bonding, so high formal charges are stabilised rather than punished as the group is descended.(iii) Ionisation enthalpies. Along any one period, ionisation enthalpy rises overall (with dips at half-filled and filled configurations). Down a column, though, d-block ionisation enthalpy increases from 3d to 4d to 5d — this is the step people misapply, carrying over the s/p-block rule that ionisation enthalpy falls down a group. It doesn't fall here because the atomic radius does not expand as much as an extra shell should allow (see point iv), so the effective nuclear charge felt by the outermost electron actually grows. Group $\displaystyle 11$ shows this clearly: first ionisation enthalpies run \(\displaystyle \mathrm{Cu}\approx 745\ \text{kJ mol}^{-1} \), \(\displaystyle \mathrm{Ag}\approx 731\ \text{kJ mol}^{-1} \), \(\displaystyle \mathrm{Au}\approx 890\ \text{kJ mol}^{-1} \) — gold's is the highest of the three, not the lowest, because the poorly-shielding 4f electrons filled in just before it leave its nucleus more exposed than expected.(iv) Atomic sizes. From the 3d to the 4d series, atomic radius increases normally, as it does anywhere a new principal shell is added. But from 4d to 5d, that expected increase is almost completely cancelled by the lanthanoid contraction: the steady, cumulative shrinkage caused by the poorly-shielding 4f electrons that fill across \(\displaystyle \mathrm{La} \) to \(\displaystyle \mathrm{Lu} \) before the 5d series even starts. The result is that 5d atoms are nearly the same size as the 4d atoms directly above them — sometimes marginally smaller. \(\displaystyle \mathrm{Zr} \) (4d, radius \(\displaystyle \approx 160\ \text{pm} \)) and \(\displaystyle \mathrm{Hf} \) (5d, radius \(\displaystyle \approx 159\ \text{pm} \)) are essentially identical in size, despite Hf carrying a whole extra filled 4f shell that "should" have made it larger. This is exactly why Zr and Hf (and similarly Nb/Ta) are so hard to separate chemically — same size, same charge density, nearly identical chemistry.Putting the four together: the first series behaves as the textbook d-block pattern, while the second and third series increasingly resemble each other rather than the first, because the lanthanoid contraction inserted between them suppresses the size increase (and so also the ionisation-enthalpy decrease) that would otherwise be expected on descending the group, while simultaneously making higher oxidation states more accessible and stable.Answer: Configurations follow \(\displaystyle (n-1)d^{1-10}ns^{0-2}\) in all three series, but 4d/5d elements deviate more often (e.g. Pd \(\displaystyle 4d^{10}5s^0\), Pt \(\displaystyle 5d^9 6s^1\)); oxidation states increase in number and the higher ones grow more stable down a group (Mn²⁺ vs. stable Re⁷⁺; Fe vs. stable Ru/Os⁸⁺); ionisation enthalpy rises from 3d→4d→5d (opposite to the s/p-block trend); and atomic radius rises from 3d→4d but then stalls from 4d→5d because of the lanthanoid contraction (Zr ≈ $\displaystyle 160$ pm ≈ Hf, $\displaystyle 159$ pm) — so the 4d and 5d elements of a group resemble each other much more closely than either resembles the 3d element above them.

When a transition-metal atom loses electrons to form a cation, the electrons leave the 4s orbital first, never the 3d — even though 4s filled first when the atom was built up. In the neutral atom \(\displaystyle 4s\) sits slightly below \(\displaystyle 3d\) in energy, so it fills first. Once the \(\displaystyle 3d\) orbitals start holding electrons, though, they drop below \(\displaystyle 4s\) in energy for that species — so ionisation always strips the outer \(\displaystyle 4s\) electrons before touching \(\displaystyle 3d\). Getting this order backwards is the single most common mistake on this question.Start from each atom's ground-state configuration (atomic number in brackets), then remove electrons \(\displaystyle 4s\) first:\(\displaystyle \mathrm{Ti}\ (Z=22):\ [\mathrm{Ar}]3d^24s^2 \ \Rightarrow\ \mathrm{Ti}^{2+}: [\mathrm{Ar}]3d^2\) — $\displaystyle 2$ electrons in 3d\(\displaystyle \mathrm{V}\ (Z=23):\ [\mathrm{Ar}]3d^34s^2 \ \Rightarrow\ \mathrm{V}^{2+}: [\mathrm{Ar}]3d^3\) — $\displaystyle 3$ electrons in 3d\(\displaystyle \mathrm{Cr}\ (Z=24):\ [\mathrm{Ar}]3d^54s^1 \ \Rightarrow\ \mathrm{Cr}^{3+}\) loses the one \(\displaystyle 4s\) electron and then two more from \(\displaystyle 3d^5\): \(\displaystyle [\mathrm{Ar}]3d^3\) — $\displaystyle 3$ electrons in 3d (chromium's neutral atom is itself an exception, \(\displaystyle 3d^54s^1\) rather than \(\displaystyle 3d^44s^2\), for the extra stability of a half-filled \(\displaystyle d\) subshell)\(\displaystyle \mathrm{Mn}\ (Z=25):\ [\mathrm{Ar}]3d^54s^2 \ \Rightarrow\ \mathrm{Mn}^{2+}: [\mathrm{Ar}]3d^5\) — $\displaystyle 5$ electrons in 3d\(\displaystyle \mathrm{Fe}\ (Z=26):\ [\mathrm{Ar}]3d^64s^2 \ \Rightarrow\ \mathrm{Fe}^{2+}: [\mathrm{Ar}]3d^6\) — $\displaystyle 6$ electrons in 3d; \(\displaystyle \mathrm{Fe}^{3+}\) loses one more, from \(\displaystyle 3d\), giving \(\displaystyle [\mathrm{Ar}]3d^5\) — $\displaystyle 5$ electrons in 3d\(\displaystyle \mathrm{Co}\ (Z=27):\ [\mathrm{Ar}]3d^74s^2 \ \Rightarrow\ \mathrm{Co}^{2+}: [\mathrm{Ar}]3d^7\) — $\displaystyle 7$ electrons in 3d\(\displaystyle \mathrm{Ni}\ (Z=28):\ [\mathrm{Ar}]3d^84s^2 \ \Rightarrow\ \mathrm{Ni}^{2+}: [\mathrm{Ar}]3d^8\) — $\displaystyle 8$ electrons in 3d\(\displaystyle \mathrm{Cu}\ (Z=29):\ [\mathrm{Ar}]3d^{10}4s^1 \ \Rightarrow\ \mathrm{Cu}^{2+}\) loses the \(\displaystyle 4s\) electron and one from \(\displaystyle 3d^{10}\): \(\displaystyle [\mathrm{Ar}]3d^9\) — $\displaystyle 9$ electrons in 3d (copper's neutral atom is also an exception, \(\displaystyle 3d^{10}4s^1\), for the extra stability of a filled \(\displaystyle d\) subshell)Now place these electrons in the five 3d orbitals of the hydrated (octahedral, \(\displaystyle [\mathrm{M}(\mathrm{H_2O})_6]^{n+}\)) ion. In an octahedral field the five otherwise-degenerate 3d orbitals split into two sets: the lower-energy \(\displaystyle t_{2g}\) set (\(\displaystyle d_{xy}, d_{yz}, d_{zx}\); $\displaystyle 3$ orbitals) and the higher-energy \(\displaystyle e_g\) set (\(\displaystyle d_{z^2}, d_{x^2-y^2}\); $\displaystyle 2$ orbitals). Water is only a weak-to-intermediate field ligand for these first-row ions, so the crystal-field splitting energy \(\displaystyle \Delta_o\) is smaller than the electron-pairing energy — the hydrated ions are high-spin. That means electrons occupy the five orbitals singly (Hund's rule) across both sets before any pairing begins; pairing starts only once a fifth electron must be added, and it fills the lower \(\displaystyle t_{2g}\) set first.

Ion	3d electrons	\(\displaystyle t_{2g}\)	\(\displaystyle e_g\)
\(\displaystyle \mathrm{Ti}^{2+}\)	$\displaystyle 2$	\(\displaystyle t_{2g}^{2}\)	\(\displaystyle e_g^{0}\)
\(\displaystyle \mathrm{V}^{2+}\)	$\displaystyle 3$	\(\displaystyle t_{2g}^{3}\)	\(\displaystyle e_g^{0}\)
\(\displaystyle \mathrm{Cr}^{3+}\)	$\displaystyle 3$	\(\displaystyle t_{2g}^{3}\)	\(\displaystyle e_g^{0}\)
\(\displaystyle \mathrm{Mn}^{2+}\)	$\displaystyle 5$	\(\displaystyle t_{2g}^{3}\)	\(\displaystyle e_g^{2}\)
\(\displaystyle \mathrm{Fe}^{2+}\)	$\displaystyle 6$	\(\displaystyle t_{2g}^{4}\)	\(\displaystyle e_g^{2}\)
\(\displaystyle \mathrm{Fe}^{3+}\)	$\displaystyle 5$	\(\displaystyle t_{2g}^{3}\)	\(\displaystyle e_g^{2}\)
\(\displaystyle \mathrm{Co}^{2+}\)	$\displaystyle 7$	\(\displaystyle t_{2g}^{5}\)	\(\displaystyle e_g^{2}\)
\(\displaystyle \mathrm{Ni}^{2+}\)	$\displaystyle 8$	\(\displaystyle t_{2g}^{6}\)	\(\displaystyle e_g^{2}\)
\(\displaystyle \mathrm{Cu}^{2+}\)	$\displaystyle 9$	\(\displaystyle t_{2g}^{6}\)	\(\displaystyle e_g^{3}\)

Reading the table row by row: up to \(\displaystyle d^3\) (\(\displaystyle \mathrm{Ti}^{2+}, \mathrm{V}^{2+}, \mathrm{Cr}^{3+}\)) the electrons go singly into the three \(\displaystyle t_{2g}\) orbitals with none in \(\displaystyle e_g\) yet. At \(\displaystyle d^5\) (\(\displaystyle \mathrm{Mn}^{2+}, \mathrm{Fe}^{3+}\)) all five orbitals now hold one electron each — three in \(\displaystyle t_{2g}\), two in \(\displaystyle e_g\) — the maximum-multiplicity, all-unpaired arrangement. Beyond \(\displaystyle d^5\) every extra electron must pair up, and pairing goes into the lower \(\displaystyle t_{2g}\) set first: \(\displaystyle \mathrm{Fe}^{2+}\) (\(\displaystyle d^6\)) pairs one \(\displaystyle t_{2g}\) electron to give \(\displaystyle t_{2g}^{4}e_g^{2}\), \(\displaystyle \mathrm{Co}^{2+}\) (\(\displaystyle d^7\)) gives \(\displaystyle t_{2g}^{5}e_g^{2}\), \(\displaystyle \mathrm{Ni}^{2+}\) (\(\displaystyle d^8\)) gives \(\displaystyle t_{2g}^{6}e_g^{2}\) (the \(\displaystyle t_{2g}\) set now fully paired), and \(\displaystyle \mathrm{Cu}^{2+}\) (\(\displaystyle d^9\)) must finally start pairing in \(\displaystyle e_g\) too, giving \(\displaystyle t_{2g}^{6}e_g^{3}\).Answer: 3d electron counts — \(\displaystyle \mathrm{Ti}^{2+}=2,\ \mathrm{V}^{2+}=3,\ \mathrm{Cr}^{3+}=3,\ \mathrm{Mn}^{2+}=5,\ \mathrm{Fe}^{2+}=6,\ \mathrm{Fe}^{3+}=5,\ \mathrm{Co}^{2+}=7,\ \mathrm{Ni}^{2+}=8,\ \mathrm{Cu}^{2+}=9\). For the hydrated (octahedral) ions, all are high-spin, with orbital occupations \(\displaystyle t_{2g}^{2}e_g^{0},\ t_{2g}^{3}e_g^{0},\ t_{2g}^{3}e_g^{0},\ t_{2g}^{3}e_g^{2},\ t_{2g}^{4}e_g^{2},\ t_{2g}^{3}e_g^{2},\ t_{2g}^{5}e_g^{2},\ t_{2g}^{6}e_g^{2},\ t_{2g}^{6}e_g^{3}\) respectively.

The claim is true, and the reason behind almost every difference is the same: 3d orbitals are compact and poorly shielded, while 4d and 5d orbitals are large, diffuse, and — for the 5d series — squeezed in size by the lanthanoid contraction that sits just before them. Once you see that one geometric fact, each "different property" below stops looking like a separate rule and becomes a consequence of it.1. Atomic and ionic radii. Going down a group, radius normally increases (more shells). It does for Sc→Y (3d→4d), but Y→La→Hf (4d→5d) barely changes size at all. Between La (Z = $\displaystyle 57$) and Hf (Z = $\displaystyle 72$) the $\displaystyle 14$ lanthanoids are filling the inner 4f subshell, which shields the nucleus very poorly. The steady, cumulative shrinkage this causes — the lanthanoid contraction — almost exactly cancels the size increase expected on adding a whole new shell. The result: Zr and Hf, Nb and Ta, Mo and W are nearly identical in radius, and hence in chemistry (which is why separating Zr from Hf is notoriously hard), while the first-row member of each triad (Ti, Nb's partner V, Mo's partner Cr) is distinctly smaller and behaves differently from both.2. Density. Density is mass divided by volume, and the lanthanoid contraction fixes the volume of a 5d atom close to that of its 4d partner while its atomic mass is far higher. So density roughly doubles from the first series to the second, and rises only a little further into the third: iron is about \(\displaystyle 7.9\ \mathrm{g\,cm^{-3}}\), but osmium and iridium — its group-mates two rows down — are around \(\displaystyle 22.6\) and \(\displaystyle 22.4\ \mathrm{g\,cm^{-3}}\), among the densest elements known. First-series metals are comparatively light.3. Enthalpy of atomization and melting point. 4d and 5d valence orbitals are larger and more diffuse than 3d orbitals, so they overlap more extensively with neighbouring atoms, giving stronger metal–metal bonding in the bulk metal. Enthalpies of atomization and melting points therefore climb sharply down each group — tungsten (5d) has the highest melting point of any metal, far above its 3d counterpart chromium — whereas the first-series metals atomize and melt far more readily.4. Ionization enthalpy and reactivity. The poor shielding from the intervening 4f electrons leaves the nucleus of a 5d atom unusually effective at holding on to its valence electrons, so first ionization enthalpies rise irregularly but markedly from 3d to 5d. This is why the heavy end of the 5d row — platinum, gold, mercury — are noble metals, resistant to oxidation and to attack by dilute acids, while several first-series metals (Fe, Mn, Zn) dissolve in dilute acids liberating \(\displaystyle \mathrm{H_2}\) with ease.5. Oxidation states. First-series metals reach high oxidation states reluctantly, and those states are strong oxidizers once formed — \(\displaystyle \mathrm{Mn}\) in \(\displaystyle \mathrm{MnO_4^-}\) is +$\displaystyle 7$, but \(\displaystyle \mathrm{MnO_4^-}\) is a powerful, easily-reduced oxidizing agent. Heavier transition metals reach even higher oxidation states and hold them stably: \(\displaystyle \mathrm{OsO_4}\) has Os in +$\displaystyle 8$, \(\displaystyle \mathrm{ReF_7}\) has Re in +$\displaystyle 7$, and both are comparatively stable, isolable compounds rather than aggressive oxidants. The larger, more diffuse 4d/5d orbitals form stronger, more covalent M–O and M–halogen bonds that can support the higher charge.6. Metal–metal bonding and cluster chemistry. Because 4d and 5d orbitals are more diffuse, they overlap with each other far more effectively than 3d orbitals do. Second- and third-row transition metals form extensive metal–metal bonded clusters — species such as \(\displaystyle \mathrm{[Re_3Cl_{12}]^{3-}}\) or \(\displaystyle \mathrm{[Mo_6Cl_8]^{4+}}\) — a type of chemistry the first series shows only rarely (a little in Cr(II) chemistry).7. Magnetic behaviour. For most 3d complexes, orbital angular momentum is largely quenched by the surrounding ligand field, so the observed magnetic moment matches the spin-only formula \(\displaystyle \mu = \sqrt{n(n+2)}\) BM (n = number of unpaired electrons) quite closely. In 4d and 5d complexes, spin–orbit coupling is much stronger, so moments deviate substantially from this simple formula, and many are diamagnetic altogether because the crystal field splitting \(\displaystyle \Delta_o\) is large enough to force full electron pairing.8. Crystal field splitting and spin state. \(\displaystyle \Delta_o\) is roughly $\displaystyle 1.5$–$\displaystyle 2$ times larger for 4d and 5d metal ions than for their 3d counterparts, because the larger d orbitals interact more strongly with ligand orbitals. Pairing energy exceeds \(\displaystyle \Delta_o\) for many 3d ions, favouring high-spin complexes; for 4d/5d ions the reverse is usually true, so low-spin complexes (and higher coordination numbers, since the bigger ion has more room) dominate.9. Abundance. First-series elements such as Fe, Ti, and Mn are among the more abundant elements in the Earth's crust, while their 4d/5d congeners (Os, Ir, Re) are among the rarest — a purely geochemical difference, but one more way the "first ten" behave unlike the thirty that follow them.Every one of these contrasts — size, density, melting point, reactivity, favoured oxidation states, tendency to metal–metal bond, and magnetic behaviour — traces back to the same cause: 3d orbitals are small and poorly-shielding, 4d/5d orbitals are large and diffuse, and the lanthanoid contraction squeezes the 5d row down to almost the same size as the 4d row above it. That is why the statement is correct.Answer: True — first-series (3d) transition elements differ from the heavier (4d, 5d) transition elements in atomic/ionic radii, density, enthalpy of atomization, ionization enthalpy, oxidation-state stability, tendency to form metal–metal bonds, and magnetic behaviour, all traceable to the compact, poorly-shielded 3d orbitals versus the larger, more diffuse 4d/5d orbitals (whose size is further fixed by the lanthanoid contraction).

The number of unpaired electrons sets the spin-only magnetic moment; when the observed value runs higher than that, the excess comes from unquenched orbital motion, not from extra unpaired spins.Step $\displaystyle 1$ — get the oxidation state and \(\displaystyle d\)-electron count of the metal in each complex.For \(\displaystyle K_4[Mn(CN)_6]\): four \(\displaystyle K^+\) balance a \(\displaystyle 4-\) charge on the complex ion, and each \(\displaystyle CN^-\) carries \(\displaystyle -1\): \[x + 6(-1) = -4 \implies x = +2 \] So the metal is \(\displaystyle Mn^{2+}\). Neutral Mn is \(\displaystyle [Ar]3d^54s^2\); the \(\displaystyle 4s\) electrons leave first on ionisation, so \(\displaystyle \mathrm{Mn^{2+}}\) is \(\displaystyle [Ar]3d^5\) — a \(\displaystyle d^5\) ion.For \(\displaystyle [Fe(H_2O)_6]^{2+}\): water is neutral, so the \(\displaystyle +2\) charge sits entirely on the metal: \(\displaystyle \mathrm{Fe^{2+}}\) is \(\displaystyle [Ar]3d^6\), a \(\displaystyle d^6\) ion.For \(\displaystyle K_2[MnCl_4]\): two \(\displaystyle K^+\) balance a \(\displaystyle 2-\) charge on the complex, and four \(\displaystyle Cl^-\) contribute \(\displaystyle -4\): \[x + 4(-1) = -2 \implies x = +2 \] Again \(\displaystyle \mathrm{Mn^{2+}}\), \(\displaystyle d^5\).Aside — get the oxidation state from the counter-ions and ligand charges first; reading it straight off the metal's "usual" valence is exactly where this kind of problem goes wrong.Step $\displaystyle 2$ — assign the spin state from the ligand and the geometry.\(\displaystyle CN^-\) is a strong-field ligand, so \(\displaystyle [Mn(CN)_6]^{4-}\) is low spin: the five \(\displaystyle d\)-electrons crowd into the lower \(\displaystyle t_{2g}\) set, \(\displaystyle t_{2g}^5e_g^0\), leaving \(\displaystyle n = 1\) unpaired electron.\(\displaystyle H_2O\) is a weak-field ligand, so \(\displaystyle [Fe(H_2O)_6]^{2+}\) is high spin: \(\displaystyle t_{2g}^4e_g^2\), giving \(\displaystyle n = 4\) unpaired electrons.\(\displaystyle [MnCl_4]^{2-}\) has only four ligands, so it is tetrahedral. The tetrahedral splitting \(\displaystyle \Delta_t\) is always too small to force pairing, so tetrahedral complexes are high spin regardless of the ligand: the configuration is \(\displaystyle e^2t_2^3\), and all \(\displaystyle n = 5\) electrons stay unpaired.Step $\displaystyle 3$ — compute the spin-only moment and compare with the given value.Formula (spin-only magnetic moment): \(\displaystyle \mu = \sqrt{n(n+2)}\) BM, where \(\displaystyle n\) is the number of unpaired electrons and BM (Bohr magneton) is the unit of magnetic moment.For \(\displaystyle [Mn(CN)_6]^{4-}\), \(\displaystyle n = 1\): \[\mu_{calc} = \sqrt{1(1+2)} = \sqrt{3} = 1.73 \text{ BM} \] Given (observed): \(\displaystyle 2.2\) BM — noticeably above \(\displaystyle 1.73\) BM.For \(\displaystyle [Fe(H_2O)_6]^{2+}\), \(\displaystyle n = 4\): \[\mu_{calc} = \sqrt{4(4+2)} = \sqrt{24} = 4.90 \text{ BM} \] Given (observed): \(\displaystyle 5.3\) BM — again above the calculated value.For \(\displaystyle [MnCl_4]^{2-}\), \(\displaystyle n = 5\): \[\mu_{calc} = \sqrt{5(5+2)} = \sqrt{35} = 5.92 \text{ BM} \] Given (observed): \(\displaystyle 5.9\) BM — matches the calculated value almost exactly.Step $\displaystyle 4$ — read the pattern.Aside — the spin-only formula assumes the electron's orbital motion contributes nothing to the magnetic moment, which is true only when the ground term has no orbital degeneracy. When that assumption fails, the observed moment sits above the spin-only number (never below), and a small excess does not mean extra unpaired electrons — you have to check whether it is a small bump (orbital contribution) or a jump all the way to the value for one more unpaired electron.In the two octahedral complexes, the odd or asymmetrically-placed electrons sit in the triply degenerate \(\displaystyle t_{2g}\) set, giving orbitally degenerate ground terms — \(\displaystyle {}^2T_{2g}\) for \(\displaystyle [Mn(CN)_6]^{4-}\) and \(\displaystyle {}^5T_{2g}\) for \(\displaystyle [Fe(H_2O)_6]^{2+}\). That residual orbital angular momentum adds to the spin moment, so both observed values (\(\displaystyle 2.2\) BM and \(\displaystyle 5.3\) BM) land above their spin-only predictions without signalling any extra unpaired electron.In \(\displaystyle [MnCl_4]^{2-}\), the high-spin \(\displaystyle d^5\) tetrahedral configuration \(\displaystyle e^2t_2^3\) has ground term \(\displaystyle {}^6A_1\) — an orbitally non-degenerate (A-type) term, so there is no orbital contribution to add. Its observed moment (\(\displaystyle 5.9\) BM) therefore lands right on the spin-only value (\(\displaystyle 5.92\) BM) for \(\displaystyle 5\) unpaired electrons.Answer: The values confirm $\displaystyle 1$, $\displaystyle 4$, and $\displaystyle 5$ unpaired electrons respectively — \(\displaystyle [Mn(CN)_6]^{4-}\) is low-spin \(\displaystyle d^5\) (\(\displaystyle t_{2g}^5\), calc. $\displaystyle 1.73$ BM, obs. $\displaystyle 2.2$ BM), \(\displaystyle [Fe(H_2O)_6]^{2+}\) is high-spin \(\displaystyle d^6\) (\(\displaystyle t_{2g}^4e_g^2\), calc. $\displaystyle 4.90$ BM, obs. $\displaystyle 5.3$ BM), and \(\displaystyle [MnCl_4]^{2-}\) is high-spin tetrahedral \(\displaystyle d^5\) (\(\displaystyle e^2t_2^3\), calc. $\displaystyle 5.92$ BM, obs. $\displaystyle 5.9$ BM); the first two exceed their spin-only values because their T-term ground states carry orbital contribution, while the third's \(\displaystyle {}^6A_1\) ground term has none, so its observed and calculated moments coincide.