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

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.