The d- and f- Block Elements

This article follows the chapter's own sequence: position and configuration (where the d- and f-blocks sit, the four transition series, the $(n-1)d\,ns$ pattern and its Cr/Cu exceptions), then the general properties of the 3d series one at a time — physical properties and atomisation enthalpy, atomic/ionic radii, ionisation enthalpies, oxidation states, the M2+/M and M3+/M2+ electrode potentials, stability of higher oxidation states, and magnetic, colour, catalytic, interstitial and alloy behaviour. A dedicated section covers the chapter's two named compounds, potassium dichromate and potassium permanganate. The article then turns to the f-block — lanthanoids (contraction, oxidation states) and actinoids (radioactivity, a wider oxidation-state range, comparison with lanthanoids) — closing with the industrial applications that make this chapter genuinely load-bearing.

Position, series and electronic configuration

The d-block occupies the broad middle section of the periodic table, flanked by the s-block and the p-block. Moving across a period, electrons fill the $(n-1)d$ orbitals — the penultimate shell — giving four rows: the 3d series (Sc to Zn, Z 21-30), the 4d series (Y to Cd, Z 39-48), the 5d series (La, then Hf to Hg, Z 57 and 72-80) and the 6d series (Ac, then Rf to Cn, Z 89 and 104-112). The f-block sits as a separate panel below: the 4f series, Ce to Lu, the lanthanoids, and the 5f series, Th to Lr, the actinoids. The general outer electronic configuration of the d-block is

where: $n$ is the principal quantum number of the outermost shell, $(n-1)d$ holds one to ten electrons, and $ns$ normally holds one or two — except palladium, $4d^{10}5s^{0}$. Chromium is $3d^{5}4s^{1}$, not the naively expected $3d^{4}4s^{2}$, and copper is $3d^{10}4s^{1}$, not $3d^{9}4s^{2}$; both trace to the extra stability of half-filled and fully filled d subshells, derived in full below. By the modern IUPAC definition, a transition element has an incompletely filled d subshell in the neutral atom or in a common ion — a sharper test than "sits in the d-block." Zinc, cadmium and mercury are $d^{10}$ in the atom and stay $d^{10}$ in their only common ion, $+2$, so none counts as a transition element despite sitting at the d-block's edge; scandium ($3d^{1}4s^{2}$) does count, because that one 3d electron is incomplete.

General physical properties: metallic character, melting points and atomisation enthalpy

Nearly all transition elements show classic metallic behaviour — high tensile strength, ductility, malleability, thermal and electrical conductivity, metallic lustre — in hcp, bcc or ccp lattices (Zn, Cd, Hg and Mn are anomalous). Their melting and boiling points are high because the $(n-1)d$ electrons join the $ns$ electrons in metallic bonding, giving more bonding electrons per atom than the neighbouring s- or p-block metals. Across a row, melting point and atomisation enthalpy rise toward a maximum near the middle and fall again, with a sharp dip at manganese. Metal-metal bond strength depends on the number of unpaired d electrons, so bonding peaks where unpairing is greatest and weakens as d orbitals pair up toward $d^{10}$; manganese's comparatively unreactive, half-filled $3d^{5}4s^{2}$ bonds less effectively than its neighbours. The 4d and 5d series also have systematically higher atomisation enthalpies than the corresponding 3d metals, which is why metal-metal bonding in compounds is far more common among the heavier transition metals.

Atomic and ionic radii: the first sighting of lanthanoid contraction

Ionic radii of the same charge decrease steadily across a series, because each added electron enters an inner d orbital that shields the outer shell only imperfectly, so effective nuclear charge rises and pulls the shell in; atomic radii show the same trend but vary less than in the s- and p-blocks. Comparing series, atomic radii grow noticeably from the 3d row to the 4d row, but the 5d row is almost the same size as the 4d row rather than growing further — because fourteen 4f electrons fill between lanthanum and hafnium, and the resulting lanthanoid contraction (derived below) shrinks the 5d atoms enough to cancel the expected increase. Zirconium (160 pm) and hafnium (159 pm) end up almost identical in radius.

Ionisation enthalpies and the special stability of d0, d5 and d10

Ionisation enthalpy rises across each series left to right, driven by the same rising effective nuclear charge, but far more gently than across an s- or p-block period, because the added and removed electrons sit in different, only partially shielding subshells ($3d$ versus $4s$). The first ionisation enthalpy rises only slightly and somewhat irregularly; the second and third rise more steeply once $4s$ electrons are gone, but with sharp breaks exactly where a $d^{5}$ configuration is created or destroyed — the second ionisation enthalpy of manganese and the third of iron are both unusually high because the ion formed ($\text{Mn}^{2+}$ or $\text{Fe}^{3+}$) is the extra-stable $3d^{5}$, and removing yet another electron sacrifices that stability. The mechanism, exchange energy among parallel-spin electrons in degenerate orbitals, is derived fully below, since it also governs oxidation-state and electrode-potential stability.

Oxidation states across the 3d series

One of the most distinctive features of transition metals is the sheer number of oxidation states a single element can show. Scandium is essentially locked into $+3$; manganese, near the middle, exhibits every state from $+2$ to $+7$; copper and zinc are largely restricted to low states because too many d electrons are left to accommodate a high one. Up to manganese, the maximum reasonably stable state equals the sum of $s$ and $d$ electrons — $\text{Ti}^{IV}$ in $\text{TiO}_2$, $\text{V}^{V}$ in $\text{VO}_2^{+}$, $\text{Cr}^{VI}$ in $\text{CrO}_4^{2-}$, $\text{Mn}^{VII}$ in $\text{MnO}_4^{-}$ — after which stability drops abruptly, so Fe, Co, Ni, Cu and Zn are typically $\text{Fe}^{II,III}$, $\text{Co}^{II,III}$, $\text{Ni}^{II}$, $\text{Cu}^{I,II}$ and $\text{Zn}^{II}$. d-block oxidation states differ from one another by one unit ($\text{V}^{II}, \text{V}^{III}, \text{V}^{IV}, \text{V}^{V}$), not the two typical of the p-block's inert-pair effect. Down a group the trend also reverses: heavier p-block elements favour lower states, but heavier d-block elements favour higher ones — molybdenum(VI) and tungsten(VI) are markedly more stable than chromium(VI), which is exactly why acidified dichromate, wanting to drop from $+6$ to the stable $+3$, is a strong oxidant while $\text{MoO}_3$ and $\text{WO}_3$ are not.

Standard electrode potentials: M2+/M and M3+/M2+

The $E^{\ominus}(\text{M}^{2+}/\text{M})$ values become less negative from titanium to copper, tracking the rising sum of the first two ionisation enthalpies. Manganese and zinc sit more negative than the trend predicts because $\text{Mn}^{2+}$ ($3d^5$) and $\text{Zn}^{2+}$ ($3d^{10}$) are unusually stable, and nickel because of its very large hydration enthalpy. Copper is the outlier with a positive $E^{\ominus}(\text{Cu}^{2+}/\text{Cu}) = +0.34\ \text{V}$: it cannot displace hydrogen from dilute acids, because the energy to atomise and ionise solid copper is not fully repaid by its hydration enthalpy. The $E^{\ominus}(\text{M}^{3+}/\text{M}^{2+})$ values tell a complementary story: low for scandium ($\text{Sc}^{3+}$, noble-gas-like, so easy to form), high for manganese ($\text{Mn}^{2+}$, $d^5$, resists further oxidation), low for iron ($\text{Fe}^{3+}$, $d^5$, easy to form this time), and low for vanadium (stable half-filled $t_{2g}$ in $\text{V}^{2+}$). These four values are a compact summary of how $d^{0}$ and $d^{5}$ stability governs which oxidation state a 3d metal prefers in solution.

Stability of higher oxidation states: the role of fluorine and oxygen

Only fluorine and oxygen reliably stabilise a metal's very highest oxidation states. Fluorine works through high lattice energy ($\text{CoF}_3$) or strong covalent bonds ($\text{VF}_5$, $\text{CrF}_6$), but has limits — manganese's $+7$ state has no simple fluoride, only $\text{MnO}_3\text{F}$. Oxygen is the stronger stabiliser because it can form multiple ($\pi$) bonds to the metal, so covalent $\text{Mn}_2\text{O}_7$ exists though the highest manganese fluoride is only $\text{MnF}_4$. As oxidation number rises, oxides grow less ionic and more acidic: $\text{Sc}_2\text{O}_3$ is basic, $\text{V}_2\text{O}_5$ is amphoteric, $\text{CrO}_3$ gives $\text{H}_2\text{CrO}_4$/$\text{H}_2\text{Cr}_2\text{O}_7$, and $\text{Mn}_2\text{O}_7$ gives the strong acid $\text{HMnO}_4$.

Magnetic properties and coloured ions

Unpaired electrons in partly filled d orbitals make most transition-metal ions paramagnetic; ions with none are diamagnetic. For first-row compounds the orbital contribution is effectively quenched, so the moment is predicted from the number of unpaired electrons alone using the spin-only formula:

where: $\mu$ is the magnetic moment in Bohr magneton (BM) and $n$ is the number of unpaired electrons. One unpaired electron gives $\mu \approx 1.73\ \text{BM}$; five (a $d^{5}$ ion such as $\text{Mn}^{2+}$ or $\text{Fe}^{3+}$) give $\mu \approx 5.92\ \text{BM}$, the largest first-row value. Measuring $\mu$ experimentally is therefore a direct way to count unpaired electrons and pin down configuration and oxidation state. Colour follows a related mechanism: visible light excites an electron from a lower to a higher d orbital, and the colour seen is complementary to the frequency absorbed. This needs a partly filled d subshell, so $d^{0}$ or $d^{10}$ ions cannot show it and are colourless — $\text{Sc}^{3+}$, $\text{Ti}^{4+}$, $\text{Cu}^{+}$ and $\text{Zn}^{2+}$ are all colourless, while $\text{V}^{3+}$ (green), $\text{Mn}^{2+}$ (pale pink), $\text{Fe}^{3+}$ (yellow) and $\text{Cu}^{2+}$ (blue) are coloured. The exact shade also depends on the ligand bound — the start of the ligand-field ideas developed fully in the coordination-compounds chapter that follows.

Complex formation, catalysis, interstitial compounds and alloys

Transition metals form unusually many complexes — $[\text{Fe}(\text{CN})_6]^{3-}$, $[\text{Cu}(\text{NH}_3)_4]^{2+}$, $[\text{PtCl}_4]^{2-}$ — because their ions are small, highly charged, and have empty or partly filled d orbitals to accept ligand electron pairs. The same variable oxidation states make them excellent catalysts: $\text{V}_2\text{O}_5$ in the Contact Process, iron in the Haber process, nickel in hydrogenation. A clean example is $\text{Fe}^{3+}$ catalysing $2\text{I}^{-} + \text{S}_2\text{O}_8^{2-} \rightarrow \text{I}_2 + 2\text{SO}_4^{2-}$: the direct reaction is slow (both reactants are negatively charged), but $\text{Fe}^{3+}$ oxidises $\text{I}^{-}$ to $\text{I}_2$ while becoming $\text{Fe}^{2+}$, which $\text{S}_2\text{O}_8^{2-}$ then reoxidises — a two-step relay avoiding the slow like-charge collision.

Interstitial compounds form when small atoms (H, C, N) sit trapped inside a metal lattice, giving non-stoichiometric formulas ($\text{TiC}$, $\text{Fe}_3\text{H}$, $\text{VH}_{0.56}$) with no normal oxidation state; they are very hard, have higher melting points than the pure metal, stay metallically conductive, and are chemically inert. Alloys form readily because transition-metal radii are usually within fifteen percent of each other — steels (Cr, V, W, Mo, Mn), brass (Cu-Zn) and bronze (Cu-Sn) are the industrially dominant examples.

Potassium dichromate and potassium permanganate

These are the two named compounds NCERT singles out, each built from an ore, isolated as a solid oxidant, and used in acidified titrations. Potassium dichromate, $\text{K}_2\text{Cr}_2\text{O}_7$, comes from chromite ore, $\text{FeCr}_2\text{O}_4$, by oxidative fusion with sodium carbonate in air to sodium chromate, acidified to sodium dichromate, then converted to the less soluble potassium salt with KCl (full sequence in the fourth derivation below). Chromate (tetrahedral $\text{CrO}_4^{2-}$) and dichromate (two corner-shared tetrahedra, $\text{Cr-O-Cr}$ angle $126^{\circ}$) interconvert with pH — acid gives orange dichromate, alkali reverses it. In acid, dichromate is a strong oxidant, $\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^{+} + 6e^{-} \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}$ ($E^{\ominus} = +1.33\ \text{V}$), oxidising iodide, tin(II), sulphide and iron(II).

Potassium permanganate, $\text{KMnO}_4$, comes from pyrolusite, $\text{MnO}_2$, fused with alkali and an oxidant (KNO$_3$ or air) to dark green potassium manganate, $\text{K}_2\text{MnO}_4$, then oxidised — commercially by electrolysis, or by disproportionation in acid ($3\text{MnO}_4^{2-} + 4\text{H}^{+} \rightarrow 2\text{MnO}_4^{-} + \text{MnO}_2 + 2\text{H}_2\text{O}$) — to purple permanganate. Its reduction depends sharply on pH: to manganate in strong alkali ($E^{\ominus} = +0.56\ \text{V}$), to brown $\text{MnO}_2$ in neutral/faint alkali ($E^{\ominus} = +1.69\ \text{V}$), and to nearly colourless $\text{Mn}^{2+}$ in acid ($\text{MnO}_4^{-} + 8\text{H}^{+} + 5e^{-} \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$, $E^{\ominus} = +1.52\ \text{V}$) — the reaction used in acidified titrations against iron(II), oxalate and hydrogen sulphide.

The lanthanoids: contraction and oxidation states

Lanthanum and the fourteen elements after it, cerium to lutetium (symbol $\text{Ln}$), form the lanthanoid series. All share a common $6s^{2}$ outer configuration, filling $4f$ with increasing irregularity, but every $\text{Ln}^{3+}$ ion is a clean $4f^{n}$, because lanthanoids overwhelmingly favour $+3$. Ionic radii shrink steadily from $\text{La}^{3+}$ to $\text{Lu}^{3+}$, the lanthanoid contraction, from imperfect shielding of one $4f$ electron by another (derived fully below). A few lanthanoids show $+2$ or $+4$, and every exception follows the same empty/half-filled/fully-filled stability logic as the 3d series: $\text{Ce}^{4+}$ ($4f^{0}$) is a slow-acting oxidant ($E^{\ominus} = +1.74\ \text{V}$); $\text{Eu}^{2+}$ ($4f^{7}$) and $\text{Yb}^{2+}$ ($4f^{14}$) are reducing agents that revert to $\text{Ln}^{3+}$; $\text{Tb}^{4+}$ ($4f^{7}$) is an oxidant. Trivalent ions other than $\text{La}^{3+}$ ($f^0$) and $\text{Lu}^{3+}$ ($f^{14}$) are paramagnetic and often coloured, from narrow $f$-$f$ transitions. Early lanthanoids behave somewhat like calcium, later ones like aluminium; all liberate hydrogen from dilute acids, burn in halogens, and form basic $\text{Ln}_2\text{O}_3$ and $\text{Ln}(\text{OH})_3$. Mischmetall (about 95% Ln, 5% Fe) goes into lighter flints and Mg alloys; mixed Ln oxides are petroleum-cracking catalysts and screen phosphors.

The actinoids: wider oxidation states and comparison with the lanthanoids

The actinoid series, thorium to lawrencium (Z 90-103), is entirely radioactive, with the heaviest members short-lived and studied only in nanogram quantities. All share $7s^{2}$ with irregular $5f$/$6d$ occupancy; because $5f$ orbitals are less buried than $4f$, they bond far more readily, giving actinoids a richer chemistry. $+3$ is the general state, but early actinoids reach unusually high maxima — Th $+4$, Pa $+5$, U $+6$, Np $+7$ — before narrowing back to $+3$ for later, more lanthanoid-like members. Actinoid contraction parallels lanthanoid contraction but is sharper, since $5f$ electrons shield each other even less effectively. Both series contract steadily in ionic radius, favour $+3$, and are silvery, reactive metals with basic hydroxides; but actinoids show a far wider oxidation-state range (especially early on), lower and less regular ionisation enthalpies, and — being radioactive — greater difficulty of study. Lanthanoid contraction remains the more chemically important effect, since it directly governs the well-studied near-identical chemistry of zirconium/hafnium and niobium/tantalum.

Applications in industry

Iron and steel remain the dominant construction materials, alloyed with Cr, Mn, V, W and Mo for hardness and corrosion resistance. $\text{TiO}_2$ is a workhorse pigment; $\text{MnO}_2$ powers dry-cell batteries; Zn and Ni-Cd power rechargeable ones. $\text{V}_2\text{O}_5$ catalyses the Contact Process, $\text{TiCl}_4$/$\text{Al}(\text{CH}_3)_3$ forms the Ziegler catalyst for polythene, iron catalyses the Haber process, nickel catalyses hydrogenation, and $\text{PdCl}_2$ catalyses the Wacker oxidation of ethyne to ethanal. $\text{AgBr}$ underlies photography, and silver, gold and copper remain the historic "coinage metals."

Board level - spin-only magnetic moment of a divalent 3d ion (extends NCERT Example 4.8)

An element has atomic number 25. Find the electronic configuration and the spin-only magnetic moment of its divalent ion in aqueous solution, and state whether the ion is coloured.

Step 1. Identify the element and its divalent ion. Atomic number 25 is manganese, ground-state configuration $[\text{Ar}]3d^{5}4s^{2}$. Forming $\text{Mn}^{2+}$ removes the two $4s$ electrons first (as it always does for d-block ions), leaving the configuration $3d^{5}$.

Step 2. Count unpaired electrons using Hund's rule. Five electrons distributed across five degenerate $3d$ orbitals occupy each orbital singly with parallel spin, so $n = 5$ unpaired electrons.

Step 3. Apply the spin-only formula: $\mu = \sqrt{n(n+2)} = \sqrt{5 \times 7} = \sqrt{35} = 5.92\ \text{BM}$.

Step 4. Decide on colour. Since $3d^{5}$ is neither $d^{0}$ nor $d^{10}$, a $d$-$d$ transition is possible, so $\text{Mn}^{2+}$ is coloured (observed as pale pink in aqueous solution). Answer: $\text{Mn}^{2+}$ has configuration $3d^{5}$, magnetic moment $5.92\ \text{BM}$, and is coloured (pale pink).

Board level - stoichiometry of potassium dichromate oxidising iron(II) (extends NCERT Exercise 4.15)

What volume of $0.100\ \text{M}$ acidified potassium dichromate solution is required to completely oxidise the $\text{Fe}^{2+}$ present in $25.0\ \text{mL}$ of $0.600\ \text{M}$ iron(II) sulphate solution?

Step 1. Write the balanced ionic equation. Dichromate's half-reaction is $\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^{+} + 6e^{-} \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}$, and iron(II)'s is $\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^{-}$. Multiplying the iron half-reaction by 6 and adding gives $\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^{+} + 6\text{Fe}^{2+} \rightarrow 2\text{Cr}^{3+} + 6\text{Fe}^{3+} + 7\text{H}_2\text{O}$, so the mole ratio is 1 mole of dichromate to 6 moles of $\text{Fe}^{2+}$.

Step 2. Compute moles of $\text{Fe}^{2+}$ available: $n(\text{Fe}^{2+}) = 0.600\ \text{mol/L} \times 0.0250\ \text{L} = 0.0150\ \text{mol}$.

Step 3. Convert to moles of dichromate required using the 1:6 ratio: $n(\text{Cr}_2\text{O}_7^{2-}) = 0.0150/6 = 2.50\times10^{-3}\ \text{mol}$.

Step 4. Convert to volume: $V = n/C = 2.50\times10^{-3}\ \text{mol} / 0.100\ \text{mol/L} = 0.0250\ \text{L}$. Answer: $25.0\ \text{mL}$ of the $0.100\ \text{M}$ dichromate solution is required.

Competitive level (JEE / NEET) - predicting colour and relative stability across a set of ions (extends NCERT Exercise 4.18)

Among $\text{Ti}^{4+}$, $\text{V}^{3+}$, $\text{Cu}^{+}$, $\text{Sc}^{3+}$, $\text{Mn}^{2+}$, $\text{Fe}^{3+}$ and $\text{Co}^{2+}$, identify which are coloured in aqueous solution, and state which single ion is most resistant to further oxidation.

Step 1. Write each ion's d-electron count. $\text{Ti}^{4+}$: $[\text{Ar}]3d^{0}$. $\text{V}^{3+}$: $[\text{Ar}]3d^{2}$. $\text{Cu}^{+}$: $[\text{Ar}]3d^{10}$. $\text{Sc}^{3+}$: $[\text{Ar}]3d^{0}$. $\text{Mn}^{2+}$: $[\text{Ar}]3d^{5}$. $\text{Fe}^{3+}$: $[\text{Ar}]3d^{5}$. $\text{Co}^{2+}$: $[\text{Ar}]3d^{7}$.

Step 2. Apply the $d^{0}$/$d^{10}$ colourless rule. $\text{Ti}^{4+}$ and $\text{Sc}^{3+}$ ($d^{0}$) and $\text{Cu}^{+}$ ($d^{10}$) have no $d$-$d$ transition available and are colourless; $\text{V}^{3+}$ ($d^{2}$, green), $\text{Mn}^{2+}$ ($d^{5}$, pale pink), $\text{Fe}^{3+}$ ($d^{5}$, pale yellow/violet depending on ligand) and $\text{Co}^{2+}$ ($d^{7}$, pink) are all coloured.

Step 3. Rank stability against further oxidation using the $d^{5}$ exchange-energy argument developed in the second derivation. Among these seven ions, $\text{Mn}^{2+}$ is the one whose configuration is already the maximally exchange-stabilised $3d^{5}$, so oxidising it further to $\text{Mn}^{3+}$ ($3d^{4}$) is energetically unfavourable — consistent with the very high, oxidising $E^{\ominus}(\text{Mn}^{3+}/\text{Mn}^{2+}) = +1.5\ \text{V}$ reported for this couple.

Step 4. State the answer clearly. Answer: $\text{Ti}^{4+}$, $\text{Cu}^{+}$ and $\text{Sc}^{3+}$ are colourless ($d^0$ or $d^{10}$); $\text{V}^{3+}$, $\text{Mn}^{2+}$, $\text{Fe}^{3+}$ and $\text{Co}^{2+}$ are coloured; and $\text{Mn}^{2+}$ is the most resistant to further oxidation because it already holds the extra-stable half-filled $3d^{5}$ configuration. Timing note: under ninety seconds once the $d^0$/$d^5$/$d^{10}$ checklist is automatic. Write only each species' outer $d^{n}$ count and scan for $n = 0$ or $n = 10$ — everything else is coloured by default. Do not waste time recalling the exact shade unless asked.

Competitive level (JEE / NEET) - permanganate titration of oxalic acid, volume back-calculation

A $10.0\ \text{mL}$ sample of an oxalic acid solution requires $16.0\ \text{mL}$ of $0.0200\ \text{M}$ acidified $\text{KMnO}_4$ for complete oxidation to $\text{CO}_2$, at $333\ \text{K}$. Find the molarity of the oxalic acid solution.

Step 1. Write the balanced ionic equation. Permanganate in acid: $\text{MnO}_4^{-} + 8\text{H}^{+} + 5e^{-} \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$. Oxalate: $\text{C}_2\text{O}_4^{2-} \rightarrow 2\text{CO}_2 + 2e^{-}$. Equalising electrons (multiply the first by 2, the second by 5) and adding gives $2\text{MnO}_4^{-} + 5\text{C}_2\text{O}_4^{2-} + 16\text{H}^{+} \rightarrow 2\text{Mn}^{2+} + 10\text{CO}_2 + 8\text{H}_2\text{O}$, a mole ratio of $2$ permanganate to $5$ oxalate.

Step 2. Compute moles of $\text{KMnO}_4$ used: $n(\text{MnO}_4^{-}) = 0.0200\ \text{mol/L} \times 0.0160\ \text{L} = 3.20\times10^{-4}\ \text{mol}$.

Step 3. Convert to moles of oxalic acid using the $2:5$ ratio: $n(\text{C}_2\text{O}_4^{2-}) = 3.20\times10^{-4} \times (5/2) = 8.00\times10^{-4}\ \text{mol}$.

Step 4. Convert to molarity in the original $10.0\ \text{mL}$ sample: $C = n/V = 8.00\times10^{-4}\ \text{mol} / 0.0100\ \text{L} = 0.0800\ \text{mol/L}$. Answer: the oxalic acid solution is $0.0800\ \text{M}$. Timing note: about a minute and a half. This titration's mole ratio is not the intuitive $1{:}1$ — always derive it from electron balance. It is self-indicating (permanganate's own colour vanishes at the endpoint) and run hot, since the reaction is slow at room temperature until autocatalysed by the $\text{Mn}^{2+}$ produced.

The single most common slip is treating "d-block element" and "transition element" as synonyms. Zinc, cadmium and mercury sit in the d-block by position but fail the IUPAC test, since their d subshell is completely full ($d^{10}$) in both atom and common ion — a distinction examiners test directly. A second error is deleting ion electrons from the wrong shell. $ns$ electrons are always removed before $(n-1)d$ electrons on ionisation, regardless of filling order in the neutral atom — so $\text{Fe}$ gives $\text{Fe}^{2+}$ as $3d^{6}$, not $3d^{4}4s^{2}$. Students often assume oxidation states differ by two, carrying the p-block's inert-pair habit across by reflex; d-block states differ by one. Relatedly, many wrongly say heavier d-block elements favour lower oxidation states "by the inert-pair effect" — the d-block trend runs the opposite way ($\text{Mo}(\text{VI})$, $\text{W}(\text{VI})$ more stable than $\text{Cr}(\text{VI})$). A further slip is quoting a metal's highest oxidation state from its chloride rather than its oxide or fluoride — only fluorine and oxygen are small/electronegative (or, for oxygen, multiply-bonding) enough to reach the very highest states. Finally, a common arithmetic error is assuming permanganate or dichromate reacts mole-for-mole with a reducing agent. The ratio must be derived from electron balance (5 electrons per $\text{MnO}_4^{-}$ in acid, 6 per $\text{Cr}_2\text{O}_7^{2-}$), and it changes if the medium changes, since permanganate's product itself changes from $\text{Mn}^{2+}$ to $\text{MnO}_2$ to $\text{MnO}_4^{2-}$.

• State the modern IUPAC definition of a transition element and correctly exclude Zn, Cd and Hg using it.
• Write the general outer electronic configuration of a d-block atom and its ion, applying the "$ns$ before $(n-1)d$" removal rule.
• Explain, using exchange energy, why chromium and copper adopt $3d^{5}4s^{1}$ and $3d^{10}4s^{1}$ rather than the naively expected configurations.
• Describe how melting point, atomisation enthalpy, atomic radius and ionic radius vary across the 3d series and explain the anomalies at manganese and zinc.
• Predict which oxidation states of a given 3d metal are most stable, using the $d^{0}$/$d^{5}$/$d^{10}$ special-stability rule and the "differ by one" spacing.
• Use the spin-only formula to calculate the magnetic moment of an ion from its number of unpaired electrons, and predict whether an ion is coloured from its $d^{n}$ configuration.
• Describe the manufacture of potassium dichromate from chromite ore and potassium permanganate from pyrolusite, and write balanced ionic equations for their oxidising reactions in acid.
• Define lanthanoid contraction, explain its cause, and connect it to the near-identical chemistry of zirconium and hafnium.
• Compare lanthanoids and actinoids on electronic configuration, oxidation states, ionic-radius contraction and chemical reactivity.
• Balance redox equations involving dichromate and permanganate and use them to solve titration and stoichiometry numericals.

Ask the tutor: Want to work through a full periodic-trends comparison table for the 3d, 4d and 5d series, or practice more dichromate/permanganate titration numericals until the electron-balancing becomes automatic?