Chapter 4 · The d- and f-Block Elements

4.1 Transition Elements: Configuration and General Properties

The transition elements fill an inner $d$ subshell. That one fact explains their configurations (with the famous exceptions Cr and Cu), the trends in size, ionisation enthalpy and $E^{\circ}$ across the 3d series, their many oxidation states, their colours and magnetism, and why they make such good catalysts and alloys.

Sodium, magnesium and aluminium each have one fixed oxidation state, make colourless compounds and are ignored by a magnet. Move ten places further along period 4 and everything changes. Manganese forms compounds in every oxidation state from $+2$ to $+7$. A solution of nickel(II) sulphate is green, copper(II) sulphate blue and cobalt(II) chloride pink. Iron(III) chloride is pulled into a strong magnetic field. Iron powder turns nitrogen and hydrogen into ammonia, and a few grams of vanadium(V) oxide make sulphuric acid by the tonne. The metals themselves are hard, strong and dense, with melting points up to $2183\ \mathrm{K}$ (vanadium) in this row and $3683\ \mathrm{K}$ (tungsten) in the next-but-one.

All of this comes from the way the $d$ orbitals fill. In this section we write the configurations of the first transition series, then follow how size, ionisation enthalpy, melting point and electrode potential change from scandium to zinc. The pattern is steady, with a few sharp exceptions, and each exception has a reason: a half-filled $d^{5}$ or a full $d^{10}$ set is especially stable. The same configurations then let us count unpaired electrons, predict magnetic moments and colours, and understand why transition metals form complexes, act as catalysts, take small atoms into their lattices and mix into alloys.

Position and Electronic Configurations

The $d$-block occupies groups 3 to 12, in the middle of the periodic table between the $s$-block (groups 1–2) and the $p$-block (groups 13–18). Across each of its rows the electrons go into the $d$ orbitals of the penultimate shell, $(n-1)d$, while the outer $ns$ orbital holds one or two electrons. There are four rows, or series: the 3d series (Sc to Zn, period 4), the 4d series (Y to Cd, period 5), the 5d series (La and Hf to Hg, period 6) and the incomplete 6d series (Ac and Rf to Cn, period 7).

📘 Definition 4.1 — d-block element and transition element

A d-block element is one in which the last electron enters an $(n-1)d$ orbital (groups 3–12). A transition element is an element whose atom, or one of its ions in a common oxidation state, has an incompletely filled $d$ subshell (from $d^{1}$ to $d^{9}$).

  • Zinc, cadmium and mercury (group 12) have the full configuration $(n-1)d^{10}ns^{2}$ in the atom and $(n-1)d^{10}$ in their common ions ($\mathrm{Zn^{2+}}$, $\mathrm{Cd^{2+}}$, $\mathrm{Hg^{2+}}$). They are $d$-block elements but are not regarded as transition elements; they are studied alongside them.
  • Copper, silver and gold have $(n-1)d^{10}ns^{1}$ atoms, but each has a common ion with a partly filled $d$ subshell ($\mathrm{Cu^{2+}}$ $3d^{9}$, $\mathrm{Au^{3+}}$ $5d^{8}$), so they are transition elements.
  • Scandium ($3d^{1}4s^{2}$) is a transition element because the atom has a partly filled $d$ subshell, even though its only common ion, $\mathrm{Sc^{3+}}$, is $d^{0}$.
📐 Rule 4.1 — Electronic configurations of the d-block
  1. The general outer configuration is $(n-1)d^{1-10}\,ns^{1-2}$. For the 3d series, $\mathrm{[Ar]\,3d^{x}\,4s^{2}}$, where $x$ rises from $1$ (Sc) to $10$ (Zn).
  2. Exceptions in the 3d series: chromium is $\mathrm{[Ar]\,3d^{5}\,4s^{1}}$ (not $3d^{4}4s^{2}$) and copper is $\mathrm{[Ar]\,3d^{10}\,4s^{1}}$ (not $3d^{9}4s^{2}$). The $3d$ and $4s$ energies are very close, so moving one electron from $4s$ to $3d$ costs little, and the half-filled $d^{5}$ or completely filled $d^{10}$ set gains extra stability (more exchange energy, a symmetrical distribution of charge).
  3. Ions: when a transition atom forms a cation, the $4s$ electrons are removed before the $3d$ electrons. Once the $3d$ orbitals hold electrons they lie below $4s$. So $\mathrm{Fe}$ $\mathrm{[Ar]\,3d^{6}\,4s^{2}}$ gives $\mathrm{Fe^{2+}}$ $\mathrm{[Ar]\,3d^{6}}$ and $\mathrm{Fe^{3+}}$ $\mathrm{[Ar]\,3d^{5}}$. A 3d ion of charge $+q$ from an atom with $Z$ electrons has $d^{\,Z-18-q}$.

The heavier series have more exceptions, because the $(n-1)d$ and $ns$ levels are even closer (for example palladium is $\mathrm{[Kr]\,4d^{10}\,5s^{0}}$). You are expected to know the 3d configurations exactly; the others are not asked as rules.

ElementScTiVCrMnFeCoNiCuZn
$Z$21222324252627282930
$3d$123556781010
$4s$2221222212
$\mathrm{M^{2+}}$—$d^{2}$$d^{3}$$d^{4}$$d^{5}$$d^{6}$$d^{7}$$d^{8}$$d^{9}$$d^{10}$
$\mathrm{M^{3+}}$$d^{0}$$d^{1}$$d^{2}$$d^{3}$$d^{4}$$d^{5}$$d^{6}$———

The exceptions Cr and Cu affect only the atoms: every ion of the same charge has the regular count, because all ions are formed by losing the $4s$ electrons first.

✏️ Example 1 — Configurations of atoms and ions
Write the electronic configurations of (a) $\mathrm{Cr}$ and $\mathrm{Cr^{3+}}$, (b) $\mathrm{Cu}$, $\mathrm{Cu^{+}}$ and $\mathrm{Cu^{2+}}$, (c) $\mathrm{Mn^{2+}}$ and $\mathrm{Co^{3+}}$. In each case give the number of $3d$ electrons.
(a) Chromium $Z=24$: $\mathrm{Cr}$ $\mathrm{[Ar]\,3d^{5}\,4s^{1}}$ (the exception). $\mathrm{Cr^{3+}}$: remove the one $4s$ electron, then two $3d$ electrons: $\mathrm{[Ar]\,3d^{3}}$. Check with Rule 4.1: $24-18-3=3$ ✓.
(b) Copper $Z=29$: $\mathrm{Cu}$ $\mathrm{[Ar]\,3d^{10}\,4s^{1}}$; $\mathrm{Cu^{+}}$ $\mathrm{[Ar]\,3d^{10}}$; $\mathrm{Cu^{2+}}$ $\mathrm{[Ar]\,3d^{9}}$.
(c) Manganese, cobalt $\mathrm{Mn}$ $\mathrm{[Ar]\,3d^{5}\,4s^{2}}$, so $\mathrm{Mn^{2+}}$ is $\mathrm{[Ar]\,3d^{5}}$. $\mathrm{Co}$ $\mathrm{[Ar]\,3d^{7}\,4s^{2}}$, so $\mathrm{Co^{3+}}$ is $\mathrm{[Ar]\,3d^{6}}$ ($27-18-3=6$ ✓).
Answer $3d$ electrons: $\mathrm{Cr}$ 5, $\mathrm{Cr^{3+}}$ 3, $\mathrm{Cu}$ 10, $\mathrm{Cu^{+}}$ 10, $\mathrm{Cu^{2+}}$ 9, $\mathrm{Mn^{2+}}$ 5, $\mathrm{Co^{3+}}$ 6. A common slip is to write $\mathrm{Mn^{2+}}$ as $\mathrm{3d^{3}\,4s^{2}}$: the $4s$ electrons always leave first.
✏️ Example 2 — Transition element or not?
Using Definition 4.1, decide whether each of the following is a transition element, and give the reason: (a) scandium, (b) zinc, (c) silver (common ion $\mathrm{Ag^{+}}$, but $\mathrm{AgF_{2}}$ contains $\mathrm{Ag^{2+}}$), (d) mercury.
(a) Sc Yes. The atom is $\mathrm{[Ar]\,3d^{1}\,4s^{2}}$: its $d$ subshell is incompletely filled, even though $\mathrm{Sc^{3+}}$ is $d^{0}$.
(b) Zn No. Atom $3d^{10}4s^{2}$ and ion $\mathrm{Zn^{2+}}$ $3d^{10}$: the $d$ subshell is full in both, and zinc shows no other oxidation state.
(c) Ag Yes. Silver is $\mathrm{[Kr]\,4d^{10}\,5s^{1}}$ and $\mathrm{Ag^{+}}$ is $4d^{10}$, but in the $+2$ state ($\mathrm{AgF_{2}}$) it is $4d^{9}$, a partly filled $d$ subshell. The same argument makes copper ($\mathrm{Cu^{2+}}$, $3d^{9}$) a transition element.
(d) Hg No. $\mathrm{[Xe]\,4f^{14}\,5d^{10}\,6s^{2}}$; its ions $\mathrm{Hg^{2+}}$ ($5d^{10}$) and $\mathrm{Hg_{2}^{2+}}$ keep a full $5d$ subshell. Group 12 metals are also softer and lower-melting than transition metals, because their $d$ electrons do not join in metallic bonding (mercury is a liquid).

Physical Properties: Metallic Character, Size and Ionisation Enthalpy

Nearly all transition elements are typical metals: hard, strong, lustrous, malleable and ductile, with high thermal and electrical conductivity. Except for zinc, cadmium, mercury and manganese, they crystallise in one of the typical metallic structures (body-centred cubic, cubic close-packed or hexagonal close-packed) at room temperature. They have high melting and boiling points and high enthalpies of atomisation, because the $(n-1)d$ electrons join the $ns$ electrons in metallic bonding: the more unpaired $d$ electrons an atom has, the stronger the bonding between atoms.

The table collects the data for the 3d series that this section uses. Interactive 1 plots every row.

PropertyScTiVCrMnFeCoNiCuZn
metallic radius / pm164147135129137126125125128137
$\Delta_{a}H^{\circ}$ / $\mathrm{kJ\,mol^{-1}}$326473515397281416425430339126
$\Delta_{i}H_{1}$ / $\mathrm{kJ\,mol^{-1}}$631656650653717762758736745906
$\Delta_{i}H_{2}$ / $\mathrm{kJ\,mol^{-1}}$1235130914141592150915611644175219581734
$\Delta_{i}H_{3}$ / $\mathrm{kJ\,mol^{-1}}$2393265728332990326029623243340235563837
melting point / K181419412183218015191811176817281358693
density / $\mathrm{g\,cm^{-3}}$3.434.16.077.197.217.88.78.98.97.1
📐 Rule 4.2 — Trends across the first transition series
  1. Melting points and enthalpies of atomisation rise to a maximum near the middle of the series (V, Cr), where the atoms have the most unpaired $d$ electrons to share in metallic bonding, and then fall. Manganese and zinc are unusually low. The 4d and 5d metals have higher enthalpies of atomisation than the 3d metals above them, which is why metal–metal bonds are much more common in compounds of the heavier transition metals.
  2. Atomic and ionic radii decrease from Sc to about Cr, stay almost constant through Fe, Co and Ni, and rise slightly at Cu and Zn. Each step adds one proton to the nucleus and one electron to an inner $3d$ orbital. A $d$ electron shields the outer electrons poorly, so the effective nuclear charge rises and the atom shrinks. In the middle of the series the extra shielding by the growing number of $d$ electrons balances the extra nuclear charge. Ions of the same charge ($\mathrm{M^{2+}}$ or $\mathrm{M^{3+}}$) also shrink steadily with increasing atomic number.
  3. Density rises from Sc ($3.43$) to Cu ($8.9\ \mathrm{g\,cm^{-3}}$): the atoms get heavier while their volume shrinks.
  4. Ionisation enthalpies increase across the series, but much less steeply than across a $p$-block row, because the added inner $d$ electrons partly shield the outer $4s$ electrons from the growing nuclear charge. The irregularities follow the stability of $d^{5}$ and $d^{10}$: $\Delta_{i}H_{2}$ is unusually high for Cr and Cu (the electron must come out of $\mathrm{Cr^{+}}$, $3d^{5}$, or $\mathrm{Cu^{+}}$, $3d^{10}$); $\Delta_{i}H_{3}$ is unusually high for Mn ($\mathrm{Mn^{2+}}$ is $3d^{5}$) and Zn ($\mathrm{Zn^{2+}}$ is $3d^{10}$), and unusually low for Fe ($\mathrm{Fe^{2+}}$, $3d^{6}$, becomes $3d^{5}$).
✏️ Example 3 — Reading the irregularities
Use the table to explain: (a) why $\Delta_{i}H_{2}$ of chromium ($1592\ \mathrm{kJ\,mol^{-1}}$) is greater than that of manganese ($1509$); (b) why $\Delta_{i}H_{3}$ of manganese ($3260$) is about $300\ \mathrm{kJ\,mol^{-1}}$ greater than that of iron ($2962$); (c) why zinc has the lowest enthalpy of atomisation and melting point in the series. (d) By how much does $\Delta_{i}H_{1}+\Delta_{i}H_{2}$ (the energy to make $\mathrm{M^{2+}}$ from a gaseous atom) differ between Cu and Zn?
(a) The second electron comes from the singly charged ion. $\mathrm{Cr^{+}}$ is $\mathrm{[Ar]\,3d^{5}}$, a stable half-filled set, so removing an electron from it costs extra energy. $\mathrm{Mn^{+}}$ is $\mathrm{[Ar]\,3d^{5}\,4s^{1}}$: the second electron is the remaining $4s$ electron, which is easier to remove.
(b) $\mathrm{Mn^{2+}}\rightarrow\mathrm{Mn^{3+}}$ breaks the half-filled $3d^{5}$ set ($d^{5}\rightarrow d^{4}$), which is difficult. $\mathrm{Fe^{2+}}\rightarrow\mathrm{Fe^{3+}}$ goes from $3d^{6}$ to the stable $3d^{5}$, which is easy. Difference $3260-2962=298\ \mathrm{kJ\,mol^{-1}}$.
(c) In zinc ($3d^{10}4s^{2}$) every $d$ orbital is full, so the $d$ electrons take no part in metallic bonding; only the two $4s$ electrons bond the atoms together. Hence $\Delta_{a}H^{\circ}=126\ \mathrm{kJ\,mol^{-1}}$ (against $339$ for Cu) and a melting point of only $693\ \mathrm{K}$.
(d) Cu: $745+1958=2703\ \mathrm{kJ\,mol^{-1}}$. Zn: $906+1734=2640\ \mathrm{kJ\,mol^{-1}}$. Copper needs $63\ \mathrm{kJ\,mol^{-1}}$ more, because its second electron is pulled out of the full $3d^{10}$ set. Together with its high enthalpy of atomisation, this makes $\mathrm{Cu\rightarrow Cu^{2+}}$ unusually difficult, which you will meet again in the $E^{\circ}$ values.
Interactive 1 — Trends Across the 3d Series

Choose a property to plot from Sc to Zn. Press Play to move the cursor along the series, or click a column. Points marked with an amber ring are the anomalies; the verdict explains each one by the configuration involved. Compare the shapes: radius falls then levels off, melting point and $\Delta_{a}H^{\circ}$ peak in the middle, $\Delta_{i}H_{3}$ zig-zags at Mn and Fe, and $E^{\circ}(\mathrm{M^{2+}/M})$ climbs to the one positive value, copper.

element · configuration—
value—
change from previous—
series range—

Oxidation States and Electrode Potentials

Because the $4s$ and $3d$ electrons have similar energies, a transition metal can lose (or share) a few or many of them. Sodium stops at $+1$ because the next electron would come from the $\mathrm{[Ne]}$ core. Manganese can go on giving up electrons until all seven of its $4s$ and $3d$ electrons are used, as in the permanganate ion, $\mathrm{MnO_{4}^{-}}$.

📘 Definition 4.2 — Variable oxidation state
An element shows variable oxidation states when it forms stable compounds in two or more oxidation states. For transition elements the different states usually differ by one unit (for example $\mathrm{Fe^{2+}}$/$\mathrm{Fe^{3+}}$, $\mathrm{Cu^{+}}$/$\mathrm{Cu^{2+}}$, $\mathrm{V^{2+}}$ to $\mathrm{V^{5+}}$), because electrons are removed one at a time from orbitals of nearly equal energy. In the $p$-block the states typically differ by two units ($\mathrm{Sn^{2+}}$/$\mathrm{Sn^{4+}}$, $\mathrm{Tl^{+}}$/$\mathrm{Tl^{3+}}$), as the $ns^{2}$ pair is either kept or used as a whole.
📐 Rule 4.3 — Patterns in the oxidation states of the 3d metals
  1. Number of states. The variety is greatest in the middle of the series: manganese shows every state from $+2$ to $+7$. Scandium has too few electrons to show more than $+3$; zinc has too many $d$ electrons, all paired, to use any of them, and shows only $+2$.
  2. Highest state. From Sc to Mn the highest oxidation state equals the total number of $4s$ and $3d$ electrons: $\mathrm{Sc}$ $+3$, $\mathrm{Ti}$ $+4$, $\mathrm{V}$ $+5$, $\mathrm{Cr}$ $+6$, $\mathrm{Mn}$ $+7$. After manganese the highest states fall again (Fe rarely beyond $+3$, Co and Ni mostly $+2$), because pairing in the $d$ orbitals and the rising nuclear charge hold the $d$ electrons more firmly. For the same reason $+2$ becomes the most common state towards the end of the series.
  3. Stabilising high states. The highest states occur in oxides, oxoanions and fluorides: $\mathrm{TiF_{4}}$, $\mathrm{VF_{5}}$, $\mathrm{CrF_{6}}$; $\mathrm{V_{2}O_{5}}$, $\mathrm{CrO_{3}}$, $\mathrm{Mn_{2}O_{7}}$, $\mathrm{CrO_{4}^{2-}}$, $\mathrm{MnO_{4}^{-}}$. Fluorine stabilises them because it is the most electronegative element and $\mathrm{F^{-}}$ is small (high lattice or bond enthalpies). Oxygen goes further than fluorine: it can form multiple bonds to the metal, so manganese reaches $+7$ in $\mathrm{Mn_{2}O_{7}}$ and $\mathrm{MnO_{4}^{-}}$ but only $+4$ in its highest fluoride, $\mathrm{MnF_{4}}$.
  4. Down a group, higher oxidation states become more stable (the opposite of the $p$-block): Mo(VI) and W(VI) compounds are stable, while Cr(VI) in $\mathrm{Cr_{2}O_{7}^{2-}}$ is a strong oxidising agent.
  5. Low states such as $0$ occur in complexes with ligands like carbon monoxide that can accept electron density back from the metal: $\mathrm{[Ni(CO)_{4}]}$, $\mathrm{[Fe(CO)_{5}]}$ (Chapter 5).

How easily a metal forms its ions in water is measured by its standard electrode potential. Converting $\mathrm{M(s)}$ into $\mathrm{M^{2+}(aq)}$ can be pictured in three steps: atomise the metal ($\Delta_{a}H^{\circ}$, endothermic), remove two electrons ($\Delta_{i}H_{1}+\Delta_{i}H_{2}$, endothermic), and hydrate the gaseous ion ($\Delta_{hyd}H^{\circ}$, strongly exothermic). The more the first two outweigh the third, the less negative (more positive) $E^{\circ}(\mathrm{M^{2+}/M})$ is.

$E^{\circ}$ / VTiVCrMnFeCoNiCuZn
$\mathrm{M^{2+}/M}$$-1.63$$-1.18$$-0.90$$-1.18$$-0.44$$-0.28$$-0.25$$+0.34$$-0.76$
$\mathrm{M^{3+}/M^{2+}}$$-0.37$$-0.26$$-0.41$$+1.57$$+0.77$$+1.97$———
📐 Rule 4.4 — Trends in standard electrode potentials
  1. $E^{\circ}(\mathrm{M^{2+}/M})$ becomes generally less negative from Ti to Cu, as $\Delta_{i}H_{1}+\Delta_{i}H_{2}$ increases: the tendency to form $\mathrm{M^{2+}}$ ions decreases across the series.
  2. Copper is the only 3d metal with a positive value ($+0.34\ \mathrm{V}$). Its high enthalpy of atomisation and high $\Delta_{i}H_{2}$ are not repaid by the hydration enthalpy of $\mathrm{Cu^{2+}}$. So copper does not liberate hydrogen from dilute non-oxidising acids (dilute $\mathrm{HCl}$ or $\mathrm{H_{2}SO_{4}}$); it dissolves only in oxidising acids such as nitric acid and hot concentrated sulphuric acid, which oxidise it.
  3. Mn, Ni and Zn are more negative than the trend predicts: Mn because $\mathrm{Mn^{2+}}$ is the stable $d^{5}$ ion, Zn because $\mathrm{Zn^{2+}}$ is the stable $d^{10}$ ion (and zinc is easy to atomise), and Ni because $\mathrm{Ni^{2+}}$ has the most negative hydration enthalpy in the series.
  4. $E^{\circ}(\mathrm{M^{3+}/M^{2+}})$ tells which of the two states is preferred in water. The large positive values for Mn ($+1.57\ \mathrm{V}$) and Co ($+1.97\ \mathrm{V}$) make $\mathrm{Mn^{3+}}$ and $\mathrm{Co^{3+}}$ strong oxidising agents: $\mathrm{Mn^{3+}}$ ($d^{4}$) gains an electron to become stable $d^{5}$. The comparatively low value for Fe ($+0.77\ \mathrm{V}$) reflects the stability of $\mathrm{Fe^{3+}}$ ($d^{5}$). The negative value for Cr ($-0.41\ \mathrm{V}$) makes $\mathrm{Cr^{2+}}$ a strong reducing agent: it loses an electron to become $\mathrm{Cr^{3+}}$ ($d^{3}$, three electrons in the lower set of $d$ orbitals in water, a very stable arrangement).

Reactivity. Apart from copper, the 3d metals have negative $E^{\circ}(\mathrm{M^{2+}/M})$ and are, in principle, oxidised by $1\ \mathrm{M}$ $\mathrm{H^{+}}$, though some react very slowly: titanium and vanadium, for example, are passive to dilute non-oxidising acids at room temperature, protected by a thin film of oxide. The same kind of film makes chromium useful as a corrosion-resistant plating.

✏️ Example 4 — Using $E^{\circ}$ values to predict behaviour
(a) Which is the stronger reducing agent, $\mathrm{Cr^{2+}}$ or $\mathrm{Fe^{2+}}$? Explain using configurations. (b) Can $\mathrm{Mn^{3+}}$ oxidise $\mathrm{Fe^{2+}}$ to $\mathrm{Fe^{3+}}$ under standard conditions? Write the equation and find $E^{\circ}_{cell}$. (c) In water, copper(I) ions disproportionate: $\mathrm{2Cu^{+}(aq)\rightarrow Cu^{2+}(aq)+Cu(s)}$. Given $E^{\circ}_{\mathrm{Cu^{+}/Cu}}=+0.52\ \mathrm{V}$ and $E^{\circ}_{\mathrm{Cu^{2+}/Cu^{+}}}=+0.15\ \mathrm{V}$, find $E^{\circ}$ for this reaction and $\Delta_{r}G^{\circ}$ ($F=96\,487\ \mathrm{C\,mol^{-1}}$).
(a) $E^{\circ}_{\mathrm{Cr^{3+}/Cr^{2+}}}=-0.41\ \mathrm{V}$ is far below $E^{\circ}_{\mathrm{Fe^{3+}/Fe^{2+}}}=+0.77\ \mathrm{V}$, so $\mathrm{Cr^{2+}}$ gives up its electron much more readily: $\mathrm{Cr^{2+}}$ is the stronger reducing agent. In configurations: $\mathrm{Cr^{2+}}$ ($d^{4}$) $\rightarrow\mathrm{Cr^{3+}}$ ($d^{3}$, very stable in water), while $\mathrm{Fe^{2+}}$ ($d^{6}$) $\rightarrow\mathrm{Fe^{3+}}$ ($d^{5}$) has less to gain.
(b) $\mathrm{Mn^{3+}(aq)+Fe^{2+}(aq)\rightarrow Mn^{2+}(aq)+Fe^{3+}(aq)}$. Cathode (reduction) $\mathrm{Mn^{3+}/Mn^{2+}}$, anode $\mathrm{Fe^{3+}/Fe^{2+}}$: $E^{\circ}_{cell}=1.57-0.77=+0.80\ \mathrm{V}\gt0$, so yes, the reaction is spontaneous.
(c) $E^{\circ}$ One $\mathrm{Cu^{+}}$ is reduced ($\mathrm{Cu^{+}(aq)+e^{-}\rightarrow Cu(s)}$, $+0.52\ \mathrm{V}$) and one is oxidised ($\mathrm{Cu^{+}(aq)\rightarrow Cu^{2+}(aq)+e^{-}}$, the reverse of the $+0.15\ \mathrm{V}$ couple): $E^{\circ}=0.52-0.15=+0.37\ \mathrm{V}$, with $n=1$.
(c) $\Delta_{r}G^{\circ}$ $\Delta_{r}G^{\circ}=-nFE^{\circ}=-1\times96\,487\times0.37=-3.57\times10^{4}\ \mathrm{J\,mol^{-1}}=-35.7\ \mathrm{kJ\,mol^{-1}}$. Negative, so $\mathrm{Cu^{+}}$ disproportionates in water. The driving force is the much more negative hydration enthalpy of the small, doubly charged $\mathrm{Cu^{2+}}$ ion, which more than repays the second ionisation enthalpy of copper. That is why most simple copper(I) compounds are insoluble solids ($\mathrm{CuCl}$, $\mathrm{Cu_{2}O}$) and copper(II) is the common state in water.

Magnetism, Colour, Complexes, Catalysis, Interstitial Compounds and Alloys

A substance placed between the poles of a strong magnet is either pulled in a little or pushed out a little. The difference tells us whether it has unpaired electrons. Each unpaired electron behaves like a tiny magnet because of its spin (and its orbital motion); paired electrons have opposite spins, and their magnetic effects cancel.

📘 Definition 4.3 — Paramagnetism, diamagnetism and the Bohr magneton

A paramagnetic substance is attracted into a magnetic field; it contains one or more unpaired electrons. A diamagnetic substance is weakly repelled by a magnetic field; all its electrons are paired. (Iron, cobalt and nickel metals show ferromagnetism, an extreme form of paramagnetism in which the moments of neighbouring atoms line up.)

Magnetic moments of atoms and ions are measured in Bohr magnetons: $1\ \mathrm{BM}=9.27\times10^{-24}\ \mathrm{A\,m^{2}}$ (equivalently $\mathrm{J\,T^{-1}}$).

📐 Rule 4.5 — Spin-only magnetic moment

For an ion of the first transition series with $n$ unpaired electrons, the magnetic moment is given closely by the spin-only formula

$$\mu=\sqrt{n(n+2)}\ \ \mathrm{BM}$$

It assumes that only the electron spins contribute; the orbital contribution is small for most 3d ions in compounds and is ignored. For a free $3d$ ion, or in water, the electrons follow Hund's rule: $d^{1}$ to $d^{5}$ have $n=1$ to $5$; $d^{6}$ to $d^{9}$ have $n=10-(\text{number of }d\text{ electrons})$, i.e. $4,3,2,1$; $d^{0}$ and $d^{10}$ have $n=0$.

$n$012345
$\mu$ / BM01.732.833.874.905.92
examples$\mathrm{Sc^{3+}}$, $\mathrm{Zn^{2+}}$$\mathrm{Ti^{3+}}$, $\mathrm{Cu^{2+}}$$\mathrm{V^{3+}}$, $\mathrm{Ni^{2+}}$$\mathrm{Cr^{3+}}$, $\mathrm{Co^{2+}}$$\mathrm{Mn^{3+}}$, $\mathrm{Fe^{2+}}$$\mathrm{Mn^{2+}}$, $\mathrm{Fe^{3+}}$

The moment rises with the number of unpaired electrons. Run backwards, a measured moment tells you $n$, and hence the $d$-electron count and often the oxidation state.

✏️ Example 5 — Magnetic moments from configurations, and back again
(a) Calculate the spin-only magnetic moments of $\mathrm{Fe^{2+}}$, $\mathrm{Mn^{2+}}$ and $\mathrm{Cu^{+}}$ (atomic numbers 26, 25, 29). (b) An aqueous ion of a 3d metal in the $+2$ state has $\mu=3.87\ \mathrm{BM}$. Which ions are possible? (c) A compound of vanadium has a magnetic moment of $1.73\ \mathrm{BM}$. What is the oxidation state of vanadium ($Z=23$) in it, assuming spin-only behaviour?
(a) Fe²⁺ $3d^{6}$: five orbitals, one pair and four single electrons, $n=4$. $\mu=\sqrt{4\times6}=\sqrt{24}=4.90\ \mathrm{BM}$.
(a) Mn²⁺, Cu⁺ $\mathrm{Mn^{2+}}$ $3d^{5}$, $n=5$: $\mu=\sqrt{35}=5.92\ \mathrm{BM}$ (the largest possible for a 3d ion). $\mathrm{Cu^{+}}$ $3d^{10}$, $n=0$: $\mu=0$, diamagnetic.
(b) $\sqrt{n(n+2)}=3.87$ gives $n(n+2)=15$, so $n=3$: a $d^{3}$ or $d^{7}$ ion. With charge $+2$: $d^{3}$ is $\mathrm{V^{2+}}$ ($23-18-2=3$) and $d^{7}$ is $\mathrm{Co^{2+}}$ ($27-18-2=7$).
(c) $1.73\ \mathrm{BM}$ means $n=1$, so vanadium is $d^{1}$: $23-18-q=1$ gives $q=4$. Vanadium is in the $+4$ state, as in the blue vanadyl ion $\mathrm{VO^{2+}}$.

Coloured ions. In an isolated ion the five $d$ orbitals have the same energy. Surrounded by water molecules or other ligands, they split into groups of slightly different energy (Chapter 5 works out the pattern). An electron can then jump from a lower to a higher $d$ orbital, a $d$–$d$ transition, by absorbing a photon whose energy matches the gap. For most 3d ions this gap corresponds to visible light. The ion absorbs one part of the spectrum, and we see the remaining light, whose colour is complementary to the colour absorbed. $\mathrm{[Ti(H_{2}O)_{6}]^{3+}}$, for example, absorbs green-yellow light (near $500\ \mathrm{nm}$) and looks purple. An ion with $d^{0}$ or $d^{10}$ has no possible $d$–$d$ transition and is colourless: $\mathrm{Sc^{3+}}$, $\mathrm{Ti^{4+}}$, $\mathrm{Cu^{+}}$, $\mathrm{Zn^{2+}}$. The colour also depends on the ligand: pale blue $\mathrm{[Cu(H_{2}O)_{4}]^{2+}}$ turns deep blue with ammonia, and green $\mathrm{[Ni(H_{2}O)_{6}]^{2+}}$ turns blue-violet with ethane-1,2-diamine.

Aqua ion$\mathrm{Sc^{3+}}$$\mathrm{Ti^{4+}}$$\mathrm{Ti^{3+}}$$\mathrm{V^{4+}}$$\mathrm{V^{3+}}$$\mathrm{Cr^{3+}}$$\mathrm{Mn^{3+}}$
$d^{n}$$d^{0}$$d^{0}$$d^{1}$$d^{1}$$d^{2}$$d^{3}$$d^{4}$
colourcolourlesscolourlesspurplebluegreenvioletviolet
Aqua ion$\mathrm{Mn^{2+}}$$\mathrm{Fe^{3+}}$$\mathrm{Fe^{2+}}$$\mathrm{Co^{2+}}$$\mathrm{Ni^{2+}}$$\mathrm{Cu^{2+}}$$\mathrm{Zn^{2+}}$
$d^{n}$$d^{5}$$d^{5}$$d^{6}$$d^{7}$$d^{8}$$d^{9}$$d^{10}$
colourpinkyellowgreenpinkgreenbluecolourless

$\mathrm{Mn^{2+}}$ and $\mathrm{Fe^{3+}}$ ($d^{5}$) are only very pale: their $d$–$d$ transitions would need an electron to reverse its spin, which makes the absorption weak. "$\mathrm{V^{4+}}$" exists in water as the vanadyl ion $\mathrm{VO^{2+}}$.

Interactive 2 — 3d Ion Explorer: Configuration, Magnetism and Colour

Choose a metal and an oxidation state. The lab builds the configuration from the atom (removing $4s$ electrons first), fills the five $3d$ boxes by Hund's rule, counts the unpaired electrons and computes the spin-only moment $\mu=\sqrt{n(n+2)}$. On the right, a colour wheel shows the light the aqua ion absorbs (outlined) and the complementary colour you see (filled). Press Play to step through the common aqua ions.

configuration—
unpaired $n$—
$\mu$ (BM)—
aqua-ion colour—
🚀 Beyond the board syllabus (JEE Main · NEET) — when the spin-only formula falls short

In JEE Main and NEET 2026 (magnetic properties of the transition elements); not required by the CBSE syllabus, and JEE Advanced lists spin-only moments only.

Measured moments are often a little larger than the spin-only values because the orbital motion of the electrons also contributes. For $\mathrm{Mn^{2+}}$ and $\mathrm{Fe^{3+}}$ ($d^{5}$) the agreement is excellent, about $5.9\ \mathrm{BM}$, since a half-filled set has no orbital contribution. For $\mathrm{Co^{2+}}$ ($d^{7}$, spin-only $3.87\ \mathrm{BM}$) values between about $4.4$ and $5.2\ \mathrm{BM}$ are typical. In JEE problems, use the spin-only formula unless the question says otherwise. For the lanthanoid ions of Section 4.3, whose $4f$ electrons are well shielded from the surroundings, the orbital contribution is large and the spin-only formula does not apply at all.

Complex formation. Transition-metal ions are small and highly charged, and they have vacant $d$ orbitals of suitable energy to accept lone pairs from molecules or anions (ligands). They therefore form a very large number of complex compounds: $\mathrm{[Fe(CN)_{6}]^{3-}}$, $\mathrm{[Fe(CN)_{6}]^{4-}}$, $\mathrm{[Cu(NH_{3})_{4}]^{2+}}$, $\mathrm{[Ni(CN)_{4}]^{2-}}$, $\mathrm{[PtCl_{4}]^{2-}}$. Chapter 5 is devoted to them.

Catalytic properties. Many transition metals and their compounds are catalysts. Two features are responsible. First, they can switch easily between oxidation states, so they can take electrons from one reactant and pass them to another. Second, metal surfaces can bond reactant molecules weakly, which brings the molecules together and weakens their bonds. Examples: vanadium(V) oxide in the contact process ($\mathrm{2SO_{2}(g)+O_{2}(g)\rightarrow2SO_{3}(g)}$), finely divided iron in the Haber process for ammonia, nickel in the hydrogenation of vegetable oils, and $\mathrm{Fe^{3+}}$ ions in the reaction between iodide and peroxodisulphate (Example 6). (Here we only use the fact that a catalyst provides a path of lower activation energy, as in Chapter 3; the detailed theory of surface catalysis is not part of this course.)

✏️ Example 6 — How iron ions catalyse a reaction between two anions
The reaction $\mathrm{S_{2}O_{8}^{2-}(aq)+2I^{-}(aq)\rightarrow2SO_{4}^{2-}(aq)+I_{2}(aq)}$ is slow, but a trace of $\mathrm{Fe^{3+}}$ speeds it up greatly. (a) Suggest why the uncatalysed reaction is slow. (b) Write two steps, each involving the iron ions, that add up to the overall reaction. (c) Use $E^{\circ}$ values ($\mathrm{S_{2}O_{8}^{2-}/SO_{4}^{2-}}$ $+2.01\ \mathrm{V}$, $\mathrm{Fe^{3+}/Fe^{2+}}$ $+0.77\ \mathrm{V}$, $\mathrm{I_{2}/I^{-}}$ $+0.54\ \mathrm{V}$) to show that both steps are feasible. (d) Would $\mathrm{Fe^{2+}}$ also act as a catalyst?
(a) Both reactants are negative ions, so they repel each other; collisions with enough energy to react are rare (high activation energy).
(b) $\mathrm{2Fe^{3+}(aq)+2I^{-}(aq)\rightarrow2Fe^{2+}(aq)+I_{2}(aq)}$ and $\mathrm{2Fe^{2+}(aq)+S_{2}O_{8}^{2-}(aq)\rightarrow2Fe^{3+}(aq)+2SO_{4}^{2-}(aq)}$. Adding them, $\mathrm{2Fe^{3+}}$ and $\mathrm{2Fe^{2+}}$ cancel and the overall equation is recovered. Each step is between oppositely charged ions, which attract.
(c) Step 1: $E^{\circ}=0.77-0.54=+0.23\ \mathrm{V}$. Step 2: $E^{\circ}=2.01-0.77=+1.24\ \mathrm{V}$. Both are positive, so both are spontaneous. The catalyst works because its $E^{\circ}$ lies between those of the two reacting couples.
(d) Yes. Starting with $\mathrm{Fe^{2+}}$, step 2 happens first and produces $\mathrm{Fe^{3+}}$, which then runs step 1; the cycle is the same. Iron's ability to switch between $+2$ and $+3$ is exactly what makes it a catalyst here.
📘 Definition 4.4 — Interstitial compounds
Interstitial compounds are formed when small atoms such as H, C, N or B are trapped in the holes (interstices) of the crystal lattice of a metal. Examples are $\mathrm{TiC}$, $\mathrm{Mn_{4}N}$, $\mathrm{Fe_{3}H}$, $\mathrm{VH_{0.56}}$ and $\mathrm{TiH_{1.7}}$. They are usually non-stoichiometric (neither typically ionic nor covalent, so the formula need not match normal valencies). Compared with the pure metal they have higher melting points, are very hard (some borides approach diamond), keep metallic conductivity, and are chemically inert. Steel and cast iron are hard because carbon atoms occupy interstitial sites in iron.
📘 Definition 4.5 — Alloy
An alloy is a blend of metals (sometimes with a non-metal such as carbon) made by mixing the components in the molten state and cooling. Transition metals form alloys very readily because their atomic radii are within about $15\%$ of one another, so atoms of one metal can take the place of atoms of another in the crystal lattice. The resulting alloys are hard and often have high melting points. Examples: ferrous alloys (steels containing Cr, V, W, Mo or Mn), stainless steel (iron with about $18\%$ chromium and $8\%$ nickel), brass (copper and zinc) and bronze (copper and tin).
⚠️ Common mistakes with transition elements

1. Removing $3d$ before $4s$. Ions lose their $4s$ electrons first: $\mathrm{Fe^{3+}}$ is $\mathrm{[Ar]\,3d^{5}}$, not $\mathrm{[Ar]\,3d^{3}\,4s^{2}}$.

2. Applying the Cr/Cu exceptions to ions. They concern atoms only. $\mathrm{Cr^{2+}}$ is $d^{4}$ and $\mathrm{Cu^{2+}}$ is $d^{9}$, as for any other ion.

3. Using total instead of unpaired electrons. In $\mu=\sqrt{n(n+2)}$, $n$ is the number of unpaired electrons: $\mathrm{Ni^{2+}}$ ($d^{8}$) has $n=2$, so $\mu=2.83\ \mathrm{BM}$, not $\sqrt{80}$.

4. Calling every $d$-block element a transition element. Zn, Cd and Hg are not; Sc, Cu, Ag and Au are.

Exercises

I
Configurations and definitions. Atomic numbers: Sc 21, Ti 22, V 23, Cr 24, Mn 25, Fe 26, Co 27, Ni 28, Cu 29, Zn 30.
  • 1 Write the electronic configurations of $\mathrm{Ti}$, $\mathrm{V^{2+}}$, $\mathrm{Cr^{2+}}$, $\mathrm{Ni^{2+}}$ and $\mathrm{Zn}$.
  • 2 Which of the following have a $3d^{5}$ configuration: $\mathrm{Mn^{2+}}$, $\mathrm{Fe^{3+}}$, $\mathrm{Cr^{+}}$, $\mathrm{Co^{2+}}$, $\mathrm{Fe^{2+}}$?
  • 3 How many $3d$ electrons are there in $\mathrm{Co^{3+}}$, $\mathrm{Cu^{2+}}$, $\mathrm{Ti^{4+}}$ and $\mathrm{V^{3+}}$?
  • 4 Copper has a completely filled $3d$ subshell in the atom, yet it is a transition element; zinc is not. Explain.
  • 5 Why is the configuration of chromium $\mathrm{[Ar]\,3d^{5}\,4s^{1}}$ rather than $\mathrm{[Ar]\,3d^{4}\,4s^{2}}$?
  • 6 Which atom of the 3d series has the largest number of unpaired electrons? How many?
  • 7 A 3d metal ion $\mathrm{M^{3+}}$ has the configuration $\mathrm{[Ar]\,3d^{5}}$. Identify M.
  • 8 A 3d metal ion $\mathrm{M^{2+}}$ has nine $3d$ electrons. Identify it, and say whether its compounds are likely to be coloured.
1 $\mathrm{Ti}$ $\mathrm{[Ar]\,3d^{2}\,4s^{2}}$; $\mathrm{V^{2+}}$ $\mathrm{[Ar]\,3d^{3}}$; $\mathrm{Cr^{2+}}$ $\mathrm{[Ar]\,3d^{4}}$; $\mathrm{Ni^{2+}}$ $\mathrm{[Ar]\,3d^{8}}$; $\mathrm{Zn}$ $\mathrm{[Ar]\,3d^{10}\,4s^{2}}$.
2 $\mathrm{Mn^{2+}}$ ($25-18-2=5$), $\mathrm{Fe^{3+}}$ ($26-18-3=5$) and $\mathrm{Cr^{+}}$ ($\mathrm{3d^{5}4s^{1}}$ loses the $4s$ electron). $\mathrm{Co^{2+}}$ is $d^{7}$ and $\mathrm{Fe^{2+}}$ is $d^{6}$.
3 $\mathrm{Co^{3+}}$ 6; $\mathrm{Cu^{2+}}$ 9; $\mathrm{Ti^{4+}}$ 0; $\mathrm{V^{3+}}$ 2.
4 In its common $+2$ state copper is $3d^{9}$, an incompletely filled $d$ subshell, so it meets the definition. Zinc is $3d^{10}$ both in the atom and in its only ion $\mathrm{Zn^{2+}}$.
5 The $3d$ and $4s$ energies are very close; moving one electron from $4s$ to $3d$ gives a half-filled $3d^{5}$ set (with $4s^{1}$), which has extra stability from exchange energy and a symmetrical charge distribution. The gain outweighs the small promotion cost.
6 Chromium: $3d^{5}4s^{1}$ has $5+1=6$ unpaired electrons (manganese, $3d^{5}4s^{2}$, has 5).
7 $Z-18-3=5$, so $Z=26$: iron, and the ion is $\mathrm{Fe^{3+}}$.
8 $Z-18-2=9$, so $Z=29$: $\mathrm{Cu^{2+}}$. Yes: $d^{9}$ allows $d$–$d$ transitions, and copper(II) compounds are typically blue or green.
II
Physical properties and trends. Use the data table of this section.
  • 1 Why do transition metals generally have high melting points and enthalpies of atomisation?
  • 2 Why is the melting point of manganese ($1519\ \mathrm{K}$) much lower than those of its neighbours chromium and iron?
  • 3 Find the decrease in metallic radius from Sc to Cr and from Cr to Ni. Explain why the second is so much smaller.
  • 4 Why are the atomic radii of Fe, Co and Ni almost the same?
  • 5 Across period 3, $\Delta_{i}H_{1}$ rises from $496$ (Na) to $1520\ \mathrm{kJ\,mol^{-1}}$ (Ar). Across the 3d series it rises only from $631$ to $906$. Explain the difference.
  • 6 Why is $\Delta_{i}H_{2}$ of copper the highest in the series?
  • 7 Arrange Sc, Ti, Cr and Zn in order of increasing enthalpy of atomisation, and explain the position of zinc.
  • 8 Calculate $\Delta_{i}H_{1}+\Delta_{i}H_{2}+\Delta_{i}H_{3}$ for Mn and for Fe. Which forms a gaseous $\mathrm{M^{3+}}$ ion more easily, and why?
1 Their atoms are held together by strong metallic bonding in which both the $ns$ and the unpaired $(n-1)d$ electrons take part. More bonding electrons mean stronger interatomic bonds.
2 Manganese ($3d^{5}4s^{2}$) holds its stable half-filled $3d^{5}$ electrons tightly, so they contribute little to metallic bonding; the metal also has an unusual, complex crystal structure. Its enthalpy of atomisation ($281\ \mathrm{kJ\,mol^{-1}}$) is correspondingly low.
3 Sc to Cr: $164-129=35\ \mathrm{pm}$. Cr to Ni: $129-125=4\ \mathrm{pm}$. At the start each added $3d$ electron shields poorly and the rising nuclear charge pulls the atom in; later the growing number of $d$ electrons shields the outer electrons enough to balance the extra protons.
4 The increase in nuclear charge from Fe to Ni is almost exactly balanced by the increased shielding (and repulsion) of the extra $3d$ electrons, so the effective nuclear charge felt by the $4s$ electrons hardly changes ($126$, $125$, $125\ \mathrm{pm}$).
5 Across period 3 the electrons enter the outer shell, which shields poorly, so $Z_{\text{eff}}$ rises steeply. Across the 3d series the electrons enter the inner $3d$ subshell, which partly shields the outer $4s$ electrons from the added nuclear charge, so $\Delta_{i}H_{1}$ rises only slowly.
6 The second electron is removed from $\mathrm{Cu^{+}}$, $\mathrm{[Ar]\,3d^{10}}$, breaking a completely filled, stable $3d$ subshell: $1958\ \mathrm{kJ\,mol^{-1}}$.
7 $\mathrm{Zn}\ (126)\lt\mathrm{Sc}\ (326)\lt\mathrm{Cr}\ (397)\lt\mathrm{Ti}\ (473)\ \mathrm{kJ\,mol^{-1}}$. Zinc's $3d^{10}$ electrons are all paired and take no part in metallic bonding, so it is lowest.
8 Mn: $717+1509+3260=5486\ \mathrm{kJ\,mol^{-1}}$. Fe: $762+1561+2962=5285\ \mathrm{kJ\,mol^{-1}}$. Iron needs $201\ \mathrm{kJ\,mol^{-1}}$ less, because its third electron comes from $d^{6}$ (giving stable $d^{5}$), while manganese's comes from the stable $d^{5}$.
III
Oxidation states and electrode potentials. Use the $E^{\circ}$ table of this section and $E^{\circ}_{\mathrm{O_{2}/H_{2}O}}=+1.23\ \mathrm{V}$ (acid).
  • 1 Why does manganese show the largest number of oxidation states in the 3d series?
  • 2 State the highest oxidation states of Ti, V, Cr and Mn, and give one compound or ion in which each is found.
  • 3 The highest fluoride of manganese is $\mathrm{MnF_{4}}$, but its highest oxide is $\mathrm{Mn_{2}O_{7}}$. Explain.
  • 4 Find the oxidation state of the transition metal in: $\mathrm{CrO_{4}^{2-}}$, $\mathrm{MnO_{4}^{2-}}$, $\mathrm{VO^{2+}}$, $\mathrm{[Fe(CN)_{6}]^{3-}}$, $\mathrm{TiCl_{4}}$, $\mathrm{Cu_{2}O}$.
  • 5 Why is $E^{\circ}(\mathrm{Cu^{2+}/Cu})$ positive? Does copper liberate hydrogen from dilute hydrochloric acid?
  • 6 Which is more stable in water, $\mathrm{Cr^{3+}}$ or $\mathrm{Mn^{3+}}$? Explain using $E^{\circ}$ values and configurations.
  • 7 Show that $\mathrm{Co^{3+}(aq)}$ can oxidise water to oxygen under standard conditions. Write the balanced equation.
  • 8 Why is $E^{\circ}(\mathrm{Zn^{2+}/Zn})$ more negative than the trend suggests, and why does zinc not form $\mathrm{Zn^{3+}}$?
  • 9 Copper(II) iodide does not exist: mixing $\mathrm{Cu^{2+}}$ and $\mathrm{I^{-}}$ gives white copper(I) iodide and iodine. Write the balanced ionic equation and identify the oxidising and reducing agents.
1 Manganese ($3d^{5}4s^{2}$) has seven $4s$ and $3d$ electrons, all of which can be lost or shared, giving every state from $+2$ to $+7$; it lies in the middle of the series, where the number of available electrons is largest while the $d$ electrons are still unpaired.
2 Ti $+4$ ($\mathrm{TiO_{2}}$, $\mathrm{TiCl_{4}}$); V $+5$ ($\mathrm{V_{2}O_{5}}$); Cr $+6$ ($\mathrm{CrO_{3}}$, $\mathrm{Cr_{2}O_{7}^{2-}}$); Mn $+7$ ($\mathrm{Mn_{2}O_{7}}$, $\mathrm{MnO_{4}^{-}}$).
3 Oxygen can form multiple ($\pi$) bonds with the metal, $\mathrm{Mn{=}O}$, so fewer oxygen atoms are needed to reach a high oxidation state; fluorine forms only single bonds, and fitting seven $\mathrm{F}$ atoms round one Mn is not possible.
4 $\mathrm{CrO_{4}^{2-}}$: $x-8=-2$, $+6$. $\mathrm{MnO_{4}^{2-}}$: $+6$. $\mathrm{VO^{2+}}$: $x-2=+2$, $+4$. $\mathrm{[Fe(CN)_{6}]^{3-}}$: $x-6=-3$, $+3$. $\mathrm{TiCl_{4}}$: $+4$. $\mathrm{Cu_{2}O}$: $+1$.
5 The energy needed to atomise copper and remove two electrons ($\Delta_{a}H^{\circ}+\Delta_{i}H_{1}+\Delta_{i}H_{2}$, both high) is not compensated by the hydration enthalpy of $\mathrm{Cu^{2+}}$. Since $E^{\circ}_{\mathrm{Cu^{2+}/Cu}}=+0.34\ \mathrm{V}\gt0=E^{\circ}_{\mathrm{H^{+}/H_{2}}}$, copper cannot reduce $\mathrm{H^{+}}$: no hydrogen is liberated.
6 $\mathrm{Cr^{3+}}$. $E^{\circ}_{\mathrm{Cr^{3+}/Cr^{2+}}}=-0.41\ \mathrm{V}$: $\mathrm{Cr^{3+}}$ ($d^{3}$) has little tendency to be reduced. $E^{\circ}_{\mathrm{Mn^{3+}/Mn^{2+}}}=+1.57\ \mathrm{V}$: $\mathrm{Mn^{3+}}$ ($d^{4}$) is readily reduced to $d^{5}$ $\mathrm{Mn^{2+}}$, so it is a strong oxidising agent and unstable in water.
7 $\mathrm{4Co^{3+}(aq)+2H_{2}O(l)\rightarrow4Co^{2+}(aq)+O_{2}(g)+4H^{+}(aq)}$; $E^{\circ}=1.97-1.23=+0.74\ \mathrm{V}\gt0$, so the reaction is spontaneous: simple $\mathrm{Co^{3+}}$ salts are not stable in water (cobalt(III) is stable only in complexes).
8 $\mathrm{Zn^{2+}}$ has the stable $3d^{10}$ configuration and zinc has a very low enthalpy of atomisation, so forming $\mathrm{Zn^{2+}(aq)}$ is easier than the trend suggests. Forming $\mathrm{Zn^{3+}}$ would mean breaking the full $3d^{10}$ set ($\Delta_{i}H_{3}=3837\ \mathrm{kJ\,mol^{-1}}$, the highest in the series), which is never repaid.
9 $\mathrm{2Cu^{2+}(aq)+4I^{-}(aq)\rightarrow2CuI(s)+I_{2}(aq)}$ (atoms: 2 Cu, 4 I each side; charge $+4-4=0$ ✓). $\mathrm{Cu^{2+}}$ is the oxidising agent ($+2\rightarrow+1$) and $\mathrm{I^{-}}$ the reducing agent ($-1\rightarrow0$). The product is often written as $\mathrm{Cu_{2}I_{2}}$; it is the same compound.
IV
Magnetism, colour, catalysis, interstitial compounds and alloys. Use the spin-only formula; Cr $52.0\ \mathrm{g\,mol^{-1}}$.
  • 1 Calculate the spin-only magnetic moments of $\mathrm{Ti^{3+}}$, $\mathrm{V^{3+}}$, $\mathrm{Cr^{3+}}$, $\mathrm{Co^{2+}}$ and $\mathrm{Ni^{2+}}$.
  • 2 Arrange $\mathrm{Mn^{2+}}$, $\mathrm{Fe^{2+}}$, $\mathrm{Cu^{2+}}$, $\mathrm{Zn^{2+}}$ and $\mathrm{Ni^{2+}}$ in order of increasing magnetic moment.
  • 3 A 3d ion $\mathrm{M^{2+}}$ has a spin-only moment of $4.90\ \mathrm{BM}$. Which two ions are possible?
  • 4 $\mathrm{Sc^{3+}}$, $\mathrm{Ti^{4+}}$ and $\mathrm{Zn^{2+}}$ salts are white, but $\mathrm{Ti^{3+}}$ salts are purple. Explain.
  • 5 Hydrated copper(II) ions absorb mainly orange-red light. What colour do their solutions appear? Explain the term you use.
  • 6 Give two reasons why transition metals and their compounds are good catalysts, and name the catalyst used in (a) the Haber process, (b) the contact process.
  • 7 What are interstitial compounds? State three of their characteristic properties.
  • 8 Kitchen engineering. A stainless-steel sink of mass $2.50\ \mathrm{kg}$ contains $18\%$ Cr and $8\%$ Ni by mass. Find the masses of Cr and Ni, and the amount (in mol) of chromium. Why do transition metals form alloys so readily?
Check: unpaired electrons from a measured magnetic moment
nearest $n$—
spin-only $\mu$ for that $n$—
possible $d^{n}$ · ions—
1 $\mathrm{Ti^{3+}}$ $d^{1}$, $n=1$: $1.73\ \mathrm{BM}$. $\mathrm{V^{3+}}$ $d^{2}$: $\sqrt{8}=2.83\ \mathrm{BM}$. $\mathrm{Cr^{3+}}$ $d^{3}$: $\sqrt{15}=3.87\ \mathrm{BM}$. $\mathrm{Co^{2+}}$ $d^{7}$, $n=3$: $3.87\ \mathrm{BM}$. $\mathrm{Ni^{2+}}$ $d^{8}$, $n=2$: $2.83\ \mathrm{BM}$.
2 $n$: $\mathrm{Zn^{2+}}$ 0, $\mathrm{Cu^{2+}}$ 1, $\mathrm{Ni^{2+}}$ 2, $\mathrm{Fe^{2+}}$ 4, $\mathrm{Mn^{2+}}$ 5. Order: $\mathrm{Zn^{2+}}\ (0)\lt\mathrm{Cu^{2+}}\ (1.73)\lt\mathrm{Ni^{2+}}\ (2.83)\lt\mathrm{Fe^{2+}}\ (4.90)\lt\mathrm{Mn^{2+}}\ (5.92\ \mathrm{BM})$.
3 $n(n+2)=24$, $n=4$: $d^{4}$ or $d^{6}$. As $\mathrm{M^{2+}}$: $\mathrm{Cr^{2+}}$ ($24-18-2=4$) or $\mathrm{Fe^{2+}}$ ($26-18-2=6$). (The check-lab above starts with this value.)
4 $\mathrm{Sc^{3+}}$ and $\mathrm{Ti^{4+}}$ are $d^{0}$ and $\mathrm{Zn^{2+}}$ is $d^{10}$: no $d$–$d$ transition is possible, so no visible light is absorbed. $\mathrm{Ti^{3+}}$ is $d^{1}$; its electron can be excited to a higher $d$ orbital by absorbing green-yellow light, and the transmitted light looks purple.
5 Blue (blue-green). We see the complementary colour, the colour that remains when the absorbed part of white light is removed; orange-red and blue (green-blue) are complementary.
6 (i) They can change oxidation state easily, accepting electrons from one reactant and passing them on; (ii) their surfaces can bond reactant molecules, bringing them together and weakening their bonds (lower activation energy). (a) Finely divided iron; (b) vanadium(V) oxide, $\mathrm{V_{2}O_{5}}$.
7 Compounds formed when small atoms (H, C, N, B) occupy holes in a metal's crystal lattice, e.g. $\mathrm{TiC}$, $\mathrm{Mn_{4}N}$, $\mathrm{Fe_{3}H}$. They are non-stoichiometric; they have higher melting points than the pure metals; they are very hard; they retain metallic conductivity; they are chemically inert (any three).
8 Cr: $0.18\times2.50=0.45\ \mathrm{kg}$; Ni: $0.08\times2.50=0.20\ \mathrm{kg}$. $n(\mathrm{Cr})=\dfrac{450\ \mathrm{g}}{52.0\ \mathrm{g\,mol^{-1}}}=8.65\ \mathrm{mol}$. Transition-metal atoms have similar radii (within about $15\%$) and similar metallic bonding, so one can replace another in the crystal lattice.
🎯 Exam corner BoardJEE MainJEE AdvancedNEET

E1. Board · 3 marks (a) Why do transition elements show variable oxidation states? (b) $\mathrm{Cr^{2+}}$ and $\mathrm{Mn^{3+}}$ both have the $d^{4}$ configuration, yet $\mathrm{Cr^{2+}}$ is reducing while $\mathrm{Mn^{3+}}$ is oxidising. Explain. (c) Which of $\mathrm{Cu^{+}}$ and $\mathrm{Cu^{2+}}$ is paramagnetic?

(a) The $(n-1)d$ and $ns$ electrons have very similar energies, so different numbers of them can be lost or shared, giving several oxidation states that usually differ by one.
(b) $\mathrm{Cr^{2+}}$ loses an electron to become $\mathrm{Cr^{3+}}$, $d^{3}$, which is very stable in water ($E^{\circ}_{\mathrm{Cr^{3+}/Cr^{2+}}}=-0.41\ \mathrm{V}$). $\mathrm{Mn^{3+}}$ gains an electron to become $\mathrm{Mn^{2+}}$, the stable half-filled $d^{5}$ ($E^{\circ}_{\mathrm{Mn^{3+}/Mn^{2+}}}=+1.57\ \mathrm{V}$).
(c) $\mathrm{Cu^{2+}}$ ($3d^{9}$, one unpaired electron). $\mathrm{Cu^{+}}$ ($3d^{10}$) is diamagnetic.
Marking tip One mark each; in (b) the two configurations $d^{3}$ and $d^{5}$ earn the mark, the $E^{\circ}$ values support it.

E2. JEE Main Numerical value. Among the aqueous ions $\mathrm{V^{2+}}$, $\mathrm{Cr^{3+}}$, $\mathrm{Mn^{2+}}$, $\mathrm{Fe^{3+}}$, $\mathrm{Co^{2+}}$, $\mathrm{Ni^{2+}}$, $\mathrm{Cu^{2+}}$ and $\mathrm{Mn^{3+}}$, the number of ions with a spin-only magnetic moment of $3.87\ \mathrm{BM}$ is ______.

Idea $3.87\ \mathrm{BM}$ means $n=3$: a $d^{3}$ or $d^{7}$ ion.
Count $\mathrm{V^{2+}}$ $d^{3}$ ✓; $\mathrm{Cr^{3+}}$ $d^{3}$ ✓; $\mathrm{Mn^{2+}}$ $d^{5}$; $\mathrm{Fe^{3+}}$ $d^{5}$; $\mathrm{Co^{2+}}$ $d^{7}$ ✓; $\mathrm{Ni^{2+}}$ $d^{8}$; $\mathrm{Cu^{2+}}$ $d^{9}$; $\mathrm{Mn^{3+}}$ $d^{4}$.
Answer $\mathbf{3}$.

E3. NEET Which ion is colourless in aqueous solution? (A) $\mathrm{Ti^{3+}}$  (B) $\mathrm{V^{3+}}$  (C) $\mathrm{Sc^{3+}}$  (D) $\mathrm{Cu^{2+}}$

Idea Colour needs a $d$–$d$ transition, i.e. $d^{1}$ to $d^{9}$. $\mathrm{Ti^{3+}}$ $d^{1}$, $\mathrm{V^{3+}}$ $d^{2}$, $\mathrm{Cu^{2+}}$ $d^{9}$, but $\mathrm{Sc^{3+}}$ is $d^{0}$.
Answer (C) $\mathrm{Sc^{3+}}$.

E4. JEE Main In which pair is the third ionisation enthalpy of the first element greater than that of the second? (A) Mn, Fe  (B) Fe, Co  (C) Cr, Mn  (D) Co, Ni

Idea $\Delta_{i}H_{3}$ is high when the electron comes out of a stable $d^{5}$ or $d^{10}$ $\mathrm{M^{2+}}$ ion and low when removal creates $d^{5}$.
Check (A) $\mathrm{Mn^{2+}}$ $d^{5}\rightarrow d^{4}$ (hard) vs $\mathrm{Fe^{2+}}$ $d^{6}\rightarrow d^{5}$ (easy): $3260\gt2962$ ✓. (B) $2962\lt3243$. (C) $2990\lt3260$. (D) $3243\lt3402$.
Answer (A).

E5. JEE Advanced One or more correct. Choose the correct statement(s). (A) $\mathrm{Cu^{+}}$ disproportionates in aqueous solution. (B) The highest oxidation state of manganese in a fluoride is $+7$. (C) $\mathrm{Cr^{2+}}$ is a stronger reducing agent than $\mathrm{Fe^{2+}}$. (D) Zn, Cd and Hg are regarded as transition elements because they are in the $d$-block.

(A) True $\mathrm{2Cu^{+}(aq)\rightarrow Cu^{2+}(aq)+Cu(s)}$, $E^{\circ}=0.52-0.15=+0.37\ \mathrm{V}$, driven by the large hydration enthalpy of $\mathrm{Cu^{2+}}$.
(B) False The highest fluoride is $\mathrm{MnF_{4}}$ ($+4$); $+7$ is reached only with oxygen ($\mathrm{Mn_{2}O_{7}}$, $\mathrm{MnO_{4}^{-}}$), which forms multiple bonds.
(C) True $E^{\circ}_{\mathrm{Cr^{3+}/Cr^{2+}}}=-0.41\ \mathrm{V}$ against $+0.77\ \mathrm{V}$ for iron.
(D) False · Answer Their $d$ subshell is full in the atom and in the common ions. Correct: (A) and (C).

E6. Board · Case study Titanium is everywhere in medicine and industry. The metal itself is used for hip joints; titanium(IV) oxide, $\mathrm{TiO_{2}}$, is the brilliant white pigment in toothpaste and paint; solutions of titanium(III) salts are purple; titanium carbide, $\mathrm{TiC}$, coats cutting tools because it is extremely hard. (a) Explain why $\mathrm{TiO_{2}}$ is white while $\mathrm{Ti^{3+}}$ salts are coloured. (2 marks) (b) Calculate the spin-only magnetic moment of $\mathrm{Ti^{3+}}$ ($Z=22$). (1 mark) (c) What type of compound is $\mathrm{TiC}$? State one property, other than hardness, that it shares with this class. (1 mark)

(a) In $\mathrm{TiO_{2}}$ titanium is $+4$, $\mathrm{Ti^{4+}}$ $\mathrm{[Ar]\,3d^{0}}$: there are no $d$ electrons, so no $d$–$d$ transition and no absorption of visible light. $\mathrm{Ti^{3+}}$ is $3d^{1}$; the electron absorbs visible (green-yellow) light to jump to a higher $d$ orbital, and the complementary colour, purple, is seen.
(b) $n=1$: $\mu=\sqrt{1\times3}=1.73\ \mathrm{BM}$.
(c) An interstitial compound (carbon atoms in the holes of the titanium lattice). It has a very high melting point (or: retains metallic conductivity; is chemically inert).
🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes

Research question. A surgeon replacing a hip can choose a stem made of a titanium alloy, a stainless steel or a cobalt–chromium alloy. What do the chemical properties of transition metals that you met in this section have to do with the choice?

Why it matters. Over a million hip and knee joints are replaced worldwide every year, in India increasingly so as people live longer. An implant must be strong, must not corrode in warm salty body fluid for twenty years or more, must not release ions that cause allergy, and should not heat up or distort an MRI image.

What to do.

  1. A femoral stem has a volume of about $40\ \mathrm{cm^{3}}$. Find its mass in Ti–6Al–4V (density $4.43\ \mathrm{g\,cm^{-3}}$), in 316L stainless steel ($8.0\ \mathrm{g\,cm^{-3}}$) and in a Co–Cr alloy ($8.3\ \mathrm{g\,cm^{-3}}$).
  2. Titanium has a very negative $E^{\circ}(\mathrm{Ti^{2+}/Ti})=-1.63\ \mathrm{V}$, yet it does not corrode in the body. Find out what the protective films on titanium and on stainless steel are made of, and explain "passivity" in two sentences.
  3. 316L steel contains about $12\%$ nickel by mass. How many grams of nickel are in the stem? Look up why nickel ions are a concern for some patients.
  4. Which of the three materials are ferromagnetic? Find out why titanium implants are preferred for patients who will need MRI scans.

Tools: Calculator; the data table of this section; open sources such as orthopaedic-materials reviews or national joint-registry reports.

Hand in: A one-page comparison table (mass, corrosion protection, allergy risk, MRI behaviour) and a recommendation for a 45-year-old active patient, with reasons tied to transition-metal chemistry (alloys, passivating oxides, magnetism).

Going further. Titanium's surface oxide is $\mathrm{TiO_{2}}$, with Ti in the $+4$ state. Explain from its configuration why this oxide is white and non-magnetic, and why bone cells can grow directly on it.

Instructor note: masses ≈ $177\ \mathrm{g}$ (Ti alloy), $320\ \mathrm{g}$ (316L), $332\ \mathrm{g}$ (Co–Cr). Passive films: $\mathrm{TiO_{2}}$ on titanium, $\mathrm{Cr_{2}O_{3}}$-rich film on stainless steel and Co–Cr; they are thin, adherent and self-repairing, so the very negative $E^{\circ}$ does not lead to corrosion. Nickel: $0.12\times320\approx38\ \mathrm{g}$; released $\mathrm{Ni^{2+}}$ is a common contact allergen. Austenitic 316L is essentially non-ferromagnetic and Co–Cr alloys are not ferromagnetic either, but titanium alloys have the lowest magnetic susceptibility of the three (only weakly paramagnetic) and give the least image distortion. $\mathrm{Ti^{4+}}$ is $d^{0}$: no $d$–$d$ transitions (white) and no unpaired electrons (diamagnetic).

🧪 Self-check quiz

0/5 correct

Q1. The electronic configuration of $\mathrm{Cr^{3+}}$ ($Z=24$) is

Q2. The spin-only magnetic moment of $\mathrm{Fe^{3+}}$ ($Z=26$) is about

Q3. Which 3d metal has a positive standard electrode potential $E^{\circ}(\mathrm{M^{2+}/M})$?

Q4. The highest oxidation state shown by manganese is reached in

Q5. Interstitial compounds such as $\mathrm{TiC}$ and $\mathrm{Fe_{3}H}$ are typically

← Chapter 4 — The d- and f-Block Elements4.2 Important Compounds: Potassium Dichromate and Potassium Permanganate →