The d- and f-Block Elements: Class 12 Chemistry · Chapter 4

Chemistry Part 1 · Companion notes, simulations and practice · Official NCERT chapter PDF ↗

Published 11 Oct 2026 · Updated 11 Oct 2026

130 min
43 practice questions
46 MCQs

What you will learn in The d- and f-Block Elements

Locate the d- and f-blocks in the periodic table and decide whether an element counts as a transition element
Write ground-state configurations of d- and f-block atoms and ions, explaining exceptions such as chromium and copper
Explain trends across the 3d series in atomisation enthalpy, size, ionisation enthalpy, oxidation states and electrode potentials
Relate magnetic moments, colour, complex formation, catalysis, interstitial compounds and alloys to partly filled d orbitals
Describe the preparation, structure and oxidising reactions of potassium dichromate and potassium permanganate, and balance their ionic equations
Compare the lanthanoids and actinoids in configuration, size, oxidation states and reactivity
Name important uses of d- and f-block elements and their compounds

Original companion text aligned to NCERT Chemistry Class 12, Chapter 4, pp. 89–117. Not affiliated with or endorsed by NCERT.

In Mehrauli, Delhi, an iron pillar has stood in the open for about sixteen centuries and has barely rusted. Iron shares the middle of the periodic table with copper, silver, gold, titanium and platinum: the metals of tools, coins, wires, bridges and jewellery. This chapter is about that middle region, the d-block, and about the two rows printed separately below the main table, the f-block.

In d-block atoms the electrons that arrive last go into a d subshell of the shell below the outermost one, the $(n - 1) d$ subshell. In f-block atoms they go even deeper, into the $4 f$ or $5 f$ subshell. Chemists call the d-block elements transition elements (संक्रमण तत्व) and the f-block elements inner transition elements. The 4f elements are the lanthanoids and the 5f elements are the actinoids.

Definition

Transition element

An element whose atom, or one of whose common ions, has an incompletely filled d subshell (the IUPAC definition).

ExampleIron: $F e$ is $3 d^{6} 4 s^{2}$ , and $F e^{3 +}$ is $3 d^{5}$ .

The name is older than our picture of electrons. Early chemists saw these metals as a bridge between the very reactive metals on the left of the table and the elements on the right. In position and in properties, they marked a 'transition' from one side to the other. Today the label is decided by electrons, as in the definition above. That test leaves out three familiar metals, zinc, cadmium and mercury, for a reason Section 4.2 makes clear; they stay in this chapter all the same, because they complete the d-series.

The f-block matters to India in a very practical way. The black sands of the Kerala and Odisha coasts contain monazite, a mineral rich in cerium, lanthanum, neodymium and the actinoid thorium. Neodymium goes into the strong magnets of motors and earphones, and thorium is the planned fuel for the last stage of India's nuclear power programme.

Before we zoom in, note how the chapter runs: first the d-block (Sections 4.1 to 4.4), with most attention on the first row, scandium to zinc; then the f-block (Sections 4.5 and 4.6); and last, uses of both (Section 4.7).

4.1 Position in the Periodic Table

Look at the periodic table and the d-block is the wide band of ten columns in the middle, groups 3 to 12, with the s-block to its left and the p-block to its right. Each period from the fourth onwards contains one row of ten d-block elements, because a d subshell holds ten electrons.

3/4The lanthanoids (4f) and actinoids (5f) are lifted out and printed below, so the table is not 32 columns wide.

Fig. 4.1The four blocks of the periodic table. The f-block rows really belong inside periods 6 and 7.NCERT p. 89

An outline of the periodic table as a grid of small squares. In the first step only the two left-hand columns (s-block) and the six right-hand columns (p-block) are shaded. In the second step the ten middle columns of periods 4 to 7 are shaded as the d-block, with rows labelled 3d, 4d, 5d and 6d and the first and last squares of the 3d row marked Sc and Zn. In the third step two separate rows of fourteen squares below the table are shaded and labelled 4f lanthanoids and 5f actinoids; star marks link them to their places in group 3 of periods 6 and 7. In the last step the group 12 column is outlined with a dashed box and a note says its d shell is always full.

3d series (period 4): scandium ( $Z = 21$ ) to zinc ( $Z = 30$ ).
4d series (period 5): yttrium ( $Z = 39$ ) to cadmium ( $Z = 48$ ).
5d series (period 6): lanthanum ( $Z = 57$ ), then hafnium ( $Z = 72$ ) to mercury ( $Z = 80$ ).
6d series (period 7): actinium ( $Z = 89$ ), then rutherfordium ( $Z = 104$ ) to copernicium ( $Z = 112$ ).

The gaps inside the 5d and 6d series are where the f-block fits. After lanthanum, fourteen electrons fill the 4f subshell (cerium to lutetium) before the 5d series resumes at hafnium. After actinium, the 5f subshell fills from thorium to lawrencium. Printing those 28 elements inside the main table would make it very wide, so they get their own two rows at the bottom.

4.2 Electronic Configurations of the d-Block Elements

Across a d-series the nuclear charge rises by one at each step, and the extra electron usually goes into the $(n - 1) d$ subshell while the outer $n s$ subshell keeps one or two electrons. The general pattern for the outer electrons is therefore

General outer configuration

(n - 1) d^{1 – 10} n s^{1 – 2}

(4.1)

Two atoms of the 3d row break this pattern (Fig. 4.2). Filling 4s before 3d predicts $3 d^{4} 4 s^{2}$ for chromium and $3 d^{9} 4 s^{2}$ for copper, but the real ground states are $3 d^{5} 4 s^{1}$ and $3 d^{10} 4 s^{1}$ . The 3d and 4s levels are so close in energy that a small bonus can tip the balance. A half-filled ( $d^{5}$ ) or completely filled ( $d^{10}$ ) set earns one: extra exchange energy and a symmetrical electron cloud. Moving one electron from 4s into 3d buys that bonus at very little cost.

Table 4.1Ground-state configurations of the 3d series (outside the argon core)NCERT Table 4.1 · p. 90

Element	$Z$	Outer configuration
Sc	21	$3 d^{1} 4 s^{2}$
Ti	22	$3 d^{2} 4 s^{2}$
V	23	$3 d^{3} 4 s^{2}$
Cr	24	$3 d^{5} 4 s^{1}$ (not $3 d^{4} 4 s^{2}$ )
Mn	25	$3 d^{5} 4 s^{2}$
Fe	26	$3 d^{6} 4 s^{2}$
Co	27	$3 d^{7} 4 s^{2}$
Ni	28	$3 d^{8} 4 s^{2}$
Cu	29	$3 d^{10} 4 s^{1}$ (not $3 d^{9} 4 s^{2}$ )
Zn	30	$3 d^{10} 4 s^{2}$

3/3Copper gains a full $3 d^{10}$ set the same way: $3 d^{10} 4 s^{1}$ .

Fig. 4.2One 4s electron moves into 3d when that completes a half-filled or filled set.NCERT p. 90

Two orbital box diagrams side by side, each with five 3d boxes and one 4s box holding electron arrows. On the left, chromium starts with four single up-arrows in 3d and a pair in 4s, labelled expected 3d4 4s2. In the second step the down-arrow from 4s jumps into the empty fifth 3d box and turns up, giving 3d5 4s1 with six unpaired electrons, labelled half-filled 3d. On the right, copper starts with four paired 3d boxes, one single, and a 4s pair; in the third step one 4s electron moves into the single 3d box, giving 3d10 4s1, labelled completely filled 3d.

In the 4d and 5d series the surprises multiply, and not all of them follow the 'half or full' rule. Palladium goes furthest: it empties its outer s subshell completely ( $4 d^{10} 5 s^{0}$ ), the only d-block atom to do so. Niobium ( $4 d^{4} 5 s^{1}$ ) and ruthenium ( $4 d^{7} 5 s^{1}$ ) obey no simple rule at all; they just show how finely the two levels are balanced. The table lists the irregular cases.

Table 4.2Irregular configurations in the 4d and 5d series: simple filling (expected) against the real ground state (actual)NCERT p. 91

Element	Expected	Actual
Nb	$4 d^{3} 5 s^{2}$	$4 d^{4} 5 s^{1}$
Mo	$4 d^{4} 5 s^{2}$	$4 d^{5} 5 s^{1}$
Ru	$4 d^{6} 5 s^{2}$	$4 d^{7} 5 s^{1}$
Rh	$4 d^{7} 5 s^{2}$	$4 d^{8} 5 s^{1}$
Pd	$4 d^{8} 5 s^{2}$	$4 d^{10} 5 s^{0}$
Ag	$4 d^{9} 5 s^{2}$	$4 d^{10} 5 s^{1}$
Pt	$5 d^{8} 6 s^{2}$	$5 d^{9} 6 s^{1}$
Au	$5 d^{9} 6 s^{2}$	$5 d^{10} 6 s^{1}$

At the right-hand end of every series stands group 12: zinc, cadmium, mercury and copernicium, all $(n - 1) d^{10} n s^{2}$ . When these metals react they give up only their s electrons, so even their ions keep a complete d set. That is the electron-based reason the IUPAC definition leaves them out, although we study them alongside their neighbours.

Look at two aqueous ions of different metals, $V^{2 +}$ and $C r^{3 +}$ . Both are violet, both have three unpaired electrons, and both are $d^{3}$ . That is no coincidence. The $(n - 1) d$ orbitals belong to an inner shell, yet their lobes reach out to the edge of the atom. There they touch the neighbouring atoms, ions and molecules, and are touched by them. So the number of d electrons, more than the identity of the metal, often decides how an ion looks and behaves.

This turns a familiar periodic-table habit on its head. In the s- and p-blocks we learn chemistry group by group, because elements in a column resemble each other. In the d-block, neighbours along a row, such as iron, cobalt and nickel, are often more alike than members of a group. They differ only by an electron tucked into the inner d subshell. Section 4.3 follows the first row and its family traits: high melting points, several oxidation states, paramagnetism, coloured ions, complexes and catalysis.

Example 4.1NCERT Example 4.1 · p. 91

Gold, yes; mercury, no

Gold ( $Z = 79$ ) has the outer configuration $5 d^{10} 6 s^{1}$ and mercury ( $Z = 80$ ) has $5 d^{10} 6 s^{2}$ . Both atoms have a full 5d subshell. Explain why gold is counted as a transition element but mercury is not.

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 3

Which is the ground-state configuration of chromium (

Z = 24

[A r] 3 d^{4} 4 s^{2}

[A r] 3 d^{6}

[A r] 3 d^{3} 4 s^{2} 4 p^{1}

[A r] 3 d^{5} 4 s^{1}

Show answer

Answer (D) $[A r] 3 d^{5} 4 s^{1}$

Moving one 4s electron into 3d gives a half-filled

3 d^{5}

set, and that bonus outweighs the small energy cost. The 'expected'

3 d^{4} 4 s^{2}

is the tempting wrong answer.

4.3 General Properties of the Transition Elements (d-Block)

From here to the end of Section 4.4 we follow the first row, scandium to zinc, and bring in the heavier rows only for contrast. The common thread is simple: almost every property we meet depends on how many d electrons an atom or ion has, and on how tightly they are held.

4.3.1 Physical Properties

Pick up a steel spoon, a copper wire or a titanium spectacle frame and you are holding a typical transition metal: it shines, it carries heat and electric current, it bends rather than snapping, and it can be drawn into wire. At room temperature most of these metals pack their atoms in one of the familiar metallic arrangements (body-centred cubic, hexagonal close-packed or cubic close-packed). Manganese adopts a complicated cubic structure of its own, zinc and cadmium use slightly distorted hexagonal packing, and mercury is not even solid.

What stands out is how tough they are. Leaving aside zinc, cadmium and mercury, they resist scratching, need very high temperatures to melt and hardly evaporate even when molten. The reason is the bonding. In an s-block metal each atom shares only its one or two s electrons with the 'sea' of electrons. A transition metal can also commit its unpaired $(n - 1) d$ electrons, and neighbouring d orbitals overlap to give extra bonding. The enthalpy of atomisation (परमाणुकरण एन्थैल्पी), $Δ_{a} H^{\circ}$ , the energy needed to turn one mole of the solid into free gaseous atoms, measures that bond strength directly.

Table 4.3Melting points and enthalpies of atomisation of the 3d metals (rounded data-table values)NCERT p. 92

Metal	m.p. / K	$Δ_{a} H^{\circ}$ / kJ mol⁻¹
Sc	1814	378
Ti	1941	473
V	2183	514
Cr	2180	397
Mn	1519	281
Fe	1811	416
Co	1768	425
Ni	1728	430
Cu	1358	337
Zn	693	130

Read the table from top to bottom. Bond strength climbs while the number of unpaired d electrons rises, and peaks near the middle of the row, where each d orbital holds about one unpaired electron. After that, electrons pair up and fewer are free to bond, so the values fall again. Manganese breaks the smooth curve: its $3 d^{5} 4 s^{2}$ arrangement holds the d electrons tightly, so they contribute less to metallic bonding. Zinc, with every d electron paired and buried, bonds only through 4s and has the lowest value of all.

The heavier rows are tougher still. Molybdenum and tungsten need about $658$ and $849 k J m o l^{- 1}$ to atomise, against $397 k J m o l^{- 1}$ for chromium above them; tungsten melts at $3695 K$ , higher than any other metal. Larger 4d and 5d orbitals overlap better, which is also why metal–metal bonds turn up so often in 4d and 5d compounds. A very large enthalpy of atomisation also makes a metal reluctant to react: platinum and gold owe part of their 'noble' character to it, as Section 4.3.5 shows.

Example 4.2NCERT Example 4.2 · p. 95

Why vanadium holds together so well

The enthalpies of atomisation of potassium, calcium and vanadium are about $89$ , $178$ and $514 k J m o l^{- 1}$ . All three are in period 4. Explain the trend.

Try it yourself first. Then reveal the solution one step at a time.

4.3.2 Variation in Atomic and Ionic Sizes of Transition Metals

Moving along a d-series, each step adds a proton to the nucleus and an electron to the $(n - 1) d$ subshell. Electrons in the same d subshell screen one another poorly, because d orbitals point in different directions. So the outer electrons feel a slowly growing effective nuclear charge and the atoms shrink, but only gradually: the metallic radius falls from $147 p m$ for titanium to $125 p m$ for nickel. Ions of the same charge shrink in the same way along the row.

Going down a group, we expect atoms to grow, since each row adds a whole new shell. From the 3d to the 4d row they do. From the 4d to the 5d row, however, the size hardly changes: the metallic radius of zirconium is $160 p m$ and that of hafnium $159 p m$ . Between lanthanum and hafnium, fourteen electrons go into the 4f subshell. 4f electrons shield even worse than d electrons, so through those fourteen elements the atoms contract steadily. This lanthanoid contraction almost exactly cancels the growth that the extra shell should have caused (Fig. 4.3).

4/4In fact they match their 4d partners: the lanthanoid contraction has cancelled the extra shell.

Fig. 4.3The 4d and 5d members of a group are almost the same size.NCERT Fig. 4.3 · p. 94

A grid of circles drawn to scale. The top row is group 4 and the bottom row group 5; columns are the 3d, 4d and 5d series. In the first step only the 3d atoms appear: titanium 147 picometres and vanadium 135 picometres. In the second step the larger 4d atoms appear: zirconium 160 and niobium 147 picometres. In the third step dashed circles in the 5d column show the larger size we would expect for hafnium and tantalum. In the last step the real hafnium (159) and tantalum (147) circles appear, the same size as the 4d atoms, with an equals sign between them.

The consequences are large. Zirconium and hafnium, or niobium and tantalum, are so alike in size and chemistry that they occur together in the same minerals and are notoriously hard to separate. The contraction also makes the 5d metals very dense: osmium and iridium, at about $22.6 g c m^{- 3}$ , are the densest elements known.

Each atom along the 3d row is a little smaller but a little heavier than the one before, so more mass is packed into every cubic centimetre. Titanium ( $4.51 g c m^{- 3}$ ) is light enough for aircraft frames and artificial hips, whereas copper reaches $8.96 g c m^{- 3}$ .

4.3.3 Ionisation Enthalpies

Ionisation enthalpies also rise along each d-series, as the nuclear charge grows, but far more gently than along a main-group period. From sodium to argon the first ionisation enthalpy triples; from scandium ( $633 k J m o l^{- 1}$ ) to zinc ( $906 k J m o l^{- 1}$ ) it rises by less than half. The new 3d electrons sit inside the 4s electron's region and screen it rather well from the extra protons, so the outermost electron feels only a slowly rising pull.

One rule saves a lot of confusion: when a 3d atom forms a cation, the 4s electrons leave first. Once the atom is charged, the 3d level drops below 4s in energy. Every $M^{2 +}$ and $M^{3 +}$ ion of the row is therefore a bare $[A r] 3 d^{n}$ ion, with no 4s electrons.

Why are the values irregular? Three factors set the energy of an electron in a subshell: its attraction to the nucleus, its repulsion by the other electrons, and the exchange energy (विनिमय ऊर्जा). It is a stabilisation that electrons with parallel spins in degenerate orbitals gain; this is the quantum reason behind Hund's rule. It grows roughly with the number of pairs of parallel spins:

Pairs of parallel spins

N_{pairs} = \frac{p (p - 1)}{2}

(4.2)

where

$p$: the number of electrons with the same spin in a set of degenerate orbitals

Definition

Exchange energy

The extra stability gained by electrons of parallel spin in a set of degenerate orbitals; it is roughly proportional to the number of parallel-spin pairs.

ExampleA $d^{5}$ ion has $5 \times 4 / 2 = 10$ pairs; a $d^{4}$ ion has only 6.

2/4Taking one electron away leaves 6 pairs: four pairs of exchange stabilisation are lost, so this ionisation is costly.

Fig. 4.4Removing an electron from

d^{5}

destroys four parallel pairs; removing one from

d^{6}

destroys none.

Five d-orbital boxes in a row. Curved lines below the boxes join every pair of up-spin electrons. In the first step all five boxes hold one up-arrow (Mn2+, d5) and there are ten connecting curves; the counter reads 10 parallel pairs. In the second step one electron is removed (Mn3+, d4); the four curves that touched it vanish and the counter reads 6, with the note 4 pairs lost. In the third step the boxes show five up-arrows plus a down-arrow in the first box (Fe2+, d6), still 10 pairs. In the last step the down-arrow is removed (Fe3+, d5); all ten curves remain and the note says none lost.

This explains the striking breaks in the data. The second ionisation enthalpy is unusually large for chromium and copper, because $C r^{+}$ ( $3 d^{5}$ ) and $C u^{+}$ ( $3 d^{10}$ ) must give up an electron from a stable set. Manganese's second value is lower than chromium's, because $M n^{+}$ ( $3 d^{5} 4 s^{1}$ ) still has its 4s electron to lose. For the third ionisation, iron ( $2957 k J m o l^{- 1}$ ) actually needs less energy than manganese ( $3248 k J m o l^{- 1}$ ), despite its extra proton: Fig. 4.4 shows why. Zinc's second value is low (it forms $d^{10}$ ), but its third is the highest in the row, as it must break that full set.

So ionisation enthalpies tell part of the story of which oxidation states are easy to reach: nickel, copper and zinc have such high third values that they seldom go beyond +2. They cannot tell the whole story, because forming an ion in a real substance also releases energy, through hydration, crystal formation or bonding to ligands; Section 4.3.5 puts these terms together. Meanwhile, the explorer below lets you compare the numbers across the row.

Interactive3d trends explorer

A bar chart of the ten 3d metals from scandium to zinc. Nine buttons choose the property plotted: enthalpy of atomisation, first, second and third ionisation enthalpies, metallic radius, the two standard electrode potentials M2+/M and M3+/M2+, density and melting point. Bars with no reliable value are shown as gaps. Tapping a bar shows the element, its value with units and, for the elements that break the trend, a one-line explanation such as the half-filled 3d set of manganese or the full 3d set of zinc. A summary line under the chart says what pattern to look for in the chosen property.

Fig. 4.5Pick a property to plot it for scandium to zinc, then tap a bar to read the value and why it stands out.

Quick check

1 of 3

Which ion needs the most energy to lose one more electron?

F e^{2 +}

(

d^{6}

)

C r^{2 +}

(

d^{4}

)

V^{2 +}

(

d^{3}

)

M n^{2 +}

(

d^{5}

)

Show answer

Answer (D) $M n^{2 +}$ ( $d^{5}$ )

Ionising

M n^{2 +}

breaks a half-filled set and loses four exchange pairs, so manganese has the highest third ionisation enthalpy of these.

F e^{2 +}

is the tempting wrong answer: it has more protons, but its sixth electron is paired and leaves easily.

4.3.4 Oxidation States

Ask a chemist to name one feature of transition metals and you will often hear 'variable oxidation states'. Vanadium, for example, is found as +2, +3, +4 and +5, and its compounds turn from violet to green to blue to yellow as the state changes. The states usually differ by one unit, because the d electrons can be removed one at a time at gently rising cost. In the p-block the steps are usually two: tin shows +2 and +4, thallium +1 and +3, because the ns² pair either stays (the inert-pair effect) or goes as a pair.

Plot the states each metal shows (Fig. 4.6) and the result looks like a hill, with manganese on top: every state from +2 to +7. Climbing the left slope is a matter of supply. A metal can at most give or share all of its 4s and 3d electrons, so the ceiling rises by one per step. It is +3 in $S c C l_{3}$ , +4 in $T i C l_{4}$ , +5 in $V_{2} O_{5}$ , +6 in $C r O_{3}$ and +7 in $K M n O_{4}$ . Scandium has only three such electrons and always uses all three, so it hardly varies at all.

Past manganese the supply keeps growing but the ceiling drops. Iron would need all eight of its outer electrons for +8 and never manages it, though its heavier relatives ruthenium and osmium do ( $R u O_{4}$ , $O s O_{4}$ ). Iron's everyday states are +2 and +3. The growing nuclear charge holds the d electrons ever more tightly, and as they pair up there are fewer half-empty orbitals left for sharing in bonds. By copper and zinc the d set is nearly or completely full, and +2 (with +1 for copper) is about all that remains.

3/3The most common states in everyday compounds are marked with squares.

Fig. 4.6Oxidation states of the 3d metals: the maximum rises to +7 at manganese, then falls.NCERT Table 4.3 · p. 96

A chart with the ten 3d metals along the bottom and oxidation states +1 to +7 up the side. In the first step filled blue dots joined by a line mark the highest state of each metal: 3, 4, 5, 6 and 7 from scandium to manganese, then 6, 4, 4, 2 and 2. A dashed grey line marks the number of 4s plus 3d electrons, which keeps rising past manganese while the highest state falls. In the second step open circles add all the other known states, filling in every value from +2 up to the maximum for titanium to manganese. In the last step orange squares mark the most common states, such as +3 and +6 for chromium and +2, +4 and +7 for manganese.

Going down a group, the d-block does the opposite of the p-block. There, heavier elements cling to lower states (lead prefers +2 to +4); here, heavier elements are more at ease in high ones. Chromium(VI) oxide, $C r O_{3}$ , is so eager to be reduced that it can set ethanol alight, yet the oxides of molybdenum(VI) and tungsten(VI) are stable, unreactive solids.

At the other extreme, a metal can even be in the zero oxidation state. Nickel tetracarbonyl, $N i (C O)_{4}$ , a volatile liquid used to purify nickel, and chromium hexacarbonyl, $C r (C O)_{6}$ , are examples. CO is a special ligand: besides giving the metal an electron pair, it takes electron density back into its own empty orbitals (it is a π-acceptor), which stops the metal from building up too much negative charge. Unit 5 explains this bonding.

Example 4.3NCERT Example 4.3 · p. 97

Why titanium stops at +4

Titanium never shows an oxidation state above +4, but manganese reaches +7. Explain.

Try it yourself first. Then reveal the solution one step at a time.

4.3.5 Trends in the $M^{2 +} / M$ Standard Electrode Potentials

The standard electrode potential $E^{\circ} (M^{2 +} / M)$ tells us how readily a metal turns into its aqueous dipositive ion. A useful way to read it is as the end result of three energy steps (Fig. 4.7): pull the solid apart into atoms ( $Δ_{a} H^{\circ}$ , uphill), strip two electrons from each atom ( $Δ_{i} H_{1} + Δ_{i} H_{2}$ , steeply uphill), and let water molecules surround the ion (hydration enthalpy $Δ_{h y d} H^{\circ}$ , steeply downhill). The more uphill the total, the harder the change, and the less negative (or more positive) $E^{\circ}$ becomes.

4/4Net: zinc +723, copper +940. The more uphill net change gives copper its positive $E^{\circ}$ .

Fig. 4.7Turning the metal into aqueous ions: copper's net step is about

217 k J m o l^{- 1}

more uphill than zinc's (values in kJ mol⁻¹).NCERT Fig. 4.4 · p. 98

Two staircase energy diagrams side by side, zinc on the left and copper on the right, drawn to the same scale. In the first step a short arrow rises from the solid metal to the gaseous atom: +130 kilojoules per mole for zinc and +337 for copper. In the second step a tall arrow rises to the gaseous 2+ ion: +2639 for zinc and +2703 for copper. In the third step a blue arrow drops to the hydrated ion: minus 2046 for zinc and minus 2100 for copper. In the last step a dashed arrow joins the solid metal to the hydrated ion: the net change is +723 for zinc and +940 for copper, so copper's final level sits higher.

The cycle explains the overall drift: along the row the two ionisation steps get dearer, so $E^{\circ}$ becomes steadily less negative. It also explains copper's oddity. Copper is the one 3d metal whose $E^{\circ} (M^{2 +} / M)$ is positive ( $+ 0.34 V$ ): it pays a lot to atomise and even more to lose its second electron from the $3 d^{10}$ set of $C u^{+}$ , and hydration does not pay that back. Put copper in dilute hydrochloric or sulphuric acid and no hydrogen bubbles appear. Dissolving it needs an acid that is itself an oxidising agent, such as nitric acid or hot, concentrated sulphuric acid; the copper then reduces the acid (to nitrogen oxides or sulphur dioxide) instead of reducing $H^{+}$ .

Three values lie below the smooth trend, more negative than their neighbours suggest:

Manganese: $M n^{2 +}$ is a half-filled $d^{5}$ ion, and manganese is easy to atomise.
Zinc: $Z n^{2 +}$ is a full $d^{10}$ ion, and zinc has the smallest enthalpy of atomisation in the row.
Nickel: the hydration step is unusually generous, because the small $N i^{2 +}$ ion holds water molecules very strongly.

4.3.6 Trends in the $M^{3 +} / M^{2 +}$ Standard Electrode Potentials

The couple $M^{3 +} / M^{2 +}$ compares the +3 and +2 states directly. A low or negative $E^{\circ}$ means the $M^{2 +}$ ion gives up an electron easily (a reducing agent); a high positive value means the $M^{3 +}$ ion grabs one eagerly (an oxidising agent). Each value can be read from the d-electron count.

Scandium: $S c^{3 +}$ has the empty $d^{0}$ (argon) configuration, so $S c^{2 +}$ hardly exists; the value is very low.
Vanadium ( $- 0.26 V$ ) and chromium ( $- 0.41 V$ ): $V^{2 +}$ and $C r^{2 +}$ are reducing. $C r^{2 +}$ ( $d^{4}$ ) is especially keen to become $C r^{3 +}$ ( $d^{3}$ ), whose three d electrons exactly half-fill the lower-energy d set in a complex (the $t_{2 g}$ set of Unit 5).
Manganese ( $+ 1.54 V$ ): $M n^{3 +}$ ( $d^{4}$ ) becomes the half-filled $d^{5}$ $M n^{2 +}$ on gaining an electron, so it is a strong oxidising agent.
Iron ( $+ 0.77 V$ ): $F e^{2 +}$ ( $d^{6}$ ) reaches half-filled $d^{5}$ by losing an electron, so the value is comparatively low.
Cobalt ( $+ 1.92 V$ ): $C o^{3 +}$ is so strongly oxidising that, in water, it slowly oxidises the water itself.
Zinc: a $Z n^{3 +}$ ion would mean breaking the $d^{10}$ set; it is unknown.

Example 4.4NCERT Example 4.4 · p. 98

Two d⁶ ions, opposite behaviour

$F e^{2 +}$ and $C o^{3 +}$ both have six d electrons. In water, $F e^{2 +}$ is a mild reducing agent but $C o^{3 +}$ is a powerful oxidising agent ( $E^{\circ}$ : $F e^{3 +} / F e^{2 +} = + 0.77 V$ , $C o^{3 +} / C o^{2 +} = + 1.92 V$ ). Explain.

Try it yourself first. Then reveal the solution one step at a time.

4.3.7 Trends in Stability of Higher Oxidation States

High oxidation states need partners that can pull electron density strongly towards themselves. The two best are fluorine and oxygen, both small and highly electronegative. Comparing the highest fluoride and the highest oxide of each metal shows how far each partner can push the metal.

Table 4.4Highest fluoride and highest oxide of each 3d metal (oxidation state of the metal in brackets)NCERT Tables 4.5, 4.6 · p. 99

Metal	Highest fluoride	Highest oxide
Ti	$T i F_{4}$ (+4)	$T i O_{2}$ (+4)
V	$V F_{5}$ (+5)	$V_{2} O_{5}$ (+5)
Cr	$C r F_{5}$ (+5)	$C r O_{3}$ (+6)
Mn	$M n F_{4}$ (+4)	$M n_{2} O_{7}$ (+7)
Fe	$F e F_{3}$ (+3)	$F e_{2} O_{3}$ (+3)
Co	$C o F_{3}$ (+3)	$C o_{3} O_{4}$ (mixed +2, +3)
Ni	$N i F_{2}$ (+2)	$N i O$ (+2)
Cu	$C u F_{2}$ (+2)	$C u O$ (+2)

Why should these two partners win? A metal in a high oxidation state carries a large positive charge. A compound holding it is stable only if a lot of energy comes back: a large lattice energy if the compound is ionic, like $C o F_{3}$ , or strong bonds if it is molecular, like $V F_{5}$ . Small, highly electronegative fluorine supplies both. Even so, the fluoride column of the table shows its limits, and the other halogens do worse:

Where fluorine stops. It matches the group number only up to vanadium: $S c F_{3}$ , $T i F_{4}$ , $V F_{5}$ . Chromium's highest fluoride is $C r F_{5}$ , and manganese's is only $M n F_{4}$ . Manganese(VII) meets fluorine only with oxygen's help, in the oxofluoride $M n O_{3} F$ . From iron onwards no simple fluoride goes beyond +3, and $C o F_{3}$ is the only halide of cobalt(III).
Heavier halogens do worse. Chlorine, bromine and iodine cannot hold vanadium at +5 on their own; vanadium(V) pairs with them only in oxohalides, $V O X_{3}$ .
Iodide and high states clash. Iodide is easily oxidised, so a strongly oxidising metal ion destroys it. Copper(II) iodide does not exist: iodide added to $C u^{2 +}$ gives white copper(I) iodide and free iodine.
Fluoride and low states clash too. Copper(I) fluoride is unknown, although copper(I) chloride, bromide and iodide are familiar solids.

The copper(II)–iodide reaction, written out below, is put to use in analysis: the iodine it releases is titrated with thiosulphate to find how much copper an alloy or ore contains.

\begin{aligned} 2 C u^{2 +} + 4 I^{-} \\ \overset{}{\to} 2 C u I (s) \\ + I_{2} \end{aligned}

(4.3)

Solid copper(I) compounds such as $C u I$ and $C u_{2} O$ are stable, yet the $C u^{+}$ ion cannot survive in water. It disproportionates: of each pair of ions, one becomes $C u^{2 +}$ and the other copper metal (Example 4.9 puts a number on it). Water is the deciding factor. A doubly charged ion attracts water molecules far more strongly than a singly charged one, and the energy released by hydrating $C u^{2 +}$ outweighs the cost of removing copper's second electron.

Oxygen goes further still. Down the oxide column the highest state keeps pace with the group number all the way to $M n_{2} O_{7}$ (+7), three units above manganese's best fluoride. Count the partners needed. A fluorine atom forms one bond, so manganese(VII) would need seven fluorine atoms crowded round one small atom. A doubly bonded oxygen accounts for two of the metal's electrons, so four oxygen atoms are enough, and they fit comfortably at the corners of a tetrahedron.

That tetrahedral $M O_{4}$ unit is the signature of the high states. It appears in vanadate ( $V O_{4}^{3 -}$ , V +5), chromate ( $C r O_{4}^{2 -}$ , Cr +6), the manganese oxoanions $M n O_{4}^{3 -}$ , $M n O_{4}^{2 -}$ and $M n O_{4}^{-}$ (+5, +6 and +7), and $M n_{2} O_{7}$ itself, which is two such tetrahedra sharing one corner oxygen. After manganese even oxygen runs out of help. Iron's highest ordinary oxide is $F e_{2} O_{3}$ , and iron(VI) exists only as the ferrate ion, $F e O_{4}^{2 -}$ , which survives for a while in strong alkali before breaking down to $F e_{2} O_{3}$ and oxygen. In acid solution, small oxocations hold high states too: $T i O^{2 +}$ (+4), blue $V O^{2 +}$ (+4) and pale yellow $V O_{2}^{+}$ (+5).

Example 4.5NCERT Example 4.5 · p. 100

Which oxoion can oxidise chloride?

In acid solution, $E^{\circ}$ is $+ 1.00 V$ for $V O_{2}^{+} / V O^{2 +}$ , $+ 1.33 V$ for $C r_{2} O_{7}^{2 -} / C r^{3 +}$ and $+ 1.51 V$ for $M n O_{4}^{-} / M n^{2 +}$ , while $E^{\circ} (C l_{2} / C l^{-}) = + 1.36 V$ . (a) Which of the three oxoions can oxidise chloride ions under standard conditions? (b) Why does oxidising power rise from vanadium to manganese?

Try it yourself first. Then reveal the solution one step at a time.

4.3.8 Chemical Reactivity and $E^{\circ}$ Values

Drop a piece of iron and a piece of titanium into cold dilute hydrochloric acid. The iron fizzes, but the titanium looks untouched, although its $E^{\circ} (M^{2 +} / M)$ ( $- 1.63 V$ ) is far more negative than iron's ( $- 0.44 V$ ). Electrode potentials answer only one question: whether a reaction can go. How fast it goes is a separate matter, and titanium, like vanadium, is protected by a thin, tough oxide film that keeps cold, non-oxidising acids away from the metal. Judged by $E^{\circ}$ alone, every first-row metal except copper should dissolve in $1 M$ $H^{+}$ . Further down the block, platinum and gold resist every single acid and dissolve only in mixtures such as aqua regia.

The $M^{3 +} / M^{2 +}$ values of Section 4.3.6 sort the ions in the same way. At one extreme, $T i^{2 +}$ , $V^{2 +}$ and $C r^{2 +}$ are reducing agents strong enough to take electrons even from $H^{+}$ , so their acidified solutions slowly give off hydrogen. At the other, $M n^{3 +}$ and $C o^{3 +}$ are the strongest oxidising agents among the aqua ions of the row. In between, $F e^{2 +}$ and $F e^{3 +}$ live together comfortably in water. For chromium(II):

\begin{aligned} 2 C r^{2 +} (a q) \\ + 2 H^{+} (a q) \\ \overset{}{\to} 2 C r^{3 +} (a q) \\ + H_{2} (g) \end{aligned}

(4.4)

Example 4.6NCERT Example 4.6 · p. 101

Putting a number on copper’s nobility

Use these rounded data (kJ mol⁻¹) to estimate the difference $E^{\circ} (C u^{2 +} / C u) - E^{\circ} (Z n^{2 +} / Z n)$ : $Δ_{a} H^{\circ}$ : Zn 130, Cu 337; $Δ_{i} H_{1} + Δ_{i} H_{2}$ : Zn 2639, Cu 2703; $Δ_{h y d} H^{\circ} (M^{2 +})$ : Zn −2046, Cu −2100. Treat $Δ G \approx Δ H$ (ignore entropy).

Try it yourself first. Then reveal the solution one step at a time.

Example 4.7NCERT Example 4.7 · p. 101

More protons, yet easier to ionise

Iron has one more proton than manganese, yet its third ionisation enthalpy is lower (Fe $2957$ , Mn $3248 k J m o l^{- 1}$ ), and $E^{\circ} (F e^{3 +} / F e^{2 +}) = + 0.77 V$ is far below $E^{\circ} (M n^{3 +} / M n^{2 +}) = + 1.54 V$ . Explain both facts.

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 2

Which of these ions will release hydrogen from a dilute acid?

F e^{3 +}

M n^{2 +}

C u^{2 +}

C r^{2 +}

Show answer

Answer (D) $C r^{2 +}$

C r^{2 +}

is a strong reducing agent (

E^{\circ} (C r^{3 +} / C r^{2 +}) = - 0.41 V

), so it can reduce

H^{+}

H_{2}

, as in (4.4). The others are already stable or are oxidising.

4.3.9 Magnetic Properties

Hang a small tube of a salt between the poles of a strong electromagnet and switch on (Fig. 4.8). Most transition metal salts are drawn into the field: they are paramagnetic (अनुचुंबकीय). A few, such as zinc salts, are nudged weakly out of it: they are diamagnetic. Iron, cobalt and nickel metals respond far more violently, because the tiny magnets of neighbouring atoms line up together; they are attracted strongly and can stay magnetised. This ferromagnetism is paramagnetism amplified by the cooperation of whole regions of atoms.

Definition

Paramagnetic

Attracted into a magnetic field, because the substance contains unpaired electrons whose magnetic moments line up with the field.

Example $C u S O_{4} \cdot 5 H_{2} O$ ( $C u^{2 +}$ is $d^{9}$ ).

Definition

Diamagnetic

Weakly repelled by a magnetic field; all its electrons are paired.

Example $Z n S O_{4} \cdot 7 H_{2} O$ ( $Z n^{2 +}$ is $d^{10}$ ).

Where does the pull come from? An unpaired electron behaves like a tiny bar magnet, partly because of its spin and partly because of its motion around the nucleus, whereas two paired electrons cancel. In compounds of the first-row metals, the ligands largely lock the orbital motion in place (chemists say it is quenched), so the spins carry almost all the magnetism. The spin-only magnetic moment then depends only on $n$ , the number of unpaired electrons:

Spin-only formula

μ = \sqrt{n (n + 2)} B M

(4.5)

where

$n$: the number of unpaired electrons
$B M$: Bohr magneton, the natural unit of an electron's magnetic moment, $9.27 \times 10^{- 24} J T^{- 1}$

Table 4.5Spin-only magnetic moments (high-spin aqua ions as examples)NCERT Table 4.7 · p. 102

$n$	$μ$ / BM	Examples
0	0	$S c^{3 +}$ , $C u^{+}$ , $Z n^{2 +}$
1	1.73	$T i^{3 +}$ , $C u^{2 +}$
2	2.83	$V^{3 +}$ , $N i^{2 +}$
3	3.87	$V^{2 +}$ , $C r^{3 +}$ , $C o^{2 +}$
4	4.90	$C r^{2 +}$ , $M n^{3 +}$ , $F e^{2 +}$
5	5.92	$M n^{2 +}$ , $F e^{3 +}$

Most first-row ions give measured moments close to these values, which makes a magnetic measurement a quick way to count unpaired electrons; the smallest non-zero moment, 1.73 BM, means exactly one. Where a measured value is higher, some orbital motion has survived: aqueous $F e^{2 +}$ shows about $5.4 B M$ and $C o^{2 +}$ between about $4.4$ and $5.2 B M$ , both above their spin-only values. The balance in Fig. 4.8 is how such moments are measured.

2/3 $M n^{2 +}$ ( $d^{5}$ , five unpaired electrons) is pulled into the field: the reading rises.

Fig. 4.8A paramagnetic sample is pulled into the field and seems to weigh more; a diamagnetic one is pushed out and seems lighter.

A digital balance at the top holds a long sample tube by a wire. The bottom of the tube hangs between the north and south poles of an electromagnet. In the first step the magnet is off and the display reads no change. In the second step dashed field lines join the poles; the tube holding a manganese(II) salt is pulled down into the field and the display reads heavier. In the third step a zinc(II) salt is used instead; the tube rises slightly out of the field and the display reads lighter.

The explorer below lets you build any 3d atom or ion, watch its electrons fill the orbital boxes by Hund's rule, and read off the unpaired electrons and the spin-only moment.

Interactived-electron explorer

An orbital box diagram with five 3d boxes and one 4s box. Ten buttons choose an element from scandium to zinc and five buttons choose a charge from 0 to +4; charges that would break into the argon core are disabled. The boxes fill with up and down arrows following Hund's rule, and the configuration is written out, including the chromium and copper exceptions for neutral atoms. Readouts give the number of d electrons, unpaired electrons, pairs of parallel spins, the spin-only magnetic moment in Bohr magnetons, whether the ion is paramagnetic or diamagnetic, and whether it can show d–d colour, with the observed colour of the common aqua ion where there is one. A note explains special cases such as half-filled and filled sets.

Fig. 4.9Choose an element from Sc to Zn and a charge. The 4s electrons leave first; then read the unpaired electrons, the spin-only moment and whether the ion should be coloured.

Example 4.8NCERT Example 4.8 · p. 102

Identify the vanadium ion

A vanadium compound has a measured magnetic moment of $2.8 B M$ . Find the number of unpaired electrons and the likely oxidation state of vanadium.

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 3

What is the spin-only magnetic moment of

N i^{2 +}

(

Z = 28

) in water, in BM?

Show answer

Answer 2.83 BM

N i^{2 +}

[A r] 3 d^{8}

: three filled orbitals and two half-filled ones, so

n = 2

and

μ = \sqrt{2 \times 4} = 2.83 B M

4.3.10 Formation of Coloured Ions

Solutions of transition metal salts look like a paint box: blue copper sulphate, green nickel chloride, pink cobalt nitrate. In an ion surrounded by water molecules (or other ligands), the five d orbitals no longer have equal energies; they split into a lower and an upper set (Unit 5 explains why). An electron in the lower set can jump to the upper set if it absorbs light of exactly the right energy, a d–d transition. For most ions that energy falls in the visible range.

We see the light that is not absorbed, so the colour of the solution is the complement of the absorbed colour (Fig. 4.10). The size of the energy gap depends on the metal, its charge and its ligands, so the same metal can change colour when its ligands change: aqueous $C u^{2 +}$ is pale blue, but add ammonia and it turns deep blue-violet as $[C u (N H_{3})_{4}]^{2 +}$ forms. An ion with no d electrons ( $S c^{3 +}$ , $T i^{4 +}$ ) or a full set ( $Z n^{2 +}$ , $C u^{+}$ ) has no possible d–d jump, so its compounds are white and its solutions colourless.

3/4The light that passes through lacks yellow-green, so the solution looks purple.

Fig. 4.10An ion absorbs one band of visible light to lift a d electron; the solution shows the remaining, complementary colour.NCERT p. 103

On the left, an energy diagram shows the d orbitals of an ion in water split into a lower set of three and an upper set of two, separated by an energy gap labelled delta. In the middle, a band of white light, drawn as six coloured stripes, enters a beaker of Ti3+ solution. In the first step the electron sits in the lower set. In the second step it jumps to the upper set, and the light leaving the beaker has lost its yellow and green stripes, labelled yellow-green absorbed. In the third step an eye on the right sees the remaining light, labelled looks purple. In the last step the beaker holds Sc3+, which has no d electron: nothing is absorbed, all six stripes pass and the eye sees it as colourless.

Table 4.6Colours of some aqua ions of the 3d series, by number of d electronsNCERT Table 4.8 · p. 103

$d^{n}$	Ion in water	Colour
$d^{0}$	$S c^{3 +}$ , $T i^{4 +}$	colourless
$d^{1}$	$T i^{3 +}$	purple
$d^{1}$	$V O^{2 +}$ (V +4)	blue
$d^{2}$	$V^{3 +}$	green
$d^{3}$	$V^{2 +}$ , $C r^{3 +}$	violet
$d^{4}$	$C r^{2 +}$	blue
$d^{5}$	$M n^{2 +}$	very pale pink
$d^{5}$	$F e^{3 +}$	yellow (from hydrolysis)
$d^{6}$	$F e^{2 +}$	pale green
$d^{7}$	$C o^{2 +}$	pink
$d^{8}$	$N i^{2 +}$	green
$d^{9}$	$C u^{2 +}$	blue
$d^{10}$	$Z n^{2 +}$	colourless

4.3.11 Formation of Complex Compounds

The red colour of blood comes from a complex. In haemoglobin, each iron(II) ion is held by four nitrogen atoms of a flat ring and a fifth from the protein, leaving a sixth position where an oxygen molecule can attach. A complex is any such unit: a central metal ion with several molecules or anions, the ligands, bonded to it, behaving as a new species with its own colour, charge and reactions. Transition metals form them in enormous numbers, for example $[C o (N H_{3})_{6}]^{3 +}$ , $[N i (C N)_{4}]^{2 -}$ , $[F e (H_{2} O)_{6}]^{2 +}$ and $[A g (N H_{3})_{2}]^{+}$ , the silver complex in Tollens' reagent; vitamin B $_{12}$ is built round a cobalt complex.

Why are transition metal ions such good hosts? Each ligand donates a pair of electrons, so the best hosts pull hard on electron pairs and have somewhere to put them. Transition metal ions are small and often highly charged, which gives them a strong pull, and they have vacant d orbitals of suitable energy to accept the pairs. Unit 5 is devoted to these compounds.

4.3.12 Catalytic Properties

Transition metal catalysts are all around you. In a car's catalytic converter, a thin film of platinum, palladium and rhodium turns carbon monoxide, unburnt fuel and nitrogen oxides into carbon dioxide, water and nitrogen. A gauze of platinum–rhodium alloy oxidises ammonia in nitric acid plants, and molybdenum and iron atoms sit at the heart of the enzyme that lets soil bacteria fix nitrogen. (Section 4.7 lists more industrial examples.) A catalyst offers a reaction an easier route, and transition metals can provide one in two ways.

A meeting place. Molecules of the reactants stick to metal atoms on the catalyst surface, forming temporary bonds that use the metal's 3d and 4s electrons. Held side by side, the reactants meet far more often, and the bonds inside them are stretched and weakened, so less activation energy is needed.
An electron shuttle. An ion that switches easily between two oxidation states can take electrons from one reactant and hand them on to another, opening a lower-energy path.

A neat example of the second kind is the oxidation of iodide by peroxodisulphate. Both reactants are negative ions that repel each other, so the direct reaction is slow. A trace of $F e^{3 +}$ speeds it up by shuttling electrons (Fig. 4.11):

overall

\begin{aligned} S_{2} O_{8}^{2 -} + 2 I^{-} \\ \overset{}{\to} 2 S O_{4}^{2 -} \\ + I_{2} \end{aligned}

the two catalysed steps

\begin{aligned} 2 F e^{3 +} + 2 I^{-} \\ \overset{}{\to} 2 F e^{2 +} + I_{2} \\ 2 F e^{2 +} + S_{2} O_{8}^{2 -} \\ \overset{}{\to} 2 F e^{3 +} \\ + 2 S O_{4}^{2 -} \end{aligned}

4/4The iron is regenerated, so a small amount keeps working.

Fig. 4.11Iron switches between +3 and +2, carrying electrons from iodide to peroxodisulphate.NCERT p. 104

In the first step two negative ions, iodide and peroxodisulphate, face each other with a double-headed arrow labelled repel: slow. In later steps a circular cycle appears with 2Fe3+ at the top and 2Fe2+ at the bottom. In the second step the left half of the circle is highlighted: iodide enters from the left and leaves as iodine as Fe3+ becomes Fe2+. In the third step the right half is highlighted: peroxodisulphate enters and leaves as two sulphate ions as Fe2+ returns to Fe3+. In the last step both halves are highlighted with the words catalyst returns in the centre.

4.3.13 Formation of Interstitial Compounds

Steel is iron with a little carbon, and those carbon atoms do not take the places of iron atoms: they are tucked into the small gaps between the packed iron atoms. Hydrogen, boron, carbon and nitrogen atoms are small enough to do this. They form interstitial compounds (अंतराकाशी यौगिक) with many transition metals: tungsten carbide ( $W C$ ), titanium nitride ( $T i N$ ), cementite ( $F e_{3} C$ ) and palladium hydride ( $P d H_{x}$ , whose $x$ can reach about 0.7).

Definition

Interstitial compound

A compound in which small atoms such as H, B, C or N sit in the holes of a metal's crystal lattice; its composition is often variable and it keeps much of the metal's character.

ExampleTungsten carbide, $W C$ , used for drill tips.

How many guests there are depends on how many holes happen to be filled. So the formulas are often fractional (as in $P d H_{0.7}$ ) and say nothing about the metal's oxidation state, and the bonding is a mixture, partly metallic and partly covalent. The guest atoms act like wedges that stop layers of metal atoms sliding over one another, which explains what these materials are good for:

they are very hard; some carbides and borides come close to diamond, which is why tungsten carbide tips cutting tools and gold-coloured $T i N$ coats drill bits;
they melt at higher temperatures than the parent metal;
they resist chemical attack;
yet they still conduct electricity like metals.

2/3Small atoms such as C or N slip into the holes, making the solid harder and higher-melting.

Fig. 4.12Small atoms fill holes (interstitial compounds); similar-sized atoms swap places (substitutional alloys).NCERT p. 104

A square array of twenty large circles represents metal atoms packed in layers. In the first step it is labelled pure metal. In the second step small dark circles appear in some of the gaps between four large atoms, labelled interstitial: small H, C or N sit in the holes, harder and inert. In the third step the small atoms are gone and four of the large circles are drawn with a dashed orange outline, labelled alloy: a similar-sized metal takes some places.

4.3.14 Alloy Formation

India's ₹1 and ₹2 coins, a stainless-steel sink and a brass diya are all alloys (मिश्रातु): metallic materials made by melting a metal together with other elements. In the simplest kind, a substitutional alloy, atoms of the added metal take over some of the host's places in the crystal at random. That works only when the two kinds of atom are close in size, roughly within 15 % in metallic radius, and transition metals, with their similar radii and bonding, mix very readily. Chromium ( $129 p m$ ) differs from iron ( $126 p m$ ) by about 2 %, and zinc ( $137 p m$ ) from copper ( $128 p m$ ) by about 7 %.

Definition

Alloy

A metallic material made by blending a metal with one or more other elements; in a substitutional alloy, atoms of similar size replace each other in the lattice.

ExampleStainless steel: iron with about 18 % chromium and 8 % nickel.

Alloying usually makes a metal harder, and often raises its melting point. Steels show how much a few per cent can do. Chromium and nickel give stainless steel, tungsten and vanadium let tool steels stay hard even when red-hot, and manganese makes the tough steel of railway crossings. Transition metals alloy with main-group metals too. Copper with zinc gives brass for lamps and fittings, and copper with tin gives bronze, the kansa bell-metal of traditional plates and bowls. A zinc-rich alloy with a little copper is the base of Bidri ware from Karnataka, inlaid with silver.

Try this

Does an alloy rust?

You’ll need

an iron nail and a piece of sandpaper
a stainless-steel spoon or a ₹1 coin
two small cups, water and salt

Predict first

Which sample will show rust first?

The stainless-steel spoon, because it contains more metalsThe iron nailBoth at the same timeNeither: salt water does not cause rust

Lightly sand an ordinary iron nail so its surface is bare metal. Take a clean stainless-steel spoon (or a ₹1 coin) as the alloy sample.
Half-fill two small cups with water and stir a teaspoon of salt into each.
Put the nail in one cup and the spoon or coin in the other, so each is half under water.
Look at both every day for four days and note any rust, the colour of the water, and where the change starts.

Table 4.7Rust watch

Day	Iron nail	Alloy
Day 1
Day 2
Day 4

Your entries are saved automatically.

4.4 Some Important Compounds of Transition Elements

Two purple and orange salts, potassium permanganate and potassium dichromate, are among the most useful oxidising agents in chemistry. Before meeting them we need one more idea and a look at the oxides they come from.

Definition

Disproportionation

A redox reaction in which one species is simultaneously oxidised and reduced, giving one product in a higher and one in a lower oxidation state.

Example $2 C u^{+}$ → $C u^{2 +}$ + $C u$

Example 4.9NCERT Example 4.9 · p. 105

Does Cu⁺ survive in water?

Given $E^{\circ} (C u^{+} / C u) = + 0.52 V$ and $E^{\circ} (C u^{2 +} / C u^{+}) = + 0.15 V$ , show that $C u^{+}$ disproportionates in water and estimate the equilibrium constant at $298 K$ .

Try it yourself first. Then reveal the solution one step at a time.

4.4.1 Oxides and Oxoanions of Metals

Most transition metals tarnish or burn when heated in oxygen, and many form several oxides; their highest oxides keep pace with the group number up to $M n_{2} O_{7}$ , as Section 4.3.7 showed. What changes along such a series is the oxide's character. Compare two oxides of manganese. $M n O$ is a green ionic solid that dissolves in acids to give $M n^{2 +}$ salts, the behaviour of a basic oxide; every first-row metal except scandium has an ionic monoxide of this kind. $M n_{2} O_{7}$ is a dark, oily covalent liquid that reacts with water to give permanganic acid, $H M n O_{4}$ , the behaviour of a non-metal oxide.

The difference lies in the charge on the metal. A metal ion with a small charge sits happily in an ionic lattice, and its oxide ions are free to pick up $H^{+}$ : the oxide is basic. A metal centre at +6 or +7 pulls the oxygen's electrons so strongly towards itself that the bonds become covalent and the metal behaves like a non-metal: the oxide is acidic. In between, oxides are amphoteric. Chromium shows the whole range: $C r O$ is basic, $C r_{2} O_{3}$ amphoteric and $C r O_{3}$ acidic, giving chromic and dichromic acids ( $H_{2} C r O_{4}$ , $H_{2} C r_{2} O_{7}$ ) with water. Vanadium does the same in three steps. $V_{2} O_{3}$ is basic, and $V_{2} O_{4}$ is weakly basic, dissolving in acids as blue $V O^{2 +}$ . $V_{2} O_{5}$ is amphoteric though mainly acidic: it gives vanadate ( $V O_{4}^{3 -}$ ) with alkalis but $V O_{2}^{+}$ with acids. Covalent bonding also shows in the low melting points of high oxides such as $C r O_{3}$ and $V_{2} O_{5}$ .

Compare

Oxide character changes with oxidation state

Low state (+2)

$M n O$ , $C r O$ , $V O$
mostly ionic
basic: dissolve in acids

Middle (+3, +4)

$C r_{2} O_{3}$ , $V_{2} O_{4}$ , $M n O_{2}$
partly covalent
amphoteric or weakly basic

High (+5 to +7)

$V_{2} O_{5}$ , $C r O_{3}$ , $M n_{2} O_{7}$
covalent, low melting
acidic: give oxoacids

Potassium dichromate

Potassium dichromate, $K_{2} C r_{2} O_{7}$ , is the chromium(VI) salt of choice in the laboratory. It is the starting material for the chromium salts used to tan leather (a big industry in Kanpur and around Chennai) and an oxidant in making dyes and other organic compounds. It is also a primary standard in volumetric analysis, because its crystals are pure, stable and do not absorb water from the air. The sodium salt dissolves far better in water and is preferred as an oxidant in organic chemistry.

Step by step

From chromite ore to potassium dichromate

Step 1 of 3

Roast the ore

Powdered chromite, $F e C r_{2} O_{4}$ , is heated with sodium carbonate in a good supply of air. Oxygen raises chromium from +3 to +6, giving soluble yellow sodium chromate, while iron ends up as insoluble $F e_{2} O_{3}$ . (Equation 4.6)

1 / 3

roasting

\begin{aligned} 4 F e C r_{2} O_{4} \\ + 8 N a_{2} C O_{3} \\ + 7 O_{2} \\ \overset{}{\to} 8 N a_{2} C r O_{4} \\ + 2 F e_{2} O_{3} \\ + 8 C O_{2} \end{aligned}

(4.6)

acidifying

\begin{aligned} 2 N a_{2} C r O_{4} \\ + 2 H^{+} \\ \overset{}{\to} N a_{2} C r_{2} O_{7} \\ + 2 N a^{+} + H_{2} O \end{aligned}

(4.7)

cation swap

\begin{aligned} N a_{2} C r_{2} O_{7} \\ + 2 K C l \\ \overset{}{\to} K_{2} C r_{2} O_{7} \\ + 2 N a C l \end{aligned}

(4.8)

Add acid to a yellow chromate solution and it turns orange; add alkali and the yellow comes back (Fig. 4.13). No electrons change hands: chromium is +6 in both ions, and the switch is an acid–base reaction in which two chromate ions join through a shared oxygen.

yellow ⇌ orange

\begin{aligned} 2 C r O_{4}^{2 -} + 2 H^{+} \\ \overset{}{⇌} C r_{2} O_{7}^{2 -} \\ + H_{2} O \end{aligned}

(4.9)

3/3 $C r_{2} O_{7}^{2 -}$ : two tetrahedra sharing a corner, with a Cr–O–Cr angle of about 126°.

Fig. 4.13Acid joins two chromate tetrahedra through one oxygen; alkali splits them again.NCERT p. 106

In the first step two separate chromate ions are drawn as a chromium atom bonded to four oxygen atoms at the corners of a tetrahedron, with dashed tetrahedron edges; each is labelled CrO4 2 minus, yellow, in alkali. In the second step the two units slide together until they meet at one oxygen; the extra oxygen of the right-hand unit floats up and fades, labelled H2O leaves, and the new ion is labelled Cr2O7 2 minus, orange on adding acid. In the last step an arc at the shared oxygen is marked 126 degrees and the label reads shared corner O: two tetrahedra, one shared oxygen.

Both sodium and potassium dichromate are strong oxidising agents in acid solution, where the orange dichromate is reduced to green $C r^{3 +}$ :

$E^\circ = +1.33\unit{V}$

\begin{aligned} C r_{2} O_{7}^{2 -} + 14 H^{+} \\ + 6 e^{-} \\ \overset{}{\to} 2 C r^{3 +} \\ + 7 H_{2} O \end{aligned}

(4.10)

To write the full equation for any reaction, take the reductant's half-equation, multiply it until it releases six electrons, add it to (4.10) and cancel what appears on both sides. Iron(II) supplies one electron per ion, so six are needed; iodide, hydrogen sulphide and tin(II) each supply two per formula unit, so three are needed. For example, with iron(II):

\begin{aligned} C r_{2} O_{7}^{2 -} + 14 H^{+} \\ + 6 F e^{2 +} \\ \overset{}{\to} 2 C r^{3 +} \\ + 6 F e^{3 +} \\ + 7 H_{2} O \end{aligned}

(4.11)

The redox combiner below does this bookkeeping for many reductants. Pick an oxidant and a reductant, and it scales and adds the half-equations, checks that atoms and charges balance, and tells you from the $E^{\circ}$ values whether the reaction should go.

InteractiveRedox combiner

Two groups of buttons. The first chooses the oxidising agent: acidified dichromate or acidified permanganate. The second chooses a reducing agent: iron(II), iodide, hydrogen sulphide, tin(II), oxalate, sulphite, nitrite, hydrogen peroxide, ethanol or chloride. The panel shows the two half-equations with their electrons, the multipliers needed to match the electrons, and the balanced overall ionic equation after cancelling hydrogen ions and water. A check line confirms that every element and the total charge balance. A last line gives the standard cell potential from the two electrode potentials and says whether the reaction is feasible under standard conditions; with chloride, dichromate is not feasible but permanganate is.

Fig. 4.14Choose dichromate or permanganate in acid and a reducing agent; the half-equations are matched for electrons and added.

Example

The breath test

Older breath-alcohol tubes contain acidified potassium dichromate on silica gel. Ethanol in the breath turns the crystals from orange to green while it is oxidised to ethanoic acid. Write the balanced ionic equation and find how many moles of dichromate react with one mole of ethanol.

Try it yourself first. Then reveal the solution one step at a time.

Potassium permanganate

Potassium permanganate, $K M n O_{4}$ , is made from pyrolusite, the mineral manganese dioxide. The route passes through green potassium manganate, in which manganese is +6.

Step by step

From pyrolusite to potassium permanganate

Step 1 of 3

Alkaline fusion

$M n O_{2}$ is fused with potassium hydroxide and an oxidising agent, air or $K N O_{3}$ . Manganese rises from +4 to +6 and the melt turns dark green with $K_{2} M n O_{4}$ . (Equation 4.12)

1 / 3

\begin{aligned} 2 M n O_{2} + 4 K O H \\ + O_{2} \\ \overset{}{\to} 2 K_{2} M n O_{4} \\ + 2 H_{2} O \end{aligned}

(4.12)

\begin{aligned} 3 M n O_{4}^{2 -} + 4 H^{+} \\ \overset{}{\to} 2 M n O_{4}^{-} \\ + M n O_{2} \\ + 2 H_{2} O \end{aligned}

(4.13)

\begin{aligned} 2 M n^{2 +} \\ + 5 S_{2} O_{8}^{2 -} \\ + 8 H_{2} O \\ \overset{}{\to} 2 M n O_{4}^{-} \\ + 10 S O_{4}^{2 -} \\ + 16 H^{+} \end{aligned}

(4.14)

Potassium permanganate forms purple crystals so dark that they look almost black. It is only moderately soluble, about 6 to 7 g in 100 g of water at room temperature. Its colour, though, is so intense that one crystal tints a whole bucket of water. Heated above about $513 K$ , the solid gives off oxygen and leaves manganate and manganese dioxide:

\begin{aligned} 2 K M n O_{4} \\ \overset{}{\to} K_{2} M n O_{4} \\ + M n O_{2} + O_{2} \end{aligned}

(4.15)

In manganate and permanganate alike, four oxygen atoms sit at the corners of a tetrahedron around manganese. The Mn–O bonds include π bonding from overlap of oxygen p orbitals with manganese d orbitals. (Solid $K M n O_{4}$ even has the same crystal structure as potassium perchlorate, whose $C l O_{4}^{-}$ ion is also tetrahedral.) The d-electron count settles the magnetism. Green manganate (Mn +6, $d^{1}$ ) has one unpaired electron and is paramagnetic. Permanganate (Mn +7, $d^{0}$ ) has none and is diamagnetic, apart from a weak paramagnetism that changes with temperature. That leaves a puzzle. A $d^{0}$ ion cannot show d–d colour, yet permanganate is intensely purple and dichromate orange. Their colour comes from a different process, charge transfer, in which light moves an electron from oxygen to the metal. Explaining it, and the weak paramagnetism, needs molecular orbital theory, which you will meet in later courses.

What permanganate turns into depends on the acidity of the solution (Fig. 4.15). The three half-equations, with their electrode potentials, show how strongly $H^{+}$ concentration steers the outcome:

acid

\begin{aligned} M n O_{4}^{-} + 8 H^{+} \\ + 5 e^{-} \\ \overset{}{\to} M n^{2 +} \\ + 4 H_{2} O \\ E^{\circ} = + 1.51 V \end{aligned}

(4.16)

towards MnO₂

\begin{aligned} M n O_{4}^{-} + 4 H^{+} \\ + 3 e^{-} \\ \overset{}{\to} M n O_{2} \\ + 2 H_{2} O \\ E^{\circ} = + 1.69 V \end{aligned}

(4.17)

strong alkali

\begin{aligned} M n O_{4}^{-} + e^{-} \\ \overset{}{\to} M n O_{4}^{2 -} \\ E^{\circ} = + 0.56 V \end{aligned}

(4.18)

4/4In strongly alkaline solution it gains only one, giving green manganate.

Fig. 4.15One oxidant, three products: permanganate takes 5, 3 or 1 electrons depending on the medium.NCERT p. 107

A permanganate ion, drawn as manganese with four oxygens and labelled +7, purple, sits on the left. In the second step an arrow labelled acid and plus 5 electrons leads to Mn2+, labelled +2, pale pink. In the third step a second arrow labelled neutral and plus 3 electrons leads to MnO2, labelled +4, brown solid. In the last step a third arrow labelled strong alkali and plus 1 electron leads to MnO4 2 minus, labelled +6, green.

Electrode potentials tell you what can happen, not how quickly. With $E^{\circ} = + 1.51 V$ , well above the $+ 1.23 V$ of the $O_{2} / H_{2} O$ couple, acidified permanganate ought to oxidise the water around it and release oxygen. A laboratory bottle of it nevertheless lasts for weeks: the reaction becomes noticeable only in hot solution or once some $M n^{2 +}$ is present to catalyse it. Oxalate titrations use both levers. The first drops of permanganate lose their colour sluggishly, so the flask is warmed to about $333 K$ ; once a little $M n^{2 +}$ has formed, each drop is decolourised at once.

Permanganate oxidises a long list of reducing agents, and the medium decides how far the manganese itself is reduced. Table 4.8, after Equation 4.21, collects the reactions you are most likely to meet. To write any of the equations, take the permanganate half-equation for that medium and the reductant's half-equation as many times as needed to cancel the electrons, then add. Two examples in acid:

iron(II)

\begin{aligned} M n O_{4}^{-} + 8 H^{+} \\ + 5 F e^{2 +} \\ \overset{}{\to} M n^{2 +} \\ + 5 F e^{3 +} \\ + 4 H_{2} O \end{aligned}

(4.19)

oxalate, warm

\begin{aligned} 2 M n O_{4}^{-} + 16 H^{+} \\ + 5 C_{2} O_{4}^{2 -} \\ \overset{}{\to} 2 M n^{2 +} \\ + 10 C O_{2} \\ + 8 H_{2} O \end{aligned}

(4.20)

In neutral or faintly alkaline solution, where the manganese stops at $M n O_{2}$ , some reductants are oxidised further than in acid. Iodide is the classic case: instead of stopping at iodine, it goes all the way to iodate.

neutral

\begin{aligned} 2 M n O_{4}^{-} + I^{-} \\ + H_{2} O \\ \overset{}{\to} 2 M n O_{2} \\ + I O_{3}^{-} \\ + 2 O H^{-} \end{aligned}

(4.21)

Table 4.8What permanganate does to common reducing agents (Mn ends as

M n^{2 +}

in acid,

M n O_{2}

in neutral solution)NCERT p. 107

Reducing agent	Medium	Oxidised to
$F e^{2 +}$ (pale green)	acid	$F e^{3 +}$ (yellow)
$I^{-}$	acid	$I_{2}$
$C_{2} O_{4}^{2 -}$	acid, warm	$C O_{2}$
$H_{2} S$	acid	sulphur (a pale yellow solid)
$S O_{3}^{2 -}$	acid	$S O_{4}^{2 -}$
$N O_{2}^{-}$	acid	$N O_{3}^{-}$
$I^{-}$	neutral	$I O_{3}^{-}$
$S_{2} O_{3}^{2 -}$	neutral	$S O_{4}^{2 -}$
$M n^{2 +}$	neutral	$M n O_{2}$

Two practical points follow for titrations. Permanganate is acidified with dilute sulphuric acid, never hydrochloric acid, because it would also oxidise chloride to chlorine (Example 4.5) and the titre would come out too large. And no indicator is needed: the first drop in excess leaves a faint, lasting pink. Outside the laboratory the same oxidising power has many uses. Dilute solutions, sold in India as lal dawa, disinfect wells and wash fruit and vegetables. Industry uses permanganate to bleach textile fibres and to take the colour out of oils, and organic chemists rely on it to oxidise alkenes, alcohols and the side chains of benzene rings.

Example

How much iron is in the tablet?

An iron supplement tablet is dissolved in dilute sulphuric acid and made up to $100.0 m L$ . A $25.0 m L$ portion needs $10.8 m L$ of $0.00500 M$ $K M n O_{4}$ to reach a faint permanent pink. Find the mass of iron(II) in the whole tablet ( $M_{Fe} = 55.85 g m o l^{- 1}$ ).

Try it yourself first. Then reveal the solution one step at a time.

Try this

Make permanganate lose its colour

You’ll need

a few crystals of potassium permanganate (sold by chemists as a disinfectant)
two clear glasses of water
a vitamin C tablet, or the juice of half a lemon
a spoon and a pair of gloves

Predict first

What will happen to the purple colour as you stir in crushed vitamin C?

It will fade, because vitamin C reduces the permanganate.It will turn deep green.It will become darker purple.Nothing will change.

Dissolve two or three crystals of potassium permanganate in a glass of water. If the colour is very dark, dilute it until it is a clear, light purple.
Pour half into a second glass and keep it aside for comparison.
Crush a vitamin C tablet. Add a pinch to the first glass, stir, and wait a minute. Add more, a pinch at a time.
Record the colour after each addition, comparing with the second glass.

Table 4.9What I saw

Glass	Colour	Cloudy?
Untouched
Little vitamin C
More vitamin C

Your entries are saved automatically.

The purple fades, sometimes through a brownish stage, until the liquid is almost colourless, while the untouched glass stays purple. Ascorbic acid gives up electrons and manganese(VII) is reduced: to nearly colourless $M n^{2 +}$ where the solution is acidic enough, or to brown $M n O_{2}$ , which makes the liquid cloudy, where it is not. This is the chemistry of a permanganate titration: the colour itself shows when all the reducing agent has been used up.

Quick check

1 of 4

Sodium hydroxide is added drop by drop to an orange solution of potassium dichromate. What do you see?

It turns green as

C r^{3 +}

forms.It turns yellow as chromate forms; chromium stays +6.It turns purple.It stays orange: pH has no effect.

Show answer

Answer (B) It turns yellow as chromate forms; chromium stays +6.

Raising the pH reverses (4.9): dichromate splits into yellow chromate. It is an acid–base change, not a reduction, so green

C r^{3 +}

does not appear.

4.5 The Lanthanoids

Below the main table sit two rows of fourteen elements each. The first follows lanthanum (cerium to lutetium) and fills the 4f subshell; the second follows actinium (thorium to lawrencium) and fills 5f. Lanthanum and actinium are so similar to the elements after them that they are always discussed together with them, and chemists use the symbol Ln for lanthanum or any lanthanoid.

The two rows behave very differently. Almost every lanthanoid compound contains the $L n^{3 +}$ ion, so moving along the row changes little except the size of that ion, a little at each step. If you wanted to find out how size alone affects chemistry, you could hardly design a better set of elements, and they are more alike than the members of any d-series. The actinoids refuse to be so tidy: from one element to the next they use different numbers of electrons in bonding, and every one of them is radioactive, which makes them hard to study. We begin with the lanthanoids.

Table 4.10Lanthanum and the lanthanoids: configurations outside the xenon core, and

r

, the radius of the

L n^{3 +}

ionNCERT Table 4.9 · p. 110

Ln (Z)	Atom	$L n^{3 +}$	$r$ / pm
La (57)	$5 d^{1} 6 s^{2}$	$4 f^{0}$	103
Ce (58)	$4 f^{1} 5 d^{1} 6 s^{2}$	$4 f^{1}$	101
Pr (59)	$4 f^{3} 6 s^{2}$	$4 f^{2}$	99
Nd (60)	$4 f^{4} 6 s^{2}$	$4 f^{3}$	98
Pm (61)	$4 f^{5} 6 s^{2}$	$4 f^{4}$	97
Sm (62)	$4 f^{6} 6 s^{2}$	$4 f^{5}$	96
Eu (63)	$4 f^{7} 6 s^{2}$	$4 f^{6}$	95
Gd (64)	$4 f^{7} 5 d^{1} 6 s^{2}$	$4 f^{7}$	94
Tb (65)	$4 f^{9} 6 s^{2}$	$4 f^{8}$	92
Dy (66)	$4 f^{10} 6 s^{2}$	$4 f^{9}$	91
Ho (67)	$4 f^{11} 6 s^{2}$	$4 f^{10}$	90
Er (68)	$4 f^{12} 6 s^{2}$	$4 f^{11}$	89
Tm (69)	$4 f^{13} 6 s^{2}$	$4 f^{12}$	88
Yb (70)	$4 f^{14} 6 s^{2}$	$4 f^{13}$	87
Lu (71)	$4 f^{14} 5 d^{1} 6 s^{2}$	$4 f^{14}$	86

The radii in the last column are for six-coordinate ions, rounded to the nearest picometre; tables based on other coordination numbers give slightly different values, but the trend is the same.

4.5.1 Electronic Configurations

Every lanthanoid atom has $6 s^{2}$ on the outside. Beneath it the 4f subshell fills from cerium to lutetium, though not perfectly regularly. Lanthanum, cerium, gadolinium and lutetium put one electron into 5d instead, keeping their 4f sets empty, nearly empty, half-filled ( $4 f^{7}$ ) or full ( $4 f^{14}$ ). The tripositive ions are far neater. Every lanthanoid forms $L n^{3 +}$ , and in that state the series runs from $4 f^{0}$ (La) to $4 f^{14}$ (Lu), one more 4f electron per element.

4.5.2 Atomic and Ionic Sizes

From lanthanum to lutetium the atoms and ions steadily shrink: the $L n^{3 +}$ radius falls from about $103$ to $86 p m$ (Fig. 4.16). The ionic radii fall smoothly; the metallic radii fall less regularly, because europium and ytterbium use only two electrons for metallic bonding and so have unusually large atoms.

Definition

Lanthanoid contraction

The gradual shrinking of the atoms and ions across the 4f series, La to Lu, because 4f electrons screen the rising nuclear charge so poorly.

Example $L a^{3 +}$ about $103 p m$ , $L u^{3 +}$ about $86 p m$ .

3/3Overall, from $L a^{3 +}$ to $L u^{3 +}$ , the radius shrinks by about 17 pm.

Fig. 4.16The radius of the

L n^{3 +}

ions falls by about 17 pm across the series (vertical axis starts at 80 pm).NCERT Fig. 4.6 · p. 109

A bar chart with the fifteen elements from lanthanum to lutetium along the bottom and ionic radius in picometres up the side, the axis starting at 80 picometres. In the first step the bars for lanthanum to gadolinium appear, each a little shorter than the one before. In the second step the bars for terbium to lutetium are added, continuing the steady fall. In the last step a dashed arrow runs from the top of the lanthanum bar, labelled 103, to the top of the lutetium bar, labelled 86, with the note about 17 picometres smaller.

Why do the ions shrink? Each step along the series adds one proton to the nucleus and one electron to the 4f subshell. A 4f electron is spread through a diffuse, many-lobed region and does a poor job of screening the other electrons from the extra proton, poorer even than a d electron does in a d-series. The effective nuclear charge therefore creeps up at every step and the whole electron cloud is drawn in.

The consequences reach well beyond the f-block:

The 5d metals are almost the same size as the 4d metals above them (zirconium $160 p m$ , hafnium $159 p m$ ), so zirconium and hafnium occur together in nature and are very hard to separate.
The lanthanoids themselves differ so little in size that separating them took decades of patient work; ion exchange and solvent extraction now make use of the small differences.
The basic character of the hydroxides $L n (O H)_{3}$ decreases from lanthanum to lutetium, because the smaller ion holds its hydroxide ions more covalently.

4.5.3 Oxidation States

Every lanthanoid forms $L n^{3 +}$ . A handful also form +2 or +4 ions, nearly always reaching an empty, half-filled or full 4f set by doing so. Because +3 is the home state, these unusual ions are keen to return to it, which makes them useful oxidising or reducing agents.

Table 4.11Lanthanoid ions outside the +3 stateNCERT p. 109

Ion	f count	Behaviour
$C e^{4 +}$	$4 f^{0}$	strong oxidant ( $E^{\circ} \approx + 1.74 V$ ); it attacks water so slowly that cerium(IV) sulphate serves as a titrant
$T b^{4 +}$	$4 f^{7}$	oxidant, as in $T b O_{2}$ and $T b F_{4}$
Pr, Nd, Dy (+4)	–	only in a few solids, such as the oxide $P r O_{2}$
$E u^{2 +}$	$4 f^{7}$	mild reductant ( $E^{\circ} (E u^{3 +} / E u^{2 +}) \approx - 0.35 V$ )
$Y b^{2 +}$	$4 f^{14}$	stronger reductant ( $E^{\circ} \approx - 1.05 V$ )
$S m^{2 +}$	$4 f^{6}$	very strong reductant ( $E^{\circ} \approx - 1.55 V$ )

Samarium is a useful warning. Its +2 ion is $4 f^{6}$ , not a special configuration, yet $S m^{2 +}$ exists (samarium(II) iodide, $S m I_{2}$ , is a reducing agent used in organic synthesis): special stability is a guide, not a law.

4.5.4 General Characteristics

Freshly cut, a lanthanoid metal is soft and silvery-white, but the shine soon fades as the surface reacts with air. They conduct heat and electricity well and adopt common metallic structures. Their properties change gradually along the row: hardness more than doubles from lanthanum to lutetium, and melting points climb from about $1190 K$ (La) to about $1930 K$ (Lu). The exceptions are europium and ytterbium, sometimes joined by samarium and thulium. Their densities and melting points are out of line with their neighbours', because they give only two electrons each to the metallic bond.

Glass-blowers protect their eyes with goggles made of neodymium-containing glass, which absorbs a narrow band of the yellow glare from hot glass. That narrowness is typical of lanthanoid colours. Many $L n^{3 +}$ ions are coloured, in crystals and in solution, because an electron can jump between 4f orbitals; but the 4f orbitals are buried so deep that ligands barely disturb them, so each ion absorbs in sharp lines rather than broad bands. With no 4f electron ( $L a^{3 +}$ ) or a full set ( $L u^{3 +}$ ) no such jump is possible, and these ions are colourless. The same count decides magnetism: every lanthanoid ion is paramagnetic except the $f^{0}$ ions ( $L a^{3 +}$ , $C e^{4 +}$ ) and the $f^{14}$ ions ( $Y b^{2 +}$ , $L u^{3 +}$ ).

Why is +3 so universal? The first two electrons cost about as much as calcium's (roughly $600$ and $1200 k J m o l^{- 1}$ ). The third costs more, but a small, triply charged ion releases so much energy in a crystal lattice or in water that the cost is repaid. The third ionisation enthalpy is lowest where losing the electron leaves a special f set: it dips at lanthanum ( $4 f^{0}$ left), gadolinium ( $4 f^{7}$ ) and lutetium ( $4 f^{14}$ ). Exchange energy, met in Section 4.3.3, is at work again, now in the f subshell.

All the lanthanoids are reactive metals. Their $E^{\circ} (L n^{3 +} / L n)$ values lie in a narrow band, about $- 2.2$ to $- 2.4 V$ (europium, at about $- 2.0 V$ , is the odd one out), close to magnesium's $- 2.37 V$ . In practice the early members react about as vigorously as calcium and the later ones more like aluminium. The table summarises their reactions. The oxides $L n_{2} O_{3}$ and hydroxides $L n (O H)_{3}$ are basic, like those of the alkaline earth metals.

Table 4.12Reactions of the lanthanoidsNCERT Fig. 4.7 · p. 111

Treated with	Conditions	Product
oxygen	burns	$L n_{2} O_{3}$
water	warm	$L n (O H)_{3}$ + $H_{2}$
dilute acids	room temperature	$L n^{3 +}$ salts + $H_{2}$
halogens	burns	$L n X_{3}$
hydrogen	gentle heating	hydrides
carbon	about 2773 K	$L n_{3} C$ , $L n_{2} C_{3}$ , $L n C_{2}$
nitrogen	heating	$L n N$
sulphur	heating	$L n_{2} S_{3}$

In India, Indian Rare Earths Limited extracts these elements from the monazite of coastal sands, and they turn up in many places. Neodymium–iron–boron magnets drive electric motors, wind turbines and earphones. Phosphors made from single lanthanoid oxides glow in screens and lamps, and mixed lanthanoid oxides help to crack petroleum into petrol. Lanthanoids are also added to alloy steels. Mischmetall is an alloy of about 95 % lanthanoid metals (mainly cerium and lanthanum) with about 5 % iron. It is the 'flint' in gas lighters and goes into magnesium alloys for shells and bullets.

4.6 The Actinoids

The actinoids run from thorium ( $Z = 90$ ) to lawrencium ( $Z = 103$ ). Every one is radioactive. Thorium and uranium have isotopes so long-lived that they survive from the formation of the Earth; the heaviest actinoids exist for hours or less and have been made only a few atoms, or at most nanograms, at a time. Everything beyond uranium is made artificially, in reactors or accelerators, which is why much of their chemistry is still being worked out. Thorium matters especially to India: the country has large reserves in monazite, and the third stage of its nuclear programme is designed to breed uranium-233 from thorium-232.

4.6.1 Electronic Configurations

Every actinoid atom has $7 s^{2}$ on the outside, but the 5f and 6d levels below it are so close in energy that electrons do not settle into 5f as tidily as they fill 4f in the lanthanoids. Thorium ( $6 d^{2} 7 s^{2}$ ) has no 5f electron at all, although cerium, the lanthanoid above it, already has one in 4f. From protactinium ( $5 f^{2} 6 d^{1} 7 s^{2}$ ) onwards 5f fills, and it is complete at nobelium ( $5 f^{14} 7 s^{2}$ ). Special f counts again shape the details: curium is $[R n] 5 f^{7} 6 d^{1} 7 s^{2}$ , keeping a half-filled 5f set by putting its next electron in 6d, just as gadolinium keeps $4 f^{7}$ with $5 d^{1}$ . Lawrencium is usually written $[R n] 5 f^{14} 6 d^{1} 7 s^{2}$ , though experiments suggest its last electron may actually sit in 7p.

The 5f orbitals have the same shapes as the 4f orbitals but are less deeply buried. 4f electrons sit inside the filled 5s and 5p shells, tucked away from neighbouring atoms. 5f electrons extend further out and can take part in bonding. Most of the differences between the two families follow from this one fact.

4.6.2 Ionic Sizes

The actinoids show an actinoid contraction: their atoms and $M^{3 +}$ ions shrink across the series, just as the lanthanoids do. The shrinkage from one element to the next is larger than in the lanthanoids, because 5f electrons shield the nuclear charge even more poorly than 4f electrons.

Definition

Actinoid contraction

The steady shrinking of actinoid atoms and $M^{3 +}$ ions as the atomic number rises; the step from one element to the next is larger than in the lanthanoid contraction because 5f electrons shield poorly.

4.6.3 Oxidation States

Plot the actinoid oxidation states (Fig. 4.17) and a hill appears again, as it did for the 3d metals. The highest state climbs from +4 at thorium through +5 at protactinium and +6 at uranium to +7 at neptunium and plutonium, then falls. From curium onwards +3 is the rule, and nobelium even prefers +2, which leaves its $5 f^{14}$ set full. Nearly all of them can also form +3 ions. The early actinoids can vary so much because their 5f, 6d and 7s levels lie close together in energy, so different numbers of electrons can be used in bonding. In water, ions in the +3 and +4 states tend to hydrolyse, and uranium(VI) and plutonium(VI) appear as the linear oxocations $U O_{2}^{2 +}$ and $P u O_{2}^{2 +}$ .

3/3From curium on, +3 dominates; Md and No also form +2 ions.

Fig. 4.17Early actinoids show many oxidation states; later ones settle at +3 (and +2 for Md and No).NCERT Table 4.11 · p. 112

A chart with the fifteen elements from actinium to lawrencium along the bottom and oxidation states +2 to +7 up the side; a shaded band marks +3. In the first step every element has a purple dot at +3, labelled all show +3 (thorium only rarely). In the second step blue dots add the higher states of the early actinoids: thorium +4, protactinium +4 and +5, uranium +4 to +6, neptunium and plutonium +4 to +7, americium +4 to +6, with the label many states, early actinoids. In the last step curium and berkelium gain +4 dots, and green rings at +2 are added for mendelevium and nobelium.

Uranium alone shows how varied this chemistry is: in acid solution you can meet it as reddish $U^{3 +}$ , green $U^{4 +}$ and yellow $U O_{2}^{2 +}$ , each with its own reactions. So the actinoids, unlike the lanthanoids, cannot be summed up by one typical state; especially in the first half of the series, each element has to be learnt on its own terms.

4.6.4 General Characteristics and Comparison with Lanthanoids

Actinoid metals are silvery, like the lanthanoids, but less uniform in other ways. Their metallic radii jump about from element to element far more than lanthanoid radii do, so the metals crystallise in an unusual variety of structures. Plutonium alone has six solid forms between room temperature and its melting point. Chemically they are even more reactive than the lanthanoids, above all as fine powders, which can catch fire in air. Here is how they treat a few reagents:

acids: hydrochloric acid dissolves them all, but nitric acid barely touches most of them, because it coats them with a protective oxide film;
water: boiling water turns them into a mixture of oxide and hydride;
non-metals: most combine with them at quite moderate temperatures;
alkalis: no reaction.

Two more contrasts follow from the less-buried 5f orbitals. First, the early actinoids lose electrons more easily than the early lanthanoids. Their ionisation enthalpies are known only roughly, but they are clearly lower. A 5f electron added early in the series stays further from the nucleus than a 4f electron would, so more of the inner electrons lie between it and the nuclear charge. It is held less firmly and is readily used in bonding. Second, their magnetism is harder to predict. The magnetic susceptibility still rises and falls roughly with the number of unpaired 5f electrons, as it does in the lanthanoids, but the values are generally lower.

Compare

Lanthanoids and actinoids side by side

Lanthanoids (4f)

outer $6 s^{2}$ ; 4f fills from Ce to Lu
steady, smooth contraction
almost only +3; a few +2 or +4
mostly non-radioactive (Pm excepted)
f orbitals buried: little f bonding
resemble each other closely

Actinoids (5f)

outer $7 s^{2}$ ; 5f and 6d compete early on
larger contraction per element
many states (up to +7) in the first half
all radioactive
5f orbitals less buried: take part in bonding
lanthanoid-like only in the second half, though neighbours still resemble each other

Why does the lanthanoid contraction get so much attention and the actinoid contraction so little? Both shrink the elements that follow them. But the elements after the lanthanoids are the familiar 5d metals, hafnium to mercury, whose sizes and chemistry the contraction reshapes. The elements after the actinoids have been made only a few atoms at a time, so there is little chemistry for the actinoid contraction to explain.

Example 4.10NCERT Example 4.10 · p. 113

Why terbium can reach +4

Terbium ( $Z = 65$ ) is $[X e] 4 f^{9} 6 s^{2}$ . Besides +3 it forms terbium(IV) compounds such as $T b O_{2}$ and $T b F_{4}$ . Explain, and predict whether terbium(IV) will act as an oxidising or a reducing agent.

Try it yourself first. Then reveal the solution one step at a time.

4.7 Some Applications of d- and f-Block Elements

Look around a kitchen and you will find the d- and f-blocks at work: a steel knife, a stainless sink, white paint on the wall, the battery in a torch and the magnet in a phone speaker. Of all these materials steel is made in by far the largest amounts, and it remains the backbone of construction. A steel plant reduces iron ore in a blast furnace and burns away surplus carbon and other impurities. It then adds carbon, together with chromium, manganese, nickel or other metals, in measured amounts to get the grade it wants.

Many compounds have one special job. Titanium dioxide, $T i O_{2}$ , is the bright white pigment in paint, paper and toothpaste. Manganese dioxide is the oxidant inside dry cells, and zinc, nickel and cadmium go into other batteries; cobalt, nickel and manganese now fill the cathodes of lithium-ion cells. Silver bromide, $A g B r$ , darkens in light, which made photographic film possible. Copper, silver and gold are still called the 'coinage metals'. Everyday coins, though, are now made of cheaper alloys such as cupronickel or stainless steel, and gold and silver coins are bought mainly by collectors. Above all, transition metals and their compounds catalyse some of the largest processes in the chemical industry:

Table 4.13Some industrial processes that use transition metal catalystsNCERT p. 114

Process	Catalyst	Change
Contact process	$V_{2} O_{5}$	$S O_{2}$ to $S O_{3}$ (for sulphuric acid)
Haber process	finely divided iron	nitrogen and hydrogen to ammonia
Hydrogenating oils	nickel	liquid oils to solid fats
Ziegler–Natta	$T i C l_{4}$ and $A l (C_{2} H_{5})_{3}$	ethene to polythene
Wacker process	$P d C l_{2}$	ethene to ethanal
Reppe synthesis	nickel complexes	ethyne to benzene and other rings

The f-block adds strong neodymium magnets, cerium-based powders for polishing glass, the phosphors of energy-saving lamps and LED screens, and the nuclear fuels thorium and uranium.

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Match each catalyst with its industrial process.

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Answer

$V_{2} O_{5}$ : Contact process
iron : Haber process
nickel : hydrogenation of oils
$P d C l_{2}$ : Wacker process

Vanadium(V) oxide makes

S O_{3}

, iron makes ammonia, nickel adds hydrogen to oils, and palladium chloride oxidises ethene to ethanal.

Recap

The d- and f-Block Elements: key points and summary

The d-block (groups 3–12) fills $(n - 1) d$ orbitals in four series; the f-block fills 4f (lanthanoids) and 5f (actinoids) (Section 4.1).
A transition element has an incomplete d subshell in its atom or a common ion; Zn, Cd and Hg do not (Section 4.2).
Outer configurations follow $(n - 1) d^{1 – 10} n s^{1 – 2}$ , with exceptions such as Cr $3 d^{5} 4 s^{1}$ and Cu $3 d^{10} 4 s^{1}$ ; ions lose ns electrons first (Section 4.2, Section 4.3.3).
Metallic bonding uses $(n - 1) d$ electrons, giving high melting points and enthalpies of atomisation that peak mid-row (Section 4.3.1).
Radii shrink slowly along a row; the lanthanoid contraction makes 4d and 5d members almost the same size (Section 4.3.2).
Ionisation enthalpies rise gently; breaks at $d^{5}$ and $d^{10}$ come from exchange energy (Section 4.3.3).
Oxidation states differ by one; the maximum equals 4s + 3d electrons up to Mn (+7), then falls (Section 4.3.4).
Electrode potentials reflect atomisation, ionisation and hydration; copper alone has a positive $E^{\circ} (M^{2 +} / M)$ (Section 4.3.5, Section 4.3.6).
Oxygen and fluorine stabilise the highest states; oxygen does better because it forms multiple bonds (Section 4.3.7).
Unpaired d electrons cause paramagnetism ( $μ = \sqrt{n (n + 2)}$ BM) and d–d transitions cause colour (Section 4.3.9, Section 4.3.10).
Transition metals form complexes, catalyse reactions, and make interstitial compounds and alloys (Section 4.3.11, Section 4.3.12, Section 4.3.14).
$K_{2} C r_{2} O_{7}$ (from chromite) and $K M n O_{4}$ (from pyrolusite) are strong oxidants whose products depend on pH (Potassium dichromate, Potassium permanganate).
Lanthanoids show mainly +3 and a steady contraction; actinoids are radioactive, show more states and contract more per element (Section 4.5, Section 4.6).
Steels, pigments, batteries, catalysts and nuclear fuels are key uses of these elements (Section 4.7).

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Quick check: MCQs on The d- and f-Block Elements

4.1 Position in the Periodic Table3 questions
4.2 Electronic Configurations of the d-Block Elements3 questions
4.3.1 Physical Properties3 questions
4.3.3 Ionisation Enthalpies3 questions
4.3.4 Oxidation States3 questions
4.3.6 Trends in the $\ce{M^3+}/\ce{M^2+}$ Standard Electrode Potentials3 questions
4.3.7 Trends in Stability of Higher Oxidation States3 questions
4.3.8 Chemical Reactivity and $E^\circ$ Values2 questions
4.3.9 Magnetic Properties3 questions
4.3.10 Formation of Coloured Ions3 questions
4.3.14 Alloy Formation4 questions
Potassium permanganate4 questions
4.5.4 General Characteristics4 questions
4.6.4 General Characteristics and Comparison with Lanthanoids3 questions
4.7 Some Applications of d- and f-Block Elements2 questions

Glossary · 15

Key terms (English–Hindi)

actinoidऐक्टिनॉयडaiktinoid: One of the fourteen 5f elements, thorium to lawrencium; all are radioactive.First seen in the chapter →
actinoid contractionऐक्टिनॉयड आकुंचनaiktinoid aakunchan: The decrease in size across the actinoids, larger per element than the lanthanoid contraction.First seen in the chapter →
alloyमिश्रातुmishraatu: A metallic blend of a metal with other elements, e.g. brass (Cu–Zn).First seen in the chapter →

d–d transition: The jump of an electron between the split d-orbital sets of a metal ion on absorbing visible light; the cause of most transition metal colours.First seen in the chapter →
diamagneticप्रतिचुंबकीयpratichumbakiya: Weakly repelled by a magnetic field; all electrons paired.First seen in the chapter →
disproportionationअसमानुपातनasamanupatan: A reaction in which one species is both oxidised and reduced, e.g. $2 C u^{+} \overset{}{\to} C u^{2 +} + C u$ .First seen in the chapter →

enthalpy of atomisationपरमाणुकरण एन्थैल्पीparmanukaran enthalpy: The enthalpy change when one mole of an element is converted into separate gaseous atoms.First seen in the chapter →
exchange energyविनिमय ऊर्जाvinimay urja: Stabilisation gained by electrons with parallel spins in degenerate orbitals, roughly proportional to the number of such pairs.First seen in the chapter →

inner transition elementआंतर संक्रमण तत्वaantar sankraman tatva: An f-block element, in which the 4f or 5f subshell is being filled.First seen in the chapter →
interstitial compoundअंतराकाशी यौगिकantarakashi yaugik: A compound in which small atoms (H, B, C, N) occupy holes in a metal lattice, e.g. $W C$ .First seen in the chapter →

lanthanoidलैन्थेनॉयडlanthenoid: One of the fourteen 4f elements, cerium to lutetium (lanthanum is studied with them).First seen in the chapter →
lanthanoid contractionलैन्थेनॉयड आकुंचनlanthenoid aakunchan: The steady decrease in size from La to Lu, caused by poor shielding by 4f electrons.First seen in the chapter →

paramagneticअनुचुंबकीयanuchumbakiya: Attracted into a magnetic field because of unpaired electrons.First seen in the chapter →

spin-only magnetic momentकेवल प्रचक्रण चुंबकीय आघूर्णkeval prachakran chumbakiya aaghurn: $μ = \sqrt{n (n + 2)}$ BM, the moment due to the spins of $n$ unpaired electrons alone.First seen in the chapter →

transition elementसंक्रमण तत्वsankraman tatva: An element with an incomplete d subshell in its atom or in a common ion; the d-block minus group 12.First seen in the chapter →

The d- and f-Block Elements: Class 12 Chemistry · Chapter 4

What you will learn in The d- and f-Block Elements

4.1 Position in the Periodic Table

4.2 Electronic Configurations of the d-Block Elements

4.3 General Properties of the Transition Elements (d-Block)

4.3.1 Physical Properties

4.3.2 Variation in Atomic and Ionic Sizes of Transition Metals

4.3.3 Ionisation Enthalpies

4.3.4 Oxidation States

4.3.5 Trends in the $M^{2 +} / M$ Standard Electrode Potentials

4.3.6 Trends in the $M^{3 +} / M^{2 +}$ Standard Electrode Potentials

4.3.7 Trends in Stability of Higher Oxidation States

4.3.8 Chemical Reactivity and $E^{\circ}$ Values

4.3.9 Magnetic Properties

4.3.10 Formation of Coloured Ions

4.3.11 Formation of Complex Compounds

4.3.12 Catalytic Properties

4.3.13 Formation of Interstitial Compounds

4.3.14 Alloy Formation

4.4 Some Important Compounds of Transition Elements

4.4.1 Oxides and Oxoanions of Metals

Potassium dichromate

Potassium permanganate

4.5 The Lanthanoids

4.5.1 Electronic Configurations

4.5.2 Atomic and Ionic Sizes

4.5.3 Oxidation States

4.5.4 General Characteristics

4.6 The Actinoids

4.6.1 Electronic Configurations

4.6.2 Ionic Sizes

4.6.3 Oxidation States

4.6.4 General Characteristics and Comparison with Lanthanoids

4.7 Some Applications of d- and f-Block Elements

The d- and f-Block Elements: key points and summary

Practice questions with answers (NCERT “Exercises”)

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Key terms (English–Hindi)

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4.1 Position in the Periodic Table

4.2 Electronic Configurations of the d-Block Elements

4.3 General Properties of the Transition Elements (d-Block)

4.3.1 Physical Properties

4.3.2 Variation in Atomic and Ionic Sizes of Transition Metals

4.3.3 Ionisation Enthalpies

4.3.4 Oxidation States

4.3.5 Trends in the MX2+/M Standard Electrode Potentials

4.3.6 Trends in the MX3+/MX2+ Standard Electrode Potentials

4.3.7 Trends in Stability of Higher Oxidation States

4.3.8 Chemical Reactivity and E∘ Values

4.3.9 Magnetic Properties

4.3.10 Formation of Coloured Ions

4.3.11 Formation of Complex Compounds

4.3.12 Catalytic Properties

4.3.13 Formation of Interstitial Compounds

4.3.14 Alloy Formation

4.4 Some Important Compounds of Transition Elements

4.4.1 Oxides and Oxoanions of Metals

Potassium dichromate

Potassium permanganate

4.5 The Lanthanoids

4.5.1 Electronic Configurations

4.5.2 Atomic and Ionic Sizes

4.5.3 Oxidation States

4.5.4 General Characteristics

4.6 The Actinoids

4.6.1 Electronic Configurations

4.6.2 Ionic Sizes

4.6.3 Oxidation States

4.6.4 General Characteristics and Comparison with Lanthanoids

4.7 Some Applications of d- and f-Block Elements

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4.3.5 Trends in the $M^{2 +} / M$ Standard Electrode Potentials

4.3.6 Trends in the $M^{3 +} / M^{2 +}$ Standard Electrode Potentials

4.3.8 Chemical Reactivity and $E^{\circ}$ Values