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Chapter 4: The d- and f-Block Elements

On a small island near Stockholm lies a quarry called Ytterby. In the late eighteenth century, workers digging there for feldspar came across an unusually heavy black mineral. A sample reached the Finnish chemist Johan Gadolin, who analysed it and reported that it contained a new “earth” — an oxide of an element nobody had yet seen. That single rock turned out to be one of the richest chemical puzzles in history.

Over the next hundred years, chemist after chemist tried to break Gadolin’s earth apart. Every time they thought they had a pure substance, careful work showed it was still a mixture. The elements hidden inside behaved so alike — same charge, almost the same size, nearly the same reactions — that ordinary chemical separation simply failed. Crystallisation had to be repeated thousands of times to tease them apart. Four of the elements finally isolated were named after that one village: yttrium, ytterbium, terbium and erbium.

Why should a whole family of elements be so stubbornly identical? And why, on the other side of the periodic table, does a single metal such as manganese behave so differently in each of its many compounds — green in one, deep purple in another, a mild ion here and a fierce oxidising agent there? Both questions lead to the same place: the quiet filling of inner d and f orbitals.

Fig. 4.0

Fig. 4.0 – Aqueous solutions of transition metals, gadolinite ore, and the d- and f-blocks.

NoteCuriosity Corner

Q1. Why are the lanthanoids so hard to separate from one another, when neighbouring s-block metals such as sodium and potassium can be told apart quite easily? Q2. Zirconium and hafnium sit in different periods of the periodic table, yet they are almost the same size and are always found together in nature. How can an element from the row above “shrink” the one below it? Q3. Copper(II) sulphate solution is blue, but zinc sulphate solution is colourless — both metals belong to the same block. Why should a completely filled set of d orbitals cost a compound its colour? Q4. Iron rusts away in moist air, while gold survives for centuries. Both are transition metals, so what decides how “noble” a metal will be? Q5. Manganese shows every oxidation state from +2 to +7, but scandium sticks to +3. Why should a metal in the middle of the series have so many more choices?

Keep these questions in mind as you read. This chapter will give you everything you need to reason your way to the answers.

4.1 Position in the Periodic Table

Two Blocks Built on Inner Orbitals

Look at a long-form periodic table and you will see a wide middle region that seems to hold the table together — the elements of groups 3 to 12. These make up the d-block. In the four long periods of the table, it is the d orbitals of these atoms that are being progressively filled with electrons.

Below the main body of the table sits a separate panel of two rows. These are the elements of the f-block, in which the 4f and 5f orbitals are progressively filled. They are printed separately only to keep the table a manageable width; chemically, they belong inside it.

The elements of the d-block are usually called transition metals, and those of the f-block are called inner transition metals. Both names will be used constantly through this Chapter, so it is worth fixing them now.

The Four Series of Transition Metals

There are mainly four series of the transition metals. The 3d series runs from scandium (Sc) to zinc (Zn). The 4d series runs from yttrium (Y) to cadmium (Cd). The 5d series contains lanthanum (La) together with the elements from hafnium (Hf) to mercury (Hg). The 6d series contains actinium (Ac) along with the elements from rutherfordium (Rf) to copernicium (Cn).

The inner transition metals form two series of their own. The 4f series runs from cerium (Ce) to lutetium (Lu) and its members are called lanthanoids. The 5f series runs from thorium (Th) to lawrencium (Lr), and these are the actinoids.

d-block (transition) vs f-block (inner transition) elements

Feature d-block f-block
Groups in the table 3–12 placed in a separate panel below the table
Orbitals being filled (n–1)d 4f and 5f
Common name transition metals inner transition metals
Series 3d, 4d, 5d, 6d 4f (lanthanoids), 5f (actinoids)
Number of series four two
What Counts as a Transition Element

The name “transition metal” was coined for a historical reason: the chemical properties of these metals were seen as transitional — a bridge between the very reactive, electropositive s-block metals on the left and the p-block elements on the right. The modern definition, however, is based squarely on electronic structure.

The word “transition” here does not mean the elements are changing into something. It simply records the nineteenth-century view that this middle block marks the passage — the transition — from the s-block to the p-block, its members showing properties intermediate between the two.

NoteDefinition

A transition metal is a metal that has an incompletely filled d subshell either in its neutral atom or in one of its ions.

Notice the two-part condition carefully, because exam questions are built on it. It is enough for the ion to have a partly filled d subshell; the neutral atom need not.

This definition has an immediate consequence. Zinc, cadmium and mercury of group 12 have a full \(d^{10}\) configuration both in the ground state of the atom and in their common oxidation states. Neither the atom nor the usual ion has a partly filled d subshell, so strictly they are not regarded as transition metals. All the same, they are the end members of the 3d, 4d and 5d series respectively, and their chemistry is studied along with that of the transition metals.

Q. If zinc is not really a transition metal, why does it keep appearing in the tables of this chapter?

A. Because it is the last element of the 3d series. Comparing zinc with the true transition metals beside it — its colourless ions, its low enthalpy of atomisation, its single oxidation state — is exactly what shows us how much the partly filled d orbitals were doing in the other elements.

Why the d and f Elements Deserve a Separate Study

It is the presence of partly filled d or f orbitals that makes transition elements behave so differently from the non-transition elements. That single structural fact is behind their variable oxidation states, their coloured ions, their magnetic behaviour, their catalytic power and their complexes — all of which are taken up later in this Chapter. For this reason, transition elements and their compounds are studied separately. This does not mean a new theory is needed: the usual theory of valence that works for the non-transition elements can be applied successfully to the transition elements too.

Much of the material world runs on this block. The precious metals silver, gold and platinum are transition metals, and so are industrially vital metals such as iron, copper and titanium. Iron gives us steel, copper carries electricity through our homes, titanium goes into aircraft and surgical implants, and platinum sits inside the catalytic converter of a car.

The Plan of This Chapter

The route ahead is worth knowing in advance. We shall first take up the electronic configuration, occurrence and general characteristics of the transition elements, with special emphasis on the trends in the properties of the first row (3d) transition metals. We shall then look at the preparation and properties of two important compounds of this block. After that we turn to certain general aspects of the inner transition metals — their electronic configurations, oxidation states and chemical reactivity.

Where the d-Block Sits

The d-block occupies the large middle section of the periodic table, flanked between the s-block and the p-block, as shown in Fig. 4.1. The d orbitals filled here belong to the penultimate energy level of the atom — that is, the shell just inside the outermost one. In an element of the 4th period, for instance, the outermost shell is the 4th, and the d orbitals receiving electrons are the 3d orbitals. Filling these penultimate d orbitals gives rise to the four rows of the transition metals: 3d, 4d, 5d and 6d.

Fig. 4.1

Fig. 4.1 – Position of the d- and f-blocks in the periodic table.

All these series of transition elements are set out in Table 4.1 in the next section, where their electronic configurations are given.

(question only — no solution) Hafnium (Z = 72) and gadolinium (Z = 64) both lie in period 6 of the periodic table. State the block to which each belongs, and name the series (3d, 4d, 5d, 6d, lanthanoid or actinoid) in which each is placed.

4.2 Electronic Configurations of the d-Block Elements

The General Pattern

In general, the electronic configuration of the outer orbitals of these elements is

\[(n-1)d^{1-10}\,ns^{1-2}\]

except for palladium, whose configuration is \(4d^{10}5s^{0}\).

Read the formula slowly. The \((n-1)\) stands for the inner d orbitals, which may hold anywhere from one to ten electrons. The \(ns\) part is the outermost s orbital, which may hold one or two electrons. So in an element of the 4th period such as iron, \(n = 4\): the 4s orbital is the outermost one, and the d orbitals being filled are the 3d orbitals, one shell inside it.

You may wonder why the outer 4s orbital gets filled before the inner 3d orbital at all. The reason, which you met while learning the Aufbau principle, is that the 4s orbital happens to lie slightly lower in energy than the 3d orbital when they are empty, so electrons enter 4s first. The gap between them, however, is very small — and that small gap is the source of everything that follows.

Why Chromium and Copper Break the Rule

The general formula has several exceptions, precisely because the energy difference between the \((n-1)d\) and \(ns\) orbitals is so little. On top of that, half-filled and completely filled sets of orbitals are relatively more stable than partly filled ones.

NotePrinciple / Law

Sets of degenerate orbitals that are exactly half-filled or completely filled have extra stability, so an electron will shift from the ns orbital into the (n–1)d orbitals when doing so produces a half-filled or fully filled d set — provided the two orbital sets are close in energy, as they are in the transition series.

A consequence of this factor shows up in the configurations of chromium and copper in the 3d series. Consider chromium, which has a \(3d^5 4s^1\) configuration instead of \(3d^4 4s^2\); the energy gap between the two sets (3d and 4s) of orbitals is small enough to prevent the electron entering the 3d orbitals — that is, the electron is not held back in 4s but moves so that the 3d set becomes exactly half-filled. Similarly in the case of copper, the configuration is \(3d^{10}4s^1\) and not \(3d^9 4s^2\), giving a completely filled 3d set.

Q. If half-filled and filled sets are so stable, why does iron not also rearrange to something like \(3d^7 4s^1\)?

A. Because \(3d^7\) is neither half-filled nor fully filled — the shift would buy no extra stability, and moving an electron out of the lower-lying 4s orbital costs energy. The rearrangement happens only at \(d^5\) (Cr) and \(d^{10}\) (Cu), where the reward is real.

Fig. 4.2

Fig. 4.2 – (a) 3d and 4s energy levels and (b) anomalous configurations of Cr and Cu.

The ground state electronic configurations of the outer orbitals of the transition elements are given in Table 4.1. Only the outermost s electrons and the \((n-1)d\) electrons are listed; the inner noble-gas core is understood.

Table 4.1: Electronic Configurations of outer orbitals of the Transition Elements (ground state)

1st Series

Sc Ti V Cr Mn Fe Co Ni Cu Zn
\(Z\) 21 22 23 24 25 26 27 28 29 30
4s 2 2 2 1 2 2 2 2 1 2
3d 1 2 3 5 5 6 7 8 10 10

2nd Series

Y Zr Nb Mo Tc Ru Rh Pd Ag Cd
\(Z\) 39 40 41 42 43 44 45 46 47 48
5s 2 2 1 1 1 1 1 0 1 2
4d 1 2 4 5 6 7 8 10 10 10

3rd Series

La Hf Ta W Re Os Ir Pt Au Hg
\(Z\) 57 72 73 74 75 76 77 78 79 80
6s 2 2 2 2 2 2 2 1 1 2
5d 1 2 3 4 5 6 7 9 10 10

4th Series

Ac Rf Db Sg Bh Hs Mt Ds Rg Cn
\(Z\) 89 104 105 106 107 108 109 110 111 112
7s 2 2 2 2 2 2 2 2 1 2
6d 1 2 3 4 5 6 7 8 10 10

Table 4.1, the ns row of each of the four series (Scan across the ns row of every series. In which columns does the ns count drop below 2, and what is the (n–1)d count in exactly those columns? State the pattern you find in one sentence, and then name the one element in the table whose ns count is 0.)

The End Members: Zn, Cd, Hg and Cn

The electronic configurations of the outer orbitals of Zn, Cd, Hg and Cn are represented by the general formula \((n-1)d^{10}ns^{2}\). The orbitals in these elements are completely filled in the ground state as well as in their common oxidation states. Therefore, as we noted while defining the term, they are not regarded as transition elements.

On what ground can you say that scandium (\(Z = 21\)) is a transition element but zinc (\(Z = 30\)) is not?

On the basis of incompletely filled 3d orbitals in case of scandium atom in its ground state (\(3d^1\)), it is regarded as a transition element. On the other hand, zinc atom has completely filled d orbitals (\(3d^{10}\)) in its ground state as well as in its oxidised state, hence it is not regarded as a transition element.

4.1 Silver atom has completely filled d orbitals (\(4d^{10}\)) in its ground state. How can you say that it is a transition element?

What Partly Filled d Orbitals Make Possible

The d orbitals of the transition elements protrude to the periphery of an atom more than the other orbitals (that is, more than the s and p orbitals). They therefore lie exposed. This has two consequences: these orbitals are more influenced by the surroundings, and they in turn affect the atoms or molecules surrounding them.

In some respects, ions of a given \(d^n\) configuration (\(n = 1 - 9\)) have similar magnetic and electronic properties, whichever metal they came from. With partly filled d orbitals, these elements exhibit certain characteristic properties: display of a variety of oxidation states, formation of coloured ions, and entering into complex formation with a variety of ligands.

The transition metals and their compounds also exhibit catalytic property and paramagnetic behaviour. All these characteristics are discussed in detail later in this Chapter; for now, simply note that every one of them can be traced back to the partly filled d set.

Reading the Block Across, Not Down

There are greater similarities in the properties of the transition elements of a horizontal row, in contrast to the non-transition elements, where a group (a vertical column) is usually the family that behaves alike. However, some group similarities also exist among the transition metals. Accordingly, we shall first study the general characteristics and their trends in the horizontal rows — particularly the 3d row — and then consider some group similarities.

Writing the Configuration of an Ion

One point that trips students up constantly: when a d-block element forms an ion, the \(ns\) electrons are removed first, before the \((n-1)d\) electrons — even though the \(ns\) orbital was filled first. So the ion, unlike the atom, has no outer s electrons left to speak of.

Write the ground-state electronic configuration of the iron atom (\(Z = 26\)) and of the \(\mathrm{Fe^{3+}}\) ion, and state the number of unpaired d electrons in the ion.

Iron has \(Z = 26\). Filling in order, the configuration is

\[\mathrm{Fe}: [\mathrm{Ar}]\,3d^6\,4s^2\]

To form \(\mathrm{Fe^{3+}}\)​, three electrons must be removed. The two 4s electrons go first, followed by one 3d electron:

\[\mathrm{Fe^{3+}}: [\mathrm{Ar}]\,3d^5\]

The five 3d electrons spread themselves singly over the five d orbitals with parallel spins (Hund’s rule), so the ion has five unpaired electrons and an exactly half-filled d set. This is why \(\mathrm{Fe^{3+}}\) turns up again and again in this chapter as a particularly stable ion.

(question only — no solution) Write the ground-state outer electronic configuration of the nickel atom (\(Z = 28\)) and of the \(\mathrm{Ni^{2+}}\) ion, and state how many unpaired 3d electrons the ion contains.

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

We will discuss the properties of elements of the first transition series only in the following sections. The 3d row is the row for which reliable data are easiest to compare, and once you understand it, the other rows fall into place.

4.3.1 Physical Properties

Nearly all the transition elements display typical metallic properties: high tensile strength, ductility, malleability, high thermal and electrical conductivity, and metallic lustre. With the exceptions of Zn, Cd, Hg and Mn, they have one or more typical metallic structures at normal temperatures — that is, their atoms pack in the close-packed or body-centred arrangements you met in the solid state.

Lattice Structures of Transition Metals

Sc Ti V Cr Mn Fe Co Ni Cu Zn
hcp (bcc) hcp (bcc) bcc bcc X (bcc, ccp) bcc (hcp) ccp (hcp) ccp ccp X (hcp)
Y Zr Nb Mo Tc Ru Rh Pd Ag Cd
hcp (bcc) hcp (bcc) bcc bcc hcp hcp ccp ccp ccp X (hcp)
La Hf Ta W Re Os Ir Pt Au Hg
hcp (ccp, bcc) hcp (bcc) bcc bcc hcp hcp ccp ccp ccp X

(bcc = body centred cubic; hcp = hexagonal close packed; ccp = cubic close packed; X = a typical metal structure).

Hardness, Melting Points and the d Electrons

The transition metals (with the exception of Zn, Cd and Hg) are very hard and have low volatility. Their melting and boiling points are high. Fig. 4.3 depicts the melting points of transition metals belonging to the 3d, 4d and 5d series; notice how each curve rises to a peak and then falls away.

Fig. 4.3

Fig. 4.3 – Trends in melting points of transition elements.

The high melting points of these metals are attributed to the involvement of a greater number of electrons from the \((n-1)d\) orbitals, in addition to the \(ns\) electrons, in the interatomic metallic bonding. In an s-block metal only the s electrons are available to glue the lattice together; in a transition metal the d electrons join in as well, and the metallic bond becomes far stronger.

In any row, the melting points of these metals rise to a maximum at \(d^5\), except for the anomalous values of Mn and Tc, and then fall regularly as the atomic number increases.

They also have high enthalpies of atomisation, which are shown in Fig. 4.4. The enthalpy of atomisation is the enthalpy needed to break one mole of the solid metal into free gaseous atoms — in effect, a direct measure of how strong the metallic bonding is.

Fig. 4.4

Fig. 4.4 – Trends in enthalpies of atomisation of transition elements.

The maxima at about the middle of each series indicate that one unpaired electron per d orbital is particularly favourable for strong interatomic interaction. In general, the greater the number of valence electrons available, the stronger is the resultant bonding. Since the enthalpy of atomisation is an important factor in determining the standard electrode potential of a metal, metals with a very high enthalpy of atomisation (that is, a very high boiling point) tend to be noble in their reactions — they resist being converted into ions. We shall return to this point when we take up electrode potentials.

Another generalisation that may be drawn from Fig. 4.4 is that the metals of the second and third series have greater enthalpies of atomisation than the corresponding elements of the first series. This is an important factor in accounting for the occurrence of much more frequent metal–metal bonding in compounds of the heavy transition metals.

Fig. 4.3, the melting-point curves of the three transition series (Read the 3d curve from Ti to Cu. At which element is the melting point highest, and at which element does the curve show an unexpected dip? Compare the peak height of the 5d curve with that of the 3d curve, and state what this says about the strength of metallic bonding in the heavier series.)

Why do the transition elements exhibit higher enthalpies of atomisation?

Because of the large number of unpaired electrons in their atoms, they have stronger interatomic interaction and hence stronger bonding between atoms, resulting in higher enthalpies of atomisation.

4.2 In the series Sc (\(Z = 21\)) to Zn (\(Z = 30\)), the enthalpy of atomisation of zinc is the lowest, i.e., \(126\ \text{kJ mol}^{-1}\). Why?

(question only — no solution) Using the lattice-structure table above, name two metals of the 3d series that crystallise in a body centred cubic lattice, and state which of Zn, Cd and Hg does not** adopt a typical metallic structure at normal temperature.**

4.3.2 Variation in Atomic and Ionic Sizes of Transition Metals

In general, ions of the same charge in a given series show a progressive decrease in radius with increasing atomic number. The reason is a tug-of-war between two effects. Each time the nuclear charge increases by unity, the new electron enters a d orbital. It may be recalled that the shielding effect of a d electron is not very effective — a d electron is poor at screening other electrons from the nucleus. Hence the net electrostatic attraction between the nuclear charge and the outermost electron increases, and the ionic radius decreases.

The same trend is observed in the atomic radii of a given series. However, the variation within a series is quite small: the added d electron does shield to some extent, so the contraction is gentle rather than dramatic (look along the metallic radii row of Table 4.2 and you will see the values crawl from 164 pm down to about 125 pm and then rise slightly again).

atomic and ionic radii of the transition metals (across a 3d series →: radius decreases, but only slightly, because each added electron goes into an inner d orbital and shields the outer electrons imperfectly, so the effective nuclear charge felt by the outer electrons rises slowly; down a group ↓: radius increases from the 3d to the 4d series as expected, but the 5d radii are almost identical to the 4d radii, because of the lanthanoid contraction.)

The Lanthanoid Contraction

An interesting point emerges when the atomic sizes of one series are compared with those of the corresponding elements in the other series. The curves in Fig. 4.5 show an increase from the first (3d) to the second (4d) series of elements, but the radii of the third (5d) series are virtually the same as those of the corresponding members of the second series.

Fig. 4.5

Fig. 4.5 – Trends in atomic radii of transition elements.

This phenomenon is associated with the intervention of the 4f orbitals, which must be filled before the 5d series of elements begins. The filling of 4f before the 5d orbital results in a regular decrease in atomic radii called lanthanoid contraction, which essentially compensates for the expected increase in atomic size with increasing atomic number.

NoteDefinition

Lanthanoid contraction is the steady decrease in atomic and ionic radii that occurs along the lanthanoid series as the 4f orbitals are progressively filled; its effect is to make the elements that follow the lanthanoids smaller than expected.

The net result of the lanthanoid contraction is that the second and the third d series exhibit similar radii (for example, Zr 160 pm, Hf 159 pm) and have very similar physical and chemical properties — much more similar than one would expect on the basis of the usual family relationship between an element and the one below it.

The factor responsible for the lanthanoid contraction is somewhat similar to that observed in an ordinary transition series and is attributed to a similar cause: the imperfect shielding of one electron by another in the same set of orbitals. However, the shielding of one 4f electron by another is even less than that of one d electron by another, and as the nuclear charge increases along the series, there is a fairly regular decrease in the size of the entire \(4f^{\,n}\) orbitals.

Q2. Zirconium and hafnium sit in different periods of the periodic table, yet they are almost the same size and are always found together in nature. How can an element from the row above “shrink” the one below it?

A. It is not the row above but the fourteen lanthanoids squeezed in before hafnium that do the shrinking. As the 4f orbitals fill, the poor shielding by 4f electrons lets the growing nuclear charge pull the whole electron cloud inward, element after element. By the time we reach hafnium, this accumulated lanthanoid contraction has cancelled out almost exactly the size increase we would expect on going down a group. Hafnium therefore ends up at 159 pm against zirconium’s 160 pm. Two ions of the same size and the same charge behave almost identically in chemical reactions, so the two metals travel together in their ores and are notoriously hard to separate.

Density Across the Series

The decrease in metallic radius, coupled with an increase in atomic mass, results in a general increase in the density of these elements. Density is mass per unit volume, so heavier atoms packed into smaller volumes must give a denser metal. Thus, from titanium (\(Z = 22\)) to copper (\(Z = 29\)), the significant increase in the density may be noted in Table 4.2 — from \(4.1\ \text{g cm}^{-3}\) to \(8.9\ \text{g cm}^{-3}\).

Table 4.2: Electronic Configurations and some other Properties of the First Series of Transition Elements

Element Sc Ti V Cr Mn Fe Co Ni Cu Zn
Atomic number 21 22 23 24 25 26 27 28 29 30
Configuration, M \(3d^14s^2\) \(3d^24s^2\) \(3d^34s^2\) \(3d^54s^1\) \(3d^54s^2\) \(3d^64s^2\) \(3d^74s^2\) \(3d^84s^2\) \(3d^{10}4s^1\) \(3d^{10}4s^2\)
Configuration, M⁺ \(3d^14s^1\) \(3d^24s^1\) \(3d^34s^1\) \(3d^5\) \(3d^54s^1\) \(3d^64s^1\) \(3d^74s^1\) \(3d^84s^1\) \(3d^{10}\) \(3d^{10}4s^1\)
Configuration, M²⁺ \(3d^1\) \(3d^2\) \(3d^3\) \(3d^4\) \(3d^5\) \(3d^6\) \(3d^7\) \(3d^8\) \(3d^9\) \(3d^{10}\)
Configuration, M³⁺ [Ar] \(3d^1\) \(3d^2\) \(3d^3\) \(3d^4\) \(3d^5\) \(3d^6\) \(3d^7\) – –
Enthalpy of atomisation, \(\Delta_a H^\circ\)/kJ mol⁻¹ 326 473 515 397 281 416 425 430 339 126
Ionisation enthalpy \(\Delta_i H^\circ\) I /kJ mol⁻¹ 631 656 650 653 717 762 758 736 745 906
\(\Delta_i H^\circ\) II /kJ mol⁻¹ 1235 1309 1414 1592 1509 1561 1644 1752 1958 1734
\(\Delta_i H^\circ\) III /kJ mol⁻¹ 2393 2657 2833 2990 3260 2962 3243 3402 3556 3837
Metallic radius M /pm 164 147 135 129 137 126 125 125 128 137
Ionic radius M²⁺ /pm – – 79 82 82 77 74 70 73 75
Ionic radius M³⁺ /pm 73 67 64 62 65 65 61 60 – –
Standard electrode potential \(E^\circ\)(M²⁺/M) /V – –1.63 –1.18 –0.90 –1.18 –0.44 –0.28 –0.25 +0.34 –0.76
Standard electrode potential \(E^\circ\)(M³⁺/M²⁺) /V – –0.37 –0.26 –0.41 +1.57 +0.77 +1.97 – – –
Density /g cm⁻³ 3.43 4.1 6.07 7.19 7.21 7.8 8.7 8.9 8.9 7.1

4.3.3 Ionisation Enthalpies

There is an increase in ionisation enthalpy along each series of the transition elements from left to right, due to an increase in nuclear charge which accompanies the filling of the inner d orbitals. Table 4.2 gives the values of the first three ionisation enthalpies of the first series of transition elements.

These values show that the successive enthalpies of these elements do not increase as steeply as in the case of non-transition elements. Also, the variation in ionisation enthalpy along a series of transition elements is much less in comparison to the variation along a period of non-transition elements. The first ionisation enthalpy, in general, increases; but the magnitude of the increase in the second and third ionisation enthalpies for the successive elements is much higher along a series.

ionisation enthalpy across the 3d series (across a period →: first ionisation enthalpy increases only gently and irregularly, because each added electron enters an inner 3d orbital that shields the 4s electrons fairly well, so the effective nuclear charge felt by the 4s electron grows slowly; second and third ionisation enthalpies rise much more steeply, and show sharp breaks wherever removing an electron would destroy a stable \(d^5\) or \(d^{10}\) arrangement.)

Why the First Ionisation Enthalpy Wanders

The irregular trend in the first ionisation enthalpy of the metals of the 3d series, though of little chemical significance, can be accounted for by considering that the removal of one electron alters the relative energies of the 4s and 3d orbitals.

You have learnt that when d-block elements form ions, the \(ns\) electrons are lost before the \((n-1)d\) electrons. As we move along the period in the 3d series, we see that the nuclear charge increases from scandium to zinc, but electrons are added to the orbital of an inner subshell, i.e., the 3d orbitals. These 3d electrons shield the 4s electrons from the increasing nuclear charge somewhat more effectively than the outer shell electrons can shield one another. Therefore, the atomic radii decrease less rapidly, and ionisation energies increase only slightly along the 3d series.

The doubly or more highly charged ions have \(d^n\) configurations with no 4s electrons. A general trend of increasing values of the second ionisation enthalpy is expected, as the effective nuclear charge increases — one d electron does not shield another electron from the influence of the nuclear charge, because d orbitals differ in direction and so the electrons in them do not get in each other’s way.

However, the trend of steady increase in second and third ionisation enthalpy breaks for the formation of \(\mathrm{Mn^{2+}}\) and \(\mathrm{Fe^{3+}}\) respectively. In both cases, the ion produced has the \(d^5\) configuration. Similar breaks occur at corresponding elements in the later transition series.

Exchange Energy: the Hidden Term

The interpretation of the variation in ionisation enthalpy for an electronic configuration \(d^n\) is as follows. The three terms responsible for the value of the ionisation enthalpy are: the attraction of each electron towards the nucleus, the repulsion between the electrons, and the exchange energy.

Exchange energy is responsible for the stabilisation of an energy state. It is approximately proportional to the total number of possible pairs of parallel spins in the degenerate orbitals — the more electrons there are with parallel spins in a set of equal-energy orbitals, the more ways they can “exchange”, and the lower the energy of the system. When several electrons occupy a set of degenerate orbitals, the lowest energy state corresponds to the maximum possible extent of single occupation of the orbitals with parallel spins (Hund’s rule).

NotePrinciple / Law

(Hund’s rule of maximum multiplicity) — In a set of degenerate orbitals, electrons occupy the orbitals singly with parallel spins before any pairing takes place, since this arrangement maximises the exchange energy and gives the lowest energy state.

The loss of exchange energy increases the stability. As the stability increases, the ionisation becomes more difficult. There is no loss of exchange energy at the \(d^6\) configuration.

Work through the two cases NCERT gives. \(\mathrm{Mn^{+}}\) has the \(3d^54s^1\) configuration and the configuration of \(\mathrm{Cr^{+}}\) is \(d^5\); therefore, the ionisation enthalpy of \(\mathrm{Mn^{+}}\) is lower than that of \(\mathrm{Cr^{+}}\). In the same way, \(\mathrm{Fe^{2+}}\) has the \(d^6\) configuration and \(\mathrm{Mn^{2+}}\) has the \(3d^5\) configuration; hence the ionisation enthalpy of \(\mathrm{Fe^{2+}}\) is lower than that of \(\mathrm{Mn^{2+}}\). In other words, we can say that the third ionisation enthalpy of Fe is lower than that of Mn — a fact you should remember, because it explains why \(\mathrm{Fe^{3+}}\) forms so readily.

Q. Why should it be easier to pull an electron out of \(\mathrm{Fe^{2+}}\) (\(d^6\)) than out of \(\mathrm{Mn^{2+}}\) (\(d^5\)), when iron has the larger nuclear charge?

A. Because of what is left behind. Removing an electron from \(d^6\) leaves the specially stable half-filled \(d^5\) set, so the process is favoured. Removing one from \(d^5\) destroys that half-filled set. Stability of the product, not just the pull of the nucleus, decides how hard ionisation is.

Forming the M²⁺ Ion

The lowest common oxidation state of these metals is +2. To form the \(\mathrm{M^{2+}}\) ions from the gaseous atoms, the sum of the first and second ionisation enthalpy is required, in addition to the enthalpy of atomisation (the metal must first be converted to free atoms).

The dominant term is the second ionisation enthalpy, which shows unusually high values for Cr and Cu, where the \(\mathrm{M^{+}}\) ions have the \(d^5\) and \(d^{10}\) configurations respectively — stable arrangements that resist losing a further electron. The value for Zn is correspondingly low, as the ionisation causes the removal of one 4s electron, which results in the formation of the stable \(d^{10}\) configuration.

The trend in the third ionisation enthalpies is not complicated by the 4s orbital factor, and shows the greater difficulty of removing an electron from the \(d^5\) (\(\mathrm{Mn^{2+}}\)) and \(d^{10}\) (\(\mathrm{Zn^{2+}}\)) ions. In general, the third ionisation enthalpies are quite high. Also, the high values for the third ionisation enthalpies of copper, nickel and zinc indicate why it is difficult to obtain an oxidation state greater than two for these elements.

Although ionisation enthalpies give some guidance concerning the relative stabilities of oxidation states, this problem is very complex and not amenable to ready generalisation. Hydration enthalpy, lattice enthalpy and the nature of the ligand all have a say, as we shall see shortly.

4.3.4 Oxidation States

One of the most notable features of the transition elements is the great variety of oxidation states these may show in their compounds. Table 4.3 lists the common oxidation states of the first row transition elements; the most common ones are printed in bold.

Table 4.3: Oxidation States of the first row Transition Metals (the most common ones are in bold types)

Sc Ti V Cr Mn Fe Co Ni Cu Zn
+3 +2 +2 +2 +2 +2 +2 +2 +1 +2
+3 +3 +3 +3 +3 +3 +3 +2
+4 +4 +4 +4 +4 +4 +4
+5 +5 +5
+6 +6 +6
+7
Why the Middle of the Series is Richest

The elements which give the greatest number of oxidation states occur in or near the middle of the series. Manganese, for example, exhibits all the oxidation states from +2 to +7.

The lesser number of oxidation states at the extreme ends stems from one of two opposite causes: either too few electrons to lose or share (Sc, Ti), or too many d electrons — hence fewer orbitals available in which to share electrons with others — for higher valence (Cu, Zn).

Thus, early in the series, scandium(II) is virtually unknown, and titanium(IV) is more stable than Ti(III) or Ti(II). At the other end, the only oxidation state of zinc is +2, and no d electrons are involved in it at all.

The maximum oxidation states of reasonable stability correspond in value to the sum of the s and d electrons up to manganese: \(\mathrm{Ti^{IV}O_2}\), \(\mathrm{V^{V}O_2^{+}}\), \(\mathrm{Cr^{VI}O_4^{2-}}\), \(\mathrm{Mn^{VII}O_4^{-}}\). After manganese, there is a rather abrupt decrease in the stability of higher oxidation states, so the typical species to follow are \(\mathrm{Fe^{II,III}}\), \(\mathrm{Co^{II,III}}\), \(\mathrm{Ni^{II}}\), \(\mathrm{Cu^{I,II}}\), \(\mathrm{Zn^{II}}\).

Q5. Manganese shows every oxidation state from +2 to +7, but scandium sticks to +3. Why should a metal in the middle of the series have so many more choices?

A. Manganese has the configuration \(3d^54s^2\) — seven electrons (five d and two s) that lie close together in energy and are all, in principle, available for losing or sharing. It can therefore stop at +2, or go all the way to +7 in \(\mathrm{MnO_4^{-}}\), where the sum of its s and d electrons is used up. Scandium, at the start of the series, has only three such electrons (\(3d^14s^2\)); once it has lost all three it reaches the noble-gas configuration of argon, and there is nothing further, and nothing in between, that is stable. Few electrons at one end, and too many crowded d electrons at the other, both narrow the choice — the middle of the series is where the options are widest.

Steps of One, Not Two

The variability of oxidation states, a characteristic of transition elements, arises out of the incomplete filling of d orbitals in such a way that their oxidation states differ from each other by unity — for example, \(\mathrm{V^{II}}\), \(\mathrm{V^{III}}\), \(\mathrm{V^{IV}}\), \(\mathrm{V^{V}}\). This is in contrast with the variability of oxidation states of non-transition elements, where oxidation states normally differ by a unit of two (think of \(\mathrm{Sn^{2+}}\) and \(\mathrm{Sn^{4+}}\), or \(\mathrm{Pb^{2+}}\) and \(\mathrm{Pb^{4+}}\)).

variable oxidation states in transition vs non-transition elements

Feature Transition (d-block) elements Non-transition (p-block) elements
Successive states differ by one unit two units
Example \(\mathrm{V^{II}}\), \(\mathrm{V^{III}}\), \(\mathrm{V^{IV}}\), \(\mathrm{V^{V}}\) \(\mathrm{Sn^{II}}\), \(\mathrm{Sn^{IV}}\)
Origin incomplete filling of d orbitals; ns and \((n-1)d\) electrons close in energy loss of p electrons, then of the s pair (inert pair effect)
Heavier member of a group favours the higher oxidation state, e.g. Mo(VI), W(VI) more stable than Cr(VI) the lower oxidation state, e.g. Pb(II) more stable than Pb(IV)

An interesting feature in the variability of oxidation states of the d-block elements is noticed among the groups (groups 4 through 10). Although in the p-block the lower oxidation states are favoured by the heavier members (due to the inert pair effect), the opposite is true in the groups of the d-block. For example, in group 6, Mo(VI) and W(VI) are found to be more stable than Cr(VI). Thus Cr(VI) in the form of dichromate in acidic medium is a strong oxidising agent, whereas \(\mathrm{MoO_3}\) and \(\mathrm{WO_3}\) are not.

Low oxidation states are found when a complex compound has ligands capable of \(\pi\)-acceptor character in addition to the \(\sigma\)-bonding. Such ligands can drain electron density away from the metal, allowing it to hold electrons it would otherwise lose. For example, in \(\mathrm{Ni(CO)_4}\) and \(\mathrm{Fe(CO)_5}\), the oxidation state of nickel and iron is zero.

Name a transition element which does not exhibit variable oxidation states.

Scandium (\(Z = 21\)) does not exhibit variable oxidation states.

4.3 Which of the 3d series of the transition metals exhibits the largest number of oxidation states and why?

Determine the oxidation number of manganese in \(\mathrm{KMnO_4}\) and in \(\mathrm{K_2MnO_4}\), and state which of the two is manganese in its highest possible oxidation state.

The oxidation number of an element in a compound is found by assigning the usual values to the other atoms and making the total add up to the overall charge. Potassium is always +1 and oxygen in these salts is –2.

For \(\mathrm{KMnO_4}\), let the oxidation number of Mn be \(x\):

\[(+1) + x + 4(-2) = 0 \quad \Rightarrow \quad x - 7 = 0 \quad \Rightarrow \quad x = +7\]

For \(\mathrm{K_2MnO_4}\):

\[2(+1) + x + 4(-2) = 0 \quad \Rightarrow \quad x - 6 = 0 \quad \Rightarrow \quad x = +6\]

Manganese has the configuration \(3d^54s^2\), so the sum of its s and d electrons is 7. Hence +7, as in \(\mathrm{KMnO_4}\) (the permanganate ion, \(\mathrm{MnO_4^{-}}\)), is the highest oxidation state manganese can reach; in \(\mathrm{K_2MnO_4}\) (the manganate ion, \(\mathrm{MnO_4^{2-}}\)) it is +6.

(question only — no solution) Determine the oxidation number of chromium in \(\mathrm{K_2Cr_2O_7}\) and in \(\mathrm{K_2CrO_4}\), and state what this tells you about the relationship between the chromate and dichromate ions.

4.3.8 Chemical Reactivity and E° Values

Transition metals vary widely in their chemical reactivity. Many of them are sufficiently electropositive to dissolve in mineral acids, although a few are ‘noble’ — that is, they are unaffected by single acids.

The metals of the first series, with the exception of copper, are relatively more reactive and are oxidised by \(1\ \mathrm{M}\ \mathrm{H^{+}}\), though the actual rate at which these metals react with oxidising agents like the hydrogen ion (\(\mathrm{H^{+}}\)) is sometimes slow. For example, titanium and vanadium, in practice, are passive to dilute non-oxidising acids at room temperature — a protective oxide film on the surface keeps the acid away, even though the reaction is thermodynamically allowed.

The \(E^\circ\) values for M²⁺/M (Table 4.2) indicate a decreasing tendency to form divalent cations across the series. This general trend towards less negative \(E^\circ\) values is related to the increase in the sum of the first and second ionisation enthalpies. It is interesting to note that the \(E^\circ\) values for Mn, Ni and Zn are more negative than expected from the general trend. Whereas the stabilities of the half-filled d subshell (\(d^5\)) in \(\mathrm{Mn^{2+}}\) and the completely filled d subshell (\(d^{10}\)) in zinc are related to their \(E^\circ\) values, for nickel the \(E^\circ\) value is related to the highest negative enthalpy of hydration.

An examination of the \(E^\circ\) values for the redox couple M³⁺/M²⁺ (Table 4.2) shows that \(\mathrm{Mn^{3+}}\) and \(\mathrm{Co^{3+}}\) ions are the strongest oxidising agents in aqueous solutions. The ions \(\mathrm{Ti^{2+}}\)​, \(\mathrm{V^{2+}}\) and \(\mathrm{Cr^{2+}}\) are strong reducing agents and will liberate hydrogen from a dilute acid, e.g.,

\[2\,\mathrm{Cr^{2+}}\text{(aq)} + 2\,\mathrm{H^{+}}\text{(aq)} \rightarrow 2\,\mathrm{Cr^{3+}}\text{(aq)} + \mathrm{H_2}\text{(g)}\]

Q4. Iron rusts away in moist air, while gold survives for centuries. Both are transition metals, so what decides how “noble” a metal will be?

A. Three energies decide it: the enthalpy of atomisation, the ionisation enthalpies, and the hydration enthalpy of the ion formed. A metal turns into an ion easily only if the energy released on hydration can pay for pulling the atom out of the metal lattice and stripping its electrons. Gold belongs to the third series, where the enthalpy of atomisation is very high (recall Fig. 4.4) and the ionisation enthalpies are large; the hydration energy cannot repay that cost, so gold stays a metal — its \(E^\circ\) is positive and it is not attacked by single acids. Iron’s atomisation and ionisation costs are far more modest, its \(E^\circ\)(Fe²⁺/Fe) is –0.44 V, and moist air is enough to oxidise it. Copper, sitting in between, cannot displace hydrogen from acids either, which is why it survives as roofing and wiring while iron must be painted or galvanised.

This is why the coinage and jewellery metals are what they are. A metal that must last for decades in a pocket or on a finger must be one whose oxidation is not worth its while. It is also why an iron water pipe corrodes and a copper one does not, and why iron is deliberately alloyed with chromium — chromium forms its own tough, invisible oxide layer that seals the surface, which is the whole principle of stainless steel.

For the first row transition metals, the \(E^\circ\) values are:

| \(E^\circ\) | V | Cr | Mn | Fe | Co | Ni | Cu |

|—|—|—|—|—|—|—|—|

| (M²⁺/M) | –1.18 | –0.91 | –1.18 | –0.44 | –0.28 | –0.25 | +0.34 |

Explain the irregularity in the above values.

The \(E^\circ\) (M²⁺/M) values are not regular, which can be explained from the irregular variation of ionisation enthalpies (\(\Delta_i H_1 + \Delta_i H_2\)) and also the sublimation enthalpies, which are relatively much less for manganese and vanadium.

Why is the \(E^\circ\) value for the \(\mathrm{Mn^{3+}}/\mathrm{Mn^{2+}}\) couple much more positive than that for \(\mathrm{Cr^{3+}}/\mathrm{Cr^{2+}}\) or \(\mathrm{Fe^{3+}}/\mathrm{Fe^{2+}}\)? Explain.

The much larger third ionisation energy of Mn (where the required change is \(d^5\) to \(d^4\)) is mainly responsible for this. This also explains why the +3 state of Mn is of little importance.

4.6 Why is the highest oxidation state of a metal exhibited in its oxide or fluoride only?

4.7 Which is a stronger reducing agent, \(\mathrm{Cr^{2+}}\) or \(\mathrm{Fe^{2+}}\)​, and why?

4.3.9 Magnetic Properties

When a magnetic field is applied to substances, mainly two types of magnetic behaviour are observed: diamagnetism and paramagnetism. Diamagnetic substances are repelled by the applied field, while paramagnetic substances are attracted to it. Substances which are attracted very strongly are said to be ferromagnetic. In fact, ferromagnetism is an extreme form of paramagnetism. Many of the transition metal ions are paramagnetic.

diamagnetism vs paramagnetism vs ferromagnetism

Feature Diamagnetic Paramagnetic Ferromagnetic
Behaviour in a magnetic field repelled attracted attracted very strongly
Unpaired electrons none (all paired) one or more many, with their spins aligned
Example from this chapter \(\mathrm{Zn^{2+}}\) (\(3d^{10}\)), \(\mathrm{Sc^{3+}}\) (\(3d^0\)) \(\mathrm{Mn^{2+}}\) (\(3d^5\)), \(\mathrm{Cu^{2+}}\) (\(3d^9\)) iron (an extreme form of paramagnetism)

Paramagnetism arises from the presence of unpaired electrons, each such electron having a magnetic moment associated with its spin angular momentum and its orbital angular momentum. For the compounds of the first series of transition metals, the contribution of the orbital angular momentum is effectively quenched and hence is of no significance. For these, the magnetic moment is determined by the number of unpaired electrons and is calculated by using the ‘spin-only’ formula:

\[\mu = \sqrt{n(n+2)}\ \text{BM}\]

where \(n\) is the number of unpaired electrons and \(\mu\) is the magnetic moment in units of Bohr magneton (BM). A single unpaired electron has a magnetic moment of 1.73 Bohr magnetons (BM).

magnetic moment \(\mu\) (Unit: Bohr magneton, BM. Only unpaired electrons count — \(n\) is the number of unpaired electrons in the ion, never the total number of d electrons. \(\mu = 0\) means diamagnetic; the larger the value of \(\mu\), the more unpaired electrons the ion carries.)

The unit is named after Niels Bohr, whose model of the atom first tied an electron’s angular momentum to a magnetic moment. The Bohr magneton is simply the natural “quantum of magnetism” that one electron contributes, so quoting \(\mu\) in BM is a way of counting unpaired electrons in disguise.

The magnetic moment increases with the increasing number of unpaired electrons. Thus, the observed magnetic moment gives a useful indication about the number of unpaired electrons present in the atom, molecule or ion — measure the magnetism, and you can work backwards to the electronic configuration. The magnetic moments calculated from the ‘spin-only’ formula and those derived experimentally for some ions of the first row transition elements are given in Table 4.7. The experimental data are mainly for hydrated ions in solution or in the solid state.

Table 4.7: Calculated and Observed Magnetic Moments (BM)

Ion Configuration Unpaired electron(s) Magnetic moment (Calculated) Magnetic moment (Observed)
\(\mathrm{Sc^{3+}}\) \(3d^0\) 0 0 0
\(\mathrm{Ti^{3+}}\) \(3d^1\) 1 1.73 1.75
\(\mathrm{Ti^{2+}}\) \(3d^2\) 2 2.84 2.76
\(\mathrm{V^{2+}}\) \(3d^3\) 3 3.87 3.86
\(\mathrm{Cr^{2+}}\) \(3d^4\) 4 4.90 4.80
\(\mathrm{Mn^{2+}}\) \(3d^5\) 5 5.92 5.96
\(\mathrm{Fe^{2+}}\) \(3d^6\) 4 4.90 5.3 – 5.5
\(\mathrm{Co^{2+}}\) \(3d^7\) 3 3.87 4.4 – 5.2
\(\mathrm{Ni^{2+}}\) \(3d^8\) 2 2.84 2.9 – 3.4
\(\mathrm{Cu^{2+}}\) \(3d^9\) 1 1.73 1.8 – 2.2
\(\mathrm{Zn^{2+}}\) \(3d^{10}\) 0 0 —

Calculate the magnetic moment of a divalent ion in aqueous solution if its atomic number is 25.

With atomic number 25, the divalent ion in aqueous solution will have the \(d^5\) configuration (five unpaired electrons). The magnetic moment, \(\mu\), is

\[\mu = \sqrt{5(5+2)} = 5.92\ \text{BM}\]

The spin-only magnetic moment of a certain divalent ion of a 3d metal is found to be 3.87 BM. Identify the ion and give its atomic number.

Put the measured value into the spin-only formula and solve for \(n\):

\[3.87 = \sqrt{n(n+2)} \quad \Rightarrow \quad n(n+2) = (3.87)^2 \approx 15\]

Trying whole numbers, \(n = 3\) gives \(3 \times 5 = 15\). So the ion has three unpaired electrons, and its configuration must be \(3d^3\).

A divalent ion \(\mathrm{M^{2+}}\) with \(3d^3\) comes from a metal that has three d electrons left after losing its two 4s electrons. From Table 4.2, the ion with \(3d^3\) is \(\mathrm{V^{2+}}\). Hence the ion is \(\mathrm{V^{2+}}\), and vanadium has atomic number 23. (Table 4.7 confirms the calculated value of 3.87 BM for \(\mathrm{V^{2+}}\).)

4.8 Calculate the ‘spin only’ magnetic moment of the \(\mathrm{M^{2+}_{(aq)}}\) ion (\(Z = 27\)).

(question only — no solution) Calculate the spin-only magnetic moment of \(\mathrm{Cr^{3+}}\) and of \(\mathrm{Cu^{2+}}\)​, and state which of the two ions would be attracted more strongly by a magnetic field.

4.3.10 Formation of Coloured Ions

When an electron from a lower energy d orbital is excited to a higher energy d orbital, the energy of excitation corresponds to the frequency of light absorbed (Chapter 5). This frequency generally lies in the visible region. The colour observed corresponds to the complementary colour of the light absorbed — the eye sees what is left over after the absorbed portion of white light has been removed.

The frequency of the light absorbed is determined by the nature of the ligand, that is, by the species attached to the metal ion. In aqueous solutions, where water molecules are the ligands, the colours of the ions observed are listed in Table 4.8. A few coloured solutions of d-block elements are illustrated in Fig. 4.8.

Fig. 4.8

Fig. 4.8 – Colours of first-row transition metal ions in aqueous solution.

Table 4.8: Colours of Some of the First Row (aquated) Transition Metal Ions

Configuration Example Colour
\(3d^0\) \(\mathrm{Sc^{3+}}\) colourless
\(3d^0\) \(\mathrm{Ti^{4+}}\) colourless
\(3d^1\) \(\mathrm{Ti^{3+}}\) purple
\(3d^1\) \(\mathrm{V^{4+}}\) blue
\(3d^2\) \(\mathrm{V^{3+}}\) green
\(3d^3\) \(\mathrm{V^{2+}}\) violet
\(3d^3\) \(\mathrm{Cr^{3+}}\) violet
\(3d^4\) \(\mathrm{Mn^{3+}}\) violet
\(3d^4\) \(\mathrm{Cr^{2+}}\) blue
\(3d^5\) \(\mathrm{Mn^{2+}}\) pink
\(3d^5\) \(\mathrm{Fe^{3+}}\) yellow
\(3d^6\) \(\mathrm{Fe^{2+}}\) green
\(3d^6\,3d^7\) \(\mathrm{Co^{3+}}\)​, \(\mathrm{Co^{2+}}\) blue, pink
\(3d^8\) \(\mathrm{Ni^{2+}}\) green
\(3d^9\) \(\mathrm{Cu^{2+}}\) blue
\(3d^{10}\) \(\mathrm{Zn^{2+}}\) colourless

Q3. Copper(II) sulphate solution is blue, but zinc sulphate solution is colourless — both metals belong to the same block. Why should a completely filled set of d orbitals cost a compound its colour?

A. The colour of a transition metal ion comes from an electron jumping from a lower-energy d orbital to a higher-energy one, absorbing exactly the visible light whose energy matches that gap; we then see the complementary colour. In \(\mathrm{Cu^{2+}}\) (\(3d^9\)) there is a vacancy in the upper set, so the jump is possible and the solution appears blue. In \(\mathrm{Zn^{2+}}\) the d orbitals are completely filled (\(3d^{10}\)) — every seat is taken, so no d–d excitation can occur, no visible light is absorbed, and the solution is colourless. The same reasoning explains why \(\mathrm{Sc^{3+}}\) (\(3d^0\)), with no d electrons to promote, is also colourless.

Q. Why is \(\mathrm{Cu^{+}}\) colourless while \(\mathrm{Cu^{2+}}\) is blue, when both come from the same metal?

A. Because \(\mathrm{Cu^{+}}\) is \(3d^{10}\) — a filled set, no d–d transition, no colour — whereas \(\mathrm{Cu^{2+}}\) is \(3d^9\), with one gap in the upper set and therefore one possible jump. Colour depends on the ion’s configuration, not on the identity of the metal.

4.3.11 Formation of Complex Compounds

Complex compounds are those in which the metal ions bind a number of anions or neutral molecules, giving complex species with characteristic properties. A few examples are: \(\mathrm{[Fe(CN)_6]^{3-}}\), \(\mathrm{[Fe(CN)_6]^{4-}}\), \(\mathrm{[Cu(NH_3)_4]^{2+}}\) and \(\mathrm{[PtCl_4]^{2-}}\). (The chemistry of complex compounds is dealt with in detail in Chapter 5.)

The transition metals form a large number of complex compounds. This is due to three things working together: the comparatively smaller sizes of the metal ions, their high ionic charges, and the availability of d orbitals for bond formation. A small, highly charged ion attracts the incoming ligands strongly, and the empty d orbitals give the ion somewhere to accept the ligands’ lone pairs.

4.3.12 Catalytic Properties

The transition metals and their compounds are known for their catalytic activity. This activity is ascribed to their ability to adopt multiple oxidation states and to form complexes. Vanadium(V) oxide (in the Contact Process), finely divided iron (in Haber’s Process), and nickel (in Catalytic Hydrogenation) are some of the examples.

Catalysts at a solid surface involve the formation of bonds between reactant molecules and atoms of the surface of the catalyst (first row transition metals utilise 3d and 4s electrons for bonding). This has the effect of increasing the concentration of the reactants at the catalyst surface, and also of weakening the bonds in the reacting molecules — the activation energy is lowered. A reaction with a lower activation energy has a much larger fraction of molecules able to cross the barrier, and so proceeds faster.

Also, because the transition metal ions can change their oxidation states, they become more effective as catalysts. For example, iron(III) catalyses the reaction between iodide and persulphate ions:

\[2\,\mathrm{I^{-}} + \mathrm{S_2O_8^{2-}} \rightarrow \mathrm{I_2} + 2\,\mathrm{SO_4^{2-}}\]

An explanation of this catalytic action can be given as:

\[2\,\mathrm{Fe^{3+}} + 2\,\mathrm{I^{-}} \rightarrow 2\,\mathrm{Fe^{2+}} + \mathrm{I_2}\]

\[2\,\mathrm{Fe^{2+}} + \mathrm{S_2O_8^{2-}} \rightarrow 2\,\mathrm{Fe^{3+}} + 2\,\mathrm{SO_4^{2-}}\]

Add the two steps and the iron cancels out: the catalyst is regenerated, exactly as a catalyst must be. The cycle is summarised in FC 4.1.

Flowchart

FC 4.1 – Catalytic cycle of Fe³⁺ in the iodide–persulphate reaction: Fe³⁺ + I⁻ → Fe²⁺ + I₂ (iron reduced, iodide oxidised) → Fe²⁺ + S₂O₈²⁻ → Fe³⁺ + SO₄²⁻ (iron re-oxidised, persulphate reduced) → back to start; net reaction shown outside the cycle: 2I⁻ + S₂O₈²⁻ → I₂ + 2SO₄²⁻

The direct, uncatalysed reaction between two negative ions, \(\mathrm{I^{-}}\) and \(\mathrm{S_2O_8^{2-}}\), is slow because the two ions repel each other. The iron(III) ion breaks the problem into two easier steps, each between oppositely charged partners. This ability of a transition metal to shuttle between two oxidation states is the same trick used by the iron in the enzymes of our own blood, and by the metal catalysts that clean the exhaust of a car.

4.3.13 Formation of Interstitial Compounds

Interstitial compounds are those which are formed when small atoms like H, C or N are trapped inside the crystal lattices of metals. The metal atoms in a crystal do not fill space completely; there are small gaps — interstices — between them, and a small enough atom can slip in.

They are usually non-stoichiometric and are neither typically ionic nor covalent; for example, TiC, \(\mathrm{Mn_4N}\), \(\mathrm{Fe_3H}\), \(\mathrm{VH_{0.56}}\) and \(\mathrm{TiH_{1.7}}\). Look at those last two formulae carefully: the fractional subscripts are not a mistake. The formulae quoted do not, of course, correspond to any normal oxidation state of the metal, because the small atoms simply occupy whatever holes are available. Because of the nature of their composition, these compounds are referred to as interstitial compounds.

NoteDefinition

An interstitial compound is a compound formed when small atoms such as hydrogen, carbon or nitrogen occupy the interstitial sites (the holes) in the crystal lattice of a metal; such compounds are usually non-stoichiometric and are neither typically ionic nor covalent.

The principal physical and chemical characteristics of these compounds are as follows:

  1. They have high melting points, higher than those of the pure metals.
  2. They are very hard; some borides approach diamond in hardness.
  3. They retain metallic conductivity.
  4. They are chemically inert.

The trapped atoms act as wedges, locking the metal layers so that they cannot slide over one another — which is precisely why the interstitial compound is harder and higher-melting than the metal it came from.

4.3.14 Alloy Formation

An alloy is a blend of metals prepared by mixing the components. Alloys may be homogeneous solid solutions in which the atoms of one metal are distributed randomly among the atoms of the other.

Such alloys are formed by atoms with metallic radii that are within about 15 per cent of each other — if the two kinds of atom are too different in size, one simply will not fit into the other’s lattice. Because of similar radii and other characteristics of transition metals, alloys are readily formed by these metals. The alloys so formed are hard and often have high melting points.

The best known are the ferrous alloys: chromium, vanadium, tungsten, molybdenum and manganese are used for the production of a variety of steels and stainless steel. Alloys of transition metals with non-transition metals, such as brass (copper–zinc) and bronze (copper–tin), are also of considerable industrial importance.

Stainless steel came out of a failed experiment. Around 1913, Harry Brearley, working in the steel city of Sheffield, was trying to find an alloy that would resist wear inside gun barrels. He added chromium to steel, was unimpressed by the result, and set the discarded samples aside. Weeks later he noticed that while the ordinary steel scraps had rusted, the chromium-bearing ones were still bright. The chromium had formed an invisible, self-repairing oxide film. Cutlery, surgical instruments and kitchen sinks have not been the same since.

What is meant by ‘disproportionation’ of an oxidation state? Give an example.

When a particular oxidation state becomes less stable relative to other oxidation states, one lower, one higher, it is said to undergo disproportionation. For example, manganese(VI) becomes unstable relative to manganese(VII) and manganese(IV) in acidic solution:

\[3\,\mathrm{Mn^{VI}O_4^{2-}} + 4\,\mathrm{H^{+}} \rightarrow 2\,\mathrm{Mn^{VII}O_4^{-}} + \mathrm{Mn^{IV}O_2} + 2\,\mathrm{H_2O}\]

4.9 Explain why \(\mathrm{Cu^{+}}\) ion is not stable in aqueous solutions.

4.4 Some Important Compounds of Transition Elements

4.4.1 Oxides and Oxoanions of Metals

These oxides are generally formed by the reaction of metals with oxygen at high temperatures. All the metals except scandium form MO oxides, which are ionic. As we have already seen, the highest oxidation number in the oxides coincides with the group number and is attained in \(\mathrm{Sc_2O_3}\) to \(\mathrm{Mn_2O_7}\). Beyond group 7, no higher oxides of iron above \(\mathrm{Fe_2O_3}\) are known. Besides the oxides, the oxocations stabilise \(\mathrm{V^{V}}\) as \(\mathrm{VO_2^{+}}\), \(\mathrm{V^{IV}}\) as \(\mathrm{VO^{2+}}\) and \(\mathrm{Ti^{IV}}\) as \(\mathrm{TiO^{2+}}\).

From Ionic and Basic to Covalent and Acidic

As the oxidation number of a metal increases, the ionic character of its oxide decreases. The reason is straightforward: a metal in a high oxidation state is small and highly charged, so it pulls the electron cloud of the oxide ion strongly towards itself and the bond becomes largely covalent.

In the case of Mn, \(\mathrm{Mn_2O_7}\) is a covalent green oil — hardly what one expects of a “metal oxide”. Even \(\mathrm{CrO_3}\) and \(\mathrm{V_2O_5}\) have low melting points, another sign of covalent, molecular character. In these higher oxides, the acidic character is predominant: they react with water to give acids and with alkalies to give salts.

Thus, \(\mathrm{Mn_2O_7}\) gives \(\mathrm{HMnO_4}\) and \(\mathrm{CrO_3}\) gives \(\mathrm{H_2CrO_4}\) and \(\mathrm{H_2Cr_2O_7}\).

\(\mathrm{V_2O_5}\) is, however, amphoteric — it can behave as an acid or a base — though mainly acidic, and it gives \(\mathrm{VO_4^{3-}}\) as well as \(\mathrm{VO_2^{+}}\) salts. In vanadium there is a gradual change from the basic \(\mathrm{V_2O_3}\) to the less basic \(\mathrm{V_2O_4}\) and to the amphoteric \(\mathrm{V_2O_5}\). \(\mathrm{V_2O_4}\) dissolves in acids to give \(\mathrm{VO^{2+}}\) salts. Similarly, \(\mathrm{V_2O_5}\) reacts with alkalies as well as acids to give \(\mathrm{VO_4^{3-}}\) and \(\mathrm{VO_4^{+}}\) respectively. The well characterised CrO is basic, but \(\mathrm{Cr_2O_3}\) is amphoteric.

acid–base character of the oxides of a transition metal (as the oxidation number of the metal rises →: the oxide changes from basic (low oxidation state, ionic, e.g. CrO, \(\mathrm{V_2O_3}\)) through amphoteric (\(\mathrm{Cr_2O_3}\), \(\mathrm{V_2O_5}\)) to acidic (high oxidation state, covalent, e.g. \(\mathrm{CrO_3}\), \(\mathrm{Mn_2O_7}\)), because a small, highly charged metal ion polarises the oxide ion strongly, the bonding turns covalent, and the oxide then reacts with water to give an oxoacid rather than a hydroxide.)

(question only — no solution) Arrange \(\mathrm{MnO}\), \(\mathrm{Mn_2O_3}\) and \(\mathrm{Mn_2O_7}\) in order of increasing acidic character, and give the reason for the order you choose.

Potassium dichromate, K₂Cr₂O₇

Potassium dichromate is a very important chemical, used in the leather industry and as an oxidant for the preparation of many azo compounds.

Dichromates are generally prepared from chromate, which in turn is obtained by the fusion of chromite ore (\(\mathrm{FeCr_2O_4}\)) with sodium or potassium carbonate in free access of air. The reaction with sodium carbonate occurs as follows:

\[4\,\mathrm{FeCr_2O_4} + 8\,\mathrm{Na_2CO_3} + 7\,\mathrm{O_2} \rightarrow 8\,\mathrm{Na_2CrO_4} + 2\,\mathrm{Fe_2O_3} + 8\,\mathrm{CO_2}\]

The yellow solution of sodium chromate is filtered and acidified with sulphuric acid to give a solution from which orange sodium dichromate, \(\mathrm{Na_2Cr_2O_7\cdot 2H_2O}\), can be crystallised.

\[2\,\mathrm{Na_2CrO_4} + 2\,\mathrm{H^{+}} \rightarrow \mathrm{Na_2Cr_2O_7} + 2\,\mathrm{Na^{+}} + \mathrm{H_2O}\]

Sodium dichromate is more soluble than potassium dichromate. The latter is therefore prepared by treating the solution of sodium dichromate with potassium chloride — the less soluble salt is the one that leaves the solution.

\[\mathrm{Na_2Cr_2O_7} + 2\,\mathrm{KCl} \rightarrow \mathrm{K_2Cr_2O_7} + 2\,\mathrm{NaCl}\]

Orange crystals of potassium dichromate crystallise out. The whole route from ore to product is set out in FC 4.2.

Flowchart

FC 4.2 – Preparation of K₂Cr₂O₇ from chromite ore: chromite ore FeCr₂O₄ → fusion with Na₂CO₃ in free access of air → sodium chromate Na₂CrO₄ (yellow solution) + Fe₂O₃ + CO₂ → filter → acidify with H₂SO₄ → sodium dichromate Na₂Cr₂O₇·2H₂O (orange) → treat with KCl → potassium dichromate K₂Cr₂O₇ crystallises (orange crystals) + NaCl remains in solution

Chromate and Dichromate: the Same Chromium, Two Ions

The chromates and dichromates are interconvertible in aqueous solution depending upon the pH of the solution. Importantly, the oxidation state of chromium in chromate and in dichromate is the same (+6) — the change is one of condensation, not of oxidation.

\[2\,\mathrm{CrO_4^{2-}} + 2\,\mathrm{H^{+}} \rightarrow \mathrm{Cr_2O_7^{2-}} + \mathrm{H_2O}\]

\[\mathrm{Cr_2O_7^{2-}} + 2\,\mathrm{OH^{-}} \rightarrow 2\,\mathrm{CrO_4^{2-}} + \mathrm{H_2O}\]

In plain words: add acid to a yellow chromate solution and it turns orange; add alkali to an orange dichromate solution and it turns yellow.

Q. If the oxidation state of chromium is +6 in both ions, why does the colour change at all when pH changes?

A. Because the species itself changes. The single tetrahedron \(\mathrm{CrO_4^{2-}}\) and the double, corner-sharing \(\mathrm{Cr_2O_7^{2-}}\) absorb visible light at slightly different energies, so one looks yellow and the other orange. Chromium is not oxidised or reduced here — two tetrahedra are simply joined together, or pulled apart.

The structures of the chromate ion, \(\mathrm{CrO_4^{2-}}\), and the dichromate ion, \(\mathrm{Cr_2O_7^{2-}}\), are shown in Fig. 4.9. The chromate ion is tetrahedral, whereas the dichromate ion consists of two tetrahedra sharing one corner, with a Cr–O–Cr bond angle of 126°.

Fig. 4.9

Fig. 4.9 – Structures of the chromate and dichromate ions.

Potassium Dichromate as an Oxidising Agent

Sodium and potassium dichromates are strong oxidising agents. The sodium salt has a greater solubility in water and is extensively used as an oxidising agent in organic chemistry. Potassium dichromate is used as a primary standard in volumetric analysis — it can be obtained pure and dry, its composition does not change on storage, so a solution of accurately known concentration can be made simply by weighing it out.

In acidic solution, its oxidising action can be represented as follows:

\[\mathrm{Cr_2O_7^{2-}} + 14\,\mathrm{H^{+}} + 6\,\mathrm{e^{-}} \rightarrow 2\,\mathrm{Cr^{3+}} + 7\,\mathrm{H_2O} \qquad (E^\circ = 1.33\ \text{V})\]

Note what this half-reaction tells you: each dichromate ion accepts six electrons, and it needs a plentiful supply of \(\mathrm{H^{+}}\) — which is why dichromate is used in acidic medium.

Thus, acidified potassium dichromate will oxidise iodides to iodine, sulphides to sulphur, tin(II) to tin(IV) and iron(II) salts to iron(III). The half-reactions of the reducing agents are noted below, each written so that it releases six electrons, to match the six taken up by dichromate:

\[6\,\mathrm{I^{-}} \rightarrow 3\,\mathrm{I_2} + 6\,\mathrm{e^{-}} \qquad ; \qquad 3\,\mathrm{Sn^{2+}} \rightarrow 3\,\mathrm{Sn^{4+}} + 6\,\mathrm{e^{-}}\]

\[3\,\mathrm{H_2S} \rightarrow 6\,\mathrm{H^{+}} + 3\,\mathrm{S} + 6\,\mathrm{e^{-}} \qquad ; \qquad 6\,\mathrm{Fe^{2+}} \rightarrow 6\,\mathrm{Fe^{3+}} + 6\,\mathrm{e^{-}}\]

The full ionic equation may be obtained by adding the half-reaction for potassium dichromate to the half-reaction for the reducing agent. For example, with iron(II):

\[\mathrm{Cr_2O_7^{2-}} + 14\,\mathrm{H^{+}} + 6\,\mathrm{Fe^{2+}} \rightarrow 2\,\mathrm{Cr^{3+}} + 6\,\mathrm{Fe^{3+}} + 7\,\mathrm{H_2O}\]

The electrons cancel because the two half-reactions were deliberately matched at six electrons each. This is the ion–electron method you met in redox chemistry, and it is the fastest way to write any dichromate or permanganate equation in an exam.

acidified potassium dichromate, \(\mathrm{K_2Cr_2O_7}/\mathrm{H^{+}}\) (A strong oxidising agent, orange in solution → needs acidic medium and a supply of \(\mathrm{H^{+}}\); accepts 6 electrons per \(\mathrm{Cr_2O_7^{2-}}\) and is reduced to green \(\mathrm{Cr^{3+}}\)​, \(E^\circ = 1.33\ \text{V}\) → used as a primary standard in volumetric analysis, to oxidise \(\mathrm{Fe^{2+}}\) to \(\mathrm{Fe^{3+}}\)​, \(\mathrm{I^{-}}\) to \(\mathrm{I_2}\), \(\mathrm{Sn^{2+}}\) to \(\mathrm{Sn^{4+}}\) and \(\mathrm{H_2S}\) to S, and as an oxidant in the leather industry and in the making of azo compounds. The orange-to-green colour change is its visual signature.)

Chromium takes its name from the Greek chroma, meaning colour — a fitting choice, since almost every chromium compound is strongly coloured: yellow chromates, orange dichromates, green chromium(III) salts, deep red \(\mathrm{CrO_3}\).

Chromium(VI) compounds are excellent oxidants but they are also toxic and carcinogenic, and effluent from tanneries and plating shops carrying \(\mathrm{Cr(VI)}\) is a serious water pollutant. Modern practice therefore reduces \(\mathrm{Cr(VI)}\) to the far less harmful \(\mathrm{Cr(III)}\) before discharge, and green-chemistry research works steadily on replacing chromium oxidants with catalytic oxidations that use air or hydrogen peroxide instead. In the laboratory, dichromate solutions must never be handled with bare hands, and the acid is always added to water — never water to acid — when the acidified solution is being made up.

Calculate the mass of iron(II) sulphate, \(\mathrm{FeSO_4}\), that can be oxidised in acidic solution by 4.9 g of potassium dichromate. (Molar masses: \(\mathrm{K_2Cr_2O_7} = 294\ \text{g mol}^{-1}\); \(\mathrm{FeSO_4} = 152\ \text{g mol}^{-1}\).)

First find the moles of dichromate:

\[n(\mathrm{K_2Cr_2O_7}) = \frac{4.9\ \text{g}}{294\ \text{g mol}^{-1}} = 0.0167\ \text{mol}\]

From the balanced ionic equation,

\[\mathrm{Cr_2O_7^{2-}} + 14\,\mathrm{H^{+}} + 6\,\mathrm{Fe^{2+}} \rightarrow 2\,\mathrm{Cr^{3+}} + 6\,\mathrm{Fe^{3+}} + 7\,\mathrm{H_2O}\]

one mole of dichromate oxidises six moles of \(\mathrm{Fe^{2+}}\) (it accepts 6 electrons, and each \(\mathrm{Fe^{2+}}\) supplies one). Therefore

\[n(\mathrm{Fe^{2+}}) = 6 \times 0.0167 = 0.100\ \text{mol}\]

\[\text{mass of } \mathrm{FeSO_4} = 0.100\ \text{mol} \times 152\ \text{g mol}^{-1} = 15.2\ \text{g}\]

So 4.9 g of potassium dichromate will oxidise 15.2 g of iron(II) sulphate.

Potassium permanganate, KMnO₄

Potassium permanganate is prepared by fusion of \(\mathrm{MnO_2}\) with an alkali metal hydroxide and an oxidising agent like \(\mathrm{KNO_3}\). This produces the dark green \(\mathrm{K_2MnO_4}\), which disproportionates in a neutral or acidic solution to give permanganate.

\[2\,\mathrm{MnO_2} + 4\,\mathrm{KOH} + \mathrm{O_2} \rightarrow 2\,\mathrm{K_2MnO_4} + 2\,\mathrm{H_2O}\]

\[3\,\mathrm{MnO_4^{2-}} + 4\,\mathrm{H^{+}} \rightarrow 2\,\mathrm{MnO_4^{-}} + \mathrm{MnO_2} + 2\,\mathrm{H_2O}\]

In the second equation manganese(VI) is simultaneously oxidised to manganese(VII) and reduced to manganese(IV) — the disproportionation you met in Solved Example 4.12.

Commercially it is prepared by the alkaline oxidative fusion of \(\mathrm{MnO_2}\), followed by the electrolytic oxidation of manganate(VI).

\[\mathrm{MnO_2} \xrightarrow{\ \text{fused with KOH, oxidised with air or } \mathrm{KNO_3}\ } \underset{\text{manganate ion}}{\mathrm{MnO_4^{2-}}}\]

\[\underset{\text{manganate}}{\mathrm{MnO_4^{2-}}} \xrightarrow{\ \text{electrolytic oxidation in alkaline solution}\ } \underset{\text{permanganate ion}}{\mathrm{MnO_4^{-}}}\]

The two industrial stages are set out in FC 4.3.

Flowchart

FC 4.3 – Preparation of KMnO₄ from pyrolusite (MnO₂): MnO₂ → fusion with KOH + oxidising agent (air or KNO₃) → green manganate ion MnO₄²⁻ (Mn in +6) → branch (a) laboratory/acidic route: add H⁺ → disproportionation → purple MnO₄⁻ + MnO₂; branch (b) commercial route: electrolytic oxidation in alkaline solution → purple permanganate ion MnO₄⁻ (Mn in +7)

In the laboratory, a manganese(II) ion salt is oxidised by peroxodisulphate to permanganate:

\[2\,\mathrm{Mn^{2+}} + 5\,\mathrm{S_2O_8^{2-}} + 8\,\mathrm{H_2O} \rightarrow 2\,\mathrm{MnO_4^{-}} + 10\,\mathrm{SO_4^{2-}} + 16\,\mathrm{H^{+}}\]

Properties of Potassium Permanganate

Potassium permanganate forms dark purple (almost black) crystals which are isostructural with those of \(\mathrm{KClO_4}\). The salt is not very soluble in water (6.4 g per 100 g of water at 293 K), but when heated it decomposes at 513 K:

\[2\,\mathrm{KMnO_4} \rightarrow \mathrm{K_2MnO_4} + \mathrm{MnO_2} + \mathrm{O_2}\]

It has two physical properties of considerable interest: its intense colour, and its diamagnetism along with a temperature-dependent weak paramagnetism. These can be explained by the use of molecular orbital theory, which is beyond the present scope.

The manganate and permanganate ions are tetrahedral; the \(\pi\)-bonding takes place by overlap of the p orbitals of oxygen with the d orbitals of manganese, as drawn in Fig. 4.10. The green manganate is paramagnetic because of one unpaired electron, but the permanganate is diamagnetic due to the absence of an unpaired electron.

Fig. 4.10

Fig. 4.10 – Structures of the manganate and permanganate ions.

Explain, by working out the electronic configuration of manganese in each ion, why \(\mathrm{MnO_4^{-}}\) is diamagnetic while \(\mathrm{MnO_4^{2-}}\) is paramagnetic, and calculate the spin-only magnetic moment of the paramagnetic ion.

First assign the oxidation number of manganese in each ion, taking oxygen as –2.

For \(\mathrm{MnO_4^{-}}\): \(x + 4(-2) = -1 \Rightarrow x = +7\).

For \(\mathrm{MnO_4^{2-}}\): \(x + 4(-2) = -2 \Rightarrow x = +6\).

Manganese is \([\mathrm{Ar}]\,3d^54s^2\), so it has seven electrons outside the argon core.

In \(\mathrm{Mn(VII)}\), all seven are used, leaving \(3d^0\) — no unpaired electrons, hence \(\mathrm{MnO_4^{-}}\) is diamagnetic.

In \(\mathrm{Mn(VI)}\), six are used, leaving \(3d^1\) — one unpaired electron, hence \(\mathrm{MnO_4^{2-}}\) is paramagnetic.

For the manganate ion, \(n = 1\):

\[\mu = \sqrt{n(n+2)} = \sqrt{1(1+2)} = \sqrt{3} = 1.73\ \text{BM}\]

(The intense purple colour of \(\mathrm{MnO_4^{-}}\), note, cannot come from a d–d transition, since it has no d electrons at all; it arises from charge transfer, which is why the salt is so deeply coloured.)

Oxidising Reactions of Permanganate

Acidified permanganate solution oxidises oxalates to carbon dioxide, iron(II) to iron(III), nitrites to nitrates and iodides to free iodine. The half-reactions of the reductants are:

\[5\,\mathrm{C_2O_4^{2-}} \rightarrow 10\,\mathrm{CO_2} + 10\,\mathrm{e^{-}}\]

\[5\,\mathrm{Fe^{2+}} \rightarrow 5\,\mathrm{Fe^{3+}} + 5\,\mathrm{e^{-}}\]

\[5\,\mathrm{NO_2^{-}} + 5\,\mathrm{H_2O} \rightarrow 5\,\mathrm{NO_3^{-}} + 10\,\mathrm{H^{+}} + 10\,\mathrm{e^{-}}\]

\[10\,\mathrm{I^{-}} \rightarrow 5\,\mathrm{I_2} + 10\,\mathrm{e^{-}}\]

The full reaction can be written by adding the half-reaction for \(\mathrm{KMnO_4}\) to the half-reaction of the reducing agent, balancing wherever necessary.

If we represent the reduction of permanganate to manganate, to manganese dioxide, and to a manganese(II) salt by half-reactions:

\[\mathrm{MnO_4^{-}} + \mathrm{e^{-}} \rightarrow \mathrm{MnO_4^{2-}} \qquad (E^\circ = +0.56\ \text{V})\]

\[\mathrm{MnO_4^{-}} + 4\,\mathrm{H^{+}} + 3\,\mathrm{e^{-}} \rightarrow \mathrm{MnO_2} + 2\,\mathrm{H_2O} \qquad (E^\circ = +1.69\ \text{V})\]

\[\mathrm{MnO_4^{-}} + 8\,\mathrm{H^{+}} + 5\,\mathrm{e^{-}} \rightarrow \mathrm{Mn^{2+}} + 4\,\mathrm{H_2O} \qquad (E^\circ = +1.52\ \text{V})\]

We can very well see that the hydrogen ion concentration of the solution plays an important part in influencing the reaction — the product is manganate in strongly alkaline conditions, \(\mathrm{MnO_2}\) in neutral or faintly alkaline conditions, and \(\mathrm{Mn^{2+}}\) in acid.

Although many reactions can be understood by consideration of redox potential, the kinetics of the reaction is also an important factor. Permanganate at \([\mathrm{H^{+}}] = 1\) should oxidise water, but in practice the reaction is extremely slow unless either manganese(II) ions are present or the temperature is raised. Thermodynamics tells us what can happen; kinetics decides what actually does happen in reasonable time.

A few important oxidising reactions of \(\mathrm{KMnO_4}\) are given below.

1. In acid solutions:

  1. Iodine is liberated from potassium iodide:

\[10\,\mathrm{I^{-}} + 2\,\mathrm{MnO_4^{-}} + 16\,\mathrm{H^{+}} \rightarrow 2\,\mathrm{Mn^{2+}} + 8\,\mathrm{H_2O} + 5\,\mathrm{I_2}\]

  1. \(\mathrm{Fe^{2+}}\) ion (green) is converted to \(\mathrm{Fe^{3+}}\) (yellow):

\[5\,\mathrm{Fe^{2+}} + \mathrm{MnO_4^{-}} + 8\,\mathrm{H^{+}} \rightarrow \mathrm{Mn^{2+}} + 4\,\mathrm{H_2O} + 5\,\mathrm{Fe^{3+}}\]

  1. Oxalate ion or oxalic acid is oxidised at 333 K:

\[5\,\mathrm{C_2O_4^{2-}} + 2\,\mathrm{MnO_4^{-}} + 16\,\mathrm{H^{+}} \rightarrow 2\,\mathrm{Mn^{2+}} + 8\,\mathrm{H_2O} + 10\,\mathrm{CO_2}\]

  1. Hydrogen sulphide is oxidised, sulphur being precipitated:

\[\mathrm{H_2S} \rightarrow 2\,\mathrm{H^{+}} + \mathrm{S^{2-}}\]

\[5\,\mathrm{S^{2-}} + 2\,\mathrm{MnO_4^{-}} + 16\,\mathrm{H^{+}} \rightarrow 2\,\mathrm{Mn^{2+}} + 8\,\mathrm{H_2O} + 5\,\mathrm{S}\]

  1. Sulphurous acid or sulphite is oxidised to a sulphate or sulphuric acid:

\[5\,\mathrm{SO_3^{2-}} + 2\,\mathrm{MnO_4^{-}} + 6\,\mathrm{H^{+}} \rightarrow 2\,\mathrm{Mn^{2+}} + 3\,\mathrm{H_2O} + 5\,\mathrm{SO_4^{2-}}\]

  1. Nitrite is oxidised to nitrate:

\[5\,\mathrm{NO_2^{-}} + 2\,\mathrm{MnO_4^{-}} + 6\,\mathrm{H^{+}} \rightarrow 2\,\mathrm{Mn^{2+}} + 5\,\mathrm{NO_3^{-}} + 3\,\mathrm{H_2O}\]

2. In neutral or faintly alkaline solutions:

  1. A notable reaction is the oxidation of iodide to iodate:

\[2\,\mathrm{MnO_4^{-}} + \mathrm{H_2O} + \mathrm{I^{-}} \rightarrow 2\,\mathrm{MnO_2} + 2\,\mathrm{OH^{-}} + \mathrm{IO_3^{-}}\]

  1. Thiosulphate is oxidised almost quantitatively to sulphate:

\[8\,\mathrm{MnO_4^{-}} + 3\,\mathrm{S_2O_3^{2-}} + \mathrm{H_2O} \rightarrow 8\,\mathrm{MnO_2} + 6\,\mathrm{SO_4^{2-}} + 2\,\mathrm{OH^{-}}\]

  1. A manganous salt is oxidised to \(\mathrm{MnO_2}\); the presence of zinc sulphate or zinc oxide catalyses the oxidation:

\[2\,\mathrm{MnO_4^{-}} + 3\,\mathrm{Mn^{2+}} + 2\,\mathrm{H_2O} \rightarrow 5\,\mathrm{MnO_2} + 4\,\mathrm{H^{+}}\]

Notice that in acid the permanganate always ends up as colourless \(\mathrm{Mn^{2+}}\)​, while in neutral or faintly alkaline solution it stops at brown \(\mathrm{MnO_2}\) — the medium, not the reducing agent, decides the destination.

Note: Permanganate titrations in the presence of hydrochloric acid are unsatisfactory, since hydrochloric acid is itself oxidised to chlorine. Some of the permanganate would then be consumed by the acid instead of by the substance being estimated, and the titre value would be wrong. Dilute sulphuric acid is used instead.

potassium permanganate, \(\mathrm{KMnO_4}\) (A powerful oxidising agent, deep purple → in acidic medium it is reduced to colourless \(\mathrm{Mn^{2+}}\) (5 electrons, \(E^\circ = +1.52\ \text{V}\)); in neutral or faintly alkaline medium it stops at brown \(\mathrm{MnO_2}\) (3 electrons, \(E^\circ = +1.69\ \text{V}\)); in strongly alkaline medium it becomes green manganate (1 electron, \(E^\circ = +0.56\ \text{V}\)) → used to oxidise \(\mathrm{Fe^{2+}}\)​, oxalate, sulphite, nitrite, iodide and \(\mathrm{H_2S}\). Acidify with dilute \(\mathrm{H_2SO_4}\), never with HCl. In a titration it needs no external indicator: the first drop of excess permanganate gives a permanent pink colour — it is a self-indicator.)

Uses: Besides its use in analytical chemistry, potassium permanganate is used as a favourite oxidant in preparative organic chemistry. Its uses for the bleaching of wool, cotton, silk and other textile fibres, and for the decolourisation of oils, are also dependent on its strong oxidising power. A very dilute solution is also familiar in first-aid and in water treatment as a disinfectant, for the same reason — it destroys organic matter by oxidising it. Handle the solid with care: it stains the skin brown, and mixing it with easily oxidisable material can start a fire.

(question only — no solution) Write the balanced ionic equation for the oxidation of oxalate ion by acidified \(\mathrm{KMnO_4}\), and calculate the volume of \(0.02\ \mathrm{M}\ \mathrm{KMnO_4}\) solution required to oxidise \(25.0\ \mathrm{cm^3}\) of \(0.05\ \mathrm{M}\) oxalic acid solution in acidic medium.

4.5 The Lanthanoids

The Inner Transition Elements (f-Block)

The f-block consists of two series: the lanthanoids (the fourteen elements following lanthanum) and the actinoids (the fourteen elements following actinium).

Because lanthanum closely resembles the lanthanoids, it is usually included in any discussion of the lanthanoids, for which the general symbol Ln is often used. Similarly, a discussion of the actinoids includes actinium besides the fourteen elements constituting the series.

The lanthanoids resemble one another more closely than do the members of ordinary transition elements in any series. They have only one stable oxidation state, and their chemistry provides an excellent opportunity to examine the effect of small changes in size and nuclear charge along a series of otherwise similar elements — a kind of chemical control experiment.

The chemistry of the actinoids is, on the other hand, much more complicated. The complication arises partly from the occurrence of a wide range of oxidation states in these elements, and partly because their radioactivity creates special problems in their study. The two series will therefore be considered separately here.

The names, symbols, electronic configurations of atomic and some ionic states, and atomic and ionic radii of lanthanum and the lanthanoids (for which the general symbol Ln is used) are given in Table 4.9.

Lanthanum takes its name from the Greek lanthanein, “to lie hidden” — an apt name, since the element stayed concealed for decades inside what chemists had believed were pure substances, and the whole family that follows it stayed hidden inside one another for a century after that.

4.5.1 Electronic Configurations

It may be noted that atoms of these elements have an electronic configuration with \(6s^2\) common, but with variable occupancy of the 4f level (Table 4.9). The differences between one lanthanoid and the next therefore lie deep inside the atom, in an orbital that is well shielded from the outside world — which is exactly why their chemistry is so alike.

However, the electronic configurations of all the tripositive ions (the most stable oxidation state of all the lanthanoids) are of the form \(4f^{\,n}\) (\(n = 1\) to 14 with increasing atomic number). In forming \(\mathrm{Ln^{3+}}\)​, the two 6s electrons and one more electron (from 5d or 4f) are lost, leaving only 4f electrons outside the xenon core.

4.5.2 Atomic and Ionic Sizes

The overall decrease in atomic and ionic radii from lanthanum to lutetium — the lanthanoid contraction — is a unique feature in the chemistry of the lanthanoids. It has far-reaching consequences in the chemistry of the third transition series of the elements.

The decrease in atomic radii (derived from the structures of the metals) is not quite regular, whereas it is regular in the \(\mathrm{M^{3+}}\) ions, as Fig. 4.11 shows.

Fig. 4.11

Fig. 4.11 – Trends in ionic radii of lanthanoids.

This contraction is, of course, similar to that observed in an ordinary transition series and is attributed to the same cause: the imperfect shielding of one electron by another in the same sub-shell. However, the shielding of one 4f electron by another is even less than that of one d electron by another. So, with the increase in nuclear charge along the series, there is a fairly regular decrease in the sizes with increasing atomic number.

atomic and ionic radii of the lanthanoids (across the series La → Lu →: both atomic and ionic radii decrease steadily, because each added electron enters the deeply buried 4f level and shields the outer electrons very poorly, so the rising nuclear charge draws the whole electron cloud inwards; the effect on the M³⁺ ions is remarkably regular, and its cumulative size is enough to cancel the expected size increase for the element that follows the series.)

The cumulative effect of the contraction of the lanthanoid series, known as lanthanoid contraction, causes the radii of the members of the third transition series to be very similar to those of the corresponding members of the second series. The almost identical radii of Zr (160 pm) and Hf (159 pm), a consequence of the lanthanoid contraction, account for their occurrence together in nature and for the difficulty faced in their separation.

Q1. Why are the lanthanoids so hard to separate from one another, when neighbouring s-block metals such as sodium and potassium can be told apart quite easily?

A. Chemical separation works only when two species differ in charge, in size, or in the readiness with which they form compounds. Sodium and potassium differ noticeably in size, so their salts differ in solubility and behaviour. The lanthanoids offer no such handle. Every one of them settles into the same stable +3 state, and the electrons that distinguish one from the next are buried in the 4f level, hidden beneath the filled outer shells and hardly involved in bonding. The lanthanoid contraction makes their ionic radii shrink so gently — a few picometres at a step — that neighbouring \(\mathrm{Ln^{3+}}\) ions are chemically almost twins. Their salts therefore have nearly the same solubility, and separating them by crystallisation demanded the same operation repeated thousands of times, which is precisely the ordeal the chemists of the Ytterby minerals endured.

Table 4.9: Electronic Configurations and Radii of Lanthanum and Lanthanoids

(Only electrons outside the [Xe] core are indicated.)

Atomic Number Name Symbol Ln Ln²⁺ Ln³⁺ Ln⁴⁺ Radius Ln /pm Radius Ln³⁺ /pm
57 Lanthanum La \(5d^16s^2\) \(5d^1\) \(4f^{\,0}\) 187 106
58 Cerium Ce \(4f^15d^16s^2\) \(4f^{\,2}\) \(4f^{\,1}\) \(4f^{\,0}\) 183 103
59 Praseodymium Pr \(4f^{\,3}6s^2\) \(4f^{\,3}\) \(4f^{\,2}\) \(4f^{\,1}\) 182 101
60 Neodymium Nd \(4f^{\,4}6s^2\) \(4f^{\,4}\) \(4f^{\,3}\) \(4f^{\,2}\) 181 99
61 Promethium Pm \(4f^{\,5}6s^2\) \(4f^{\,5}\) \(4f^{\,4}\) 181 98
62 Samarium Sm \(4f^{\,6}6s^2\) \(4f^{\,6}\) \(4f^{\,5}\) 180 96
63 Europium Eu \(4f^{\,7}6s^2\) \(4f^{\,7}\) \(4f^{\,6}\) 199 95
64 Gadolinium Gd \(4f^{\,7}5d^16s^2\) \(4f^{\,7}5d^1\) \(4f^{\,7}\) 180 94
65 Terbium Tb \(4f^{\,9}6s^2\) \(4f^{\,9}\) \(4f^{\,8}\) \(4f^{\,7}\) 178 92
66 Dysprosium Dy \(4f^{\,10}6s^2\) \(4f^{\,10}\) \(4f^{\,9}\) \(4f^{\,8}\) 177 91
67 Holmium Ho \(4f^{\,11}6s^2\) \(4f^{\,11}\) \(4f^{\,10}\) 176 89
68 Erbium Er \(4f^{\,12}6s^2\) \(4f^{\,12}\) \(4f^{\,11}\) 175 88
69 Thulium Tm \(4f^{\,13}6s^2\) \(4f^{\,13}\) \(4f^{\,12}\) 174 87
70 Ytterbium Yb \(4f^{\,14}6s^2\) \(4f^{\,14}\) \(4f^{\,13}\) 173 86
71 Lutetium Lu \(4f^{\,14}5d^16s^2\) \(4f^{\,14}5d^1\) \(4f^{\,14}\) – – –

4.5.3 Oxidation States

In the lanthanoids, La(II) and Ln(III) compounds are the predominant species. However, occasionally +2 and +4 ions in solution or in solid compounds are also obtained.

This irregularity (as in the ionisation enthalpies) arises mainly from the extra stability of an empty, half-filled or filled f subshell — the same principle that made \(d^5\) and \(d^{10}\) so special in the transition series, now applied to \(f^{\,0}\), \(f^{\,7}\) and \(f^{\,14}\).

Thus, the formation of \(\mathrm{Ce^{IV}}\) is favoured by its noble gas configuration (\(f^{\,0}\)), but it is a strong oxidant, reverting to the common +3 state. The \(E^\circ\) value for \(\mathrm{Ce^{4+}}/\mathrm{Ce^{3+}}\) is +1.74 V, which suggests that it can oxidise water. However, the reaction rate is very slow, and hence Ce(IV) is a good analytical reagent — a reminder, once again, that a favourable electrode potential does not by itself guarantee a fast reaction.

Pr, Nd, Tb and Dy also exhibit the +4 state, but only in the oxides, \(\mathrm{MO_2}\).

\(\mathrm{Eu^{2+}}\) is formed by losing the two s electrons, and its \(f^{\,7}\) configuration accounts for the formation of this ion. However, \(\mathrm{Eu^{2+}}\) is a strong reducing agent, changing to the common +3 state. Similarly \(\mathrm{Yb^{2+}}\), which has the \(f^{\,14}\) configuration, is a reductant. \(\mathrm{Tb^{IV}}\) has half-filled f orbitals and is an oxidant. The behaviour of samarium is very much like europium, exhibiting both +2 and +3 oxidation states.

Q. If the \(f^{\,7}\) configuration of \(\mathrm{Eu^{2+}}\) is so stable, why is \(\mathrm{Eu^{2+}}\) still a reducing agent?

A. Because stability is always relative. The half-filled \(f^{\,7}\) set is stable enough for \(\mathrm{Eu^{2+}}\) to exist at all — most lanthanoids offer no +2 ion worth speaking of. But the +3 state is the one the whole family settles into, and its stability still wins. So \(\mathrm{Eu^{2+}}\) survives long enough to be used, and readily gives up an electron to reach \(\mathrm{Eu^{3+}}\) — which is what a reducing agent does.

Write the electronic configuration of the \(\mathrm{Eu^{2+}}\) ion and calculate its spin-only magnetic moment. (Europium has \(Z = 63\).)

From Table 4.9, the europium atom is \([\mathrm{Xe}]\,4f^{\,7}6s^2\). The \(\mathrm{Eu^{2+}}\) ion is formed by losing the two 6s electrons:

\[\mathrm{Eu^{2+}}: [\mathrm{Xe}]\,4f^{\,7}\]

The 4f subshell has seven orbitals. Following Hund’s rule, the seven electrons occupy them singly with parallel spins, so the ion has seven unpaired electrons (\(n = 7\)) and an exactly half-filled f set.

\[\mu = \sqrt{n(n+2)} = \sqrt{7(7+2)} = \sqrt{63} = 7.94\ \text{BM}\]

The large moment is a direct consequence of the half-filled \(f^{\,7}\) arrangement, which is also the reason the +2 state of europium exists at all.

(question only — no solution) Using Table 4.9, write the electronic configuration of the \(\mathrm{Gd^{3+}}\) ion and of the \(\mathrm{Lu^{3+}}\) ion, and state which of the two is paramagnetic and why.

4.5.4 General Characteristics

All the lanthanoids are silvery white soft metals and tarnish rapidly in air. The hardness increases with increasing atomic number, samarium being steel hard. Their melting points range between 1000 and 1200 K, but samarium melts at 1623 K. They have a typical metallic structure and are good conductors of heat and electricity. Density and other properties change smoothly, except for Eu and Yb, and occasionally for Sm and Tm.

Colour and Magnetic Behaviour

Many trivalent lanthanoid ions are coloured, both in the solid state and in aqueous solutions. The colour of these ions may be attributed to the presence of f electrons. Neither \(\mathrm{La^{3+}}\) nor \(\mathrm{Lu^{3+}}\) ion shows any colour, but the rest do so — and that is no accident, since these two are precisely the \(f^{\,0}\) and \(f^{\,14}\) ions, with no possibility of an f–f electron jump.

However, the absorption bands are narrow, probably because of the excitation within the f level, which is well shielded from the surroundings and therefore barely disturbed by the ligands.

The lanthanoid ions other than the \(f^{\,0}\) type (\(\mathrm{La^{3+}}\) and \(\mathrm{Ce^{4+}}\)) and the \(f^{\,14}\) type (\(\mathrm{Yb^{2+}}\) and \(\mathrm{Lu^{3+}}\)) are all paramagnetic, since all the others have unpaired f electrons.

Ionisation Enthalpies

The first ionisation enthalpies of the lanthanoids are around \(600\ \text{kJ mol}^{-1}\), and the second about \(1200\ \text{kJ mol}^{-1}\), comparable with those of calcium. This is why the early members of the series behave chemically rather like calcium.

A detailed discussion of the variation of the third ionisation enthalpies indicates that exchange enthalpy considerations (as in the 3d orbitals of the first transition series) appear to impart a certain degree of stability to empty, half-filled and completely filled orbitals of the f level. This is indicated by the abnormally low value of the third ionisation enthalpy of lanthanum, gadolinium and lutetium — the three elements whose \(\mathrm{M^{3+}}\) ions come out as \(f^{\,0}\), \(f^{\,7}\) and \(f^{\,14}\) respectively.

Chemical Behaviour

In their chemical behaviour, in general, the earlier members of the series are quite reactive, similar to calcium; but with increasing atomic number, they behave more like aluminium.

Values of \(E^\circ\) for the half-reaction

\[\mathrm{Ln^{3+}}\text{(aq)} + 3\,\mathrm{e^{-}} \rightarrow \mathrm{Ln}\text{(s)}\]

are in the range of –2.2 to –2.4 V, except for Eu, for which the value is –2.0 V. This is, of course, a small variation — another sign of how alike the members of this family are. Such large negative values mean the metals are strongly electropositive and lose electrons readily.

The metals combine with hydrogen when gently heated in the gas. The carbides \(\mathrm{Ln_3C}\), \(\mathrm{Ln_2C_3}\) and \(\mathrm{LnC_2}\) are formed when the metals are heated with carbon. They liberate hydrogen from dilute acids and burn in halogens to form halides. They form oxides \(\mathrm{M_2O_3}\) and hydroxides \(\mathrm{M(OH)_3}\). The hydroxides are definite compounds, not just hydrated oxides. They are basic, like alkaline earth metal oxides and hydroxides.

Their general reactions are depicted in Fig. 4.12 and collected in FC 4.4.

Fig. 4.12

Fig. 4.12 – Chemical reactions of the lanthanoids.

Flowchart

FC 4.4 – Reactions of a lanthanoid metal Ln: Ln → burnt in O₂ → Ln₂O₃; Ln → with dilute acid → H₂ liberated; Ln → with halogens → LnX₃; Ln → with H₂O → Ln(OH)₃ + H₂; Ln → heated with C (2773 K) → LnC₂ (also Ln₃C, Ln₂C₃); Ln → heated with N → LnN; Ln → heated with S → Ln₂S₃; Ln → gently heated in H₂ gas → hydride

Uses of the Lanthanoids

The best single use of the lanthanoids is for the production of alloy steels for plates and pipes. A well known alloy is mischmetall, which consists of a lanthanoid metal (~ 95%) and iron (~ 5%) and traces of S, C, Ca and Al.

A good deal of mischmetall is used in Mg-based alloys to produce bullets, shells and lighter flint. Mixed oxides of lanthanoids are employed as catalysts in petroleum cracking. Some individual Ln oxides are used as phosphors in television screens and similar fluorescing surfaces.

The spark you strike from a cigarette lighter is lanthanoid chemistry at work: the “flint” is mischmetall, and the shavings scraped off by the steel wheel are so reactive that they ignite in air on the spot. The same family also supplies the strongest permanent magnets known — the neodymium–iron–boron magnets inside earphones, hard-disk drives, electric-vehicle motors and wind turbines. Because these metals are scarce, awkward to extract and hard to purify, recovering them from discarded electronics is now an active area of green chemistry.

The word mischmetall is simply German for “mixed metal” — an honest name, since it is what you get when the lanthanoids are reduced together without going to the trouble of separating them. For most alloying purposes, the mixture works as well as the pure metals, and it costs a fraction as much.

4.6 The Actinoids

The actinoids include the fourteen elements from Th to Lr. The names, symbols and some properties of these elements are given in Table 4.10.

The actinoids are radioactive elements, and the earlier members have relatively long half-lives. The later ones have half-life values ranging from a day down to 3 minutes for lawrencium (\(Z = 103\)). The later members could be prepared only in nanogram quantities. These facts render their study more difficult: an element that vanishes while you are weighing it, and which must be handled behind shielding, does not lend itself to ordinary laboratory chemistry.

Table 4.10: Some Properties of Actinium and Actinoids

(Only electrons outside the [Rn] core are indicated.)

Atomic Number Name Symbol M M³⁺ M⁴⁺ Radius M³⁺ /pm Radius M⁴⁺ /pm
89 Actinium Ac \(6d^17s^2\) \(5f^{\,0}\) 111
90 Thorium Th \(6d^27s^2\) \(5f^{\,1}\) \(5f^{\,0}\) 99
91 Protactinium Pa \(5f^{\,2}6d^17s^2\) \(5f^{\,2}\) \(5f^{\,1}\) 96
92 Uranium U \(5f^{\,3}6d^17s^2\) \(5f^{\,3}\) \(5f^{\,2}\) 103 93
93 Neptunium Np \(5f^{\,4}6d^17s^2\) \(5f^{\,4}\) \(5f^{\,3}\) 101 92
94 Plutonium Pu \(5f^{\,6}7s^2\) \(5f^{\,5}\) \(5f^{\,4}\) 100 90
95 Americium Am \(5f^{\,7}7s^2\) \(5f^{\,6}\) \(5f^{\,5}\) 99 89
96 Curium Cm \(5f^{\,7}6d^17s^2\) \(5f^{\,7}\) \(5f^{\,6}\) 99 88
97 Berkelium Bk \(5f^{\,9}7s^2\) \(5f^{\,8}\) \(5f^{\,7}\) 98 87
98 Californium Cf \(5f^{\,10}7s^2\) \(5f^{\,9}\) \(5f^{\,8}\) 98 86
99 Einsteinium Es \(5f^{\,11}7s^2\) \(5f^{\,10}\) \(5f^{\,9}\) – –
100 Fermium Fm \(5f^{\,12}7s^2\) \(5f^{\,11}\) \(5f^{\,10}\) – –
101 Mendelevium Md \(5f^{\,13}7s^2\) \(5f^{\,12}\) \(5f^{\,11}\) – –
102 Nobelium No \(5f^{\,14}7s^2\) \(5f^{\,13}\) \(5f^{\,12}\) – –
103 Lawrencium Lr \(5f^{\,14}6d^17s^2\) \(5f^{\,14}\) \(5f^{\,13}\) – –

The naming of this series reads like a map of the outer solar system followed by a roll-call of physicists. Uranium was named after the planet Uranus; the two elements discovered just beyond it were accordingly named neptunium and plutonium, after the next planets outward. Later members honour people and places: einsteinium, fermium, mendelevium, berkelium, californium.

4.6.1 Electronic Configurations

All the actinoids are believed to have the electronic configuration of \(7s^2\) and variable occupancy of the 5f and 6d subshells. The fourteen electrons are formally added to 5f, though not in thorium (\(Z = 90\)); but from Pa onwards, the 5f orbitals fill until they are complete at element 103.

The irregularities in the electronic configurations of the actinoids, like those in the lanthanoids, are related to the stabilities of the \(f^{\,0}\), \(f^{\,7}\) and \(f^{\,14}\) occupancies of the 5f orbitals. Thus, the configurations of Am and Cm are \([\mathrm{Rn}]\,5f^{\,7}7s^2\) and \([\mathrm{Rn}]\,5f^{\,7}6d^17s^2\) — in curium, an extra electron prefers to sit alone in a 6d orbital rather than disturb the half-filled 5f set.

Although the 5f orbitals resemble the 4f orbitals in the angular part of their wave-function, they are not as buried as the 4f orbitals. Hence 5f electrons can participate in bonding to a far greater extent. This single structural difference is the key to almost everything that separates the actinoids from the lanthanoids: electrons that can take part in bonding can also be lost in many different numbers, which is why the actinoids show so wide a range of oxidation states.

4.6.2 Ionic Sizes

The general trend seen in the lanthanoids is observable in the actinoids as well. There is a gradual decrease in the size of the atoms or \(\mathrm{M^{3+}}\) ions across the series. This may be referred to as the actinoid contraction, like the lanthanoid contraction.

The contraction is, however, greater from element to element in this series, resulting from the poor shielding by 5f electrons.

NoteDefinition

Actinoid contraction is the steady decrease in the size of the atoms and \(\mathrm{M^{3+}}\) ions along the actinoid series as the 5f orbitals fill; because 5f electrons shield the nuclear charge poorly, the contraction from one element to the next is greater than in the lanthanoids.

4.6.3 Oxidation States

There is a greater range of oxidation states in the actinoids, which is in part attributed to the fact that the 5f, 6d and 7s levels are of comparable energies. When three sets of orbitals lie so close together, electrons can be drawn from any of them, and many different numbers of electrons can be removed for a similar energy cost. The known oxidation states of the actinoids are listed in Table 4.11.

Table 4.11: Oxidation States of Actinium and Actinoids

Ac Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr
3 3 3 3 3 3 3 3 3 3 3 3 3 3
4 4 4 4 4 4 4 4
5 5 5 5 5
6 6 6 6
7 7

The actinoids show in general the +3 oxidation state. The elements in the first half of the series frequently exhibit higher oxidation states. For example, the maximum oxidation state increases from +4 in Th to +5, +6 and +7 respectively in Pa, U and Np, but decreases in the succeeding elements (Table 4.11).

The actinoids resemble the lanthanoids in having more compounds in the +3 state than in the +4 state. However, +3 and +4 ions tend to hydrolyse — that is, they are so small and highly charged that they pull apart the water molecules around them, giving acidic solutions and often precipitating hydroxides.

Because the distribution of oxidation states among the actinoids is so uneven, and so different for the former and the later elements, it is unsatisfactory to review their chemistry in terms of oxidation states.

Using Table 4.11, state the maximum oxidation state exhibited by Th, Pa, U and Np, and account for the fact that the maximum rises to +7 and then falls again in the later actinoids.

From Table 4.11, the maximum oxidation states are:

Th: +4 Pa: +5 U: +6 Np: +7

Each of these equals the total number of electrons available outside the radon core — for uranium, \(5f^{\,3}6d^17s^2\) gives \(3 + 1 + 2 = 6\), so +6 is its maximum; for neptunium, \(5f^{\,4}6d^17s^2\) gives \(4 + 1 + 2 = 7\).

These high states are possible only because the 5f, 6d and 7s levels lie at comparable energies in the early actinoids, so all these electrons can be lost or shared.

Beyond neptunium, as the nuclear charge grows, the 5f electrons are pulled in and held more tightly. They are no longer freely available for bonding, so the maximum oxidation state falls, and the later actinoids settle into the +3 state, like the lanthanoids.

(question only — no solution) Determine the oxidation number of uranium in the uranyl ion \(\mathrm{UO_2^{2+}}\) and in \(\mathrm{UF_6}\), and state whether either value is the maximum oxidation state of uranium.

4.6.4 General Characteristics and Comparison with Lanthanoids

The actinoid metals are all silvery in appearance, but display a variety of structures. The structural variability is obtained due to irregularities in metallic radii, which are far greater than in the lanthanoids.

Chemical Reactivity

The actinoids are highly reactive metals, especially when finely divided. The action of boiling water on them, for example, gives a mixture of oxide and hydride, and combination with most non-metals takes place at moderate temperatures.

Hydrochloric acid attacks all these metals, but most are only slightly affected by nitric acid, owing to the formation of protective oxide layers — the metal makes its own coat and hides behind it. Alkalies have no action on them.

Magnetic Properties

The magnetic properties of the actinoids are more complex than those of the lanthanoids. Although the variation in the magnetic susceptibility of the actinoids with the number of unpaired 5f electrons is roughly parallel to the corresponding results for the lanthanoids, the latter have higher values.

Ionisation Enthalpies

It is evident from the behaviour of the actinoids that the ionisation enthalpies of the early actinoids, though not accurately known, are lower than for the early lanthanoids.

This is quite reasonable. It is to be expected that when 5f orbitals are beginning to be occupied, they will penetrate less into the inner core of electrons. The 5f electrons will therefore be more effectively shielded from the nuclear charge than the 4f electrons of the corresponding lanthanoids. Because the outer electrons are less firmly held, they are more easily removed — and they are available for bonding in the actinoids.

Q. The 5f electrons are better shielded from the nucleus, yet the actinoid contraction is greater than the lanthanoid contraction. Is that not a contradiction?

A. No — two different shieldings are involved. The 5f electrons are well shielded by the inner core from the nucleus, so they are loosely held and easy to remove, giving low ionisation enthalpies. But a 5f electron is very poor at shielding other electrons from the nucleus. So as each new 5f electron is added, it hardly offsets the extra nuclear charge, and the contraction from one element to the next is larger than in the lanthanoids.

How the Two Series Compare

A comparison of the actinoids with the lanthanoids, with respect to the different characteristics discussed above, reveals that behaviour similar to that of the lanthanoids is not evident until the second half of the actinoid series. However, even the early actinoids resemble the lanthanoids in showing close similarities with each other, and in gradual variation in properties which do not entail a change in oxidation state.

The lanthanoid and actinoid contractions have extended effects on the sizes, and therefore on the properties, of the elements succeeding them in their respective periods. The lanthanoid contraction is more important, because the chemistry of the elements succeeding the actinoids is much less known at the present time.

lanthanoids vs actinoids

Feature Lanthanoids Actinoids
Orbitals being filled 4f 5f
Common configuration \(4f^{\,n}\,5d^{0-1}\,6s^2\) \(5f^{\,n}\,6d^{0-1}\,7s^2\)
Principal oxidation state +3; occasionally +2 and +4 +3; but +4, +5, +6 and +7 are common in the early members
Reason for range of states 4f electrons buried and unavailable 5f, 6d and 7s levels of comparable energy; 5f electrons available for bonding
Contraction along the series lanthanoid contraction actinoid contraction — greater from element to element (poorer shielding by 5f)
Radioactivity only promethium is radioactive all members are radioactive
Magnetic behaviour simpler; higher susceptibility values more complex; lower values
Ionisation enthalpies of early members higher lower
Action of acids/alkalies react with dilute acids attacked by HCl; only slightly attacked by \(\mathrm{HNO_3}\) (protective oxide layer); no action of alkalies

The actinoids are the elements of the nuclear age. Uranium and plutonium are the fuels of nuclear reactors, where the energy released by fission is used to raise steam and generate electricity; thorium, abundant in the monazite sands of India’s coastline, is being developed as a fuel of the future. Because every one of these elements is radioactive, they are handled remotely, behind shielding, in sealed enclosures — the hazard here is not chemical reactivity but ionising radiation, and the waste they leave behind must be isolated for a very long time.

Name a member of the lanthanoid series which is well known to exhibit the +4 oxidation state.

Cerium (\(Z = 58\)).

4.10 Actinoid contraction is greater from element to element than lanthanoid contraction. Why?

4.7 Some Applications of d- and f-Block Elements

Metals of Construction and of the Battery

Iron and steels are the most important construction materials we have. Their production is based on three operations: the reduction of iron oxides, the removal of impurities, and the addition of carbon and of alloying metals such as Cr, Mn and Ni. Carbon hardens the iron; chromium resists corrosion; manganese and nickel toughen the steel — each addition is made for a reason.

Some compounds are manufactured for special purposes, such as TiO for the pigment industry and \(\mathrm{MnO_2}\) for use in dry battery cells. The battery industry also requires Zn and Ni/Cd. Zinc forms the case and the anode of the ordinary dry cell, while nickel–cadmium cells were long the standard rechargeable battery.

The Coinage Metals

The elements of Group 11 — copper, silver and gold — are still worthy of being called the coinage metals, although Ag and Au are now restricted to collection items. Contemporary UK ‘copper’ coins are copper-coated steel, and the ‘silver’ UK coins are a Cu/Ni alloy. The reason these three metals were chosen for coins in the first place is the one we met in Section 4.3.8: they resist oxidation, so a coin survives decades of handling without corroding away.

Catalysts of the Chemical Industry

Many of the metals and/or their compounds are essential catalysts in the chemical industry. Recall from Section 4.3.12 why they are so good at the job: a transition metal can slip between oxidation states, and its surface atoms have partly filled d orbitals with which to grip and weaken the bonds of the molecules that land on them, as sketched in Fig. 4.13.

Fig. 4.13

Fig. 4.13 – Heterogeneous catalysis on a transition metal surface.

\(\mathrm{V_2O_5}\) catalyses the oxidation of \(\mathrm{SO_2}\) in the manufacture of sulphuric acid. \(\mathrm{TiCl_4}\) with \(\mathrm{Al(CH_3)_3}\) forms the basis of the Ziegler catalysts used to manufacture polyethylene (polythene).

Iron catalysts are used in the Haber process for the production of ammonia from \(\mathrm{N_2}/\mathrm{H_2}\) mixtures. Nickel catalysts enable the hydrogenation of fats to proceed — this is the reaction that converts liquid vegetable oils into the semi-solid fats used in cooking.

In the Wacker process, the oxidation of ethyne to ethanal is catalysed by \(\mathrm{PdCl_2}\). Nickel complexes are useful in the polymerisation of alkynes and other organic compounds such as benzene.

Finally, the photographic industry relies on the special light-sensitive properties of AgBr, which darkens where light falls on it — the basis of the photographic film.

some industrial catalysts from the d-block and what they do

Catalyst Process it drives What is produced
\(\mathrm{V_2O_5}\) Contact process — oxidation of \(\mathrm{SO_2}\) sulphuric acid
Finely divided iron Haber process — \(\mathrm{N_2} + \mathrm{H_2}\) ammonia
Nickel catalytic hydrogenation hardened fats from oils
\(\mathrm{TiCl_4}\) with \(\mathrm{Al(CH_3)_3}\) (Ziegler catalysts) polymerisation polyethylene (polythene)
\(\mathrm{PdCl_2}\) Wacker process ethanal
Nickel complexes polymerisation of alkynes and other organic compounds polymers, benzene

Vanadium(V) oxide, \(\mathrm{V_2O_5}\), catalyses the oxidation of \(\mathrm{SO_2}\) to \(\mathrm{SO_3}\). Explain, in terms of the oxidation states of vanadium, how a transition metal oxide is able to act as a catalyst here.

Vanadium in \(\mathrm{V_2O_5}\) is in the +5 state. A catalyst of this type works by taking part in the reaction and then being regenerated.

The oxide first hands over oxygen to sulphur dioxide, oxidising it to sulphur trioxide; in doing so, vanadium is itself reduced from +5 to a lower oxidation state (+4):

\[\mathrm{SO_2} + \mathrm{V_2O_5} \rightarrow \mathrm{SO_3} + \mathrm{V_2O_4}\]

The reduced oxide is then re-oxidised by the oxygen of the air, returning vanadium to +5:

\[2\,\mathrm{V_2O_4} + \mathrm{O_2} \rightarrow 2\,\mathrm{V_2O_5}\]

The catalyst has come back unchanged, and the net reaction is simply the oxidation of \(\mathrm{SO_2}\) by \(\mathrm{O_2}\). The whole cycle is possible only because vanadium has two oxidation states, +5 and +4, that lie close in energy — the defining property of a transition element. A main-group oxide with a single fixed oxidation state could not do this.

Two of these catalysts quietly feed the world. The iron catalyst of the Haber process fixes atmospheric nitrogen into ammonia, from which almost all nitrogen fertilisers are made; without it, a large part of humanity could not be fed. The nickel catalyst that hardens vegetable oils gave us margarine and vanaspati. But industry has also had to learn caution: partial hydrogenation over nickel produces trans fats, now known to be harmful to health, and manufacturers have had to redesign the process to avoid them. A catalyst decides not only how fast a reaction goes, but what exactly comes out of it.

The Ziegler catalysts are named after Karl Ziegler, who in the 1950s found that a mixture of a titanium halide with an aluminium alkyl would knit ethene molecules into long polythene chains at ordinary pressure — where the older process had needed extreme pressures. Working with Giulio Natta, whose name is joined to his in the term Ziegler–Natta catalysis, he opened the way to the cheap plastics that surround us today. Both shared the Nobel Prize in Chemistry for the work.

(question only — no solution) Name the d-block catalyst used in each of the following, and state the oxidation state of the metal in the catalyst where a definite compound is named: (i) the manufacture of sulphuric acid, (ii) the manufacture of ammonia, (iii) the hardening of vegetable oils, (iv) the manufacture of polythene.

4.8 Summary

The Two Blocks
  1. The d-block, consisting of Groups 3–12, occupies the large middle section of the periodic table. In these elements the inner d orbitals are progressively filled.
  2. The f-block is placed outside, at the bottom of the periodic table. In the elements of this block, the 4f and 5f orbitals are progressively filled.
  3. Corresponding to the filling of the 3d, 4d and 5d orbitals, three series of transition elements are well recognised.
General Characteristics of the Transition Elements
  1. All the transition elements exhibit typical metallic properties: high tensile strength, ductility, malleability, thermal and electrical conductivity, and metallic character.
  2. Their melting and boiling points are high. This is attributed to the involvement of the \((n-1)d\) electrons, in addition to the \(ns\) electrons, resulting in strong interatomic bonding.
  3. In many of these properties, the maxima occur at about the middle of each series. This indicates that one unpaired electron per d orbital is a particularly favourable configuration for strong interatomic interaction.
  4. Successive ionisation enthalpies do not increase as steeply as in the main group elements with increasing atomic number. Hence the loss of a variable number of electrons from the \((n-1)d\) orbitals is not energetically unfavourable.
  5. The involvement of the \((n-1)d\) electrons gives these elements their distinct character. Thus, in addition to variable oxidation states, they exhibit paramagnetic behaviour, catalytic properties, and a tendency to form coloured ions, interstitial compounds and complexes.
  6. The magnetic moment of a first-row transition metal ion is obtained from the spin-only formula, where \(n\) is the number of unpaired electrons:

\[\mu = \sqrt{n(n+2)}\ \text{BM}\]

Chemical Behaviour
  1. The transition elements vary widely in their chemical behaviour. Many of them are sufficiently electropositive to dissolve in mineral acids, although a few are ‘noble’. Of the first series, with the exception of copper, all the metals are relatively reactive.
  2. The transition metals react with a number of non-metals — oxygen, nitrogen, sulphur and the halogens — to form binary compounds.
  3. The first series transition metal oxides are generally formed from the reaction of the metals with oxygen at high temperatures. These oxides dissolve in acids and bases to form oxometallic salts.
Two Important Compounds
  1. Potassium dichromate (\(\mathrm{K_2Cr_2O_7}\)) is prepared from the chromite ore by fusion with alkali in the presence of air, and acidifying the extract.
  2. Potassium permanganate (\(\mathrm{KMnO_4}\)) is prepared from pyrolusite ore (\(\mathrm{MnO_2}\)).
  3. Both the dichromate and the permanganate ions are strong oxidising agents.

Flowchart

FC 4.5 – The two preparations side by side. (a) K₂Cr₂O₇: chromite ore FeCr₂O₄ → fusion with Na₂CO₃ in air → Na₂CrO₄ (yellow) → acidify with H₂SO₄ → Na₂Cr₂O₇ (orange) → add KCl → K₂Cr₂O₇. (b) KMnO₄: pyrolusite MnO₂ → alkaline oxidative fusion with KOH + air/KNO₃ → K₂MnO₄ (green, Mn +6) → electrolytic oxidation in alkaline solution (or disproportionation in acid) → KMnO₄ (purple, Mn +7).

The Inner Transition Elements
  1. The two series of inner transition elements, the lanthanoids and the actinoids, constitute the f-block of the periodic table.
  2. With the successive filling of the inner 4f orbitals, there is a gradual decrease in the atomic and ionic sizes of these metals along the series — the lanthanoid contraction. This has far-reaching consequences in the chemistry of the elements succeeding them, notably the close similarity in size between the second and third transition series.
  3. Lanthanum and all the lanthanoids are rather soft white metals. They react easily with water to give solutions containing +3 ions.
  4. The principal oxidation state of the lanthanoids is +3, although +4 and +2 oxidation states are also exhibited by some of them occasionally.
  5. The chemistry of the actinoids is more complex, in view of their ability to exist in different oxidation states. Furthermore, many of the actinoid elements are radioactive, which makes the study of these elements rather difficult.
Applications
  1. There are many useful applications of the d- and f-block elements and their compounds. Notable among them are their use in the varieties of steels, as catalysts, in complexes and in organic syntheses.

4.9 NCERT Questions

  1. Write down the electronic configuration of:
  1. \(\mathrm{Cr^{3+}}\) (ii) \(\mathrm{Pm^{3+}}\) (iii) \(\mathrm{Cu^{+}}\) (iv) \(\mathrm{Ce^{4+}}\) (v) \(\mathrm{Co^{2+}}\) (vi) \(\mathrm{Lu^{2+}}\) (vii) \(\mathrm{Mn^{2+}}\) (viii) \(\mathrm{Th^{4+}}\)

  1. Why are \(\mathrm{Mn^{2+}}\) compounds more stable than \(\mathrm{Fe^{2+}}\) towards oxidation to their +3 state?

  2. Explain briefly how the +2 state becomes more and more stable in the first half of the first row transition elements with increasing atomic number.

  3. To what extent do the electronic configurations decide the stability of oxidation states in the first series of the transition elements? Illustrate your answer with examples.

  4. What may be the stable oxidation state of the transition element with the following d electron configurations in the ground state of their atoms: \(3d^3\), \(3d^5\), \(3d^8\) and \(3d^4\)?

  5. Name the oxometal anions of the first series of the transition metals in which the metal exhibits the oxidation state equal to its group number.

  6. What is lanthanoid contraction? What are the consequences of lanthanoid contraction?

  7. What are the characteristics of the transition elements and why are they called transition elements? Which of the d-block elements may not be regarded as the transition elements?

  8. In what way is the electronic configuration of the transition elements different from that of the non-transition elements?

  9. What are the different oxidation states exhibited by the lanthanoids?

  10. Explain giving reasons:

    1. Transition metals and many of their compounds show paramagnetic behaviour.
    2. The enthalpies of atomisation of the transition metals are high.
    3. The transition metals generally form coloured compounds.
    4. Transition metals and many of their compounds act as good catalysts.

  11. What are interstitial compounds? Why are such compounds well known for transition metals?

  12. How is the variability in oxidation states of transition metals different from that of the non-transition metals? Illustrate with examples.

  13. Describe the preparation of potassium dichromate from iron chromite ore. What is the effect of increasing pH on a solution of potassium dichromate?

  14. Describe the oxidising action of potassium dichromate and write the ionic equations for its reaction with:

  1. iodide (ii) iron(II) solution and (iii) \(\mathrm{H_2S}\)

  1. Describe the preparation of potassium permanganate. How does the acidified permanganate solution react with (i) iron(II) ions (ii) \(\mathrm{SO_2}\) and (iii) oxalic acid? Write the ionic equations for the reactions.

  2. For \(\mathrm{M^{2+}}/\mathrm{M}\) and \(\mathrm{M^{3+}}/\mathrm{M^{2+}}\) systems, the \(E^\circ\) values for some metals are as follows:

Couple \(E^\circ\) Couple \(E^\circ\)
\(\mathrm{Cr^{2+}}/\mathrm{Cr}\) –0.9 V \(\mathrm{Cr^{3+}}/\mathrm{Cr^{2+}}\) –0.4 V
\(\mathrm{Mn^{2+}}/\mathrm{Mn}\) –1.2 V \(\mathrm{Mn^{3+}}/\mathrm{Mn^{2+}}\) +1.5 V
\(\mathrm{Fe^{2+}}/\mathrm{Fe}\) –0.4 V \(\mathrm{Fe^{3+}}/\mathrm{Fe^{2+}}\) +0.8 V

Use this data to comment upon: a. the stability of \(\mathrm{Fe^{3+}}\) in acid solution as compared to that of \(\mathrm{Cr^{3+}}\) or \(\mathrm{Mn^{3+}}\)​, and b. the ease with which iron can be oxidised as compared to a similar process for either chromium or manganese metal.

  1. Predict which of the following will be coloured in aqueous solution: \(\mathrm{Ti^{3+}}\)​, \(\mathrm{V^{3+}}\)​, \(\mathrm{Cu^{+}}\)​, \(\mathrm{Sc^{3+}}\)​, \(\mathrm{Mn^{2+}}\)​, \(\mathrm{Fe^{3+}}\) and \(\mathrm{Co^{2+}}\). Give reasons for each.

  2. Compare the stability of the +2 oxidation state for the elements of the first transition series.

  3. Compare the chemistry of the actinoids with that of the lanthanoids with special reference to:

  1. electronic configuration (ii) atomic and ionic sizes (iii) oxidation state (iv) chemical reactivity.

  1. How would you account for the following:

    1. Of the \(d^4\) species, \(\mathrm{Cr^{2+}}\) is strongly reducing while manganese(III) is strongly oxidising.
    2. Cobalt(II) is stable in aqueous solution, but in the presence of complexing reagents it is easily oxidised.
    3. The \(d^1\) configuration is very unstable in ions.

  2. What is meant by ‘disproportionation’? Give two examples of disproportionation reactions in aqueous solution.

  3. Which metal in the first series of transition metals exhibits the +1 oxidation state most frequently, and why?

  4. Calculate the number of unpaired electrons in the following gaseous ions: \(\mathrm{Mn^{3+}}\)​, \(\mathrm{Cr^{3+}}\)​, \(\mathrm{V^{3+}}\) and \(\mathrm{Ti^{3+}}\)​. Which one of these is the most stable in aqueous solution?

  5. Give examples and suggest reasons for the following features of transition metal chemistry:

    1. The lowest oxide of a transition metal is basic, the highest is amphoteric/acidic.
    2. A transition metal exhibits its highest oxidation state in oxides and fluorides.
    3. The highest oxidation state is exhibited in the oxoanions of a metal.

  6. Indicate the steps in the preparation of:

  1. \(\mathrm{K_2Cr_2O_7}\) from chromite ore. (ii) \(\mathrm{KMnO_4}\) from pyrolusite ore.

  1. What are alloys? Name an important alloy which contains some of the lanthanoid metals. Mention its uses.

  2. What are inner transition elements? Decide which of the following atomic numbers are the atomic numbers of the inner transition elements: 29, 59, 74, 95, 102, 104.

  3. The chemistry of the actinoid elements is not so smooth as that of the lanthanoids. Justify this statement by giving some examples from the oxidation states of these elements.

  4. Which is the last element in the series of the actinoids? Write the electronic configuration of this element. Comment on the possible oxidation state of this element.

  5. Use Hund’s rule to derive the electronic configuration of the \(\mathrm{Ce^{3+}}\) ion, and calculate its magnetic moment on the basis of the ‘spin-only’ formula.

  6. Name the members of the lanthanoid series which exhibit the +4 oxidation state and those which exhibit the +2 oxidation state. Try to correlate this type of behaviour with the electronic configurations of these elements.

  7. Compare the chemistry of the actinoids with that of the lanthanoids with reference to:

  1. electronic configuration (ii) oxidation states and (iii) chemical reactivity.

  1. Write the electronic configurations of the elements with the atomic numbers 61, 91, 101 and 109.

  2. Compare the general characteristics of the first series of the transition metals with those of the second and third series metals in the respective vertical columns. Give special emphasis on the following points:

  1. electronic configurations (ii) oxidation states (iii) ionisation enthalpies and (iv) atomic sizes.

  1. Write down the number of 3d electrons in each of the following ions: \(\mathrm{Ti^{2+}}\)​, \(\mathrm{V^{2+}}\)​, \(\mathrm{Cr^{3+}}\)​, \(\mathrm{Mn^{2+}}\)​, \(\mathrm{Fe^{2+}}\)​, \(\mathrm{Fe^{3+}}\)​, \(\mathrm{Co^{2+}}\)​, \(\mathrm{Ni^{2+}}\) and \(\mathrm{Cu^{2+}}\)​. Indicate how you would expect the five 3d orbitals to be occupied for these hydrated ions (octahedral).

  2. Comment on the statement that the elements of the first transition series possess many properties different from those of the heavier transition elements.

  3. What can be inferred from the magnetic moment values of the following complex species?

Example Magnetic Moment (BM)
\(\mathrm{K_4[Mn(CN)_6]}\) 2.2
\(\mathrm{[Fe(H_2O)_6]^{2+}}\) 5.3
\(\mathrm{K_2[MnCl_4]}\) 5.9

4.10 Check Your Concepts

  1. Zinc, cadmium and mercury are placed in the d-block, yet they are not counted as transition metals. State the IUPAC definition of a transition metal and use it to explain this exclusion. Why, then, is zinc still discussed alongside the 3d series?

  2. Assertion: Chromium has the ground-state configuration \(3d^54s^1\) and not \(3d^44s^2\). Reason: The 3d and 4s orbitals are very close in energy, and a half-filled d set has extra stability. Decide whether the assertion and the reason are each correct, and whether the reason correctly explains the assertion.

  3. Arrange the following in increasing order of the property named, and give a reason for the order in each case:

    1. \(\mathrm{Sc}\), \(\mathrm{Ti}\), \(\mathrm{Cr}\), \(\mathrm{Zn}\) — enthalpy of atomisation
    2. \(\mathrm{VO_2^{+}}\), \(\mathrm{Cr_2O_7^{2-}}\), \(\mathrm{MnO_4^{-}}\) — oxidising power
    3. \(\mathrm{Ti}\), \(\mathrm{Cu}\), \(\mathrm{Fe}\) — density

  4. “Two tetrahedra sharing one corner.” Identify the ion described, name the ion it turns into when alkali is added, and explain why the oxidation state of the metal does not change even though the colour of the solution does.

  5. Why does the melting point of the 3d series rise to a maximum near the middle of the row and then fall? Account for the anomalously low value of manganese in this trend.

  6. Explain why permanganate titrations are carried out with dilute sulphuric acid and not with hydrochloric acid. Write the equation for the unwanted reaction that would take place with HCl.

  7. \(\mathrm{Cu^{+}}\) has a \(3d^{10}\) configuration, which should be stable, yet copper(I) salts disproportionate in water. Write the equation for the disproportionation and explain, in terms of hydration enthalpy, why \(\mathrm{Cu^{2+}}\)(aq) wins.

  8. Complete and balance the following ionic equations, and in each case state the oxidation state of the metal in the product:

    1. \(\mathrm{Cr_2O_7^{2-}} + \mathrm{H^{+}} + \mathrm{Sn^{2+}} \rightarrow\)
    2. \(\mathrm{MnO_4^{-}} + \mathrm{H^{+}} + \mathrm{NO_2^{-}} \rightarrow\)
    3. \(\mathrm{MnO_4^{-}} + \mathrm{H_2O} + \mathrm{I^{-}} \rightarrow\)

  9. A student is given three aqueous solutions labelled X, Y and Z. X is colourless, Y is pink and Z is blue. All three contain first-row transition metal ions in the +2 state. Using Table 4.8, suggest a possible identity for each ion and justify your choice on the basis of the d-electron count.

  10. Assertion: The second and third transition series have almost the same atomic radii. Reason: The lanthanoid contraction cancels the expected increase in size on going from the second to the third series. Decide whether the assertion and the reason are each correct, and whether the reason correctly explains the assertion. Give one chemical consequence of this fact.

  11. Why is the third ionisation enthalpy of manganese much higher than that of iron, even though iron has the larger nuclear charge? Use your answer to explain why \(\mathrm{Mn^{3+}}\) is a strong oxidising agent while \(\mathrm{Fe^{3+}}\) is not.

  12. Match the following:

Column I Column II
a. \(\mathrm{VH_{0.56}}\) i. Primary standard in volumetric analysis
b. Mischmetall ii. Interstitial compound
c. \(\mathrm{K_2Cr_2O_7}\) iii. Ziegler catalyst component
d. \(\mathrm{TiCl_4}\) iv. Lighter flint

  1. Explain each of the following observations:

    1. The highest fluoride of manganese is \(\mathrm{MnF_4}\), but its highest oxide is \(\mathrm{Mn_2O_7}\).
    2. \(\mathrm{Cr(VI)}\) in acidic medium is a strong oxidising agent, but \(\mathrm{MoO_3}\) and \(\mathrm{WO_3}\) are not.
    3. The oxidation state of nickel in \(\mathrm{Ni(CO)_4}\) is zero.

  2. Interstitial compounds such as TiC are harder and higher-melting than the pure metal, yet they still conduct electricity. Explain both facts in terms of what happens inside the metal lattice.

  3. Ce(IV) has an \(E^\circ\) value of +1.74 V for the \(\mathrm{Ce^{4+}}/\mathrm{Ce^{3+}}\) couple, which suggests it should oxidise water — yet it is used as an analytical reagent in aqueous solution. Resolve this apparent contradiction.

  4. Why are the ionisation enthalpies of the early actinoids lower than those of the early lanthanoids, while the actinoid contraction is nevertheless greater than the lanthanoid contraction? Distinguish carefully between the two kinds of shielding involved.

  5. State, with reasons, whether each of the following ions is coloured or colourless in aqueous solution: \(\mathrm{Zn^{2+}}\)​, \(\mathrm{Cu^{2+}}\)​, \(\mathrm{Ti^{4+}}\)​, \(\mathrm{V^{3+}}\)​, \(\mathrm{La^{3+}}\)​, \(\mathrm{Nd^{3+}}\)​.

  6. Iron(III) catalyses the reaction between iodide and persulphate ions. Write the two steps of the catalytic cycle, show that they add up to the overall reaction, and explain why the uncatalysed reaction is slow.

  7. Titanium and vanadium are strongly electropositive, yet both are passive to dilute non-oxidising acids at room temperature. Explain this apparent contradiction between the \(E^\circ\) values and the observed behaviour.

  8. Predict the product formed when acidified \(\mathrm{KMnO_4}\) solution reacts with each of the following, and state the colour change you would observe: (i) \(\mathrm{FeSO_4}\) solution, (ii) oxalic acid at 333 K, (iii) \(\mathrm{H_2S}\).

4.11 Practice With Numericals

  1. Calculate the spin-only magnetic moment of each of the following ions and identify which one would be attracted most strongly by a magnetic field: \(\mathrm{Ti^{3+}}\)​, \(\mathrm{Cr^{3+}}\)​, \(\mathrm{Fe^{3+}}\)​, \(\mathrm{Ni^{2+}}\)​.

  2. The spin-only magnetic moment of a divalent ion of a 3d metal is 4.90 BM. Determine the number of unpaired electrons, write the electronic configuration of the ion, and identify it.

  3. Determine the oxidation number of the transition metal in each of the following: \(\mathrm{K_2Cr_2O_7}\), \(\mathrm{KMnO_4}\), \(\mathrm{K_2MnO_4}\), \(\mathrm{FeCr_2O_4}\), \(\mathrm{VO_2^{+}}\) and \(\mathrm{[Fe(CN)_6]^{4-}}\).

  4. Calculate the mass of potassium dichromate required to oxidise 7.6 g of iron(II) sulphate in acidic solution. (Molar masses: \(\mathrm{K_2Cr_2O_7} = 294\ \text{g mol}^{-1}\); \(\mathrm{FeSO_4} = 152\ \text{g mol}^{-1}\).)

  5. In an acidic medium, \(25.0\ \mathrm{cm^3}\) of an iron(II) solution required \(20.0\ \mathrm{cm^3}\) of \(0.02\ \mathrm{M}\ \mathrm{KMnO_4}\) for complete oxidation. Write the balanced ionic equation and calculate the molarity of the iron(II) solution.

  6. Balance the following equation by the ion-electron (half-reaction) method in acidic medium, and state how many electrons are transferred per mole of dichromate: \[\mathrm{Cr_2O_7^{2-}} + \mathrm{I^{-}} + \mathrm{H^{+}} \rightarrow \mathrm{Cr^{3+}} + \mathrm{I_2} + \mathrm{H_2O}\]

  7. Calculate the percentage of chromium by mass in potassium dichromate, \(\mathrm{K_2Cr_2O_7}\). (Atomic masses: K = 39, Cr = 52, O = 16.)

  8. The solubility of potassium permanganate is 6.4 g per 100 g of water at 293 K. Calculate the mass of water required to dissolve 16.0 g of \(\mathrm{KMnO_4}\) completely at this temperature.

  9. Write the electronic configuration of \(\mathrm{Eu^{3+}}\) and of \(\mathrm{Yb^{2+}}\)​, and calculate the spin-only magnetic moment of each. State which of the two is diamagnetic.

  10. Using the data of Table 4.4, calculate the total energy required to convert one mole of gaseous chromium atoms into gaseous \(\mathrm{Cr^{2+}}\) ions, and compare it with the corresponding value for manganese. Comment on how this difference is reflected in their \(E^\circ\)(M²⁺/M) values.