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Chapter 8: Transition and Inner Transition Elements

In the early 1940s the periodic table had an awkward gap in its logic. Uranium was then the heaviest element known in nature, and chemists had quietly slotted thorium, protactinium and uranium into the transition-metal rows — thorium below hafnium, uranium below tungsten — as though they were simply very heavy d-block metals. The arrangement looked tidy, but the chemistry of the newly made elements neptunium and plutonium refused to fit it.

The American chemist Glenn Seaborg suggested something far bolder. What if these heavy elements were not d-block metals at all, but the beginning of a second inner transition series, one in which electrons were entering 5f orbitals just as they enter 4f orbitals in the lanthanoids? That meant pulling a whole set of elements out of the main body of the table and placing them in a new row underneath it.

Senior chemists warned him not to publish. Rearranging the established periodic table, they said, would damage his standing as a scientist. Seaborg later remarked, with some humour, that he had no scientific reputation to lose — so he published anyway.

He was right. The actinoid concept became the framework that guided the discovery of element after element, and decades later element 106 was named seaborgium while he was still alive to see it.

Diagram

Fig. 8.0 – A periodic table outline with the d-block shaded in one colour and the two f-block rows (lanthanoids and actinoids) shaded in another and shown pulled out below the main table, with an arrow indicating their true position after La and Ac.

NoteCuriosity Corner

Q1. Manganese shows six different oxidation states in its compounds, while calcium — a metal in the very same period — shows only one. What is it about d-block atoms that allows such variety? Q2. A solution of cobalt chloride in water is pink, but on adding concentrated hydrochloric acid it turns deep blue. The metal ion is the same in both. So what else must be controlling the colour? Q3. Iron ore is reduced in the blast furnace using coke, yet limestone — which contains no iron at all — is also charged into the furnace. What job could it possibly be doing? Q4. Lanthanoid ions get steadily smaller from lanthanum to lutetium, even though an electron is being added at every step. How can adding electrons make an ion shrink? Q5. Actinoids show oxidation states ranging from +2 right up to +8, while lanthanoids stay almost entirely at +3. Why should two f-block series that look so similar behave so differently?

You will not be able to answer these questions yet. As you work through this chapter, watch for the ideas that let you reason each one out — every question above is answered somewhere in the pages that follow.

In which block of the modern periodic table are the transition and inner transition elements placed?

8.1 Introduction

The transition elements belong to the d block of the periodic table. What marks them out is not simply where they sit, but what their atoms are doing electronically.

As per IUPAC convention, a transition metal atom has an incomplete d-subshell, or it gives cations with an incomplete d subshell. The second half of that sentence matters as much as the first, and we shall see in Section 8.3 that it decides some borderline cases.

NoteDefinition

A transition element is one whose atom has an incomplete d subshell, or which forms at least one cation with an incomplete d subshell.

They exhibit properties between those of s and p block elements — neither as reactive and electropositive as the metals of groups 1 and 2, nor as varied and non-metallic as the elements on the right of the table.

Where the four series run

The transition elements of the modern periodic table appear as groups 3 to 12, or as four long periods. In each element of a series, the (n-1)d orbital is successively filled, where ‘n’ is the ultimate or valence shell. In other words, the electron being added goes not into the outermost shell but into the d subshell of the shell just inside it.

Q. If the d electrons are being added to the shell inside the valence shell, why does that subshell get filled at this point at all?

A. Because of the order in which orbital energies rise. The 4s orbital lies slightly lower in energy than the 3d, so it fills first — this is why calcium is \(\mathrm{4s^2}\) with an empty 3d. Once 4s is full, the next lowest vacancy is 3d, and electrons begin entering it even though it belongs to the shell inside. The same pattern repeats one shell higher for the 4d, 5d and 6d series.

The four series are marked out as follows. The 3d series is comprised of elements from scandium (Z = 21) to zinc (Z = 30). The 4d series has elements from yttrium (Z = 39) to cadmium (Z = 48). The 5d series runs from lanthanum (Z = 57) to mercury (Z = 80), without those from cerium to lutetium. The 6d series has actinium to copernicium, without those from thorium to lawrencium.

Those two exclusions are not arbitrary: the elements left out are the lanthanoids and the actinoids, which fill f orbitals rather than d orbitals and are treated separately from Section 8.10 onwards. All four series are set out together in Table 8.1.

Table 8.1 Shows the four transition series elements

d series  Group 3 4 5 6 7 8 9 10 11 12
3d Sc(21) Ti(22) V(23) Cr(24) Mn(25) Fe(26) Co(27) Ni(28) Cu(29) Zn(30)
4d Y(39) Zr(40) Nb(41) Mo(42) Tc(43) Ru(44) Rh(45) Pd(46) Ag(47) Cd(48)
5d La(57) Hf(72) Ta(73) W(74) Re(75) Os(76) Ir(77) Pt(78) Au(79) Hg(80)
6d Ac(89) Rf(104) Db(105) Sg(106) Bh(107) Hs(108) Mt(109) Ds(110) Rg(111) Cn(112)

The 5d series is described as running from lanthanum (Z = 57) to mercury (Z = 80). That span covers 24 elements, yet Table 8.1 shows only ten. Account for the difference.

First count the whole span. From Z = 57 to Z = 80 inclusive,

\[80 - 57 + 1 = 24\ \text{elements}\]

Now identify what has been excluded. The definition states that the elements from cerium to lutetium are left out. Cerium is Z = 58 and lutetium is Z = 71, so the excluded block contains

\[71 - 58 + 1 = 14\ \text{elements}\]

Subtracting,

\[24 - 14 = 10\ \text{elements}\]

which is exactly the number shown in the 5d row of Table 8.1 — lanthanum followed by hafnium through mercury.

The fourteen elements removed are the lanthanoids. They are set aside because in them the 4f subshell, not the 5d, is being filled. Notice that a transition series must contain ten elements, because the five d orbitals can hold ten electrons between them.

The general electronic configuration of transition elements is therefore written as

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

The ranges in the exponents are worth reading carefully. The d count runs from 1 to 10 across a series, and the s count is given as 1 to 2 rather than simply 2, because in a few elements one of the s electrons shifts into the d subshell. Section 8.3.1 examines why.

The name “transition” records an old and useful idea about the periodic table’s shape. These elements form a bridge across the middle of the table, between the strongly electropositive metals of the s block on the left and the non-metals of the p block on the right, and their properties shift gradually from one character to the other. The name describes the job they do in the table rather than any property of the metals themselves.

Using Table 8.1, state the transition series and the group to which the element of atomic number 41 belongs, and write the valence-shell configuration this element would be expected to have from the general formula \(\mathrm{(n-1)d^{1-10}\ ns^{1-2}}\).

Questions from this section are usually short and definitional — state the IUPAC criterion for a transition element, name the four series with their ranges, or write the general electronic configuration. The criterion is worth learning word for word, because the exceptions discussed in Section 8.3 are decided by it.

8.2 Position in the periodic table

The transition elements are placed in the periods 4 to 7 and groups 3 to 12, and these constitute the 3d, 4d, 5d and 6d series. Their position in the modern periodic table is shown in Fig. 8.1.

Diagram

Fig. 8.1 – The modern periodic table with the blocks colour-coded and labelled: s-block on the left (groups 1 and 2), p-block on the right (groups 13 to 18), d-block occupying the centre (groups 3 to 12, periods 4 to 7), and the f-block (lanthanoids and actinoids) drawn as two rows detached below the main table, with an arrow linking them to their place after La and Ac; a key marks alkali metal, alkaline earth, transition metal, basic metal, metalloid, nonmetal, halogen, noble gas, lanthanoid and actinoid.

They are placed at the centre, with the s block on one side and the p block on the other. This central position is the whole point of the name met in Section 8.1: the electropositivity, reactivity and other properties show a gradual change from the s block to the p block through those of the d block elements.

Read across period 4 in Fig. 8.1 and the shift is easy to follow. Potassium and calcium on the left lose electrons violently and are stored under oil; the transition metals in the middle are hard, comparatively unreactive, and useful as structural materials; and by the time we reach bromine and krypton on the right the elements are no longer metals at all. The d block is where one kind of behaviour hands over to the other.

Q. Why does the d block occupy exactly ten groups, from 3 to 12?

A. Because there are five d orbitals in a subshell, and each can hold two electrons of opposite spin. Ten electrons must therefore be accommodated before the d subshell is complete, and since each successive element adds one electron, ten elements are needed. The width of a block in the periodic table is always simply the number of electrons its subshell can hold — two for the s block, six for the p block, ten for the d block and fourteen for the f block.

metallic and electropositive character across a long period (across a period →: falls steadily from the s block, through the d block, to the p block; because effective nuclear charge rises across the period and the atoms hold their electrons progressively more tightly, so the readiness to lose electrons — and with it the metallic character — declines. The d block occupies the middle of that change, which is why its properties lie between those of the two blocks on either side.)

s block vs d block vs p block

Feature s block d block p block
Groups 1 and 2 3 to 12 13 to 18
Subshell being filled ns (n-1)d np
Number of groups in the block 2 10 6
Character of the elements Strongly metallic and electropositive Metallic, but far less reactive Ranges from metalloid to non-metal
Position in the table Left edge Centre Right side

Fig. 8.1 – position of the blocks in the modern periodic table (Using the figure alone, answer the following. Which periods contain d-block elements, and which is the first period to do so? Which groups form the left and right boundaries of the d block? Where are the lanthanoids and actinoids printed, and between which two elements of the main table do they actually belong?)

The element tungsten has atomic number 74. Using Table 8.1 and Fig. 8.1, state its group, its period, and the transition series to which it belongs.

This section supplies the framework for everything that follows, and questions on it are normally one-liners — which groups and periods make up the d block, or why the transition elements are described as lying between the s and p blocks in their properties.

8.3 Electronic configuration

The 3d series begins with Sc (Z = 21) and ends with Zn (Z = 30). To see why the twenty-first electron goes where it does, it helps to start one step earlier.

Argon, Ar, is the noble gas preceding the 3d series, and its electronic configuration is \(\mathrm{1s^2\,2s^2\,2p^6\,3s^2\,3p^6}\). Calcium (Z = 20), belonging to the ‘s’ block of the 4th period, has the electronic configuration \(\mathrm{1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,4s^2}\) — the two extra electrons have gone into the 4s orbital, not the 3d.

With 4s now full, the next available orbital is 3d. Hence the 21st electron in scandium (Z = 21) enters the available 3d orbital. The electronic configuration of Sc is written as \(\mathrm{1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,3d^1\,4s^2}\), or can also be represented as \(\mathrm{[Ar]\,3d^1\,4s^2}\).

The condensed form on the right is the one used throughout this chapter. Replacing the first eighteen electrons by the symbol [Ar] is simply a shorthand for the noble gas core, and it keeps attention on the electrons that actually take part in chemistry.

The electronic configuration of the elements of the 3d series is given in Table 8.2.

Table 8.2 Electronic configuration of 3d series of d-block elements

Elements Symbol At. No. Expected electronic configuration Observed electronic configuration
Scandium Sc 21 \(\mathrm{[Ar]\,3d^1\,4s^2}\) \(\mathrm{[Ar]\,3d^1\,4s^2}\)
Titanium Ti 22 \(\mathrm{[Ar]\,3d^2\,4s^2}\) \(\mathrm{[Ar]\,3d^2\,4s^2}\)
Vanadium V 23 \(\mathrm{[Ar]\,3d^3\,4s^2}\) \(\mathrm{[Ar]\,3d^3\,4s^2}\)
Chromium Cr 24 \(\mathrm{[Ar]\,3d^4\,4s^2}\) \(\mathrm{[Ar]\,3d^5\,4s^1}\)
Manganese Mn 25 \(\mathrm{[Ar]\,3d^5\,4s^2}\) \(\mathrm{[Ar]\,3d^5\,4s^2}\)
Iron Fe 26 \(\mathrm{[Ar]\,3d^6\,4s^2}\) \(\mathrm{[Ar]\,3d^6\,4s^2}\)
Cobalt Co 27 \(\mathrm{[Ar]\,3d^7\,4s^2}\) \(\mathrm{[Ar]\,3d^7\,4s^2}\)
Nickel Ni 28 \(\mathrm{[Ar]\,3d^8\,4s^2}\) \(\mathrm{[Ar]\,3d^8\,4s^2}\)
Copper Cu 29 \(\mathrm{[Ar]\,3d^9\,4s^2}\) \(\mathrm{[Ar]\,3d^{10}\,4s^1}\)
Zinc Zn 30 \(\mathrm{[Ar]\,3d^{10}\,4s^2}\) \(\mathrm{[Ar]\,3d^{10}\,4s^2}\)

Read the two right-hand columns against each other and eight of the ten agree exactly. Only chromium and copper differ, and those two cases are taken up in Section 8.3.1.

Write the full and the condensed electronic configuration of vanadium (Z = 23), and state how many electrons occupy the 3d subshell.

Fill the orbitals in order of increasing energy, exactly as was done for scandium above.

The first 18 electrons complete the argon arrangement, \(\mathrm{1s^2\,2s^2\,2p^6\,3s^2\,3p^6}\).

Electrons 19 and 20 enter the 4s orbital, since 4s lies lower in energy than 3d.

The remaining electrons, \(23 - 20 = 3\), enter the 3d orbitals.

Hence the full configuration is

\[\mathrm{1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,3d^3\,4s^2}\]

and in condensed form,

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

The 3d subshell holds 3 electrons, which agrees with the entry for vanadium in Table 8.2. Note the convention followed in writing the answer: the orbitals are listed in order of shell number, so 3d is written before 4s even though 4s was filled first.

Which d-block elements are not transition elements

Since zinc has a completely filled (n-1)d orbital in the ground state (\(\mathrm{3d^{10},\,4s^2}\)) and also in its common oxidation state \(+2\) (\(\mathrm{3d^{10}}\)), it is not regarded as a transition element. On the same ground, cadmium and mercury from the 4d and 5d series are not considered as transition elements.

Copper looks like a similar case but is not. Copper in the elementary state (\(\mathrm{3d^{10}\,4s^1}\)) contains filled 3d orbitals, but in the \(+2\) oxidation state it has a partly filled 3d orbital (\(\mathrm{3d^9}\)); hence copper is a transition element.

This is where the second half of the IUPAC criterion from Section 8.1 earns its place. An element qualifies if either the atom or one of its cations has an incomplete d subshell. Zinc fails on both counts; copper fails the first test but passes the second.

zinc, cadmium and mercury versus copper

Feature Zn, Cd, Hg Cu
d subshell in the neutral atom Completely filled Completely filled
Common oxidation state \(+2\) \(+2\)
d subshell in that ion Completely filled, \(\mathrm{3d^{10}}\) for \(\mathrm{Zn^{2+}}\) Partly filled, \(\mathrm{3d^9}\) for \(\mathrm{Cu^{2+}}\)
Satisfies the IUPAC criterion? No Yes
Regarded as a transition element? No Yes

Zinc and copper are neighbours in the 3d series and both have a completely filled 3d subshell in the free atom. Explain, using the IUPAC criterion, why copper is classed as a transition element and zinc is not.

Write the configurations of the atoms and of the common ions of each.

Zinc: the atom is \(\mathrm{[Ar]\,3d^{10}\,4s^2}\). Its common ion is formed by losing the two 4s electrons:

\[\mathrm{Zn^{2+}} = \mathrm{[Ar]\,3d^{10}}\]

Copper: the atom is \(\mathrm{[Ar]\,3d^{10}\,4s^1}\). Its \(+2\) ion is formed by losing the single 4s electron and one 3d electron:

\[\mathrm{Cu^{2+}} = \mathrm{[Ar]\,3d^{9}}\]

Now apply the criterion. A transition element must have an incomplete d subshell in the atom, or must give a cation with an incomplete d subshell.

For zinc, both the atom and the \(\mathrm{Zn^{2+}}\) ion have a complete \(\mathrm{3d^{10}}\) subshell, so zinc satisfies neither part of the criterion and is not a transition element.

For copper, the atom has a complete subshell, but \(\mathrm{Cu^{2+}}\) has \(\mathrm{3d^9}\) — one electron short of full. Copper therefore satisfies the second part of the criterion and is a transition element.

The same reasoning excludes cadmium and mercury, which lie directly below zinc and behave in the same way.

Write the condensed electronic configuration of the element with atomic number 26 and of its \(+3\) ion, and state the number of 3d electrons in each.

The general electronic configuration of the four series of d-block elements of the periodic table can be represented as given below:

  1. 3d series : \(\mathrm{[Ar]\,3d^{1-10}\,4s^2}\)

  2. 4d series : \(\mathrm{[Kr]\,4d^{1-10}\,5s^{0-2}}\)

  3. 5d series : \(\mathrm{[Xe]\,4f^{14}}\) (except La) \(\mathrm{5d^{1-10}\,6s^2}\)

  4. 6d series : \(\mathrm{[Rn]\,5f^{14}}\) (except Ac) \(\mathrm{6d^{1-10}\,7s^2}\)

The bracketed exceptions are worth noticing. Lanthanum and actinium each begin their series before any f electrons have been added, which is why the \(\mathrm{4f^{14}}\) and \(\mathrm{5f^{14}}\) parts do not apply to them.

Fill in the blanks with the correct outer electronic configurations.

2nd series

Y Zr Nb Mo Tc Ru Rh Pd Ag Cd
Z 39 40 41 42 43 44 45 46 47 48
valence shell electronic configuration \(\mathrm{4d^1 5s^2}\) \(\mathrm{4d^2 5s^2}\) \(\mathrm{4d^4 5s^1}\) \(\mathrm{4d^6 5s^1}\) \(\mathrm{4d^8 5s^1}\) \(\mathrm{4d^{10} 5s^0}\) \(\mathrm{4d^{10} 5s^2}\)

3rd series

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

4th series

Ac Rf Db Sg Bh Hs Mt Ds Rg Cn
Z 89 104 105 106 107 108 109 110 111 112
valence shell electronic configuration \(\mathrm{6d^1 7s^2}\) \(\mathrm{6d^2 7s^2}\) \(\mathrm{6d^4 7s^2}\) \(\mathrm{6d^5 7s^2}\) \(\mathrm{6d^6 7s^2}\) \(\mathrm{6d^7 7s^2}\) \(\mathrm{6d^8 7s^2}\) \(\mathrm{6d^{10} 7s^2}\)

From the general formula for the 4d series, write the configuration palladium (Z = 46) would be expected to have, compare it with the observed configuration given in the table above, and state how many electrons have moved out of the 5s orbital.

8.3.1 Electronic configuration of chromium and copper

Table 8.2 indicates that the expected electronic configuration of chromium (Z = 24) differs from the observed configuration. The same is true of copper (Z = 29), and both departures have a single explanation.

This can be explained on the basis of the concept of additional stability associated with the completely filled and half filled subshells.

NotePrinciple / Law

Any subshell having a half filled or a completely filled electronic configuration has extra stability.

Half filled means one electron in each orbital of the subshell, all with parallel spins; completely filled means every orbital holding its full pair. Both arrangements are more stable than a partly filled subshell of intermediate occupancy, and an atom will rearrange its outermost electrons to reach one of them if the cost of doing so is small.

The general electronic configuration of the elements of the 3d series is \(\mathrm{3d^{1-10}\,4s^2}\), with the exceptions of Cr and Cu. The 3d and 4s orbitals are close in energy, and in order to gain extra stability the last electron, instead of occupying the 4s orbital, occupies the 3d orbital. This assigns Cr the \(\mathrm{3d^5,\,4s^1}\) configuration and Cu the \(\mathrm{3d^{10},\,4s^1}\) configuration.

Follow the logic for each element in turn. Chromium would be expected to be \(\mathrm{3d^4\,4s^2}\); shifting one 4s electron into 3d gives \(\mathrm{3d^5\,4s^1}\), in which the d subshell is exactly half filled. Copper would be expected to be \(\mathrm{3d^9\,4s^2}\); the same shift gives \(\mathrm{3d^{10}\,4s^1}\), in which the d subshell is completely filled. The comparison is drawn in Fig. 8.1a.

Diagram

Fig. 8.1a – Orbital box diagrams for chromium and copper, expected against observed. For Cr: the expected row shows five 3d boxes with one pair and three single arrows, beside a paired 4s box; the observed row shows five 3d boxes each with a single upward arrow, beside a 4s box with one arrow. For Cu: the expected row shows five 3d boxes with four pairs and one single arrow, beside a paired 4s box; the observed row shows five fully paired 3d boxes beside a 4s box with one arrow. An arrow marked “one 4s electron shifts to 3d” links each expected row to its observed row.

The shift is only possible because the two orbitals lie so close in energy. Promoting an electron across a wide energy gap would cost far more than the stability gained, which is why no such rearrangement occurs in, say, manganese or iron.

Q. If the 4s orbital fills before 3d, why is it the 4s electrons that are removed first when a transition metal forms an ion?

A. Because the energy order changes once the 3d orbitals are occupied. In the empty-3d atoms of potassium and calcium the 4s orbital is the lower of the two, so it fills first. As soon as d electrons are present, the 3d orbitals are pulled below 4s in energy, and 4s becomes the outermost and highest-energy orbital of the atom. An electron is always removed from the highest-energy occupied orbital, so the 4s electrons leave first. This rule is used constantly from Section 8.4 onwards.

These two exceptions are among the most frequently examined facts in the whole chapter. Questions ask for the observed configuration of chromium or copper, or set an assertion–reason pair in which the assertion states the anomalous configuration and the reason states the extra stability of half filled and completely filled subshells. Write the expected configuration first, then the observed one, and name the subshell that has been completed or half completed — that sequence earns full marks.

8.4 Oxidation states of first transition series

One of the most striking features of the transition elements is the sheer variety of oxidation states they show in their compounds. Recall that the oxidation state of an atom in a compound is the charge it would carry if every bond it formed were counted as fully ionic. A representative metal such as calcium is locked into a single state, \(+2\), in practically everything it forms. Manganese, by contrast, turns up in six different oxidation states.

Table 8.3 lists the common oxidation states of the first row transition elements, set beside the outer electronic configuration of each atom, because the configuration is what makes the variety possible.

Table 8.3: Oxidation states of first transition series elements

Elements Outer electronic configuration Oxidation states Elements Outer electronic configuration Oxidation states
Sc \(\mathrm{3d^1\,4s^2}\) +1, +2, +3 Fe \(\mathrm{3d^6\,4s^2}\) +2, +3, +4, +5, +6
Ti \(\mathrm{3d^2\,4s^2}\) +2, +3, +4 Co \(\mathrm{3d^7\,4s^2}\) +2, +3, +4, +5
V \(\mathrm{3d^3\,4s^2}\) +2, +3, +4, +5 Ni \(\mathrm{3d^{10}\,4s^2}\) +2, +3, +4
Cr \(\mathrm{3d^5\,4s^1}\) +2, +3, +4, +5, +6 Cu \(\mathrm{3d^{10}\,4s^1}\) +1, +2
Mn \(\mathrm{3d^5\,4s^2}\) +2, +3, +4, +5, +6, +7 Zn \(\mathrm{3d^{10}\,4s^2}\) +2

Which of the first transition series elements shows the maximum number of oxidation states and why? Which elements in the 4d and 5d series will show maximum number of oxidation states?

Why these elements have a choice

The variety comes from an accident of energy. In a transition element the 4s and the 3d subshells lie very close together in energy. Once the two 4s electrons have been pulled off, the 3d electrons are not far behind — removing one costs only a little more energy, and the compound formed can easily pay that cost through bond formation.

In an s-block metal there is no such option. Below the valence s electrons lies a completed noble-gas shell held tightly by the nucleus, so the metal stops at one fixed oxidation state.

Two configurations in Table 8.3 are worth fixing in memory now, because they run through the whole chapter: chromium is \(\mathrm{[Ar]\,3d^5\,4s^1}\) and copper is \(\mathrm{[Ar]\,3d^{10}\,4s^1}\), not the \(\mathrm{3d^4\,4s^2}\) and \(\mathrm{3d^9\,4s^2}\) you might expect. The half-filled and completely filled d subshells are the more stable arrangements.

Vanadium was named after Vanadis, a name of the Norse goddess of beauty, because the element’s compounds appear in such a striking range of colours. Those colours are simply vanadium sitting in its different oxidation states — a piece of chemistry hidden inside the element’s name.

Building the ions step by step

Loss of 4s and 3d electrons progressively leads to the formation of ions, and this is what generates the series of oxidation states. Loss of one 4s electron leads to the formation of the \(\mathrm{M^{+}}\) ion, and loss of two 4s electrons gives an \(\mathrm{M^{2+}}\) ion, while loss of unpaired 3d electrons along with the 4s electrons gives \(\mathrm{M^{3+}}\), \(\mathrm{M^{4+}}\) ions and so on. The order of removal is summarised in FC 8.1, and worked out element by element in Table 8.4.

Flowchart

FC 8.1 – Successive electron loss in a 3d metal: neutral atom M (3dⁿ 4s²) → lose one 4s electron → M⁺ → lose second 4s electron → M²⁺ (4s empty, only 3d electrons left) → begin removing unpaired 3d electrons → M³⁺ → M⁴⁺ → higher states; branch: removal stops early once only paired 3d electrons remain, as in Zn²⁺ (3d¹⁰).

Table 8.4: Electronic configuration of various ions of 3d elements

Species Sc (21) Ti (22) V (23) Cr (24) Mn (25) Fe (26) Co (27) Ni (28) Cu (29) Zn (30)
M \(\mathrm{3d^1\,4s^2}\) \(\mathrm{3d^2\,4s^2}\) \(\mathrm{3d^3\,4s^2}\) \(\mathrm{3d^5\,4s^1}\) \(\mathrm{3d^5\,4s^2}\) \(\mathrm{3d^6\,4s^2}\) \(\mathrm{3d^7\,4s^2}\) \(\mathrm{3d^8\,4s^2}\) \(\mathrm{3d^{10}\,4s^1}\) \(\mathrm{3d^{10}\,4s^2}\)
\(\mathrm{M^{+}}\) \(\mathrm{3d^1\,4s^1}\) \(\mathrm{3d^2\,4s^1}\) \(\mathrm{3d^3\,4s^1}\) \(\mathrm{3d^5}\) \(\mathrm{3d^5\,4s^1}\) \(\mathrm{3d^6\,4s^1}\) \(\mathrm{3d^7\,4s^1}\) \(\mathrm{3d^8\,4s^1}\) \(\mathrm{3d^{10}\,4s^0}\) \(\mathrm{3d^{10}\,4s^1}\)
\(\mathrm{M^{2+}}\) \(\mathrm{3d^1}\) \(\mathrm{3d^2}\) \(\mathrm{3d^3}\) \(\mathrm{3d^4}\) \(\mathrm{3d^5}\) \(\mathrm{3d^6}\) \(\mathrm{3d^7}\) \(\mathrm{3d^8}\) \(\mathrm{3d^9}\) \(\mathrm{3d^{10}}\)
\(\mathrm{M^{3+}}\) [Ar] \(\mathrm{3d^1}\) \(\mathrm{3d^2}\) \(\mathrm{3d^3}\) \(\mathrm{3d^4}\) \(\mathrm{3d^5}\) \(\mathrm{3d^6}\) \(\mathrm{3d^7}\) – –

Read Table 8.4 downwards for any one element and a useful pattern appears: by the time the \(\mathrm{M^{2+}}\) ion is reached, the 4s orbital is empty in every case. Whatever electrons remain are 3d electrons, and it is these that decide how much further the element can go.

Q. Zinc is a d-block element, so why is it stuck at only one oxidation state, \(+2\)?

A. Zinc is \(\mathrm{3d^{10}\,4s^2}\). Losing the two 4s electrons gives \(\mathrm{Zn^{2+}}\) with a completely filled \(\mathrm{3d^{10}}\) shell, and there are no unpaired d electrons left to remove cheaply. Breaking into that full, stable subshell costs far too much energy, so zinc goes no further.

Write the valence-shell electronic configuration of \(\mathrm{Cr^{3+}}\) and of \(\mathrm{Fe^{3+}}\), and state the number of unpaired d electrons in each.

Start from the neutral atoms. Chromium is \(\mathrm{[Ar]\,3d^5\,4s^1}\) and iron is \(\mathrm{[Ar]\,3d^6\,4s^2}\).

For \(\mathrm{Cr^{3+}}\), three electrons are removed: first the single 4s electron, then two 3d electrons.

\[\mathrm{Cr}\ \mathrm{[Ar]\,3d^5\,4s^1} \rightarrow \mathrm{Cr^{3+}}\ \mathrm{[Ar]\,3d^3}\]

The three 3d electrons occupy three separate d orbitals, so there are 3 unpaired electrons.

For \(\mathrm{Fe^{3+}}\), the two 4s electrons go first, followed by one 3d electron.

\[\mathrm{Fe}\ \mathrm{[Ar]\,3d^6\,4s^2} \rightarrow \mathrm{Fe^{3+}}\ \mathrm{[Ar]\,3d^5}\]

This leaves a half-filled d subshell with 5 unpaired electrons. Both answers agree with the \(\mathrm{M^{3+}}\) row of Table 8.4. Notice the rule being used: in a cation the 4s electrons are always removed before the 3d electrons.

Write the electronic configuration of \(\mathrm{Mn^{6+}}\), \(\mathrm{Mn^{4+}}\), \(\mathrm{Fe^{4+}}\), \(\mathrm{Co^{5+}}\), \(\mathrm{Ni^{2+}}\).

Unpaired electrons decide the count

From Table 8.3 it is clear that as the number of unpaired electrons in the 3d orbitals increases, the number of oxidation states shown by the element also increases. Scandium has only one unpaired electron, and it shows three oxidation states. Manganese, with 5 unpaired d electrons, shows six different oxidation states.

The logic is direct. Each unpaired d electron is one more electron that can be removed at a modest energy cost, and each such removal opens up one more oxidation state.

Sc vs Mn vs Zn — unpaired d electrons against the number of oxidation states

Feature Sc Mn Zn
Outer configuration \(\mathrm{3d^1\,4s^2}\) \(\mathrm{3d^5\,4s^2}\) \(\mathrm{3d^{10}\,4s^2}\)
Unpaired d electrons 1 5 0
Oxidation states shown +1, +2, +3 +2 to +7 +2 only
Number of states 3 6 1
Why the middle of the series wins

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

Manganese sits at the peak for a simple arithmetical reason. Its outer configuration is \(\mathrm{3d^5\,4s^2}\), a total of seven electrons in the 3d and 4s orbitals, and every one of them is available for removal — five unpaired d electrons plus the two 4s electrons. That upper limit of \(+7\) is exactly what is reached in the permanganate ion, \(\mathrm{MnO_4^{-}}\).

Beyond the middle of the series the d electrons begin to pair up. Pulling an electron out of a pair costs extra energy, so the range of accessible oxidation states narrows again as we move towards nickel, copper and zinc.

number of oxidation states across the first transition series (across the period →: rises from Sc to a maximum at Mn, then falls away to Zn; because the number of unpaired 3d electrons rises to five at \(\mathrm{3d^5}\) and then drops as electrons start pairing in the second half of the series, and paired electrons are much harder to remove)

Q1. Manganese shows six different oxidation states in its compounds, while calcium — a metal in the very same period — shows only one. What is it about d-block atoms that allows such variety?

A. In calcium the only removable electrons are the two 4s electrons; below them lies a tightly held, complete argon core. Manganese has an incomplete 3d subshell whose energy is very close to that of 4s, holding five unpaired electrons. These d electrons can be removed one after another at a small extra energy cost, so manganese can stop at \(+2\), \(+3\), \(+4\), \(+5\), \(+6\) or \(+7\) depending on what it is reacting with — while calcium has no such choice.

Using its ground-state electronic configuration, work out the highest oxidation state vanadium can show, and write the valence-shell configuration of the vanadium ion in that state.

The oxidation state of a transition metal decides not only its colour but also how dangerous it is. Chromium in the \(+6\) state is toxic and is treated as a carcinogen, whereas chromium in the \(+3\) state is far less hazardous. Tanneries and electroplating units therefore chemically reduce chromium(VI) in their effluent to chromium(III) before the water is released, so that the metal leaves the plant in its safer oxidation state. The element is the same throughout; only the number of electrons it has lost has changed.

Application-based questions from this section usually take one of three forms in board and entrance papers: write the configuration of a given ion, order elements by the number of oxidation states they show, or explain in assertion–reason format why a particular element reaches an unusually high state. In every case the working starts the same way — write the neutral atom’s configuration first, then remove 4s before 3d.

8.5 Physical properties of first transition series

All transition elements are metals, and they show the properties we normally associate with metals. They are hard, lustrous, malleable and ductile, and they form alloys with other metals. Lustrous simply means they have a shining surface that reflects light; malleable means they can be beaten into thin sheets without cracking; ductile means they can be drawn out into wires. They are also good conductors of heat and electricity.

Structure at ordinary temperature

Except Zn, Cd, Hg and Mn, all the other transition elements have one or more typical metallic structures at ambient temperature — that is, at the ordinary temperature of the surroundings, without heating or cooling. In these regular close-packed arrangements each atom is surrounded by a large number of neighbours, which is one reason a metal holds together so firmly.

Hardness, volatility and melting points

These transition metals, with the exception of Zn, Cd and Hg, are very hard and have low volatility. Low volatility means the metal does not turn into vapour easily; its atoms are reluctant to leave the solid.

They possess high melting and boiling points. The variation of melting point across the three transition series is plotted in Fig. 8.2, and the values are strikingly high — well above those of the s-block metals of the same periods. A high melting point is direct evidence that the atoms in the metallic lattice are held to one another very strongly indeed.

Diagram

Fig. 8.2 – Graph of melting point (M.p./10³ K) against atomic number for the 3d, 4d and 5d transition series, each series drawn as a separate labelled curve with element symbols marked (Ti…Cu, Zr…Ag, Hf…Au), showing the rise to a maximum in each series and the dips at Mn and Tc.

Fig. 8.2 – melting points of the three transition series (Compare the three curves. Which of the three series lies highest overall, and roughly where along each series does the curve reach its peak? Which two elements sit noticeably below the smooth curve of their own series?)

Q. Zinc, cadmium and mercury are all d-block elements, so why are they soft and easily vaporised while the rest of the block is hard and stubborn?

A. All three have a completely filled d subshell in the metal, so their d electrons are locked into stable pairs and take almost no part in binding one atom to the next. Only the two outer s electrons are left to do the job, and the resulting metallic bond is weak — hence soft metals, low melting points and high volatility. Mercury takes this to the extreme and is liquid at room temperature.

Zn, Cd, Hg versus the other transition metals

Feature Zn, Cd, Hg Other transition metals
d subshell in the metal Completely filled Incomplete, with unpaired d electrons
Hardness Soft Very hard
Volatility Comparatively high Low
Melting and boiling points Low High
Typical metallic structure at ambient temperature Absent (Mn is also an exception here) Present, one or more forms

The symbol for mercury, Hg, comes from its Latin name hydrargyrum, built from the Greek words for water and silver — a silver that flows like water. The name records exactly what puzzled early chemists about it: a metal that behaves like a liquid.

Mercury’s liquid range and even expansion once made it the standard filling for thermometers and barometers in every school laboratory. It is also a cumulative poison, and its vapour is hazardous to breathe, which is why a broken mercury thermometer is never swept up like ordinary glass. Under the international Minamata Convention on Mercury, mercury-filled thermometers and many other mercury-added products are being phased out, and school laboratories now use alcohol-filled or digital thermometers instead. The property that made mercury useful and the property that made it dangerous belong to the same element.

Sodium and chromium are both metals, yet a knife blade can be made from chromium steel while sodium can be cut with a knife. Using the physical properties described in this section, list three measurable properties in which the two metals would differ most sharply, and state in each case which metal shows the higher value.

In board and entrance papers this section is usually tested through short reasoning questions — why transition metals are used for tools and machine parts, or which d-block elements are the exceptions to hardness and high melting point. The exception list Zn, Cd, Hg (with Mn added for metallic structure) is worth learning precisely, because questions are often framed around it.

8.7 Compounds of Mn and Cr (\(\mathrm{KMnO_4}\) and \(\mathrm{K_2Cr_2O_7}\))

Two compounds of the first transition series are met so often in the laboratory that they deserve a section to themselves. Both \(\mathrm{KMnO_4}\) and \(\mathrm{K_2Cr_2O_7}\) are strong oxidising agents — that is, both are eager to accept electrons from other substances, and in doing so they are themselves reduced.

Their oxidising power comes directly from what we studied in Section 8.4. Manganese in \(\mathrm{KMnO_4}\) is in its highest oxidation state, \(+7\), and chromium in \(\mathrm{K_2Cr_2O_7}\) is in the \(+6\) state. A metal already stripped of that many electrons has a strong pull to take some back, and the drop to a lower, more comfortable oxidation state is what drives every reaction in this section.

8.7.1 Preparation of potassium permanganate

i. Chemical oxidation

When a finely divided manganese dioxide (\(\mathrm{MnO_2}\)) is heated strongly with a fused mass of caustic potash (KOH) and an oxidising agent, potassium chlorate (\(\mathrm{KClO_3}\)), dark green potassium manganate, \(\mathrm{K_2MnO_4}\), is formed.

\[3\,\mathrm{MnO_2} + 6\,\mathrm{KOH} + \mathrm{KClO_3} \xrightarrow{\ \Delta\ } 3\,\mathrm{K_2MnO_4} + \mathrm{KCl} + 3\,\mathrm{H_2O}\]

Manganese has been pushed from \(+4\) in \(\mathrm{MnO_2}\) up to \(+6\) in the manganate ion, with the chlorate supplying the oxidising power for the step.

In neutral or acidic medium the green potassium manganate disproportionates to \(\mathrm{KMnO_4}\) and \(\mathrm{MnO_2}\).

NoteDefinition

Disproportionation is a reaction in which the same element in a single oxidation state is simultaneously oxidised to a higher oxidation state and reduced to a lower one.

\[3\,\mathrm{K_2MnO_4} + 4\,\mathrm{CO_2} + 2\,\mathrm{H_2O} \rightarrow 2\,\mathrm{KMnO_4} + \mathrm{MnO_2} + 4\,\mathrm{KHCO_3}\]

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

The liquid is filtered through glass wool or sintered glass and evaporated until crystallisation occurs. Potassium permanganate forms small crystals which are almost black in appearance.

Show that the second reaction above is a disproportionation, by tracking the oxidation number of manganese.

Work out the oxidation number of Mn in each manganese-containing species. Take oxygen as \(-2\) throughout.

In \(\mathrm{MnO_4^{2-}}\): \(x + 4(-2) = -2\), so \(x = +6\).

In \(\mathrm{MnO_4^{-}}\): \(x + 4(-2) = -1\), so \(x = +7\).

In \(\mathrm{MnO_2}\): \(x + 2(-2) = 0\), so \(x = +4\).

So manganese begins entirely at \(+6\). Part of it ends at \(+7\) (oxidised, having lost an electron) and part at \(+4\) (reduced, having gained two electrons). One element, one starting oxidation state, two different fates — which is precisely the definition of disproportionation. The stoichiometry confirms the electron book-keeping: 2 manganese atoms are oxidised, losing 1 electron each, while 1 manganese atom is reduced, gaining 2 electrons.

ii. Electrolytic oxidation

In electrolytic oxidation, an alkaline solution of manganate ion is electrolysed between iron electrodes separated by a diaphragm, as sketched in Fig. 8.4e. The overall reaction is as follows:

\[2\,\mathrm{K_2MnO_4} + \mathrm{H_2O} + [\mathrm{O}] \rightarrow 2\,\mathrm{KMnO_4} + 2\,\mathrm{KOH}\]

Diagram

Fig. 8.4e – Electrolytic cell for oxidation of manganate: a vessel containing alkaline green manganate solution, two iron electrodes labelled anode and cathode, a porous diaphragm dividing the two compartments, oxygen bubbles shown at the anode and purple permanganate forming in the anode compartment.

The oxygen evolved at the anode converts manganate to permanganate. The diaphragm matters here: it keeps the freshly made permanganate away from the cathode, where it would simply be reduced back again.

The solution is filtered and evaporated to get deep purple black coloured crystals of \(\mathrm{KMnO_4}\). The two routes to the compound are compared side by side in FC 8.5.

Flowchart

FC 8.5 – Preparation of potassium permanganate: finely divided MnO₂ → fused with KOH + KClO₃, heated → dark green K₂MnO₄ (manganate, Mn in +6) → branch A, chemical route: acidify or pass CO₂ → disproportionation → KMnO₄ + MnO₂ → filter through glass wool → evaporate → almost black crystals; branch B, electrolytic route: alkaline manganate solution electrolysed between iron electrodes with a diaphragm → O₂ at anode oxidises manganate → KMnO₄ → filter → evaporate → deep purple-black crystals.

Chromium takes its name from the Greek word chroma, meaning colour, and the reason becomes obvious the moment you handle its compounds: chromate solutions are yellow, dichromate orange, chromium(III) salts green, and chromium(II) blue. The element was named for the one property that no chemist working with it could miss.

8.7.2 Chemical properties of \(\mathrm{KMnO_4}\)

The medium in which permanganate is used decides how far manganese is reduced, and therefore which product appears. This is worth grasping before looking at the individual reactions, because it organises the whole subsection.

Q. Why does permanganate end up as \(\mathrm{Mn^{2+}}\) in acid but as \(\mathrm{MnO_2}\) in a neutral or weakly alkaline medium?

A. In acidic medium there is a plentiful supply of \(\mathrm{H^{+}}\) ions to combine with the oxygen released from the permanganate ion, forming water. That lets the reduction run all the way down to \(\mathrm{Mn^{2+}}\), a five-electron change. Without that supply of protons the oxygen cannot be carried away as readily, the reduction stops after three electrons, and manganese settles at \(+4\) as insoluble brown \(\mathrm{MnO_2}\).

\(\mathrm{KMnO_4}\) in acidic versus neutral or weakly alkaline medium

Feature Acidic medium Neutral or weakly alkaline medium
Manganese product \(\mathrm{Mn^{2+}}\) \(\mathrm{MnO_2}\)
Oxidation state of Mn after reaction \(+2\) \(+4\)
Electrons gained per Mn 5 3
Appearance of product Very pale pink, effectively colourless in solution Brown insoluble solid
Example from the text \(\mathrm{I^{-}}\) oxidised to \(\mathrm{I_2}\) \(\mathrm{I^{-}}\) oxidised further to \(\mathrm{IO_3^{-}}\)
a. In acidic medium

The oxidizing reactions of \(\mathrm{KMnO_4}\) in acidic medium are as follows.

i. Oxidation of iodide to iodine:

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

ii. Oxidation of \(\mathrm{Fe^{2+}}\) to \(\mathrm{Fe^{3+}}\):

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

iii. Oxidation of \(\mathrm{H_2S}\):

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

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

iv. Oxidation of oxalic acid:

\[2\,\mathrm{MnO_4^{-}} + 5\,\mathrm{H_2C_2O_4} + 6\,\mathrm{H^{+}} \rightarrow 2\,\mathrm{Mn^{2+}} + 10\,\mathrm{CO_2} + 8\,\mathrm{H_2O}\]

In the reaction of acidified \(\mathrm{KMnO_4}\) with \(\mathrm{Fe^{2+}}\), identify the oxidising agent and the reducing agent, and show how the balancing ratio of 1 : 5 follows from the change in oxidation number.

First assign oxidation numbers to the two metals on both sides.

Manganese: \(+7\) in \(\mathrm{MnO_4^{-}}\) falls to \(+2\) in \(\mathrm{Mn^{2+}}\). A fall in oxidation number means reduction, so permanganate is the oxidising agent, and each manganese atom gains 5 electrons.

Iron: \(+2\) in \(\mathrm{Fe^{2+}}\) rises to \(+3\) in \(\mathrm{Fe^{3+}}\). A rise means oxidation, so \(\mathrm{Fe^{2+}}\) is the reducing agent, and each iron atom loses 1 electron.

For the equation to balance, electrons lost must equal electrons gained:

\[5 \times (1\ \text{electron lost per Fe}) = 1 \times (5\ \text{electrons gained per Mn})\]

Hence 5 iron(II) ions are needed for every 1 permanganate ion, exactly the ratio written in the balanced equation above. The \(8\,\mathrm{H^{+}}\) and \(4\,\mathrm{H_2O}\) then take care of the oxygen and hydrogen balance.

How many moles of oxalic acid, \(\mathrm{H_2C_2O_4}\), are oxidised by one mole of \(\mathrm{KMnO_4}\) in acidic medium? Support your answer by working out the change in oxidation number of both carbon and manganese.

b. In neutral or weakly alkaline medium

i. Oxidation of iodide \(\mathrm{I^{-}}\) to iodate \(\mathrm{IO_3^{-}}\):

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

ii. Thiosulphate (\(\mathrm{S_2O_3^{2-}}\)) is oxidised to sulphate (\(\mathrm{SO_4^{2-}}\)):

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

iii. Manganous salt is oxidised to \(\mathrm{MnO_2}\):

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

The third reaction is worth a second look. Permanganate at \(+7\) meets manganous ion at \(+2\), and both end up as \(\mathrm{MnO_2}\) at \(+4\) — the reverse of disproportionation, where two different oxidation states of one element meet in the middle.

potassium permanganate, \(\mathrm{KMnO_4}\) (A powerful oxidising agent, deep purple in solution → needs an acidic medium if the full five-electron reduction to \(\mathrm{Mn^{2+}}\) is wanted, and works in neutral or weakly alkaline medium to give brown \(\mathrm{MnO_2}\) → used to oxidise iodide, iron(II), hydrogen sulphide and oxalic acid, and as its own indicator in titrations, since the purple colour disappears the instant it is reduced and reappears at the end point.)

8.7.3 Uses of \(\mathrm{KMnO_4}\)

  • An antiseptic.
  • For the unsaturation test in the laboratory.
  • In volumetric analysis of reducing agents.
  • For detecting halides in qualitative analysis.
  • Powerful oxidising agent in laboratory and industry.

Municipal water treatment plants dose raw water with small quantities of potassium permanganate before filtration. Dissolved iron(II) and manganese(II), which would otherwise stain plumbing and give water a metallic taste, are oxidised to insoluble oxides that can simply be filtered out, and the same treatment destroys many of the organic compounds responsible for musty odours. Because the permanganate is fully consumed in the process, what reaches the tap contains none of it. The very intensity of the purple colour is useful here too — a faint pink tinge in the treated water is an immediate visual warning that too much has been added.

8.7.4 \(\mathrm{K_2Cr_2O_7}\) : Preparation of potassium dichromate

In industrial production, finely powdered chromite ore (\(\mathrm{FeOCr_2O_3}\)) is heated with anhydrous sodium carbonate (\(\mathrm{Na_2CO_3}\)) and a flux of lime in air in a reverberatory furnace. A flux is a substance added to combine with unwanted material and carry it away; you will meet the idea again in Section 8.9.

\[4(\mathrm{FeO.Cr_2O_3}) + 8\,\mathrm{Na_2CO_3} + 7\,\mathrm{O_2} \xrightarrow{\ \Delta\ } 8\,\mathrm{Na_2CrO_4} + 2\,\mathrm{Fe_2O_3} + 8\,\mathrm{CO_2}\]

Sodium chromate (\(\mathrm{Na_2CrO_4}\)) formed in this reaction is then extracted with water and treated with concentrated sulphuric acid to get sodium dichromate and hydrated sodium sulphate:

\[2\,\mathrm{Na_2CrO_4} + \mathrm{H_2SO_4} \rightarrow \mathrm{Na_2Cr_2O_7} + 2\,\mathrm{NaCl} + \mathrm{Na_2Cr_2O_7.H_2O}\]

Addition of potassium chloride to a concentrated solution of sodium dichromate precipitates the less soluble orange-red coloured potassium dichromate, \(\mathrm{K_2Cr_2O_7}\).

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

The last step works purely on solubility. Both salts are present in the same solution, but potassium dichromate is the less soluble of the two, so it is the one that crystallises out and can be filtered off. The full route is summarised in FC 8.6.

Flowchart

FC 8.6 – Preparation of potassium dichromate: chromite ore FeO·Cr₂O₃, finely powdered → heated in air with Na₂CO₃ and lime flux in a reverberatory furnace → sodium chromate Na₂CrO₄ (yellow) + Fe₂O₃ + CO₂ → extract with water → treat with concentrated H₂SO₄ → sodium dichromate Na₂Cr₂O₇ (orange) → add KCl to the concentrated solution → less soluble K₂Cr₂O₇ crystallises out as orange-red crystals.

8.7.5 Chemical properties of \(\mathrm{K_2Cr_2O_7}\)

i. Oxidation of \(\mathrm{I^{-}}\) from an aqueous solution of KI by acidified \(\mathrm{K_2Cr_2O_7}\) gives \(\mathrm{I_2}\). Potassium dichromate is reduced to chromic sulphate. The liberated \(\mathrm{I_2}\) turns the solution brown.

\[\mathrm{K_2Cr_2O_7} + 6\,\mathrm{KI} + 7\,\mathrm{H_2SO_4} \rightarrow 4\,\mathrm{K_2SO_4} + \mathrm{Cr_2(SO_4)_3} + 7\,\mathrm{H_2O} + 3\,\mathrm{I_2}\]

ii. When \(\mathrm{H_2S}\) gas is passed through acidified \(\mathrm{K_2Cr_2O_7}\) solution, \(\mathrm{H_2S}\) is oxidised to a pale yellow precipitate of sulphur. Simultaneously potassium dichromate is reduced to chromic sulphate, which is reflected as a colour change of the solution from orange to green.

\[\mathrm{K_2Cr_2O_7} + 4\,\mathrm{H_2SO_4} + 3\,\mathrm{H_2S} \rightarrow \mathrm{K_2SO_4} + \mathrm{Cr_2(SO_4)_3} + 7\,\mathrm{H_2O} + 3\,\mathrm{S}\]

In both reactions chromium falls from \(+6\) in the dichromate ion to \(+3\) in chromic sulphate, a three-electron gain per chromium atom. That colour change from orange to green is the visual signal every student should learn to recognise: it means dichromate has done its work as an oxidising agent.

potassium dichromate, \(\mathrm{K_2Cr_2O_7}\) (A strong oxidising agent, orange-red in the solid and orange in solution → needs an acidic medium, usually dilute sulphuric acid → oxidises iodide to iodine and hydrogen sulphide to sulphur, and is reduced to green chromium(III), so the orange-to-green colour change acts as a built-in indicator that the reaction has taken place.)

\(\mathrm{KMnO_4}\) vs \(\mathrm{K_2Cr_2O_7}\)

Feature \(\mathrm{KMnO_4}\) \(\mathrm{K_2Cr_2O_7}\)
Oxidation state of the metal Mn, \(+7\) Cr, \(+6\)
Colour of the crystals Deep purple, almost black Orange-red
Colour in solution Deep purple Orange
Product after reduction in acid \(\mathrm{Mn^{2+}}\), effectively colourless \(\mathrm{Cr^{3+}}\) as chromic sulphate, green
Electrons gained per metal atom in acid 5 3
Prepared from \(\mathrm{MnO_2}\) (pyrolusite route) Chromite ore, \(\mathrm{FeO.Cr_2O_3}\)

The orange-to-green change of acidified dichromate was the basis of the first practical breathalysers. A driver blew through a tube containing crystals coated with acidified potassium dichromate; ethanol vapour in the breath reduced the orange chromium(VI) to green chromium(III), and the length of the green band gave a rough measure of how much alcohol was present. Modern instruments use fuel-cell or infrared sensors instead, which are far more precise and avoid handling a chromium(VI) compound at the roadside — but the chemistry of that first device is exactly the reaction written above.

Calculate the mass of \(\mathrm{KMnO_4}\) required to prepare \(250\ \text{mL}\) of a \(0.02\ \text{M}\) solution. (Molar mass of \(\mathrm{KMnO_4} = 158\ \text{g mol}^{-1}\).)

Questions from this section are usually set as complete-the-reaction, balance-the-equation, or colour-change items — for instance, naming the product formed when permanganate acts in an alkaline medium, or explaining why an orange solution turns green. In every case, start by writing the oxidation number of manganese or chromium on both sides; the rest of the answer follows from that one step.

8.8 Common properties of d block elements

Having examined the first transition series in detail, we can now gather the findings into a single working list. What follows applies to the d block as a whole, and it is worth reading as a checklist: every entry has already been explained somewhere in this chapter, and each one traces back to the partially filled (n-1)d subshell.

Physical properties

  1. All d block elements are lustrous and shining.

  2. They are hard and have high density.

  3. Have high melting and boiling points.

  4. Are good electrical and thermal conductors.

  5. Have high tensile strength and malleability. Tensile strength is the pull a material can withstand before it snaps, which is why these metals are chosen for cables, girders and machine parts.

  6. Can form alloys with transition and non transition elements.

  7. Many metals and their compounds are paramagnetic.

  8. Most of the metals are efficient catalysts.

Chemical properties

  1. All d block elements are electropositive metals.

  2. They exhibit variable valencies and form colored salts and complexes.

  3. They are good reducing agents — they part with electrons to other species, which is the natural behaviour of a metal that has several accessible oxidation states.

  4. They form insoluble oxides and hydroxides.

  5. Iron, cobalt, copper, molybdenum and zinc are biologically important metals.

  6. They catalyse biological reactions.

The last two entries are easy to skim past, yet they describe chemistry running inside your body at this moment. Iron sits at the centre of haemoglobin, where it binds oxygen in the lungs and releases it in the tissues — a job it can do because it can hold oxygen reversibly without being permanently oxidised. Cobalt is the metal at the heart of vitamin B₁₂. Zinc is built into hundreds of enzymes, among them carbonic anhydrase, which converts carbon dioxide to bicarbonate fast enough for your blood to carry it away. Molybdenum sits in nitrogenase, the enzyme with which certain soil bacteria convert atmospheric nitrogen into a form plants can use — the biological equivalent of the Haber process, running at ordinary temperature and pressure. In each case the metal is doing exactly what it does in a laboratory catalyst: shifting oxidation state to move electrons from one molecule to another.

Differences between the three series

Although most properties exhibited by d block elements are similar, the elements of the first row differ from the second and third rows in the stabilization of higher oxidation states in their compounds.

For example, Mo(V) and W(VI) compounds are more stable than Cr(VI) and Mn(VII). The highest oxidation state for elements of the first row is \(+7\), and in the case of the 3rd row a \(+8\) oxidation state is reached, as in \(\mathrm{RuO_4}\) and \(\mathrm{OsO_4}\).

Q. Why should the heavier transition metals hold on to high oxidation states more comfortably than the first-row metals?

A. The 4d and 5d orbitals are larger and more diffused than 3d orbitals, so they reach further out from the nucleus and overlap more effectively with the orbitals of the surrounding atoms. Electrons in these orbitals are drawn into bonding more readily, and the energy released by forming those extra bonds pays for the cost of stripping the metal down to a high oxidation state. In the compact 3d orbitals of the first row, that pay-off is smaller, so high oxidation states such as Cr(VI) and Mn(VII) exist but are far more eager to be reduced.

3d vs 4d vs 5d series

Feature 3d series (first row) 4d series (second row) 5d series (third row)
Highest oxidation state reached \(+7\) Reaches \(+8\), as in \(\mathrm{RuO_4}\) Reaches \(+8\), as in \(\mathrm{OsO_4}\)
Stability of the highest state Low; Cr(VI) and Mn(VII) are strong oxidising agents Higher; Mo(V) compounds are more stable than Cr(VI) Highest; W(VI) compounds are more stable than Cr(VI)
Size of the d orbitals Compact Larger and more diffused Largest and most diffused
Ionisation enthalpies Lowest of the three Similar to the 3d series Much higher, because of poorly shielding filled 4f orbitals

stability of high oxidation states down a group of transition elements (down a group ↓: the higher oxidation states become progressively more stable — Cr(VI) is a strong oxidising agent, Mo(V) and W(VI) compounds are comparatively stable; because the larger, more diffused 4d and 5d orbitals are more readily involved in bonding, so the energy released in bond formation offsets the energy spent in removing so many electrons)

The highest oxidation state shown by any element of the first transition series is \(+7\), reached by manganese. Iron has one more electron in its outer shells than manganese. Explain, using electronic configurations, why iron does not therefore show \(+8\).

Write the outer configurations first.

Manganese: \(\mathrm{3d^5\,4s^2}\) — a total of 7 electrons in the 3d and 4s orbitals, all of them removable, giving a maximum of \(+7\).

Iron: \(\mathrm{3d^6\,4s^2}\) — a total of 8 electrons, which on the arithmetic alone would suggest \(+8\).

The arithmetic is not the whole story. Beyond manganese, the nuclear charge has increased by one more unit while the added electron has gone into the poorly shielding 3d subshell. The effective nuclear charge felt by the d electrons is therefore higher in iron, and they are held more tightly. Removing all eight would cost more energy than any compound could recover through bond formation.

In addition, the sixth d electron in iron must pair up in an already occupied d orbital, and pulling an electron out of a pair is not the difficulty here — the difficulty is that after \(+6\) the remaining electrons are simply too strongly bound. Table 8.3 confirms this: iron’s highest listed oxidation state is \(+6\), not \(+8\). Only in the larger 4d and 5d elements, with their more diffused orbitals, does \(+8\) actually become attainable.

Osmium is named from the Greek osme, meaning smell. Osmium tetroxide, the very \(\mathrm{OsO_4}\) in which the metal reaches the \(+8\) oxidation state, is volatile and has a sharp, penetrating odour — unusual behaviour for a metal compound, and unmistakable enough that the element was named for it.

Assign the oxidation number of the metal in \(\mathrm{RuO_4}\) and in \(\mathrm{OsO_4}\), showing your working. Then name the element of the first transition series that comes closest to this value, and write the formula of the ion in which it does so.

This section is a favourite source of “give the similarities and differences between the 3d, 4d and 5d series” questions, and of assertion–reason items pairing a statement about Cr(VI) with one about W(VI). Answer them by naming the property first and the reason second, and the reason will nearly always be orbital size and effective nuclear charge.

8.9 Extraction of metals

How are metals found in nature? Name two salts of metals that are found in nature.

Most metals are found in the earth’s crust in the form of their salts, such as carbonates, sulphates, sulphides and oxides. A reactive metal cannot survive as the free element for long; oxygen, water, carbon dioxide and sulphur in the environment combine with it over geological time, and what we dig up is the resulting compound.

A few metals are nonreactive and occur in the free state in the earth’s crust — for example, silver, gold and platinum. These are exactly the metals that resist corrosion, which is also why they have been prized for ornaments and coinage for thousands of years.

Minerals and ores

Not every rock containing a metal is worth mining, and chemistry makes a careful distinction between the two words used here.

NoteDefinition

A mineral is a naturally occurring substance found in the earth’s crust containing inorganic salts, solids, siliceous matter and so on.

The mineral which contains a high percentage of the metal, and from which the metal can be extracted economically, is called an ore.

NoteDefinition

An ore is a mineral from which the metal can be extracted economically, because the percentage of metal it contains is high enough to be worth processing.

Every ore is a mineral, but not every mineral is an ore. The deciding factor is partly chemical and partly economic — a mineral too poor in metal, or too difficult to reduce, is left in the ground. Table 8.9 lists the important minerals of three transition metals and shows which of them are actually worked as ores.

Table 8.9 : List of minerals and ores of some transition metals

Metals Mineral Ore
Iron Haematite \(\mathrm{Fe_2O_3}\)
Magnetite \(\mathrm{Fe_3O_4}\)
Limonite \(2\mathrm{Fe_2O_3}\cdot3\mathrm{H_2O}\)
Iron pyrites \(\mathrm{FeS_2}\)
Siderite \(\mathrm{FeCO_3}\)		 Haematite
Copper Chalcopyrite \(\mathrm{CuFeS_2}\)
Chalcocite \(\mathrm{Cu_2S}\)
Cuprite \(\mathrm{Cu_2O}\)		 Chalcopyrite
Chalcocite
Zinc Zinc blende \(\mathrm{ZnS}\)
Zincite \(\mathrm{ZnO}\)
Calamine \(\mathrm{ZnCO_3}\)		 Zinc blende

Iron extraction has been known to Indians since 700 BC, and Indian blacksmiths also understood the thermo-mechanical process for forging — heating and hammering metal to change its properties, not merely its shape. Archaeological studies have turned up iron objects belonging to the middle iron age at excavated sites. The most famous survivor is the iron pillar at Delhi, which the textbook records as 1300 years old and free of rust to this day, an achievement in corrosion resistance that modern metallurgists have studied closely.

8.9.1 Metallurgy

Commercial extraction of metals from their ores is called metallurgy. Different methods are used for their extraction, depending on the nature of a metal and its ore.

NoteDefinition

Metallurgy is the commercial extraction of a metal from its ore.

The choice of method is not arbitrary. A metal that is easy to reduce can be smelted with cheap carbon; a metal that clings to its electrons must be prised free by electricity. The three broad routes are set out in FC 8.7.

Flowchart

FC 8.7 – Branches of metallurgy: metallurgy → branch 1, pyrometallurgy (ore reduced at high temperature using C, H₂ or Al); → branch 2, hydrometallurgy (metal recovered from an aqueous solution of its salt using a suitable reducing agent); → branch 3, electrometallurgy (metal obtained by electrolytic reduction of the molten, fused compound).

a. Pyrometallurgy: A process in which the ore is reduced to metal at high temperature using reducing agents like carbon, hydrogen, aluminium, etc., is called pyrometallurgy. The prefix comes from heat, and heat is exactly what this route depends on.

b. Hydrometallurgy: The process of extracting metals from the aqueous solution of their salts using a suitable reducing agent is called hydrometallurgy. Here the metal is first brought into solution and then displaced from it.

c. Electrometallurgy: A process in which metal is extracted by electrolytic reduction of a molten (fused) metallic compound is called electrometallurgy. Electrons are supplied directly by the current, which is why this route is reserved for metals no chemical reducing agent can handle.

pyrometallurgy vs hydrometallurgy vs electrometallurgy

Feature Pyrometallurgy Hydrometallurgy Electrometallurgy
State of the ore during reduction Solid, at high temperature Dissolved in aqueous solution Molten (fused) compound
Source of electrons Chemical reducing agent — C, \(\mathrm{H_2}\), Al Suitable chemical reducing agent in solution Electric current
Energy input Heat Chemical, at ordinary temperature Electrical
Suited to Metals whose oxides are readily reduced by carbon Metals recoverable from solution Highly reactive metals that resist chemical reduction
Steps involved in the process of extraction

Concentration. After mining the ore from the earth’s crust, it is subjected to concentration. In this step, impurities termed as gangue are removed from the ore and the ore gets concentrated.

NoteDefinition

Gangue is the sand, mud and other unwanted impurities which remain mixed with the ore deposit.

During the process of concentration, the ore is separated from the gangue material using different methods such as washing, hydraulic classification, magnetic separation, froth floatation, etc. Each of these exploits a physical difference between ore and gangue — density in washing and hydraulic classification, magnetic behaviour in magnetic separation, and the tendency of a sulphide surface to cling to oil and froth in froth floatation.

The method chosen for concentration depends upon the nature of the ore.

8.9.2 Extraction of Iron from Haematite ore using Blast furnace

Iron is the fourth most abundant element in the earth’s crust, which is one reason it became the working metal of civilisation.

The haematite ore that is mined is not pure iron oxide. Its composition is:

\[\mathrm{Fe_2O_3} + \mathrm{SiO_2} + \mathrm{Al_2O_3} + \text{phosphates}\]

where the silica, alumina and phosphates together make up the gangue. Everything the furnace does, apart from reducing the oxide, is aimed at getting rid of that gangue.

Iron is extracted from haematite by its reduction using coke and limestone. Carbon from the coke is converted to carbon monoxide, and carbon and carbon monoxide together reduce \(\mathrm{Fe_2O_3}\) to metallic iron.

Limestone acts as flux; it combines with the gangue material to form molten slag. Slag is the fusible waste product so formed, and being lighter than molten iron it floats clear and can be drawn off separately.

The overall industrial route, from ore to pure metal, is laid out in FC 8.8, and extraction of iron from haematite ore involves the following steps.

Flowchart

FC 8.8 – General route from ore to pure metal: iron ore → crushing and grinding → concentration, by magnetic separation / gravity separation / leaching / froth flotation → reduction, by coke / by CO / by heat / by Al / by electrolysis → refining, by liquification / distillation / oxidation / electro-refining → pure iron.

i. Concentration

The powdered ore is washed in a powerful current of water introduced into the hydraulic classifier. The lighter gangue particles are separated and carried away by the stream, and the concentrated ore, being denser, is collected at the bottom.

ii. Roasting

The concentrated ore is heated in a current of air. The sulfur and arsenic impurities present in the ore get converted into their oxides and escape as vapour, so a chemical cleaning has taken place before any reduction begins. Ferrous oxide in the ore is converted to \(\mathrm{Fe_2O_3}\).

\[4\,\mathrm{FeO} + \mathrm{O_2} \rightarrow 2\,\mathrm{Fe_2O_3}\]

The roasted ore is converted into lumps by sintering. Lumps are essential, because powder would pack tightly and block the upward flow of gas through the furnace.

iii. Reduction (Smelting)

This step is carried out in a blast furnace. The blast furnace is a tall cylindrical steel tower which is lined with refractory bricks — bricks that survive very high temperatures without melting or reacting.

The height of a typical blast furnace is 25 m and its diameter varies between 5 and 10 m. The furnace works on the counter current principle, where the charge comes down and hot gases move up the tower. This opposed movement is what makes the furnace efficient: the descending solid is warmed by the very gases that have already done their work lower down. The furnace is comprised of 3 parts — 1. Hearth, 2. Bosh and 3. Stack, as shown in Fig. 8.5.

Diagram

Fig. 8.5 – Vertical section of a blast furnace, labelled from top to bottom: burden charging point and cup-and-cone arrangement, stack (250°C, CO + N₂ leaving, indirect reduction, swelling and softening), bosh (1200°C, coke solution, direct reduction, primary slag formation), tuyeres (1700°C, hot blast air entering), 1500°C raceway region, slag tap hole, and hearth (1350°C, pig iron tap hole); the principal reaction of each zone written alongside.

The charge containing ore and limestone is introduced into the furnace through a cup and cone arrangement. In this arrangement the cone enables uniform distribution of charge and the cup prevents the loss of gases. A blast of preheated air is introduced into the furnace below the bosh. The charge and hot air come in contact with each other and various reactions take place.

limestone, \(\mathrm{CaCO_3}\), in the blast furnace (Decomposes on heating to quicklime, CaO → needs the high temperature of the slag-formation zone, around 1200 K → the CaO produced combines with the acidic gangue oxides, silica and alumina, to give a molten slag that floats on the iron and can be tapped off separately. It is charged in with the ore purely to remove waste, not because it contains any iron.)

Fig. 8.5 – the blast furnace in section (Read the temperatures marked on the diagram. Which part of the furnace is hottest, and which is coolest? At what height is the hot air blown in, and at what height does the charge enter? Explain how these two answers together account for the counter current principle described in the text.)

Reactions in the blast furnace

There are different temperature zones in the blast furnace. The temperature goes on increasing from top to bottom in the furnace. At the top, the temperature is 500 K. The maximum temperature of the furnace is 2000 K above the tuyers. There are 3 temperature zones in the furnace.

  1. Zone of combustion — combustion of coke with \(\mathrm{O_2}\) in the air.
  2. Zone of reduction — reduction of \(\mathrm{Fe_2O_3}\) to metallic iron.
  3. Zone of slag formation — formation of slag by reaction of gangue with limestone.

The chemistry of each zone, together with the fusion and refining stages that follow, is set out below and summarised in FC 8.9.

Flowchart

FC 8.9 – Zones of the blast furnace, from the top downwards: charge enters at top (500 K, moisture lost) → zone of reduction, 22–25 m near the top (~900 K, Fe₂O₃ + 3CO → 2Fe + 3CO₂, spongy iron formed) → zone of slag formation, 20 m (1200 K, CaCO₃ → CaO + CO₂; CaO + SiO₂ → CaSiO₃) → zone of fusion, 15 m (MnO₂ and Ca₃(PO₄)₂ reduced to Mn and P; spongy iron melts and absorbs C, Si, Mn, P, S) → zone of combustion, 5–10 m from the bottom (~2000 K, C + ½O₂ → CO, highly exothermic) → hearth: molten iron collects at the bottom, lighter slag floats above it, both tapped through separate outlets; hot gases N₂, CO, CO₂ rise and leave at the top, and are used to preheat the incoming air blast.

1. Zone of combustion. This is 5 - 10 m from the bottom. The hot air blown through the tuyers reacts with coke from the charge to form CO.

\[\mathrm{C} + \frac{1}{2}\,\mathrm{O_2} \rightarrow \mathrm{CO}, \qquad \Delta H = -220\ \text{kJ}\]

enthalpy change, ΔH (SI unit: joule, J, though kilojoule, kJ, is the practical unit here. Sign convention: ΔH negative means heat is released to the surroundings, so the reaction is exothermic; ΔH positive means heat is absorbed and the reaction is endothermic. The value \(-220\) kJ above therefore tells you at a glance that this reaction heats the furnace rather than cooling it.)

The reaction is highly exothermic; thus the temperature of this zone is around 2000 K. Some of the CO formed dissociates to form finely divided carbon.

\[2\,\mathrm{CO} \rightarrow 2\,\mathrm{C} + \mathrm{O_2}\]

The hot gas rich in CO rises upwards in the blast furnace. The charge coming down gets heated and reacts with CO. Thus CO acts as a fuel and also a reducing agent — it carries heat upward and strips oxygen from the ore on the way.

2. Zone of Reduction (22-25 m near the top). Here, the temperature is around 900 K. \(\mathrm{Fe_2O_3}\) is reduced to spongy iron by CO.

\[\mathrm{Fe_2O_3} + 3\,\mathrm{CO} \rightarrow 2\,\mathrm{Fe} + 3\,\mathrm{CO_2}\]

Some amount of \(\mathrm{Fe_2O_3}\) is reduced to iron by carbon.

\[\mathrm{Fe_2O_3} + 3\,\mathrm{C} \rightarrow 2\,\mathrm{Fe} + 3\,\mathrm{CO}\]

The iron formed here is called spongy because it is porous — it has been produced in the solid state, at a temperature well below the melting point of iron, so it never becomes liquid at this stage.

In the reaction \(\mathrm{Fe_2O_3} + 3\,\mathrm{CO} \rightarrow 2\,\mathrm{Fe} + 3\,\mathrm{CO_2}\), identify the oxidising agent and the reducing agent, and state the change in oxidation number of each element involved.

Assign oxidation numbers on both sides, taking oxygen as \(-2\).

Iron in \(\mathrm{Fe_2O_3}\): \(2x + 3(-2) = 0\), so \(x = +3\). In the free metal Fe, the oxidation number is \(0\). Iron therefore falls from \(+3\) to \(0\) — it is reduced, so \(\mathrm{Fe_2O_3}\) is the oxidising agent.

Carbon in CO: \(x + (-2) = 0\), so \(x = +2\). In \(\mathrm{CO_2}\): \(x + 2(-2) = 0\), so \(x = +4\). Carbon rises from \(+2\) to \(+4\) — it is oxidised, so CO is the reducing agent.

Each iron atom gains 3 electrons and each carbon atom loses 2. For 2 iron atoms that is 6 electrons gained, and for 3 carbon atoms 6 electrons lost — which is exactly why the equation balances with the coefficients 1 : 3 : 2 : 3.

3. Zone of slag formation (20 m unit). The gangue present in the ore is converted to slag. This slag can be used for making road foundation. The temperature of this zone is 1200 K.

The gangue contains silica, alumina and phosphates. Removal of this gangue is effected by adding lime-stone in the charge, which acts as flux. Limestone decomposes to give CaO (quick lime).

\[\mathrm{CaCO_3} \xrightarrow{\ \Delta\ } \mathrm{CaO} + \mathrm{CO_2}\]

CaO combines with gangue to form molten slag of calcium silicate and calcium aluminate.

\[\mathrm{CaO} + \mathrm{SiO_2} \rightarrow \mathrm{CaSiO_3}\]

\[12\,\mathrm{CaO} + 2\,\mathrm{Al_2O_3} \rightarrow 4\,\mathrm{Ca_3AlO_3} + 3\,\mathrm{O_2}\]

Notice the chemical logic: silica and alumina are acidic oxides, and quicklime is a basic oxide, so they combine readily. That is the whole reason a substance containing no iron at all is shovelled into an iron furnace.

Q3. Iron ore is reduced in the blast furnace using coke, yet limestone — which contains no iron at all — is also charged into the furnace. What job could it possibly be doing?

A. Limestone is the flux. In the heat of the furnace it decomposes to quicklime, CaO, which then combines with the acidic gangue oxides in the ore — silica and alumina — to form a molten slag of calcium silicate and calcium aluminate. The slag is lighter than molten iron, so it floats on top and is tapped off through a separate hole, carrying the earthy impurities out of the furnace with it. Without limestone the gangue would stay mixed with the iron and the product would be hopelessly impure.

The slag drawn off a blast furnace was once simply dumped. Today it is rapidly cooled with water, ground to a fine powder, and blended into cement — ground granulated blast-furnace slag is a standard ingredient of the cement used in dams, bridges and marine structures, where it gives concrete better resistance to chemical attack. Producing cement clinker releases a great deal of carbon dioxide, so every tonne of slag used in its place lowers the carbon footprint of construction. A waste stream from one industry has become a raw material for another, which is exactly what green chemistry sets out to achieve.

4. Zone of fusion (15 m ht). \(\mathrm{MnO_2}\) and \(\mathrm{Ca_3(PO_4)_2}\) present in the iron ore are reduced to Mn and P. Some of the silica is also reduced to Si.

The spongy iron coming down in the furnace melts and absorbs impurities like C, Si, Mn, P and S. This molten iron collects at the bottom in the furnace. The slag, which is lighter, floats on the surface of molten iron. Molten slag and iron are collected through separate outlets.

Molten iron is poured into moulds. These solid blocks are called pigs. This iron contains about 4% of carbon. When pig iron is remelted, run into moulds and cooled, it becomes cast iron.

The waste gases containing \(\mathrm{N_2}\), CO and \(\mathrm{CO_2}\) escape through the outlet at the top. These hot gases are used for preheating the blast of air, so the heat they carry is recovered rather than lost up the chimney.

5. Refining. Pure iron can be obtained by electrolytic refining of impure iron, or by the other methods given in the flow chart. The choice of extraction technique is governed by the following factors:

  1. Nature of ore.
  2. Availability and cost of reducing agent; generally cheap coke is used.
  3. Availability of hydraulic power.
  4. Purity of product (metal) required.
  5. Value of byproducts — for example, \(\mathrm{SO_2}\) obtained during roasting of sulphide ores is vital for the manufacture of \(\mathrm{H_2SO_4}\).

Knowledge of the electrochemical series provides solutions to many problems, because it tells us in advance which reducing agent is strong enough to displace a given metal.

Table 8.10 : Summary of reactions taking place in blast furnace at different temperature zones

Temp K Changes taking place Reaction
500 loss of moisture from ore —
900 Reduction of ore by CO \(\mathrm{Fe_2O_3} + 3\,\mathrm{CO} \rightarrow 2\,\mathrm{Fe} + 3\,\mathrm{CO_2}\)
1200 Decomposition of lime \(\mathrm{CaCO_3} \xrightarrow{\ \Delta\ } \mathrm{CaO} + \mathrm{CO_2}\)
1500 Reduction of ore by C \(\mathrm{Fe_2O_3} + 3\,\mathrm{C} \rightarrow 2\,\mathrm{Fe} + 3\,\mathrm{CO}\)
2000 Slag formation \(\mathrm{CaO} + \mathrm{SiO_2} \rightarrow \mathrm{CaSiO_3}\)
\(12\,\mathrm{CaO} + 2\,\mathrm{Al_2O_3} \rightarrow 4\,\mathrm{Ca_3AlO_3} + 3\,\mathrm{O_2}\)

Calculate the mass of iron that can be obtained from 1000 kg of haematite ore containing 80% \(\mathrm{Fe_2O_3}\) by mass, assuming the reduction is complete. (Atomic masses: Fe = 56, O = 16)

First find how much of the ore is actually iron oxide.

\[\text{Mass of }\mathrm{Fe_2O_3} = \frac{80}{100} \times 1000 = 800\ \text{kg}\]

Next, work out what fraction of \(\mathrm{Fe_2O_3}\) is iron. Its molar mass is

\[M(\mathrm{Fe_2O_3}) = 2 \times 56 + 3 \times 16 = 112 + 48 = 160\ \text{g mol}^{-1}\]

Of this, 112 units are iron, so the mass fraction of iron is

\[\frac{112}{160} = 0.70\]

Therefore the mass of iron obtainable is

\[0.70 \times 800 = 560\ \text{kg}\]

The 20% of the ore that is gangue contributes nothing to this figure — it leaves the furnace in the slag. Note also that the answer needs no balanced equation; the mass ratio inside the formula \(\mathrm{Fe_2O_3}\) is enough, because all the iron in the ore ends up as metal.

Calculate the mass of quicklime, CaO, produced by the complete decomposition of 100 kg of pure limestone, \(\mathrm{CaCO_3}\). (Atomic masses: Ca = 40, C = 12, O = 16)

Commercial forms of iron

Iron leaves the furnace as one material but reaches the market as three, distinguished mainly by how much carbon they contain. The three commercial forms are shown in FC 8.10.

Flowchart

FC 8.10 – Commercial forms of iron: iron → branch 1, cast iron (about 4% carbon, hard and brittle); → branch 2, wrought iron (less than 0.2% carbon, very soft); → branch 3, steel (0.2 to 2% carbon, neither too hard nor too soft).

Iron melts at a very high temperature (1800 K). On addition of carbon its melting point decreases, depending upon the percentage of carbon. The mechanical properties of steel can be modified by the addition of small amounts of suitable elements such as manganese, chromium, sulfur, nickel, etc. These elements are called alloying elements, and such steels are alloy steels.

Differences between cast iron, wrought iron and steel

Feature Cast iron Wrought iron Steel
Nature Hard and brittle Very soft Neither too hard nor too soft
Carbon content Contains 4% carbon Contains less than 0.2% carbon Contains 0.2 to 2% carbon
Uses Making pipes, manufacturing automotive parts, pots, pans, utensils Making pipes, bars for stay bolts, engine bolts and rivetts Buildings infrastructure, tools, ships, automobiles, weapons etc.

Read the table across and a single variable explains everything: carbon content. More carbon means more interstitial atoms wedged into the iron lattice, hence greater hardness but also greater brittleness — which is precisely the behaviour of interstitial compounds described in Section 8.6.7.

Iron possesses a high degree of magnetism below 1042 K. This is known as ferromagnetism, the property we met earlier in Section 8.6.4; above that temperature the ordered alignment of electron spins breaks down and iron loses its permanent magnetism.

Explain why the charge is fed into a blast furnace at the top while the blast of hot air is introduced near the bottom, and state what would go wrong if the two were interchanged.

Questions from this section are commonly set as “name the zones of the blast furnace and write the reactions occurring in them”, as differences between the three commercial forms of iron, or as short numericals on the yield of metal from an ore of stated purity. Learning the zones in order of increasing temperature, together with one reaction for each, covers most of what is asked.

8.10 Inner Transition (f-block) Elements: Lanthanoids and Actinoids

Elements whose f orbitals get filled up by electrons are called f block elements. These elements are placed separately at the bottom of the periodic table. They are a subset of the 6th and 7th periods — that is, they are not a separate part of the table at all, but two long stretches of those two periods, lifted out and set below for convenience.

NoteDefinition

f block elements are the elements in whose atoms the f orbitals are progressively filled with electrons.

Why are f-block elements called inner transition metals? Are there any similarities between transition and inner transition metals?

Why “inner” transition elements

Since the f orbital lies much inside the d orbital, in relation to the transition metals the f block elements are called inner transition elements. The electrons being added go into a subshell buried two shells below the outermost one, as Fig. 8.5a shows, so the filling happens deeper inside the atom than in an ordinary transition element.

Diagram

Fig. 8.5a – Simple radial sketch of an f-block atom: concentric shaded regions labelled from the nucleus outwards as 4f (innermost of the three shown, drawn well inside), then 5d, then 6s as the outermost, with a caption arrow indicating that the electron being added enters the innermost 4f region while the outer 6s shell stays unchanged.

NoteDefinition

Inner transition elements are those elements in which the last electron enters an f subshell, which lies well inside the d subshell of the same atom.

These elements have 1 to 14 electrons in their f orbital, 0 or 1 in the penultimate energy level, and 2 electrons in the outermost orbital. The penultimate level is simply the shell just inside the outermost one.

The two series

The lanthanoids are characterized by gradual filling up of 4f orbitals, and the actinides by the 5f orbitals. There are 14 elements filling the f orbital in each series.

The relative energies of the nd and (n-1)f orbitals are very similar and are sensitive to electronic configurations. Because the two sets of orbitals lie so close in energy, a small change in the number of electrons present can decide which of the two an incoming electron chooses — which is why the observed configurations of these elements are not always the ones simple rules would predict.

Why does each inner transition series contain exactly 14 elements, no more and no fewer?

The number follows directly from the number of orbitals available in an f subshell.

An f subshell consists of 7 orbitals. By Pauli’s exclusion principle, each orbital can hold a maximum of 2 electrons, of opposite spin.

\[\text{Maximum electrons in an f subshell} = 7 \times 2 = 14\]

Filling one f subshell therefore takes 14 electrons, and since each successive element in the series adds exactly one electron to that subshell, exactly 14 elements are needed to complete it. The lanthanoids fill the 4f subshell and the actinoids the 5f subshell, so each series contains 14 elements. The same argument, applied to the 5 d orbitals, explains why a transition series contains 10 elements.

Q. If the lanthanoids and actinoids really belong to periods 6 and 7, why are they printed separately at the bottom of the periodic table?

A. Purely for convenience of printing and reading. Inserting 14 extra elements into each of those two periods would make the table roughly twice as wide as it is, and the columns would no longer line up with the groups above. Pulling the two series out and placing them below keeps the main body of the table compact, while their true position — immediately after lanthanum and actinium — is understood.

Lanthanum, the element that gives the lanthanoids their name, is named from the Greek lanthanein, meaning “to lie hidden”. It was found concealed within a cerium mineral, and the name turned out to be an accurate description of the whole family: these elements occur together, resemble one another closely, and stayed hidden from chemists far longer than most of the periodic table.

The story of how these elements came to occupy their present position in the periodic table — and of the chemist who insisted on putting them there against the advice of his colleagues — is told in the opening pages of this chapter.

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

Feature Transition elements Inner transition elements
Subshell being filled (n-1)d (n-2)f
Depth of the subshell in the atom Penultimate shell One shell deeper still, well inside the d orbitals
Number of elements in one series 10 14
Position in the periodic table Main body, between the s and p blocks Placed separately at the bottom
Periods involved 4th to 7th 6th and 7th only
Names of the series 3d, 4d, 5d, 6d Lanthanoids (4f), actinoids (5f)

Cerium has atomic number 58 and thorium has atomic number 90. State which f subshell is being filled in each, name the series to which each belongs, and give the period of the periodic table in which each element actually sits.

Questions on this section are usually short and definitional — why the f block is called “inner”, how many elements each series contains and why, and one or two similarities between transition and inner transition metals. Answer the “how many” questions by counting orbitals rather than by memorising the number.

8.11 Properties of f-block elements

  1. Properties are similar to d block elements.

  2. Electrons are added to f subshells of the (n-2) level.

  3. Placed between the (n-1)d and ns block elements.

These three statements say, in compressed form, that the f-block elements behave like metals of the transition type but have their newly added electrons tucked two shells below the surface. The rest of this section works out what that arrangement means in practice.

Are the rare earths really rare?

Lanthanoids begin with atomic number 57 and end at 71. Although, historically, lanthanoids are termed as rare earth elements, they are fairly abundant in the earth’s crust. For example, thulium is found in more abundance than silver (\(4.5 \times 10^{-5}\) versus \(0.79 \times 10^{-5}\) percent by mass).

The name rare earth elements was coined because of the difficulty in extracting them economically in pure form from other lanthanoids having similar chemical properties. The problem was never scarcity but separation — the fourteen elements are so alike chemically that ordinary methods could not tell them apart.

Now, due to newer separation methods like ion exchange resins, the separation of these elements has become easier and more economical.

Separating one lanthanoid from another was, for over a century, one of the hardest routine problems in inorganic chemistry; early workers repeated fractional crystallisations thousands of times to obtain a single element in reasonable purity. Ion-exchange chromatography, developed in the middle of the twentieth century, changed that completely: a resin column holds the slightly different \(\mathrm{Ln^{3+}}\) ions with slightly different strengths, and a suitable eluting agent washes them off one at a time, in order of ionic size. The tiny difference in radius that makes these elements so hard to distinguish chemically is precisely what the column exploits.

Four elements — yttrium, ytterbium, terbium and erbium — all take their names from Ytterby, a single small quarry village in Sweden, because the minerals first found there proved to contain a whole tangle of these look-alike elements. Europium, by contrast, is named after the continent of Europe. No other patch of ground has given its name to so much of the periodic table.

Physical nature of the lanthanoid metals

These metals are soft with moderate densities of about \(7\ \text{g cm}^{-3}\). They have high melting points (~1000 °C) and boiling points (~3000 °C).

Similar to groups 1 and 2, lanthanoids in the metallic state are very reactive and resemble alkali and alkaline earth metals in their reactivities more than they resemble transition metals. For example, they react with water to give the metal hydroxide and hydrogen gas.

\[2\,\mathrm{M}\text{(s)} + 6\,\mathrm{H_2O}\text{(l)} \rightarrow 2\,\mathrm{M(OH)_3}\text{(s)} + 3\,\mathrm{H_2}\text{(g)}\]

The energy required to break up the metal lattice is the heat of atomization.

NoteDefinition

Heat of atomization is the energy required to break up the metal lattice and convert the metal into free gaseous atoms.

Lanthanoids have a lower heat of atomization than transition metals. This is because, with d electrons available for bonding, transition metals are much harder and require a high heat of atomization. The buried 4f electrons of a lanthanoid contribute far less to holding the lattice together, so the metal is softer and its atoms are more easily separated.

Europium and ytterbium have the lowest enthalpies of vaporization and the largest atomic radii of the lanthanoids, and resemble barium. These two elements resemble alkaline earth elements; they dissolve in liquid ammonia to give blue conducting solutions — the same striking behaviour shown by sodium and the other alkali metals.

lanthanoids vs alkali and alkaline earth metals vs transition metals

Feature Alkali / alkaline earth metals Lanthanoids Transition metals
Hardness Soft Soft Hard
Heat of atomization Low Low High
Reactivity of the metal Very high Very high Moderate
Reaction with water Gives hydroxide and \(\mathrm{H_2}\) Gives \(\mathrm{M(OH)_3}\) and \(\mathrm{H_2}\) Generally slow or no reaction
Characteristic oxidation state \(+1\) or \(+2\) \(+3\) Variable
Oxidation states of the lanthanoids

Although the common oxidation state for lanthanoids is \(+3\), the \(+2\) oxidation state is also important. They all form stable oxides of the type \(\mathrm{M_2O_3}\), where M is the metal ion.

\(\mathrm{Eu^{2+}}\) and \(\mathrm{Yb^{2+}}\) are the most stable dipositive metal ions. Higher oxidation states are unusual for lanthanoids, with the only exception of cerium, which forms a stable \(+4\) species.

Q. Why should \(+3\) be the characteristic oxidation state of almost every lanthanoid, when the transition metals of Section 8.4 showed such variety?

A. A lanthanoid loses its two outermost 6s electrons together with one more from the 5d or 4f level, which brings it to \(+3\) and no further. The remaining 4f electrons lie deep inside the atom, screened by the outer 5s and 5p shells, and are therefore not available for bonding at all. In a transition metal the d electrons sit in the penultimate shell, close enough to the surface to be removed one after another — which is exactly why variable oxidation states are the rule there and the exception here.

Cerium is the one lanthanoid that forms a stable \(+4\) species. Using electronic configurations, explain why.

The observed ground-state configuration of cerium is \(\mathrm{[Xe]\,4f^1\,5d^1\,6s^2}\).

Removing four electrons — the two 6s electrons, the single 5d electron and the single 4f electron — gives

\[\mathrm{Ce} \rightarrow \mathrm{Ce^{4+}} + 4\,\mathrm{e^{-}}, \qquad \mathrm{Ce^{4+}} = \mathrm{[Xe]\,4f^0}\]

The resulting ion has an empty 4f subshell and the electronic arrangement of xenon, a noble gas. An empty subshell together with a noble-gas core is a particularly stable arrangement, so cerium is willing to pay the energy cost of removing a fourth electron.

No other lanthanoid gains anything comparable by going to \(+4\), because in every other case the fourth electron would have to be dragged out of a partly filled 4f subshell, leaving a less stable arrangement behind. That is why \(\mathrm{Ce^{4+}}\) stands almost alone. Note also what this predicts about chemistry: \(\mathrm{Ce^{4+}}\) readily accepts an electron to return to the common \(+3\) state, which makes cerium(IV) a useful oxidising agent.

Explain, using electronic configurations, why \(\mathrm{Eu^{2+}}\) and \(\mathrm{Yb^{2+}}\) are the most stable dipositive lanthanoid ions.

Write the ground-state configurations of the two metals.

Europium: \(\mathrm{[Xe]\,4f^7\,6s^2}\). Removing only the two 6s electrons gives

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

which is a half-filled 4f subshell — seven orbitals with one electron each.

Ytterbium: \(\mathrm{[Xe]\,4f^{14}\,6s^2}\). Removing the two 6s electrons gives

\[\mathrm{Yb^{2+}} = \mathrm{[Xe]\,4f^{14}}\]

which is a completely filled 4f subshell.

Half-filled and completely filled subshells carry extra stability, exactly as they did for chromium and copper in Section 8.4. In both these cases stopping at \(+2\) lands the ion on one of those specially stable arrangements, so there is no incentive to remove a third electron. For every other lanthanoid, stopping at \(+2\) would leave an ordinary partly filled 4f subshell, and going on to \(+3\) is preferred instead.

Size, coordination number and basicity

Their ionic radii decrease from 117 pm for La to 100 pm for Lu. This is because the f orbitals do not shield the outer 5s and 5p electrons effectively, leading to an increase in effective nuclear charge and a decrease in the ionic size.

Such large ions have a higher coordination number, that varies from 6 (most common) to 9, 10 and up to 12 in some cases. Coordination number here means the number of species attached directly to the central metal ion — a big ion simply has room around it for more neighbours. For example, the hydrated lanthanum ion is a nonahydrate, \([\mathrm{La(H_2O)_9}]^{3+}\), drawn in Fig. 8.5b.

Diagram

Fig. 8.5b – Structure of the nonahydrated lanthanum ion [La(H₂O)₉]³⁺: a central La³⁺ ion with nine water molecules arranged around it, each shown attached through its oxygen atom, with the 3+ charge marked on the whole ion.

All the lanthanoids form hydroxides of the general formula \(\mathrm{Ln(OH)_3}\), where Ln represents any element of the lanthanoid series. These are ionic and basic.

Since the ionic size decreases from \(\mathrm{La^{3+}}\) to \(\mathrm{Lu^{3+}}\), the basicity of the hydroxides decreases. \(\mathrm{La(OH)_3}\) is the strongest base while \(\mathrm{Lu(OH)_3}\) is the weakest base.

The reason is worth spelling out. A smaller ion of the same charge holds the hydroxide ion more tightly, so it releases \(\mathrm{OH^{-}}\) less readily into solution — and it is the release of \(\mathrm{OH^{-}}\) that makes a hydroxide basic. The steady shrinking of the ion across the series therefore produces a steady weakening of basic strength.

basic strength of \(\mathrm{Ln(OH)_3}\) across the lanthanoid series (across the series → from \(\mathrm{La(OH)_3}\) to \(\mathrm{Lu(OH)_3}\): basic strength decreases steadily; because the \(\mathrm{Ln^{3+}}\) ion becomes progressively smaller while its charge stays at \(+3\), so it grips the \(\mathrm{OH^{-}}\) ions more strongly and gives them up less readily — strongest base \(\mathrm{La(OH)_3}\), weakest base \(\mathrm{Lu(OH)_3}\))

Arrange \(\mathrm{La(OH)_3}\), \(\mathrm{Nd(OH)_3}\) and \(\mathrm{Lu(OH)_3}\) in increasing order of basic strength, and justify the order you give in one sentence.

Taking the ionic radius of \(\mathrm{La^{3+}}\) as 117 pm and that of \(\mathrm{Lu^{3+}}\) as 100 pm, calculate the percentage decrease in ionic radius across the lanthanoid series.

Compounds and colour

Lanthanoids react with nitrogen and halogens to give nitrides and halides of the formulae LnN and \(\mathrm{LnX_3}\) respectively. While doing so, lanthanoids lose their outermost 3 electrons to form a stable compound in the \(+3\) oxidation state.

When lanthanoids are heated at elevated temperatures (~2800 K) with carbon, the carbides with general formula \(\mathrm{LnC_2}\) are obtained.

In the \(+3\) oxidation state, many of the lanthanoids are coloured, mostly green, pink and yellow. This is attributed to the electronic transitions among the f orbitals — f–f transitions, the counterpart of the d–d transitions that coloured the transition metal ions in Section 8.6.5.

Like transition metals the electronic spectra of lanthanoids, however, do not get affected by different ligands. This is a genuine point of difference, and it follows from everything said above: the 4f orbitals are buried so deep inside the atom that an approaching ligand barely reaches them. The absorption bands therefore stay in almost the same place whatever the ligand — which is why a lanthanoid salt keeps its colour where a cobalt salt, as we saw, changes from pink to blue.

Board and entrance questions from this section tend to ask why \(\mathrm{Ce^{4+}}\), \(\mathrm{Eu^{2+}}\) or \(\mathrm{Yb^{2+}}\) is unusually stable, or to compare the basic strengths of two lanthanoid hydroxides. Both types are answered the same way — write the configuration or compare the ionic radius, and let the reasoning follow from there.

8.12 Properties of Lanthanoids

  1. Soft metals with a silvery white colour and moderate densities of ~\(7\ \text{g cm}^{-3}\). Colour and brightness reduce on exposure to air, because a dull oxide layer forms on the fresh surface.

  2. Good conductors of heat and electricity.

  3. Except promethium (Pm), all are non-radioactive in nature.

  4. The atomic and ionic radii decrease from lanthanum (La) to lutetium (Lu). This is known as lanthanoid contraction.

  5. Binding to water is common, that is, \(\mathrm{H_2O}\) is often found in the products when they are isolated from aqueous solutions.

  6. Coordination numbers are usually greater than 6, typically 8, 9 and so on, with values up to 12 found.

  7. The lanthanoids are strongly paramagnetic. Gadolinium becomes ferromagnetic below 16 °C (Curie point). The other heavier lanthanoids — terbium, dysprosium, holmium, erbium, thulium and ytterbium — become ferromagnetic at much lower temperatures.

  8. Magnetic and optical properties are largely independent of environment, giving similar spectra in the gas, solution and solid states.

Promethium, the one radioactive member of the series, is named after Prometheus, the Titan of Greek myth who stole fire from the gods and gave it to human beings. The name was chosen deliberately: the element is not found in any appreciable quantity in nature and had to be obtained from nuclear reactions, so it was, in a sense, taken from a fire that was not meant to be handled.

Table 8.11 : Electronic configuration and atomic ionic radii of lanthanoids

Element Symbol Atomic number Expected (ground state) Observed (ground state) (+2 oxidation state) (+3 oxidation state) Atomic radii, pm Ionic radii (Ln⁺³, 6-coordinate), pm
Lanthanum La 57 \(\mathrm{[Xe]4f^0 5d^1 6s^2}\) \(\mathrm{[Xe]4f^0 5d^1 6s^2}\) \(\mathrm{4f^0,\ 5d^1}\) \(\mathrm{4f^0}\) 187 103
Cerium Ce 58 \(\mathrm{[Xe]4f^2 6s^2}\) \(\mathrm{[Xe]4f^1 5d^1 6s^2}\) \(\mathrm{4f^2}\) \(\mathrm{4f^1}\) 183 102
Praseodymium Pr 59 \(\mathrm{[Xe]4f^3 6s^2}\) \(\mathrm{[Xe]4f^3 6s^2}\) \(\mathrm{4f^3}\) \(\mathrm{4f^2}\) 182 99
Neodymium Nd 60 \(\mathrm{[Xe]4f^4 6s^2}\) \(\mathrm{[Xe]4f^4 6s^2}\) \(\mathrm{4f^4}\) ………… 181 98.3
Promethium Pm 61 \(\mathrm{[Xe]4f^5 6s^2}\) \(\mathrm{[Xe]4f^5 6s^2}\) \(\mathrm{4f^5}\) ………… 181 97
Samarium Sm 61 \(\mathrm{[Xe]4f^6 6s^2}\) \(\mathrm{[Xe]4f^6 6s^2}\) \(\mathrm{4f^6}\) \(\mathrm{4f^5}\) 180 95.8
Europium Eu 63 \(\mathrm{[Xe]4f^7 6s^2}\) \(\mathrm{[Xe]4f^7 6s^2}\) \(\mathrm{4f^7}\) \(\mathrm{4f^6}\) 199 94.7
Gadolinium Gd 64 \(\mathrm{[Xe]4f^8 6s^2}\) \(\mathrm{[Xe]4f^7 5d^1 6s^2}\) \(\mathrm{4f^7,\ 5d^1}\) ………… 178 93.8
Terbium Tb 65 \(\mathrm{[Xe]4f^9 6s^2}\) \(\mathrm{[Xe]4f^9 6s^2}\) \(\mathrm{4f^9}\) \(\mathrm{4f^8}\) 177 92.3
Dysprosium Dy 66 \(\mathrm{[Xe]4f^{10} 6s^2}\) \(\mathrm{[Xe]4f^{10} 6s^2}\) \(\mathrm{4f^{10}}\) \(\mathrm{4f^9}\) 176 91.2
Holmium Ho 67 \(\mathrm{[Xe]4f^{11} 6s^2}\) \(\mathrm{[Xe]4f^{11} 6s^2}\) \(\mathrm{4f^{11}}\) \(\mathrm{4f^{10}}\) 175 90.1
Erbium Er 68 \(\mathrm{[Xe]4f^{12} 6s^2}\) \(\mathrm{[Xe]4f^{12} 6s^2}\) \(\mathrm{4f^{12}}\) ………… 174 89
Thulium Tm 69 \(\mathrm{[Xe]4f^{13} 6s^2}\) \(\mathrm{[Xe]4f^{13} 6s^2}\) \(\mathrm{4f^{13}}\) \(\mathrm{4f^{12}}\) 173 88
Ytterbium Yb 70 \(\mathrm{[Xe]4f^{14} 6s^2}\) \(\mathrm{[Xe]4f^{14} 6s^2}\) \(\mathrm{4f^{14}}\) ………… – 86.8
Lutetium Lu 71 \(\mathrm{[Xe]4f^{14} 5d^1 6s^2}\) \(\mathrm{[Xe]4f^{14} 5d^1 6s^2}\) \(\mathrm{4f^{14},\ 5d^1}\) \(\mathrm{4f^{14}}\) – 86.1

8.12.1 Electronic configuration

The electronic configuration of lanthanoids is \(\mathrm{[Xe]\,4f^{0-14}\,5d^{0-2}\,6s^2}\). This is because

\[\mathrm{1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,4s^2\,3d^{10}\,4p^6\,5s^2\,4d^{10}\,5p^6}\]

is the electronic configuration of xenon, and we can simplify the electronic configuration of lanthanoids by writing \(\mathrm{[Xe]\,4f^{0-14}\,5d^{0-2}\,6s^2}\). Using the noble gas symbol in this way saves writing out 54 electrons that take no part in the chemistry.

The lanthanoids involve the gradual filling of f-orbitals. The energies of the 5d and 4f orbitals are very close, and this is the source of every irregularity in the series.

For lanthanum, 4f is slightly higher in energy than 5d. Lanthanum therefore has the electronic configuration \(\mathrm{[Xe]\,6s^2\,5d^1}\) and not \(\mathrm{[Xe]\,6s^2\,4f^1}\) — the electron simply goes into whichever orbital is lower at that point.

Gadolinium (Gd) and lutetium (Lu) have a \(\mathrm{5d^1}\) electron, which leaves the f-orbital half-filled and full-filled respectively, and this renders them extra stability. Compare their “expected” and “observed” columns in Table 8.11 to see the difference.

The electronic configurations of lanthanoids have variable occupancy in the 4f (0 to 14) orbitals. This can be noticed from Table 8.11. The number of electrons in the 6s orbitals remains constant in the ground state — every lanthanoid, without exception, has \(\mathrm{6s^2}\).

The valence shell electronic configuration of these elements can thus be represented as:

\[\mathrm{(n-2)f^{\,0,\,2-14}\ (n-1)d^{\,0,1,2}\ ns^2}\]

The electronic distribution in different orbitals of the elements in their ground and excited states is shown in Table 8.11.

Fill in the blanks in Table 8.11.

Write the expected and the observed ground-state electronic configuration of gadolinium, and explain in one or two sentences why the observed one is preferred.

Ionization Enthalpies

What is ionization enthalpy? Some elements have variable oxidation states and some have only two. Can this be justified on the basis of their ionization enthalpies?

The ionization enthalpies of lanthanoids are given in Table 8.12.

Table 8.12: First, second, third and fourth ionization enthalpies of lanthanoids in kJ/mol

Lanthanoid IE₁ IE₂ IE₃
La 538.1 1067 1850.3
Ce 528.0 1047 1949
Pr 523.0 1018 2086
Nd 530.0 1034 2130
Pm 536.0 1052 2150
Sm 543.0 1068 2260
Eu 547.0 1085 2400
Gd 592.0 1170 1990
Tb 564.0 1112 2110
Dy 572.0 1126 2200
Ho 581.0 1139 2200
Er 589.0 1151 2190
Tm 596.7 1163 2284
Yb 603.4 1175 2415
Lu 523.5 1340 2022

Read the table across any one row and the familiar pattern appears: each successive electron is harder to remove than the one before, because it is being pulled away from an ion that is already positively charged. Read it down a column and the values stay comparatively flat, which is exactly why almost every lanthanoid settles on the same \(+3\) oxidation state.

Which of the following will have the highest fourth ionization enthalpy, \(\mathrm{IE_4}\) — La, Gd or Lu?

First write the electronic configuration of the element or ion concerned. Then check whether any unpaired electrons are present, because the energy required to remove such an electron will be less than the energy required to remove an electron from an electron pair. Also compare the energies of the orbitals occupied by those electrons, since it will be easier to remove an electron from an orbital that is lower in energy than from one that is higher in energy. Remember too that the first ionization enthalpy generally decreases across the period.

Applying that method to the three species, the fourth electron in each case must come out of the tripositive ion:

\[\mathrm{La^{3+}} = \mathrm{[Xe]}, \qquad \mathrm{Gd^{3+}} = \mathrm{[Xe]\,4f^7}, \qquad \mathrm{Lu^{3+}} = \mathrm{[Xe]\,4f^{14}}\]

For gadolinium the fourth electron would break into a half-filled 4f subshell, and for lutetium into a completely filled one — both stable arrangements, so both cost a great deal of energy. For lanthanum, however, \(\mathrm{La^{3+}}\) has already reached the electronic arrangement of xenon, and the fourth electron would have to be pulled out of a complete noble gas core, which lies much lower in energy than any 4f orbital.

Hence lanthanum has the highest fourth ionization enthalpy of the three.

8.12.2 Oxidation state

The \(+3\) oxidation state is common to all these elements, and arises when 2 electrons of the s-subshell and one from the d- or f-subshell are removed.

The 4f electrons are strongly screened by the inner 5s and 5p electrons. Thus, 4f electrons are not involved in the bonding, and once the metal has reached \(+3\) there is nothing further that can be removed at reasonable cost.

Besides this, some lanthanoids show oxidation states \(+2\) and \(+4\). They are formed in the case of \(\mathrm{f^0}\), \(\mathrm{f^7}\) and \(\mathrm{f^{14}}\) configurations, or of ions resulting in those configurations.

For example: \(\mathrm{Ce^{4+}}\) (\(\mathrm{f^0}\)); \(\mathrm{Eu^{2+}}\) and \(\mathrm{Tb^{4+}}\) (\(\mathrm{f^7}\)); \(\mathrm{Yb^{2+}}\) (\(\mathrm{f^{14}}\)). Refer to Table 8.11.

Terbium is the new case in this list. Its configuration is \(\mathrm{[Xe]\,4f^9\,6s^2}\), and removing four electrons leaves \(\mathrm{Tb^{4+}} = \mathrm{4f^7}\), the same half-filled subshell that makes \(\mathrm{Eu^{2+}}\) so stable. Notice how a single rule — aim for \(\mathrm{f^0}\), \(\mathrm{f^7}\) or \(\mathrm{f^{14}}\) — accounts for every exception to the \(+3\) rule in the whole series.

8.12.3 Colour and Spectra

Some trivalent ions (\(\mathrm{M^{3+}}\)) are coloured in the solid state as well as in solution. The colour of a lanthanoid ion is due to f–f transitions, which correspond to energy in the visible region of the electromagnetic spectrum.

The colour of ions having nf electrons is about the same as those having \((14 - n)\)f electrons, where n is an integer from 1 to 13. In other words, an ion counting up from an empty f subshell and an ion counting down from a full one behave alike, because what governs the transition is the number of gaps as much as the number of electrons.

Ln ion No. of f-electrons Colour
\(\mathrm{Pr^{3+}}\) \(\mathrm{4f^2}\) green \((14 - n)\) f-electrons \(= 14 - 2 = 12\)
\(\mathrm{Tm^{3+}}\) \(\mathrm{4f^{12}}\) green n f-electrons \(= 12\)
\(\mathrm{Nd^{3+}}\) \(\mathrm{4f^3}\) pink \((14 - n)\) f-electrons \(= 14 - 3 = 11\)
\(\mathrm{Er^{3+}}\) \(\mathrm{4f^{11}}\) pink nf-electrons \(= 11\)

Read the table in pairs. \(\mathrm{Pr^{3+}}\) with 2 f-electrons and \(\mathrm{Tm^{3+}}\) with 12 are both green; \(\mathrm{Nd^{3+}}\) with 3 and \(\mathrm{Er^{3+}}\) with 11 are both pink. In each pair the two numbers add up to 14.

8.12.4 Atomic and ionic radii (Lanthanoid Contraction)

As we move along the lanthanoid series, there is a decrease in atomic and ionic radii, as Fig. 8.6 shows. This steady decrease in the atomic and ionic radii is called lanthanoid contraction.

NoteDefinition

Lanthanoid contraction is the steady decrease in atomic and ionic radii of the elements of the lanthanoid series on moving from lanthanum to lutetium.

Diagram

Fig. 8.6 – Graph of ionic radius of Ln³⁺ (pm, from 85 to 105 on the vertical axis) plotted against the lanthanoid elements La, Ce, Pr, Nd, Pm, Sm, Eu, Gd, Tb, Dy, Ho, Er, Tm, Yb, Lu along the horizontal axis, the points joined by a line that falls steadily from left to right.

As we move from one element to another, the nuclear charge increases by one unit and one electron is added. The new electrons are added to the same inner 4f subshell.

Thus the 4f electrons shield each other from the nuclear charge poorly, owing to their diffused nature. With increasing atomic number and nuclear charge, the effective nuclear charge experienced by each 4f electron increases. As a result, the whole 4f electron shell contracts at each successive element.

Notice that the argument is the same one used for the transition series in Section 8.6.1 — poor shielding by a diffused subshell, rising effective nuclear charge, shrinking size. The effect is more pronounced here because f orbitals shield even more poorly than d orbitals do.

Q4. Lanthanoid ions get steadily smaller from lanthanum to lutetium, even though an electron is being added at every step. How can adding electrons make an ion shrink?

A. Because of where the electron goes. At each step the nucleus gains one proton and the atom gains one electron, but the new electron enters the inner 4f subshell rather than the outer shell. The 4f orbitals are diffused and shield very poorly, so the added electron fails to cancel the extra nuclear charge as far as the outer electrons are concerned. The effective nuclear charge felt by the outer shell therefore creeps up at every step, and the whole electron cloud is pulled in a little tighter. Fourteen small contractions, one after another, add up to the substantial shrinkage from La to Lu that we call lanthanoid contraction.

atomic and ionic radii across the lanthanoid series (across the series → from La to Lu: both atomic and ionic radii decrease steadily; because each added electron enters the poorly shielding, diffused 4f subshell while the nuclear charge rises by one unit at every step, so the effective nuclear charge increases and the 4f shell — and the atom with it — contracts at each successive element)

Fig. 8.6 – ionic radii of lanthanoids in the +3 oxidation state (Read the first and last points on the graph and state the total fall in ionic radius across the series. Is the fall smooth from one element to the next, or are there steps? Using Table 8.11, check your reading against the tabulated values for La³⁺ and Lu³⁺.)

Lanthanoid contraction has one famous consequence far outside the f block. Zirconium and hafnium sit one below the other in the periodic table, and hafnium should be distinctly the larger of the two — but the fourteen lanthanoids come in between, and their contraction cancels the expected increase almost exactly. The two metals end up with nearly identical sizes and, as a result, nearly identical chemistry, so they occur together and are extremely difficult to separate. This matters a great deal in the nuclear industry: zirconium hardly absorbs neutrons and is used to clad reactor fuel rods, while hafnium absorbs them strongly and is used in control rods. The two metals do exactly opposite jobs in the same reactor, and telling them apart is a direct practical cost of lanthanoid contraction.

Magnetic behaviour of the lanthanoids

In Section 8.6.4 we learnt about the magnetic behaviour of transition metal complexes. The lanthanoids raise the same question, but the answer is not quite the same.

Do you think that lanthanoid complexes would show magnetism? Can you calculate the spin only magnetic moment of lanthanoid complexes using the same formula that you used for transition metal complexes? Calculate the spin only magnetic moment of \(\mathrm{La^{3+}}\), compare the value with that given in Table 8.13, and state whether it is the same or different.

Table 8.13: Effective magnetic moments of lanthanoids in +3 oxidation state

Ln Ln³⁺ oxidation state No. of unpaired electrons Observed magnetic moment, μ_eff B.M
La \(\mathrm{4f^0}\) 0 0
Ce \(\mathrm{4f^1}\) 1 2.3-2.5
Pr \(\mathrm{4f^2}\) 2 3.4-3.6
Nd \(\mathrm{4f^3}\) 3 3.5-3.6
Pm \(\mathrm{4f^4}\) 4 –
Sm \(\mathrm{4f^5}\) 5 1.4-1.7
Eu \(\mathrm{4f^6}\) 6 3.3-3.5
Gd \(\mathrm{4f^7}\) 7 7.9-8.0
Tb \(\mathrm{4f^8}\) 6 9.5-9.8
Dy \(\mathrm{4f^9}\) 5 10.4-10.6
Ho \(\mathrm{4f^{10}}\) 4 10.4-10.7
Er \(\mathrm{4f^{11}}\) 3 9.4-9.6
Tm \(\mathrm{4f^{12}}\) 2 7.1-7.6
Yb \(\mathrm{4f^{13}}\) 1 4.3-4.9
Lu \(\mathrm{4f^{14}}\) 0 0

Q. The spin-only formula worked well for the first transition series. Why does it fail so badly for most lanthanoids?

A. Because the orbital contribution is not quenched here. In a 3d metal the d orbitals lie near the surface of the ion, where approaching ligands lock the orbital motion in place and leave only spin to contribute. The 4f orbitals lie deep inside the atom, well shielded from the ligands, so the orbital angular momentum survives and adds its own share to the magnetic moment. Compare Dy³⁺ in Table 8.13: five unpaired electrons would give 5.92 BM on the spin-only formula, yet the observed value is above 10 BM.

Calculate the spin-only magnetic moment of \(\mathrm{Gd^{3+}}\) and compare it with the observed value in Table 8.13.

Gadolinium has the observed configuration \(\mathrm{[Xe]\,4f^7\,5d^1\,6s^2}\). Removing three electrons — the two 6s and the one 5d — gives

\[\mathrm{Gd^{3+}} = \mathrm{[Xe]\,4f^7}\]

The seven f electrons occupy the seven f orbitals singly, so \(n = 7\). Substituting in the spin-only formula:

\[\mu = \sqrt{n(n+2)} = \sqrt{7 \times 9} = \sqrt{63} = 7.94\ \text{BM}\]

Table 8.13 gives the observed value for \(\mathrm{Gd^{3+}}\) as 7.9–8.0 BM, so in this one case the calculated and observed values agree closely.

The agreement is not an accident. With a half-filled 4f subshell the orbital contributions of the seven electrons cancel one another out, leaving spin alone to determine the moment — the same situation the spin-only formula assumes. For lanthanoids with a partly filled subshell, as the sidebar above explains, no such cancellation occurs and the formula fails.

Calculate the spin-only magnetic moment of \(\mathrm{Sm^{3+}}\) (\(\mathrm{4f^5}\)), compare your answer with the observed value listed in Table 8.13, and state whether the spin-only formula is adequate in this case.

Examination questions from this section cluster around three skills: writing the configuration of a lanthanoid or its ion, explaining a stable \(+2\) or \(+4\) state in terms of \(\mathrm{f^0}\), \(\mathrm{f^7}\) or \(\mathrm{f^{14}}\), and defining lanthanoid contraction together with one of its consequences. All three reward the same habit of starting from the electronic configuration.

8.13 Applications

The lanthanoid compounds are present in every household. They are inside the colour television tube.

When electrons are bombarded on certain mixed lanthanoid compounds, they emit visible light over a small wavelength range. Therefore, the inside surface of a television tube or computer monitor is coated with tiny patches of three different lanthanoid compositions to give three colours that make the colour image, as shown in Fig. 8.6a.

Diagram

Fig. 8.6a – Enlarged view of a small area of a television screen: the inner surface shown coated with a repeating pattern of three kinds of tiny phosphor patches labelled red, green and blue, with an electron beam arrow striking one patch and light being emitted from it.

For example, the mixed oxide of europium and yttrium, \((\mathrm{Eu,Y})_2\mathrm{O_3}\), releases an intense red colour when bombarded with high energy electrons. The three primary colours, mixed in different proportions by varying the strength of the electron beam, are enough to build every shade in the picture.

Q. Why are lanthanoid compounds, of all things, chosen to make the light in a display screen?

A. Because they emit over a very narrow band of wavelengths, giving a pure, sharply defined colour rather than a smeared one. The reason goes back to the buried 4f orbitals of Section 8.12: shielded by the outer 5s and 5p shells, they are barely disturbed by whatever surrounds the ion, so the energy of an f–f transition is almost fixed. What is a nuisance for a chemist trying to separate these elements turns out to be exactly the property a display engineer wants.

Lasers, optical fibres and magnets

The optoelectronics applications use lanthanoid ions as active ions in luminescent materials — that is, the lanthanoid ion is the centre that actually emits the light, while the surrounding solid merely holds it in place.

The most notable application is the Nd:YAG laser (Nd:YAG = neodymium doped yttrium aluminium garnet). Erbium-doped fibre amplifiers are significant devices in optical-fibre communication systems, where they boost a light signal that has weakened over a long stretch of cable without ever converting it back into an electrical signal.

Lanthanoids are also used in hybrid cars, superconductors and permanent magnets.

three lanthanoid-based devices

Feature Colour display screen Nd:YAG laser Erbium-doped fibre amplifier
Lanthanoid used Europium, with yttrium Neodymium Erbium
Host material Mixed oxide \((\mathrm{Eu,Y})_2\mathrm{O_3}\) Yttrium aluminium garnet Glass of the optical fibre
What excites the ion High energy electrons Light from a pump source Light from a pump source
What it delivers Visible light of one pure colour An intense, narrow laser beam A strengthened optical signal
Where it is met Television tubes, computer monitors Industrial cutting, medical and research lasers Long-distance optical-fibre communication

Neodymium takes its name from the Greek for “new twin”, and praseodymium from “green twin”. The two were originally believed to be a single element, named didymium — Greek for twin — until careful work showed that the supposed element was in fact a pair of them. The names preserve the moment the twin was split, and the colour of the salts that gave one of them away.

Diagram

Fig. 8.6b – A panel of eight small labelled photographs of inner transition and postactinoid elements: promethium, europium, actinium, berkelium, bohrium, nobelium, mendelevium and seaborgium, each captioned with its name.

Neodymium is the basis of the strongest permanent magnets in ordinary use, and they are almost certainly within a metre of you as you read this — in the earphone driver, the vibration motor of a phone, the read head of a hard disk, and, on a much larger scale, in the generators of wind turbines and the drive motors of hybrid and electric cars. Because supplies of these elements are concentrated in a few mining regions, and because separating them is energy-intensive, there is now serious effort worldwide to recover lanthanoids from discarded electronics rather than mine them afresh. Recycling a magnet from a scrapped hard drive avoids the entire separation problem described in Section 8.11 — the elements are already separated.

With the help of the internet, find out the applications of the elements listed below and share this information with your friends: lanthanum, ytterbium, erbium, praseodymium, samarium, promethium.

Explain, using what you have learnt about f–f transitions, why lanthanoid ions are preferred over transition metal ions as the light-emitting centres in display screens and lasers.

Questions from this section are usually straightforward recall — name two applications of lanthanoids, or state the use of a named compound — but they are worth more marks when the answer connects the application to the property behind it, such as narrow-band emission or strong permanent magnetism.

8.14 Actinoids

The last row of elements in the periodic table is the actinoid series. It begins at thorium (Z = 90) and ends at lawrencium (Z = 103).

Most of these elements are not found in nature. They are all radioactive and man-made — they have to be built up in nuclear reactions, atom by atom, rather than dug out of the ground.

Two of them are exceptions, and the reason is a matter of time. The half-lives of the isotopes of thorium (Th-232 = \(1.4 \times 10^{10}\) years) and uranium (U-238 = \(4.5 \times 10^9\) years) are so long that these elements still exist in rocks on earth. Half-life is the time in which half of a radioactive sample decays, so an element with a half-life comparable to the age of the earth has not yet had time to disappear.

The long lived isotopes, such as thorium, protactinium, uranium, neptunium, plutonium and americium, are studied in more detail, simply because they survive long enough for their chemistry to be examined.

Q. Why should every element in this series be radioactive, when the lanthanoids just above them are almost all stable?

A. Because of the size of the nucleus. Each actinoid nucleus holds 90 or more protons, all repelling one another electrically, and beyond a certain point the short-range forces binding the nucleus together can no longer hold that repulsion in check. The nucleus becomes unstable and breaks down by radioactive decay. Lanthanoid nuclei, with roughly 57 to 71 protons, are still small enough to remain stable — which is why promethium is the only radioactive member of that series.

Physical nature and reactivity

These elements have high densities (~15-20 g cm⁻³), high melting points (~1000 °C) and high boiling points (~3000 °C).

Actinoids are less reactive than lanthanoids. For example, they react with hot, but not cold, water to give the hydroxide and hydrogen gas — the same reaction the lanthanoids carry out at ordinary temperature.

Uranium was named after the planet Uranus, which had been discovered only a few years before the element itself. When the next two members of the series were made artificially, the naming followed the solar system outwards: element 93 became neptunium, after Neptune, and element 94 plutonium, after Pluto. Three elements in a row carry the order of the outer planets.

Oxidation states of the actinoids

Unlike lanthanoids, they exhibit a range of oxidation numbers in their compounds, which varies from \(+2\) to \(+8\). The most common oxidation numbers of the actinoids are shown in Fig. 8.7.

Diagram

Fig. 8.7 – Bar chart of the most common oxidation numbers of the actinoids: oxidation number 1 to 6 on the vertical axis, elements Ac, Th, Pa, U, Np, Pu, Am, Cm, Bk, Cf, Es, Fm, Md, No, Lr along the horizontal axis, shaded blocks marking the states each element commonly shows — a wide spread of states for the early members, narrowing to +3 for the later ones.

Fig. 8.7 – common oxidation numbers of the actinoids (Read across the chart. Which element shows the widest range of oxidation numbers, and which oxidation state is shown by the largest number of elements? At roughly which element does the spread of states begin to narrow, and how does that compare with the point at which the 5f shell begins to fill regularly?)

As can be seen from Fig. 8.7, the most stable oxidation state in actinoids is \(+3\), but the highest common oxidation states of the early actinoids reflect the loss of all outer electrons — behaviour much closer to that of the transition metals than of the lanthanoids.

The reason lies in orbital energies. A ready loss of 5f electrons by the early actinoids indicates that these electrons are much closer in energy to the 7s and 6d electrons than the 4f electrons are to the 6s and 5d electrons in the lanthanoids. The 5f and 6d orbitals expand, as they are partially shielded from the nuclear attraction by the 7s electrons. As a result, all three sets of orbitals — 6d, 5f and 7s — have very similar energies.

For Th, Pa and Np the difference in energy levels is small, so electrons occupy either 6d or 5f orbitals. Further along the actinoid series the 5f orbitals are appreciably lower in energy, and thus from Pu onwards the 5f shell gets filled in a regular way.

Q5. Actinoids show oxidation states ranging from \(+2\) right up to \(+8\), while lanthanoids stay almost entirely at \(+3\). Why should two f-block series that look so similar behave so differently?

A. Because of how deeply the f electrons are buried. In a lanthanoid the 4f orbitals lie well inside the atom and are strongly screened by the 5s and 5p shells, so the 4f electrons take no part in bonding; only the two 6s electrons and one more are available, and the element stops at \(+3\). In an actinoid the 5f, 6d and 7s orbitals all lie at very similar energies, because the 5f and 6d orbitals expand outwards when partially shielded by the 7s electrons. The 5f electrons are therefore within reach and can be lost along with the 6d and 7s electrons, giving the early actinoids a wide range of oxidation states — behaviour much more like a transition metal than a lanthanoid.

Electronic configuration

The electronic configuration of actinoids is \(\mathrm{[Rn]\,5f^{0-14}\,6d^{0-2}\,7s^2}\), where Rn stands for the electronic configuration of radon.

For example, uranium has the electronic configuration \(\mathrm{[Rn]\,7s^2\,5f^3\,6d^1}\). The formation of the \(+6\) oxidation state corresponds to an electronic configuration of \(\mathrm{[Rn]}\).

Similar to lanthanoids, loss of s and d electrons occurs before f electrons in the formation of \(3+\) ions. The full set of ground-state configurations and radii is given in Table 8.14.

Uranium has the electronic configuration \(\mathrm{[Rn]\,7s^2\,5f^3\,6d^1}\). Show that the \(+6\) oxidation state of uranium corresponds exactly to the radon core, and state what this tells us about why \(+6\) is common for uranium.

Count the electrons lying outside the radon core.

\[7s^2 \rightarrow 2\ \text{electrons}, \qquad 5f^3 \rightarrow 3\ \text{electrons}, \qquad 6d^1 \rightarrow 1\ \text{electron}\]

\[\text{Total outside the core} = 2 + 3 + 1 = 6\]

Removing all six leaves

\[\mathrm{U} \rightarrow \mathrm{U^{6+}} + 6\,\mathrm{e^{-}}, \qquad \mathrm{U^{6+}} = \mathrm{[Rn]}\]

which is the electronic arrangement of the noble gas radon.

This explains why \(+6\) is a common and stable state for uranium: all six removable electrons lie in three sets of orbitals — 7s, 6d and 5f — that are close together in energy, so none of them is especially hard to remove, and the ion left behind has a complete noble gas core. Compare this with a lanthanoid, where the f electrons lie far below the outer shell and stripping the atom this far is simply not possible.

Using its ground-state configuration from Table 8.14, work out the maximum oxidation state that thorium can show, and write the electronic configuration of the ion in that state.

early actinoids versus later actinoids

Feature Early actinoids (Th, Pa, U, Np) Later actinoids (Pu onwards)
Relative energies of 5f, 6d and 7s Very close, so electrons occupy either 6d or 5f 5f appreciably lower in energy
Filling of the 5f shell Irregular Regular, one electron at a time
Range of oxidation states Wide, up to \(+6\) and beyond Narrow, mostly \(+3\)
Resemblance Behave much like transition metals Behave much like lanthanoids

Table 8.14 Electronic configuration of actinoids and their ionic radii in +3 oxidation state

Element Symbol Atomic number Ground state configuration +3 oxidation state *Atomic radii, pm *Ionic radii (Ac³⁺), pm
Actinium Ac 89 \(\mathrm{[Rn]5f^0 6d^1 7s^2}\) \(\mathrm{5f^0}\) 203 126
Thorium Th 90 \(\mathrm{[Rn]5f^0 6d^2 7s^2}\) \(\mathrm{5f^1}\) 180 –
Protactinium Pa 91 \(\mathrm{[Rn]5f^2 6d^1 7s^2}\) \(\mathrm{5f^2}\) 162 118
Uranium U 92 \(\mathrm{[Rn]5f^3 6d^1 7s^2}\) \(\mathrm{5f^3}\) 153 118
Neptunium Np 93 \(\mathrm{[Rn]5f^4 6d^1 7s^2}\) \(\mathrm{5f^4}\) 150 116
Plutonium Pu 94 \(\mathrm{[Rn]5f^6 6d^0 7s^2}\) \(\mathrm{5f^5}\) 162 115
Americium Am 95 \(\mathrm{[Rn]5f^7 6d^0 7s^2}\) \(\mathrm{5f^6}\) 173 114
Curium Cm 96 \(\mathrm{[Rn]5f^7 6d^1 7s^2}\) \(\mathrm{5f^7}\) 174 112
Berkelium Bk 97 \(\mathrm{[Rn]5f^9 6d^0 7s^2}\) \(\mathrm{5f^8}\) 170 110
Californium Cf 98 \(\mathrm{[Rn]5f^{10} 6d^0 7s^2}\) \(\mathrm{5f^9}\) 186 109
Einsteinium Es 99 \(\mathrm{[Rn]5f^{11} 6d^0 7s^2}\) \(\mathrm{5f^{10}}\) 186 98
Fermium Fm 100 \(\mathrm{[Rn]5f^{12} 6d^0 7s^2}\) \(\mathrm{5f^{11}}\) 198 91
Mendelevium Md 101 \(\mathrm{[Rn]5f^{13} 6d^0 7s^2}\) \(\mathrm{5f^{12}}\) 194 90
Nobelium No 102 \(\mathrm{[Rn]5f^{14} 6d^0 7s^2}\) \(\mathrm{5f^{13}}\) 197 95
Lawrencium Lr 103 \(\mathrm{[Rn]5f^{14} 6d^1 7s^2}\) \(\mathrm{5f^{14}}\) 171 88
Actinoid contraction

The ionic radius decreases as we move across the series, which is known as actinoid contraction. This is attributed to the poor shielding offered by f electrons.

NoteDefinition

Actinoid contraction is the steady decrease in ionic radius on moving across the actinoid series, caused by the poor shielding of nuclear charge by the 5f electrons.

The cause is exactly the one established for the lanthanoids in Section 8.12.4: each added 5f electron shields the outer electrons poorly, so the effective nuclear charge climbs steadily and the ion is drawn in tighter at every step. Fig. 8.8 places the two contractions side by side, and shows that the actinoid curve runs parallel to the lanthanoid one.

Diagram

Fig. 8.8 – Graph of ionic radius r (pm) against element for four series plotted together: actinoid 3+, lanthanoid 3+, actinoid 4+ and lanthanoid 4+, with the lanthanoid elements La to Lu labelled along the top of the horizontal axis and the actinoid elements Ac to Lr along the bottom, all four curves falling steadily from left to right.

ionic radius across the actinoid series (across the series → from Ac to Lr: ionic radius decreases steadily — actinoid contraction; because each successive element adds a proton to the nucleus while the added electron enters the inner 5f subshell, which shields poorly, so the effective nuclear charge felt by the outer electrons increases at every step)

The half-life of U-238 is \(4.5 \times 10^9\) years. What fraction of an original sample of U-238 will remain undecayed after \(1.35 \times 10^{10}\) years?

Board and entrance questions on this section usually ask why actinoids show more oxidation states than lanthanoids, what actinoid contraction is, or which actinoids occur naturally. All three are answered from the same two facts: the closeness in energy of the 5f, 6d and 7s orbitals, and the poor shielding by f electrons.

8.15 Properties of Actinoids

  1. Similar to lanthanoids, they appear silvery white in colour.

  2. These are highly reactive radioactive elements.

  3. They experience a decrease in the atomic and ionic radii from actinium (Ac) to lawrencium (Lr), known as actinoid contraction.

  4. They usually exhibit the \(+3\) oxidation state. Elements of the first half of the series usually exhibit higher oxidation states.

Q. Section 8.14 said actinoids are less reactive than lanthanoids, yet point ii here calls them highly reactive. Which is it?

A. Both statements are true, because they are comparisons against different standards. Measured against most metals — and certainly against the transition metals — the actinoids are very reactive indeed; their electronegativity is low and they combine readily with water, oxygen and the halogens. Measured against the lanthanoids specifically, they are the less reactive of the two f-block series, which is why an actinoid needs hot water to give its hydroxide where a lanthanoid reacts with cold. “Highly reactive” and “less reactive than lanthanoids” describe the same behaviour from two different vantage points.

Using the ionic radii in Table 8.14, calculate the total decrease and the percentage decrease in ionic radius on going from \(\mathrm{Ac^{3+}}\) to \(\mathrm{Lr^{3+}}\).

From Table 8.14, the ionic radius of \(\mathrm{Ac^{3+}}\) is 126 pm and that of \(\mathrm{Lr^{3+}}\) is 88 pm.

Total decrease in ionic radius:

\[126 - 88 = 38\ \text{pm}\]

Percentage decrease, taken with respect to the starting value:

\[\frac{38}{126} \times 100 = 30.2\%\]

So the ion shrinks by nearly a third of its original radius over the course of the series. This single figure is what the phrase “actinoid contraction” summarises, and the reason, as established in Section 8.14, is the poor shielding offered by the 5f electrons.

Using the ionic radii of \(\mathrm{La^{3+}}\) and \(\mathrm{Lu^{3+}}\) given in Table 8.11, calculate the percentage decrease in ionic radius across the lanthanoid series. Compare your answer with the value obtained above for the actinoids, and state which of the two contractions is the greater.

Comparing the two f-block series

The four properties listed above are best understood alongside the lanthanoids, since the two series run parallel to one another in some respects and diverge sharply in others. The table below sets out the position.

Similarities and differences between lanthanoids and actinoids

Similarities Differences
Both the series show a \(+3\) oxidation state Lanthanoids show a maximum oxidation state of \(+4\) while actinoids show oxidation states of \(+3\), \(+4\), \(+5\), \(+6\) and \(+7\)
In both the series, the f-orbitals are filled gradually Lanthanoids do not form complexes easily. Actinoids have a greater tendency to form complexes with ligands such as thioethers
Ionic radii of the elements in both series decrease with an increase in atomic number All lanthanoids are non-radioactive except promethium, but actinoids are radioactive in nature
The electronegativity of all the elements in both the series is low and they are said to be highly reactive Lanthanoids do not form oxocations, but actinoids form oxocations such as \(\mathrm{UO^{+}}\), \(\mathrm{PuO^{+}}\), \(\mathrm{NpO_2^{+}}\)
The nitrates, perchlorates and sulphates of all the elements are soluble while the hydroxides, fluorides and carbonates are insoluble Most of the lanthanoids are colourless in nature whereas the actinoids are coloured ions

Read the right-hand column and a single cause runs through it. Every difference — the wider range of oxidation states, the readiness to form complexes, the existence of oxocations — follows from the fact established in Section 8.14 that the 5f orbitals of an actinoid lie close in energy to the 6d and 7s orbitals and are therefore available for bonding, while the 4f orbitals of a lanthanoid are buried and are not. Radioactivity is the one difference that comes from the nucleus rather than the electron cloud.

The f block against the rest of the periodic table

Widening the comparison one step further shows where the lanthanoids sit between the s-block metals on one side and the transition metals on the other.

Table 8.15 : Some comparison between s - block, Lanthanoids and Transition Metals

s - block Metals Lanthanoids Transition Metals
Essentially monovalent - show group (n+) oxidation state Essentially in (+3) oxidation state (+2/+4 for certain configurations) Show variable oxidation states
Periodic trends dominated by effective nuclear charge at noble gas configuration Lanthanoid contraction of \(\mathrm{Ln^{3+}}\) Size changes of \(\mathrm{M^{n+}}\), less marked
Similar properties for a given group Similar properties Substantial changes in properties
Always ‘hard’ (O, X, N donors, preferably negatively charged) Always ‘hard’ (O, X, N donors, preferably negatively charged) heavier metals (increasingly from Fe-Cu) may show a ‘soft’ character
No ligand field effects Insignificant ligand field effects Substantial ligand field effects
Poor coordination properties (C.N. determined by size) High coordination numbers (C.N. determined by size) Coordination number 6 is typical maximum (many exceptions)
Flexibility in geometry Flexibility in geometry Fixed geometries (ligand field effects)
No magnetism Show magnetism Show magnetism

The middle column tells the story of this chapter’s second half. In oxidation state and in geometry the lanthanoids behave much like s-block metals; in magnetism they behave like transition metals; and in one respect — the near absence of ligand field effects — they behave like nothing else, because their f orbitals are shielded from the ligands altogether.

Several actinoids carry the names of the scientists who made the field possible. Curium honours Marie and Pierre Curie, whose work on radioactivity opened the subject; einsteinium and fermium were named for Albert Einstein and Enrico Fermi. Mendelevium, further along the row, is named for Dmitri Mendeleev, whose periodic table was the framework into which all of these elements eventually had to be fitted.

Americium, a man-made actinoid, sits inside a great many ordinary smoke detectors. A minute quantity of americium-241 emits alpha particles that ionise the air in a small chamber, allowing a tiny current to flow; smoke particles entering the chamber disrupt the ionised air, the current falls, and the alarm sounds. The quantity is small and sealed, and the alpha particles cannot penetrate the plastic casing — but the same radioactivity that makes these elements hazardous in bulk is exactly what makes the device work. It is also why smoke detectors carry disposal instructions rather than being thrown in with household waste.

Questions on this section are usually set as “give the similarities and differences between lanthanoids and actinoids”, worth several marks. Learn at least three of each from the table above, and be ready to give the reason — orbital energies for the chemical differences, nuclear size for the radioactivity.

8.16 Applications of actinoids

We have seen that the half-lives of the natural thorium and uranium isotopes are so long that we get very negligible radiation from these elements. Because so few nuclei decay in any given second, the radiation coming from a sample is correspondingly small — which is exactly why these two actinoids, alone in the series, can be handled in ordinary situations.

We find them in everyday use. For example, Th(IV) oxide, \(\mathrm{ThO_2}\) with 1% \(\mathrm{CeO_2}\), was used as a major source of indoor lighting before incandescent lamps came into existence, only because these oxides convert heat energy from burning natural gas to an intense light.

Even today, there is a great demand for these lights for outdoor camping, where a gas lantern with such a mantle gives a bright white light from a flame that would otherwise be dim and yellow.

Q. The gas flame itself is barely luminous. How does a thin mantle of thorium and cerium oxide make it blaze?

A. The flame supplies heat, not light. Most of the energy of a burning gas leaves as heat and as invisible infrared radiation. The mesh of \(\mathrm{ThO_2}\) with a little \(\mathrm{CeO_2}\) is heated white-hot by that flame and converts the heat energy into an intense emission concentrated in the visible region. The oxides are chosen because they survive the temperature without melting and because they emit far more strongly in the visible than an ordinary hot solid would.

The gas mantle was the invention of Carl Auer von Welsbach, an Austrian chemist working in the late nineteenth century, who found that a fabric mesh soaked in a solution of thorium and cerium salts, then burnt away to leave the bare oxides, would glow brilliantly in a gas flame. His mantle lit streets and homes across Europe for a generation, and its inventor is the same chemist who separated didymium into praseodymium and neodymium — the rare earths turn up twice in one career.

Uranium and the yellow cake

Uranium is another actinoid which is in great demand, as it is used in nuclear reactors. One of the extraction methods for uranium has a very interesting chemistry, summarised in FC 8.11.

The ore containing U(IV) oxide, \(\mathrm{UO_2}\), is first treated with Fe(III) ion to give U(VI) oxide, \(\mathrm{UO_3}\):

\[\mathrm{UO_2}\text{(s)} + \mathrm{H_2O}\text{(l)} \rightarrow \mathrm{UO_3}\text{(s)} + 2\,\mathrm{H^{+}}\text{(aq)} + 2\,\mathrm{e^{-}}\]

\[\mathrm{Fe^{3+}}\text{(aq)} + \mathrm{e^{-}} \rightarrow \mathrm{Fe^{2+}}\text{(aq)}\]

Addition of \(\mathrm{H_2SO_4}\) to this solution produces uranyl sulphate containing the \(\mathrm{UO_2^{2+}}\) cation:

\[\mathrm{UO_3}\text{(s)} + \mathrm{H_2SO_4}\text{(aq)} \rightarrow \mathrm{UO_2SO_4}\text{(aq)} + \mathrm{H_2O}\text{(l)}\]

After purification, ammonia is added to the solution, giving a bright yellow precipitate of ammonium diuranate, \((\mathrm{NH_4})_2\mathrm{U_7O_7}\):

\[2\,\mathrm{UO_2SO_4}\text{(aq)} + 6\,\mathrm{NH_3}\text{(aq)} + 3\,\mathrm{H_2O}\text{(l)} \rightarrow (\mathrm{NH_4})_2\mathrm{U_7O_7}\text{(s)} + 2\,(\mathrm{NH_4})_2\mathrm{SO_4}\text{(aq)}\]

This yellow cake is the marketable form of uranium — the material in which uranium is bought, sold and shipped before any further processing.

Flowchart

FC 8.11 – Extraction of uranium to yellow cake: ore containing UO₂, uranium in the +4 state → oxidised by Fe(III), which is itself reduced to Fe(II) → UO₃, uranium now in the +6 state → treated with H₂SO₄ → soluble uranyl sulphate UO₂SO₄ containing the UO₂²⁺ cation → purification → ammonia added → bright yellow precipitate of ammonium diuranate → filtered and dried → yellow cake, the marketable form of uranium.

In the first step of the uranium extraction above, show by assigning oxidation numbers that uranium is oxidised, identify the oxidising agent, and state how many iron(III) ions are needed per uranium atom.

Take oxygen as \(-2\) in both oxides.

In \(\mathrm{UO_2}\): \(x + 2(-2) = 0\), so \(x = +4\).

In \(\mathrm{UO_3}\): \(x + 3(-2) = 0\), so \(x = +6\).

Uranium therefore rises from \(+4\) to \(+6\) — an increase in oxidation number, which is oxidation. Each uranium atom releases 2 electrons, exactly as the first half-equation shows.

Those electrons have to go somewhere, and the second half-equation shows where. Iron falls from \(+3\) in \(\mathrm{Fe^{3+}}\) to \(+2\) in \(\mathrm{Fe^{2+}}\), gaining 1 electron per ion. Iron(III) is therefore the oxidising agent, and since each uranium atom supplies 2 electrons while each iron(III) ion accepts only 1,

\[\text{2 Fe}^{3+}\ \text{ions are needed for every 1 uranium atom.}\]

Notice the purpose of the whole step: uranium(IV) oxide is insoluble, but once oxidised to the \(+6\) state it can be brought into solution as the uranyl ion, and only a dissolved uranium compound can be purified and precipitated in the steps that follow.

Determine the oxidation number of uranium in uranyl sulphate, \(\mathrm{UO_2SO_4}\), showing your working, and state whether it agrees with the oxidation state reached in the first step of the extraction.

Questions on this section are short and usually factual — name one everyday application of thorium, or state what yellow cake is — but the extraction sequence is also a convenient source of oxidation-number and complete-the-reaction questions, so it repays working through the four equations once with oxidation numbers written above each uranium atom.

8.17 Postactinoid elements

You have seen that elements with atomic number greater than 92 are called ‘Transuranium’ elements — literally, those lying beyond uranium, which is the heaviest element found in nature in any appreciable quantity.

NoteDefinition

Transuranium elements are the elements with atomic number greater than 92, that is, those beyond uranium in the periodic table.

Elements from atomic number 93 to 103 are now included in the actinoid series, and those from 104 to 118 are called postactinoid elements.

NoteDefinition

Postactinoid elements are the elements with atomic numbers from 104 to 118, which follow the actinoid series in the periodic table.

The postactinoid elements that are known so far are transition metals — their electrons enter the 6d subshell, so they continue the fourth transition series rather than the f block. They are included as postactinoids because, similar to actinoid elements, they can be synthesized in nuclear reactions.

So far, nine postactinoid elements have been synthesized. It is difficult to study their chemistry owing to their short half-lives. For example, element 112 has a half-life of only \(2.8 \times 10^{-4}\) seconds.

Q. If an element exists for less than a thousandth of a second, how can anyone determine its chemistry at all?

A. With enormous difficulty, and only a few atoms at a time. These elements are made one atom at a time in an accelerator, and the chemistry must be carried out on that single atom within moments of its creation — usually by seeing whether it sticks to a surface, or travels through a column, in the same way a lighter member of its group does. A conclusion about “the chemistry of element 112” may rest on the behaviour of a handful of individual atoms, which is why so little is known with confidence.

What little is known

With half-lives of milliseconds, only a little is known about the chemistry of these elements.

Rutherfordium forms a chloride, \(\mathrm{RfCl_4}\), similar to zirconium and hafnium in the \(+4\) oxidation state. This is exactly what the periodic table predicts, since all three elements belong to the same group.

Dubnium resembles both the group 5 transition metal niobium(V) and the actinoid protactinium(V) — a double resemblance that reflects its position at the meeting point of the two families.

Rutherfordium forms the chloride \(\mathrm{RfCl_4}\). Determine the oxidation state of rutherfordium in this compound, and explain why zirconium and hafnium form the corresponding chlorides.

Chlorine in a metal chloride is assigned an oxidation number of \(-1\). Let the oxidation number of rutherfordium be \(x\). Since \(\mathrm{RfCl_4}\) is a neutral molecule,

\[x + 4(-1) = 0, \qquad \text{so}\ x = +4\]

Rutherfordium is therefore in the \(+4\) oxidation state.

Zirconium and hafnium form \(\mathrm{ZrCl_4}\) and \(\mathrm{HfCl_4}\) for the same reason: all three elements lie in group 4 of the periodic table and therefore have the same number of electrons available in their outermost shells. Losing all of them gives the \(+4\) state in every case.

This is a useful demonstration of what the periodic table is for. Nobody could accumulate enough rutherfordium to run ordinary experiments on it, yet its position in the table allowed chemists to predict what its chloride would look like before it was ever examined — and the prediction held.

actinoids vs postactinoid elements

Feature Actinoids Postactinoid elements
Atomic numbers 89 to 103 104 to 118
Subshell being filled 5f 6d
Block of the periodic table f block (inner transition) d block (transition metals)
How obtained Two occur naturally; the rest are made in nuclear reactions All are made in nuclear reactions
Half-lives Range from very long (Th, U) to short Extremely short, down to milliseconds
Extent of chemistry known Studied in reasonable detail for the long-lived members Very little known

Traditionally, no element was named after a still-living scientist; the honour was reserved for those who could no longer be embarrassed by it. That principle was set aside when element 106 was named seaborgium, and the newer names in this part of the table now honour people and places together — rutherfordium for Ernest Rutherford, and dubnium for Dubna, the Russian town whose laboratory produced several of these elements.

The half-life of element 112 is \(2.8 \times 10^{-4}\) seconds. Calculate the time required for a freshly prepared sample to fall to one-eighth of its initial amount.

This closing section is rarely the subject of long questions, but short factual ones do appear — what transuranium elements are, which atomic numbers the postactinoids cover, and why their chemistry is so poorly known. The answer to the last of these is always the same: half-lives too short, and far too few atoms to work with.

8.18 Summary

  1. Transition elements show a great variety of oxidation states because the 4s and 3d subshells lie very close in energy, so 3d electrons can be removed after the 4s electrons at only a small extra energy cost.

  2. Ions are built by removing the 4s electrons first and the 3d electrons afterwards, giving \(\mathrm{M^{+}}\), \(\mathrm{M^{2+}}\), \(\mathrm{M^{3+}}\), \(\mathrm{M^{4+}}\) and so on.

  3. The number of oxidation states increases with the number of unpaired 3d electrons. Scandium, with one unpaired electron, shows three oxidation states; manganese, with five, shows six.

  4. The greatest number of oxidation states occurs in or near the middle of the series. Manganese shows \(+2\) to \(+7\), the sum of its \(\mathrm{3d^5\,4s^2}\) electrons.

  5. Two configurations are exceptions worth remembering: chromium is \(\mathrm{[Ar]\,3d^5\,4s^1}\) and copper is \(\mathrm{[Ar]\,3d^{10}\,4s^1}\), because half-filled and completely filled d subshells are more stable.

  6. All transition elements are metals: hard, lustrous, malleable, ductile, alloy-forming and good conductors of heat and electricity, with high melting and boiling points and low volatility.

  7. Zn, Cd and Hg are the exceptions to hardness, low volatility and high melting point; Zn, Cd, Hg and Mn are the exceptions to having a typical metallic structure at ambient temperature.

  8. Atomic radii decrease gradually across a transition series, because the nuclear charge rises by one unit at each step while the added d electrons, being diffused and poorly penetrating, screen it weakly, so the effective nuclear charge increases.

  9. Ionic radii follow the same trend. For ions of the same charge the radius falls across the series (\(\mathrm{Cr^{2+}}\) 82 pm, \(\mathrm{Cu^{2+}}\) 73 pm), and for one element the radius falls as the oxidation state rises, from \(\mathrm{M^{2+}}\) to \(\mathrm{M^{3+}}\).

  10. Ionic radii of transition elements are smaller than those of representative elements of the same period.

  11. Ionization enthalpies of transition elements lie between those of the s-block and p-block, so these elements are less electropositive than groups 1 and 2. Lower oxidation states give ionic compounds and higher oxidation states give covalent compounds.

  12. For any one element \(\mathrm{IE_1}\), \(\mathrm{IE_2}\) and \(\mathrm{IE_3}\) increase substantially, while across a series the variation is only slight.

  13. Ionization enthalpies of the third transition series are much higher than those of the first and second, because filled 4f orbitals shield poorly and the valence electrons feel a greater nuclear attraction.

  14. Metallic character arises from low ionisation enthalpies and vacant d orbitals; the hardness indicates additional covalent character from unpaired (n-1)d electrons. Nearly all transition metals adopt hcp, ccp or bcc lattices.

  15. Melting points increase up to the \(\mathrm{d^5}\) configuration, with Cr, Mo and W highest in their series; Mn and Tc are anomalous, and after \(\mathrm{d^5}\) the melting point decreases regularly.

  16. Substances attracted into a magnetic field are paramagnetic, those repelled are diamagnetic, and those very strongly attracted are ferromagnetic. Paramagnetism and ferromagnetism come from unpaired electrons; fully paired spins give diamagnetism. Fe, Co and Ni are ferromagnetic.

  17. For the first transition series the orbital contribution is quenched, so the spin-only formula applies:

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

where \(n\) is the number of unpaired electrons. One unpaired electron gives \(\mu = 1.73\) BM. In the second and third series the orbital contribution is significant and the formula fails.

  1. Transition metal compounds are coloured because ligands split the five degenerate d orbitals by a small energy gap; absorption of visible light causes a d–d transition, and the colour seen is complementary to the colour absorbed. Ions with \(\mathrm{d^0}\) or \(\mathrm{d^{10}}\) configurations, such as \(\mathrm{Ti^{4+}}\) and \(\mathrm{Cu^{+}}\), are colourless.

  2. Colour depends on the presence of unpaired d electrons, d–d transitions, the nature of the ligands and the geometry of the complex. Pink octahedral \([\mathrm{Co(H_2O)_6}]^{2+}\) becomes deep blue tetrahedral \([\mathrm{CoCl_4}]^{2-}\) with concentrated hydrochloric acid.

  3. Colour may also arise from charge transfer, as in the intense purple of \(\mathrm{MnO_4^{-}}\), where an electron moves momentarily from oxygen to manganese.

  4. Transition metals and their compounds are good homogeneous and heterogeneous catalysts, because they move easily between oxidation states. In homogeneous catalysis the metal ion forms unstable intermediates; in heterogeneous catalysis the metal provides a surface.

  5. Important examples include \(\mathrm{MnO_2}\) for the decomposition of \(\mathrm{KClO_3}\), Mo/Fe in Haber’s process, Co-Th alloy in the Fischer Tropsch process, finely divided Ni for hydrogenation of ethene at 140 °C, platinised asbestos in the contact process, and Fe-Cr for the reaction of carbon monoxide with steam.

  6. Interstitial compounds form when small atoms of hydrogen, carbon or nitrogen occupy gaps in the metal lattice. They are hard, conduct heat and electricity, resemble the parent metal chemically, melt higher and are less dense than it, and include the inert, diamond-hard metallic carbides and the strongly reducing transition metal hydrides.

  7. Steel and cast iron are interstitial compounds of carbon and iron; the trapped carbon lowers malleability and ductility but raises tenacity.

  8. Alloys form when atoms of one metal are distributed randomly in the lattice of another, which happens readily when the radii and properties are similar. Ferrous alloys contain iron in the largest proportion; non-ferrous alloys combine a transition metal other than iron with a non transition element.

  9. Useful alloys include bronze (Cu and Sn) for statues and medals, cupra-nickel (Cu and Ni) for marine machinery, stainless steels for aircraft fuselage, nichrome (Ni and Cr, 80 : 20) for gas turbine engines, and titanium alloys for high-temperature, high-stress applications.

  10. \(\mathrm{KMnO_4}\) and \(\mathrm{K_2Cr_2O_7}\) are strong oxidising agents, with manganese at \(+7\) and chromium at \(+6\).

  11. Potassium permanganate is prepared by fusing \(\mathrm{MnO_2}\) with KOH and \(\mathrm{KClO_3}\) to give green \(\mathrm{K_2MnO_4}\), which disproportionates in neutral or acidic medium to \(\mathrm{KMnO_4}\) and \(\mathrm{MnO_2}\); alternatively, alkaline manganate is oxidised electrolytically between iron electrodes separated by a diaphragm.

\[3\,\mathrm{MnO_2} + 6\,\mathrm{KOH} + \mathrm{KClO_3} \xrightarrow{\ \Delta\ } 3\,\mathrm{K_2MnO_4} + \mathrm{KCl} + 3\,\mathrm{H_2O}\]

  1. In acidic medium \(\mathrm{KMnO_4}\) is reduced to \(\mathrm{Mn^{2+}}\), oxidising iodide to iodine, \(\mathrm{Fe^{2+}}\) to \(\mathrm{Fe^{3+}}\), \(\mathrm{H_2S}\) to sulphur and oxalic acid to carbon dioxide. In neutral or weakly alkaline medium it is reduced only to \(\mathrm{MnO_2}\), oxidising iodide to iodate, thiosulphate to sulphate and manganous salts to \(\mathrm{MnO_2}\).

  2. Potassium permanganate is used as an antiseptic, for the unsaturation test, in volumetric analysis of reducing agents, for detecting halides in qualitative analysis, and as a powerful oxidising agent.

  3. Potassium dichromate is made by roasting chromite ore with sodium carbonate and lime in air to give sodium chromate, converting this to sodium dichromate with concentrated sulphuric acid, and precipitating the less soluble potassium salt with potassium chloride.

  4. Acidified \(\mathrm{K_2Cr_2O_7}\) oxidises iodide to iodine and \(\mathrm{H_2S}\) to sulphur, and is itself reduced to green chromic sulphate — the orange-to-green change that signals the reaction.

  5. The d block elements as a whole are lustrous, hard, dense, high-melting conductors of high tensile strength that form alloys, are often paramagnetic and are efficient catalysts; chemically they are electropositive, show variable valency, form coloured salts and complexes, act as reducing agents, form insoluble oxides and hydroxides, and include biologically important metals that catalyse biological reactions.

  6. The first row differs from the second and third in the stability of higher oxidation states: Mo(V) and W(VI) compounds are more stable than Cr(VI) and Mn(VII), and the highest state is \(+7\) in the first row but \(+8\) in \(\mathrm{RuO_4}\) and \(\mathrm{OsO_4}\).

  7. Most metals occur as carbonates, sulphates, sulphides and oxides; silver, gold and platinum occur free. A mineral is any naturally occurring substance containing the metal, while an ore is a mineral rich enough for economical extraction.

  8. Metallurgy is the commercial extraction of a metal from its ore, by pyrometallurgy (reduction at high temperature with C, \(\mathrm{H_2}\) or Al), hydrometallurgy (reduction in aqueous solution) or electrometallurgy (electrolytic reduction of the fused compound).

  9. Concentration removes the gangue — sand, mud and unwanted impurities — by washing, hydraulic classification, magnetic separation or froth floatation, the method depending on the nature of the ore.

Flowchart

FC 8.12 – The chapter’s extraction sequence in condensed form: ore mined → concentration, gangue removed → roasting, sulphur and arsenic impurities driven off as oxides and FeO converted to Fe₂O₃ → reduction in the blast furnace, where coke and CO reduce Fe₂O₃ while limestone acts as flux and carries the gangue away as slag → molten iron tapped as pig iron → refining → pure iron; commercial forms branch off as cast iron, wrought iron and steel.

  1. Haematite ore contains \(\mathrm{Fe_2O_3}\) together with \(\mathrm{SiO_2}\), \(\mathrm{Al_2O_3}\) and phosphates as gangue. Iron is obtained by concentration in a hydraulic classifier, roasting in a current of air, and reduction in a blast furnace with coke and limestone.

  2. The blast furnace is a 25 m steel tower of 5 to 10 m diameter, lined with refractory bricks, made up of hearth, bosh and stack, and worked on the counter current principle, the charge descending as hot gases rise.

  3. Its zones run from 500 K at the top to 2000 K above the tuyers: combustion of coke to CO, reduction of \(\mathrm{Fe_2O_3}\) to spongy iron by CO and by carbon, slag formation from limestone and gangue, and fusion, in which molten iron absorbs C, Si, Mn, P and S.

\[\mathrm{Fe_2O_3} + 3\,\mathrm{CO} \rightarrow 2\,\mathrm{Fe} + 3\,\mathrm{CO_2} \qquad \mathrm{CaO} + \mathrm{SiO_2} \rightarrow \mathrm{CaSiO_3}\]

  1. Molten iron is cast into pigs containing about 4% carbon; remelting gives cast iron. Cast iron is hard and brittle with 4% carbon, wrought iron very soft with less than 0.2%, and steel intermediate with 0.2 to 2% carbon.

  2. The choice of extraction technique depends on the nature of the ore, the availability and cost of the reducing agent, the availability of hydraulic power, the purity required and the value of the byproducts.

  3. f block elements are those in which f orbitals are filled. They lie at the bottom of the periodic table as part of periods 6 and 7, and are called inner transition elements because the f orbital lies well inside the d orbital. Each series has 14 elements, since an f subshell of 7 orbitals holds 14 electrons.

  4. Lanthanoids run from atomic number 57 to 71 and fill the 4f orbitals; actinoids fill the 5f orbitals. The general configurations are \(\mathrm{[Xe]\,4f^{0-14}\,5d^{0-2}\,6s^2}\) and \(\mathrm{[Rn]\,5f^{0-14}\,6d^{0-2}\,7s^2}\).

  5. The rare earths are not actually rare — thulium is more abundant than silver — and the name reflects the old difficulty of separating chemically similar elements, a problem now solved by ion exchange resins.

  6. Lanthanoid metals are soft, silvery white, good conductors, non-radioactive except promethium, of moderate density and low heat of atomization, and reactive enough to give hydroxides and hydrogen with water.

\[2\,\mathrm{M}\text{(s)} + 6\,\mathrm{H_2O}\text{(l)} \rightarrow 2\,\mathrm{M(OH)_3}\text{(s)} + 3\,\mathrm{H_2}\text{(g)}\]

  1. The characteristic oxidation state is \(+3\), because 4f electrons are strongly screened by 5s and 5p electrons and take no part in bonding. States of \(+2\) and \(+4\) appear only where they give an \(\mathrm{f^0}\), \(\mathrm{f^7}\) or \(\mathrm{f^{14}}\) configuration, as in \(\mathrm{Ce^{4+}}\), \(\mathrm{Eu^{2+}}\), \(\mathrm{Tb^{4+}}\) and \(\mathrm{Yb^{2+}}\).

  2. Lanthanoids form \(\mathrm{M_2O_3}\) oxides, \(\mathrm{Ln(OH)_3}\) hydroxides, LnN nitrides, \(\mathrm{LnX_3}\) halides and \(\mathrm{LnC_2}\) carbides. The basicity of the hydroxides falls from \(\mathrm{La(OH)_3}\), the strongest base, to \(\mathrm{Lu(OH)_3}\), the weakest, as the ionic size decreases.

  3. Coordination numbers are high, commonly 6 but often 8, 9 or more, as in the nonahydrate \([\mathrm{La(H_2O)_9}]^{3+}\).

  4. Colour arises from f–f transitions, and ions with \(n\) f-electrons have about the same colour as those with \((14 - n)\). Unlike transition metals, lanthanoid spectra are almost unaffected by the ligands, because the 4f orbitals are shielded from them.

  5. Lanthanoids are strongly paramagnetic, with gadolinium ferromagnetic below its Curie point of 16 °C. The spin-only formula agrees with experiment only for \(\mathrm{Gd^{3+}}\), since elsewhere the orbital contribution is not quenched.

  6. Lanthanoid contraction is the steady decrease in atomic and ionic radii from lanthanum to lutetium, caused by the poor shielding of the increasing nuclear charge by the diffused 4f electrons.

  7. Lanthanoid compounds are used as phosphors in colour television tubes and monitors, as in \((\mathrm{Eu,Y})_2\mathrm{O_3}\) for red, in the Nd:YAG laser, in erbium-doped fibre amplifiers, and in hybrid cars, superconductors and permanent magnets.

  8. Actinoids are radioactive, mostly man-made, dense, high-melting elements, less reactive than lanthanoids, reacting with hot but not cold water. Thorium and uranium survive in rocks only because their half-lives are so long.

  9. Actinoids show oxidation states from \(+2\) to \(+8\), with \(+3\) the most stable, because the 5f, 6d and 7s orbitals lie close in energy. For Th, Pa and Np electrons occupy either 6d or 5f, while from Pu onwards the 5f shell fills regularly.

  10. Actinoid contraction is the steady decrease in ionic radius across the series, again from poor shielding by f electrons.

  11. Compared with lanthanoids, actinoids show a wider range of oxidation states, form complexes and oxocations such as \(\mathrm{UO^{+}}\), \(\mathrm{PuO^{+}}\) and \(\mathrm{NpO_2^{+}}\) more readily, are all radioactive, and give coloured ions.

  12. Thorium(IV) oxide with 1% \(\mathrm{CeO_2}\) converts the heat of a gas flame into intense light and is still used in camping lanterns. Uranium is oxidised from \(\mathrm{UO_2}\) to \(\mathrm{UO_3}\) by iron(III), dissolved as uranyl sulphate and precipitated with ammonia as the yellow cake in which it is traded.

  13. Elements beyond atomic number 92 are transuranium elements; those from 93 to 103 belong to the actinoid series and those from 104 to 118 are postactinoid elements, which are transition metals made in nuclear reactions. Their half-lives are so short — element 112 lasts \(2.8 \times 10^{-4}\) seconds — that little of their chemistry is known, though rutherfordium forms \(\mathrm{RfCl_4}\) like zirconium and hafnium, and dubnium resembles both niobium(V) and protactinium(V).

8.19 Objective Questions

  1. Which one of the following is diamagnetic?
    1. \(\mathrm{Cr^{2+}}\)
    2. \(\mathrm{Fe^{3+}}\)
    3. \(\mathrm{Cu^{2+}}\)
    4. \(\mathrm{Sc^{3+}}\)
  2. Most stable oxidation state of Titanium is
    1. \(+2\)
    2. \(+3\)
    3. \(+4\)
    4. \(+5\)
  3. Components of Nichrome alloy are
    1. Ni, Cr, Fe
    2. Ni, Cr, Fe, C
    3. Ni, Cr
    4. Cu, Fe
  4. Most stable oxidation state of Ruthenium is
    1. \(+2\)
    2. \(+4\)
    3. \(+8\)
    4. \(+6\)
  5. Stable oxidation states for chromium are
    1. \(+2, +3\)
    2. \(+3, +4\)
    3. \(+4, +5\)
    4. \(+3, +6\)
  6. Electronic configuration of Cu and \(\mathrm{Cu^{+1}}\)
    1. \(\mathrm{3d^{10},\,4s^0}\); \(\mathrm{3d^9,\,4s^0}\)
    2. \(\mathrm{3d^9,\,4s^1}\); \(\mathrm{3d^9 4s^0}\)
    3. \(\mathrm{3d^{10},\,4s^1}\); \(\mathrm{3d^{10},\,4s^0}\)
    4. \(\mathrm{3d^8,\,4s^1}\); \(\mathrm{3d^{10},\,4s^0}\)
  7. Which of the following have \(\mathrm{d^0s^0}\) configuration?
    1. \(\mathrm{Sc^{3+}}\)
    2. \(\mathrm{Ti^{4+}}\)
    3. \(\mathrm{V^{5+}}\)
    4. all of the above
  8. Magnetic moment of a metal complex is 5.9 B.M. Number of unpaired electrons in the complex is
    1. 2
    2. 3
    3. 4
    4. 5
  9. In which of the following series are all the elements radioactive in nature?
    1. Lanthanoids
    2. Actinoids
    3. d-block elements
    4. s-block elements
  10. Which of the following sets of ions contain only paramagnetic ions?
  1. \(\mathrm{Sm^{3+}}\), \(\mathrm{Ho^{3+}}\), \(\mathrm{Lu^{3+}}\)
  2. \(\mathrm{La^{3+}}\), \(\mathrm{Ce^{3+}}\), \(\mathrm{Sm^{3+}}\)
  3. \(\mathrm{La^{3+}}\), \(\mathrm{Eu^{3+}}\), \(\mathrm{Gd^{3+}}\)
  4. \(\mathrm{Ce^{3+}}\), \(\mathrm{Eu^{3+}}\), \(\mathrm{Yb^{3+}}\)
  1. Which actinoid, other than uranium, occurs in significant amount naturally?
  1. Thorium
  2. Actinium
  3. Protactinium
  4. Plutonium
  1. The flux added during extraction of Iron from hematite is
  1. Silica
  2. Calcium carbonate
  3. Sodium carbonate
  4. Alumina

8.20 Check Your Concepts

  1. Assertion: Zinc is not regarded as showing the properties typical of a transition element. Reason: The \(\mathrm{Zn^{2+}}\) ion has a completely filled 3d subshell. State whether the assertion and the reason are true, and whether the reason correctly explains the assertion.

  2. Manganese shows six oxidation states while nickel shows three. Account for this difference in terms of electronic configuration.

  3. The atomic radii of cobalt and nickel are almost identical, and copper is slightly larger than both. Explain why the steady decrease in atomic radius does not continue all the way to the end of the first transition series.

  4. Why are the ionic radii of transition elements smaller than those of representative elements of the same period?

  5. Predict, with reasons, whether an aqueous solution of each of the following would be coloured or colourless: \(\mathrm{V^{3+}}\), \(\mathrm{Ti^{4+}}\), \(\mathrm{Cu^{+}}\), \(\mathrm{Cu^{2+}}\).

  6. The permanganate ion is intensely purple although manganese in it has no d electrons left. Explain how this colour arises, and name two other ions whose colour is explained in the same way.

  7. A pink aqueous solution of cobalt chloride turns deep blue on adding concentrated hydrochloric acid. Name the two species responsible for the two colours, state the geometry of each, and explain why the colour changes.

  8. The spin-only formula predicts the magnetic moments of first-series transition metal ions well, but fails for most lanthanoid ions. Explain why, and name the one lanthanoid ion for which it works.

  9. Complete the following reactions and state the oxidation state of manganese in each product: (i) \(\mathrm{MnO_4^{-}}\) with \(\mathrm{Fe^{2+}}\) in acidic medium; (ii) \(\mathrm{MnO_4^{-}}\) with \(\mathrm{I^{-}}\) in neutral or weakly alkaline medium.

  10. When hydrogen sulphide gas is passed through acidified potassium dichromate solution, the colour changes from orange to green. Write the balanced equation, name the precipitate formed, and explain the colour change in terms of oxidation states.

  11. Green potassium manganate is said to disproportionate in acidic medium. Define disproportionation and show, using oxidation numbers, that this reaction fits the definition.

  12. Match the catalyst in column A with the process in column B.

Column A Column B
a. \(\mathrm{MnO_2}\) i. Contact process
b. Mo/Fe ii. Hydrogenation of ethene
c. Platinised asbestos iii. Decomposition of \(\mathrm{KClO_3}\)
d. Finely divided Ni iv. Fischer Tropsch process
e. Co-Th alloy v. Haber’s process

  1. Distinguish between homogeneous and heterogeneous catalysis, and state which type is illustrated by the hydrogenation of oils to margarine.

  2. Assertion: Interstitial compounds have higher melting points than the parent metal. Reason: Small atoms occupying the interstitial sites prevent the metal layers from sliding over one another. State whether each statement is true, and whether the reason explains the assertion.

  3. Explain why the addition of carbon to iron lowers its malleability and ductility but increases its tenacity.

  4. Match the alloy in column A with its typical use in column B.

Column A Column B
a. Bronze i. Gas turbine engines
b. Nichrome ii. Marine condenser tubes
c. Cupra-nickel iii. Statues, medals and trophies
d. Stainless steel iv. Fire proof bulkheads in high speed flight
e. Titanium alloy v. Outer fuselage of ultra-high speed aircraft

  1. Limestone is charged into the blast furnace although it contains no iron. Explain its function, write the two reactions it takes part in, and state why it is called a flux rather than a reducing agent.

  2. Why is the roasted ore sintered into lumps before it is charged into the blast furnace? What difficulty would arise if powdered ore were charged directly?

  3. Distinguish between a mineral and an ore, giving one example of each from the transition metals.

  4. Arrange \(\mathrm{Sc^{3+}}\), \(\mathrm{Ti^{3+}}\), \(\mathrm{Cr^{3+}}\) and \(\mathrm{Mn^{2+}}\) in increasing order of spin-only magnetic moment, and justify the order you have written without performing any calculation.

  5. Arrange \(\mathrm{La(OH)_3}\), \(\mathrm{Gd(OH)_3}\) and \(\mathrm{Lu(OH)_3}\) in decreasing order of basic strength and explain the order in terms of ionic size.

  6. \(\mathrm{Pr^{3+}}\) and \(\mathrm{Tm^{3+}}\) are both green, and \(\mathrm{Nd^{3+}}\) and \(\mathrm{Er^{3+}}\) are both pink, although the two members of each pair have very different numbers of f electrons. Explain the relationship between the pairs.

  7. Zirconium and hafnium are extremely difficult to separate from each other. Name the effect responsible for this, explain its cause, and state why the separation matters in the nuclear industry.

  8. Assertion: Actinoids show a much wider range of oxidation states than lanthanoids. Reason: In actinoids the 5f, 6d and 7s orbitals are close in energy. State whether each statement is true, and whether the reason explains the assertion.

  9. Lanthanoid spectra are almost unaffected by a change of ligand, whereas transition metal complexes change colour readily when the ligand is changed. Explain this difference, and state one practical use that depends on it.

  10. Very little is known about the chemistry of the postactinoid elements. Give two reasons, and explain how chemists were nevertheless able to predict that rutherfordium would form a tetrachloride.

  11. Give uses of \(\mathrm{KMnO_4}\).

  12. Why are salts of \(\mathrm{Sc^{3+}}\), \(\mathrm{Ti^{4+}}\), \(\mathrm{V^{5+}}\) colourless?

  13. Which steps are involved in the manufacture of potassium dichromate from chromite ore?

  14. What are the stable oxidation states of plutonium, cerium, manganese and europium?

  15. Write the electronic configuration of chromium and copper.

  16. Why is nobelium the only actinoid with \(+2\) oxidation state?

  17. Explain with the help of a balanced chemical equation why the solution of Ce(IV) is prepared in acidic medium.

  18. What is meant by ‘shielding of electrons’ in an atom?

  19. The atomic number of an element is 90. Is this element diamagnetic or paramagnetic?

  20. Explain the trends in atomic radii of d block elements.

  21. Name the different zones in the blast furnace. Write the reactions taking place in them.

  22. What are the differences between cast iron, wrought iron and steel?

  23. Iron exhibits \(+2\) and \(+3\) oxidation states. Write their electronic configuration. Which will be more stable? Why?

  24. Give the similarities and differences in elements of the 3d, 4d and 5d series.

  25. Explain trends in ionisation enthalpies of d block elements.

  26. What is meant by diamagnetic and paramagnetic metal? Give one example of a diamagnetic and a paramagnetic transition metal and lanthanoid metal.

  27. Why are the ground-state electronic configurations of gadolinium and lawrencium different from those expected?

  28. Write the steps involved in a metallurgical process.

  29. Cerium and terbium behave as good oxidising agents in the \(+4\) oxidation state. Explain.

  30. Europium and ytterbium behave as good reducing agents in the \(+2\) oxidation state. Explain.

8.21 Practice With Numericals

  1. Calculate the spin-only magnetic moment of \(\mathrm{Fe^{3+}}\) and of \(\mathrm{Ni^{2+}}\), and state which of the two is the more strongly paramagnetic.

  2. The spin-only magnetic moment of a first-series metal ion is found to be 2.84 BM. Determine the number of unpaired electrons present, and name one ion from the first transition series that fits this value.

  3. Calculate the spin-only magnetic moment of \(\mathrm{Ti^{3+}}\), \(\mathrm{V^{3+}}\) and \(\mathrm{Cr^{2+}}\), and arrange the three ions in increasing order of magnetic moment.

  4. Using the \(\mathrm{M^{3+}}\) radii given in Table 8.5, calculate the percentage decrease in ionic radius on going from \(\mathrm{Sc^{3+}}\) (73 pm) to \(\mathrm{Ni^{3+}}\) (60 pm).

  5. Calculate the mass of iron that can be obtained from 2500 kg of haematite ore containing 60% \(\mathrm{Fe_2O_3}\) by mass, assuming complete reduction. (Fe = 56, O = 16)

  6. Calculate the percentage of iron by mass in haematite, \(\mathrm{Fe_2O_3}\), and in magnetite, \(\mathrm{Fe_3O_4}\), and state which mineral is the richer source of the metal. (Fe = 56, O = 16)

  7. How many moles of \(\mathrm{Fe^{2+}}\) can be oxidised by 0.02 mol of \(\mathrm{KMnO_4}\) in acidic medium? Show your reasoning from the change in oxidation number.

  8. Calculate the mass of calcium silicate, \(\mathrm{CaSiO_3}\), formed when 60 kg of silica reacts completely with excess quicklime in the blast furnace. (Ca = 40, Si = 28, O = 16)

  9. Determine the oxidation number of the metal in each of the following: \(\mathrm{K_2Cr_2O_7}\), \(\mathrm{K_2MnO_4}\), \(\mathrm{Cr_2(SO_4)_3}\) and \(\mathrm{UO_2SO_4}\).

  10. Balance the following equation by the change-in-oxidation-number method:

\[\mathrm{Fe_2O_3} + \mathrm{C} \rightarrow \mathrm{Fe} + \mathrm{CO}\]

  1. The half-life of Th-232 is \(1.4 \times 10^{10}\) years. What fraction of an original sample of Th-232 remains undecayed after \(4.2 \times 10^{10}\) years?

  2. Calculate the mass of \(\mathrm{K_2Cr_2O_7}\) required to prepare 500 mL of a 0.05 M solution. (Molar mass of \(\mathrm{K_2Cr_2O_7} = 294\ \text{g mol}^{-1}\))

  3. What is the oxidation state of Manganese in (i) \(\mathrm{MnO_4^{2-}}\) (ii) \(\mathrm{MnO_4^{-}}\)?

  4. Balance the following equations:

\[\text{(i)}\quad \mathrm{KMnO_4} + \mathrm{H_2C_2O_4} + \mathrm{H_2SO_4} \rightarrow \mathrm{MnSO_4} + \mathrm{K_2SO_4} + \mathrm{H_2O} + \mathrm{O_2}\]

\[\text{(ii)}\quad \mathrm{K_2Cr_2O_7} + \mathrm{KI} + \mathrm{H_2SO_4} \rightarrow \mathrm{K_2SO_4} + \mathrm{Cr_2(SO_4)_3} + 7\,\mathrm{H_2O} + 3\,\mathrm{I_2}\]