8. The d- and f- block Elements

Iron, copper, silver and gold are among the transition elements that have played important roles in the development of human civilisation. The inner transition elements such as Th, Pa and U are proving excellent sources of nuclear energy in modern times.

The d-block of the periodic table contains the elements of the groups 3-12 in which the d orbitals are progressively filled in each of the four long periods. The elements constituting the f -block are those in which the 4 f and 5 f orbitals are progressively filled in the latter two long periods; these elements are formal members of group 3 from which they have been taken out to form a separate f-block of the periodic table. The names transition metals and inner transition metals are often used to refer to the elements of d-and f-blocks respectively.

There are mainly three series of the transition metals, 3d series (Sc to Zn), 4d series (Y to Cd) and 5d series (La to Hg, omitting Ce to Lu). The fourth 6d series which begins with Ac is still incomplete. The two series of the inner transition metals, (4f and 5f) are known as lanthanoids and actinoids respectively.

Strictly speaking, a transition element is defined as the one which has incompletely filled d orbitals in its ground state or in any one of its oxidation states. Zinc, cadmium and mercury of group 12 have full d10 configuration in their ground state as well as in their common oxidation states and hence, are not regarded as transition metals. However, being the end members of the three transition series, their chemistry is studied along with the chemistry of the transition metals.
The presence of partly filled d or f orbitals in their atoms sets the study of the transition elements and their compounds apart from that of the main group elements. However, the usual theory of valence as applicable to the main group elements can also be applied successfully to the transition elements.

Various precious metals such as silver, gold and platinum and industrially important metals like iron, copper and titanium form part of the transition metals.

In this Unit, besides introduction, we shall first deal with the electronic configuration, occurrence and general characteristics of the transition elements with special emphasis on the trends in the properties of the first row (3d) transition metals and the preparation and properties of some important compounds. This will be followed by consideration of certain general aspects such as electronic configurations, oxidation states and chemical reactivity of the inner transition metals.

THE TRANSITION ELEMENTS (d-BLOCK)

8.1 Position in the Periodic Table

The d–block occupies the large middle section flanked by s– and p– blocks in the periodic table. The very name ‘transition’ given to the elements of d-block is only because of their position between s– and p– block elements. The d–orbitals of the penultimate energy level in their atoms receive electrons giving rise to the three rows of the transition metals, i.e., 3d, 4d and 5d. The fourth row of 6d is still incomplete. These series of the transition elements are shown in Table 8.1.

8.2 Electronic Configurations of the d-Block Elements

In general the electronic configuration of these elements is (n-1)d1–10ns1–2 . The (n–1) stands for the inner d orbitals which may have one to ten electrons and the outermost ns orbital may have one or two electrons. However, this generalisation has several exceptions because of very little energy difference between (n-1)d and ns orbitals. Furthermore, half and completely filled sets of orbitals are relatively more stable. A consequence of this factor is reflected in the electronic configurations of Cr and Cu in the 3d series. Consider the case of Cr, for example, which has 3d54s1 instead of 3d44s2 ; the energy gap between the two sets (3d and 4s) of orbitals is small enough to prevent electron entering the 3d orbitals. Similarly in case of Cu, the configuration is 3d104s1 and not 3d94s2 . The outer electronic configurations of the transition elements are given in Table 8.1.

Table 8.1: Outer Electronic Configurations of the Transition Elements (ground state)
1st Series
Sc Ti V Cr Mn Fe Co Ni Cu Zn
Z 21 22 23 24 25 21 22 23 24 25
4s 2 2 2 1 2 2 2 2 1 2
3d 1 2 3 5 5 6 7 8 10 10
2nd Series
Y Zr Nb Mo Tc Ru Rh Pd Ag Cd
Z 39 40 41 42 43 44 45 46 47 48
5s 2 2 1 1 1 1 1 0 1 2
3d 1 2 4 5 6 7 8 10 10 10
3rd Series
La Hf Ta W Re Os Ir Pt Au Hg
Z 57 72 73 74 75 76 77 78 79 80
6s 2 2 2 2 2 2 2 1 1 2
5d 1 2 3 4 5 6 7 9 10 10
4th Series
Ac Rf Db Sg Bh Hs Mt Ds Rg Uub
Z 89 104 105 106 107 108 109 110 111 112
7s 2 2 2 2 2 2 2 2 1 2
6d 1 2 3 4 5 6 7 8 10 10

The electronic configurations of Zn, Cd and Hg are represented by the general formula (n-1)d10 ns2. The orbitals in these elements are completely filled in the ground state as well as in their common oxidation states. Therefore, they are not regarded as transition elements.

The d orbitals of the transition elements project to the periphery of an atom more than the other orbitals (i.e., s and p), hence, they are more influenced by the surroundings as well as affecting the atoms or molecules surrounding them. In some respects, ions of a given dn configuration (n = 1 – 9) have similar magnetic and electronic properties. With partly filled d orbitals these elements exhibit certain characteristic properties such as display of a variety of oxidation states, formation of coloured ions and entering into complex formation with a variety of ligands.

The transition metals and their compounds also exhibit catalytic property and paramagnetic behaviour. All these characteristics have been discussed in detail later in this Unit.

There are greater horizontal similarities in the properties of the transition elements in contrast to the main group elements. However, some group similarities also exist. We shall first study the general characteristics and their trends in the horizontal rows (particularly 3d row) and then consider some group similarities.

Example 8.1
On what ground can you say that scandium (Z = 21) is a transition element but zinc (Z = 30) is not?
Solution
On the basis of incompletely filled 3d orbitals in case of scandium atom in its ground state (3d1 ), it is regarded as a transition element. On the other hand, zinc atom has completely filled d orbitals (3d10 ) in its ground state as well as in its oxidised state, hence it is not regarded as a transition element.

Intext Question
8.1 Silver atom has completely filled d orbitals (4d10) in its ground state. How can you say that it is a transition element?

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

8.3.1 Physical Properties

Nearly all the transition elements display typical metallic properties such as high tensile strength, ductility malleability, high thermal and electrical conductivity and metallic lustre. With the exceptions of Zn, Cd, Hg and Mn, they have one or more typical metallic structures at normal temperatures.

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

The transition metals (with the exception of Zn, Cd and Hg) are very much hard and have low volatility. Their melting and boiling points are high. Fig. 8.1 depicts the melting points of the 3d, 4d and 5d transition metals. The high melting points of these metals are attributed to the involvement of greater number of electrons from (n-1)d in addition to the ns electrons in the interatomic metallic bonding. In any row the melting points of these metals rise to a maximum at d5 except for anomalous values of Mn and Tc and fall regularly as the atomic number increases. They have high enthalpies of atomisation which are shown in Fig. 8.2. The maxima at about the middle of each series indicate that one unpaired electron per d orbital is particularly favourable for strong interatomic interaction. In general, greater the number of valence electrons, stronger is the resultant bonding. Since the enthalpy of atomisation is an important factor in determining the standard electrode potential of a metal, metals with very high enthalpy of atomisation (i.e., very high boiling point) tend to be noble in their reactions (see later for electrode potentials).

Another generalisation that may be drawn from Fig. 8.2 is that the metals of the second and third series have greater enthalpies of atomisation than the corresponding elements of the first series; this is an important factor in accounting for the occurrence of much more frequent metal – metal bonding in compounds of the heavy transition metals.

8.3.2 Variation in Atomic and Ionic Sizes of Transition Metals

In general, ions of the same charge in a given series show progressive decrease in radius with increasing atomic number. This is because the new electron enters a d orbital each time the nuclear charge increases by unity. It may be recalled that the shielding effect of a d electron is not that effective, hence the net electrostatic attraction between the nuclear charge and the outermost electron increases and the ionic radius decreases. The same trend is observed in the atomic radii of a given series. However, the variation within a series is quite small. An interesting point emerges when atomic sizes of one series are compared with those of the corresponding elements in the other series. The curves in Fig. 8.3 show an increase from the first (3d) to the second (4d) series of the elements but the radii of the third (5d) series are virtually the same as those of the corresponding members of the second series. This phenomenon is associated with the intervention of the 4f orbitals which must be filled before the 5d series of elements begin. The filling of 4f before 5d orbital results in a regular decrease in atomic radii called Lanthanoid contraction which essentially compensates for the expected increase in atomic size with increasing atomic number. The net result of the lanthanoid contraction is that the second and the third d series exhibit similar radii (e.g., Zr 160 pm, Hf 159 pm) and have very similar physical and chemical properties much more than that expected on the basis of usual family relationship.

The factor responsible for the lanthanoid contraction is somewhat similar to that observed in an ordinary transition series and is attributed to similar cause, i.e., the imperfect shielding of one electron by another in the same set of orbitals. However, the shielding of one 4f electron by another is less than that of one d electron by another, and as the nuclear charge increases along the series, there is fairly regular decrease in the size of the entire 4fn orbitals.

The decrease in metallic radius coupled with increase in atomic mass results in a general increase in the density of these elements. Thus, from titanium (Z = 22) to copper (Z = 29) the significant increase in the density may be noted elements (Table 8.2).

Table 8.2: Electronic Configurations and some other Properties of the First Series of Transition Elements
Element Sc Ti V Cr Mn Fe Co Ni Cu Zn
Atomic number 21 22 23 24 25 26 27 28 29 30
Electronic configuration
M 3d14s2 3d24s2 3d34s2 3d54s1 3d54s2 3d64s2 3d74s2 3d84s2 3d104s1 3d104s2
M+ 3d14s1 3d24s1 3d34s1 3d5 3d54s1 3d64s1 3d74s1 3d84s1 3d10 3d104s1
M2+ 3d1 3d2 3d3 3d4 3d5 3d6 3d7 3d8 3d9 3d10
M3+ [Ar] 3d1 3d2 3d3 3d4 3d5 3d6 3d7 3d- 3d-
Enthalpy of atomisation, ΔaHΘ/kJ mol-1
326 473 515 397 281 416 425 430 339 126
Ionisation enthalpy/ ΔiHΘ/kJ mol-1
Metallic/ionic M 164 147 135 129 137 126 125 125 128 137
radii/pm M2+ - - 79 82 82 77 74 70 73 75
M3+ 73 67 64 62 65 65 61 60 - -
Standard electrode M2+/M - -1.63 -1.18 -0.90 -1.18 -0.44 -0.28 -0.25 +0.34 -0.76
potential Eθ/V M3+/M2+ - -0.37 -0.26 -0.41 +1.57 +0.77 +1.97 - - -
Density/g cm-3 3.43 4.1 6.07 7.19 7.21 7.8 8.7 8.9 8.9 7.1

Example 8.2
Why do the transition elements exhibit higher enthalpies of atomisation?
Solution
Because of large number of unpaired electrons in their atoms they have stronger interatomic interaction and hence stronger bonding between atoms resulting in higher enthalpies of atomisation.

8.3.3 Ionisation Enthalpies

Due to an increase in nuclear charge which accompanies the filling of the inner d orbitals, there is an increase in ionisation enthalpy along each series of the transition elements from left to right. However, many small variations occur. Table 8.2 gives the values for the first three ionisation enthalpies of the first row elements. These values show that the successive enthalpies of these elements do not increase as steeply as in the main group elements. Although the first ionisation enthalpy, in general, increases, the magnitude of the increase in the second and third ionisation enthalpies for the successive elements, in general, is much higher.

The irregular trend in the first ionisation enthalpy of the 3d metals, though of little chemical significance, can be accounted for by considering that the removal of one electron alters the relative energies of 4s and 3d orbitals. So the unipositive ions have dn configurations with no 4s electrons. There is thus, a reorganisation energy accompanying ionisation with some gains in exchange energy as the number of electrons increases and from the transference of s electrons into d orbitals. There is the generally expected increasing trend in the values as the effective nuclear charge increases. However, the value of Cr is lower because of the absence of any change in the d configuration and the value for Zn higher because it represents an ionisation from the 4s level. The lowest common oxidation state of these metals is +2. To form the M2+ ions from the gaseous atoms, the sum of the first and second ionisation energies is required in addition to the enthalpy of atomisation for each element. The dominant term is the second ionisation enthalpy which shows unusually high values for Cr and Cu where the d5 and d10 configurations of the M+ ions are disrupted, with considerable loss of exchange energy. The value for Zn is correspondingly low as the ionisation consists of the removal of an electron which allows the production of the stable d10 configuration. The trend in the third ionisation enthalpies is not complicated by the 4s orbital factor and shows the greater difficulty of removing an electron from the d5 (Mn2+ ) and d10 (Zn2+ ) ions superimposed upon the general increasing trend. In general, the third ionisation enthalpies are quite high and there is a marked break between the values for Mn2+ and Fe2+ . Also the high values for copper, nickel and zinc indicate why it is difficult to obtain oxidation state greater than two for these elements.

Although ionisation enthalpies give some guidance concerning the relative stabilities of oxidation states, this problem is very complex and not amendable to ready generalisation.

8.3.4 Oxidation States

One of the notable features of a transition element is the great variety of oxidation states it may show in its compounds. Table 8.3 lists the common oxidation states of the first row transition elements.

Table 8.3: Oxidation States of the first row Transition Metals (the most common ones are in bold types)
Sc Ti V Cr Mn Fe Co Ni Cu Zn
+2 +2 +2 +2 +2 +2 +2 +1 +2
+3 +3 +3 +3 +3 +3 +3 +3 +2
+4 +4 +4 +4 +4 +4 +4
+5 +5 +5
+6 +6 +6
+7

The elements which give the greatest number of oxidation states occur in or near the middle of the series. Manganese, for example, exhibits all the oxidation states from +2 to +7. The lesser number of oxidation states at the extreme ends stems from either too few electrons to lose or share (Sc, Ti) or too many d electrons (hence fewer orbitals available in which to share electrons with others) for higher valence (Cu, Zn). Thus, early in the series scandium(II) is virtually unknown and titanium (IV) is more stable than Ti(III) or Ti(II). At the other end, the only oxidation state of zinc is +2 (no d electrons are involved). The maximum oxidation states of reasonable stability correspond in value to the sum of the s and d electrons upto manganese (TiIVO2, VVO2+, CrV1O42–, MnVIIO4–) followed by a rather abrupt decrease in stability of higher oxidation states, so that the typical species to follow are FeII,III, CoII,III, NiII, CuI,II, ZnII.

The variability of oxidation states, a characteristic of transition elements, arises out of incomplete filling of d orbitals in such a way that their oxidation states differ from each other by unity, e.g., VII , VIII , VIV , VV . This is in contrast with the variability of oxidation states of non transition elements where oxidation states normally differ by a unit of two.

An interesting feature in the variability of oxidation states of the d–block elements is noticed among the groups (groups 4 through 10). Although in the p–block the lower oxidation states are favoured by the heavier members (due to inert pair effect), the opposite is true in the groups of d-block. For example, in group 6, Mo(VI) and W(VI) are found to be more stable than Cr(VI). Thus Cr(VI) in the form of dichromate in acidic medium is a strong oxidising agent, whereas MoO3 and WO3 are not.
Low oxidation states are found when a complex compound has ligands capable of π-acceptor character in addition to the σ-bonding. For example, in Ni(CO)4 and Fe(CO)5, the oxidation state of nickel and iron is zero.

Example 8.3
Name a transition element which does not exhibit variable oxidation states.
Solution
Scandium (Z = 21) does not exhibit variable oxidation states.

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

Table 8.4 contains the thermochemical parameters related to the transformation of the solid metal atoms to M2+ ions in solution and their standard electrode potentials. The observed values of Eθ and those calculated using the data of Table 8.4 are compared in Fig. 8.4.

The unique behaviour of Cu, having a positive EΘ, accounts for its inability to liberate H2 from acids. Only oxidising acids (nitric and hot concentrated sulphuric) react with Cu, the acids being reduced. The high energy to transform Cu(s) to Cu2+(aq) is not balanced by its hydration enthalpy. The general trend towards less negative EΘ values across the series is related to the general increase in the sum of the first and second ionisation enthalpies. It is interesting to note that the value of E for Mn, Ni and Zn are more negative than expected from the trend.

Example 8.4
Why is Cr2+ reducing and Mn3+ oxidising when both have d4 configuration.
Solution
Cr2+ is reducing as its configuration changes from d4 to d3 , the latter having a half-filled t2g level (see Unit 9) . On the other hand, the change from Mn2+ to Mnd3+ results in the half-filled (d ) configuration which has extra stability.

Intext Question
8.4 The EΘ(M2+/M) value for copper is positive (+0.34V). What is possibly the reason for this? (Hint: consider its high ΔaHΘ and low ΔhydHΘ)

Table 8 .4: Thermochemical data (kJ mol ) for the first row Transition Elements and the Standard Electrode Potentials for the II Reduction of M to M.

Element (M) ΔiHΘ(M) ΔiH1Θ ΔiH2Θ ΔhydHΘ(M2+) EΘ/V
Ti 469 661 1310 -1866 -1.63
V 515 648 1370 -1895 -1.18
Cr 398 653 1590 -1925 -0.90
Mn 279 716 1510 -1862 -1.18
Fe 418 762 1560 -1998 -0.44
Co 427 757 1640 -2079 -0.28
Ni 431 736 1750 -2121 -0.25
Cu 339 745 1960 -2121 0.34
Zn 130 908 1730 -2059 -0.76

The stability of the half-filled d sub-shell in Mn2+ and the completely filled d10 configuration in Zn2+ are related to their EΘ values, whereas EΘ for Ni is related to the highest negative ΔhydHΘ .

8.3.6 Trends in the M3+/M2+ Standard Electrode Potentials

An examination of the EΘ (M3+/M2+) values (Table 8.2) shows the varying trends. The low value for Sc3+ reflects the stability of Sc which has a noble gas configuration. The highest value for Zn is due to the removal of an electron from the stable d10 configuration of Zn2+ . The comparatively high value for Mn shows that Mn2+ (d5 ) is particularly stable, whereas comparatively low value for Fe shows the extra stability of Fe3+ (d5 ). The comparatively low value for V is related to the stability of V2+ (half-filled t2g level, Unit 9).

An examination of the EΘ (M3+/M2+ ) values (Table 8.2) shows the varying trends. The low value for Sc reflects the stability of Sc which has a noble gas configuration. The highest value for Zn is due to the removal of an electron from the stable d10 configuration of Zn2+ . The comparatively high value for Mn shows that Mn2+ (d5 ) is particularly stable, whereas comparatively low value for Fe shows the extra stability of Fe3+ (d5 ). The comparatively low value for V is related to the stability of V2+ (half-filled t2g level, Unit 9).

8.3.7 Trends in Stability of Higher Oxidation States

Table 8.5 shows the stable halides of the 3d series of transition metals. The highest oxidation numbers are achieved in TiX4 (tetrahalides), VF5 and CrF6. The +7 state for Mn is not represented in simple halides but MnO3F is known, and beyond Mn no metal has a trihalide except FeX3 and CoF3. The ability of fluorine to stabilise the highest oxidation state is due to either higher lattice energy as in the case of CoF3, or higher bond enthalpy terms for the higher covalent compounds, e.g., VF5 and CrF6.
Although VV is represented only by VF5, the other halides, however, undergo hydrolysis to give oxohalides, VOX3. Another feature of fluorides is their instability in the low oxidation states e.g., VX2 (X = CI, Br or I)

key: X = F →I; X1 = F → Br; XII = F, Cl; XIII =Cl→I

+1       CuXIII

Table 8.5: Formulas of Halides of 3d Metals
Oxidation Number
+6 CrF6
+5 VF5 CrF5
+4 TiX4 VX14 CrX4 MnF4
+3 TiX3 VX3 CrX3 MnF3 FeXI3 CoF3
+2 TiX2III VX2 CrX2 MnX2 FeX2 CoX2 NiX2 CuX2II ZnX2

and the same applies to CuX. On the other hand, all CuII halides are known except the iodide. In this case, Cu2+ oxidises I to I2:
2Cu2+ + 4I → Cu2I2 ( s ) + I2
However, many copper (I) compounds are unstable in aqueous solution and undergo disproportionation.
2Cu+ → Cu2+ + Cu
The stability of Cu2+ (aq) rather than Cu+ (aq) is due to the much more negative ΔhydHΘ of Cu2+ (aq) than Cu+ , which more than compensates for the second ionisation enthalpy of Cu.
The ability of oxygen to stabilise the highest oxidation state is demonstrated in the oxides. The highest oxidation number in the oxides (Table 8.6) coincides with the group number and is attained in Sc2O3 to Mn2O7. Beyond Group 7, no higher oxides of Fe above Fe2O3, are known, although ferrates (VI)(FeO4)2–, are formed in alkaline media but they readily decompose to Fe2O3 and O2. Besides the oxides, oxocations stabilise Vv as VO2+, VIV as VO2+ and TiIV as TiO2+. The ability of oxygen to stabilise these high oxidation states exceeds that of fluorine. Thus the highest Mn fluoride is MnF4 whereas the highest oxide is Mn2O7. The ability of oxygen to form multiple bonds to metals explains its superiority. In the covalent oxide Mn2O7, each Mn is tetrahedrally surrounded by O’s including a Mn–O–Mn bridge. The tetrahedral [MO4]n- ions are known for VV, CrVl, MnV, MnVl and MnVII.

Table 8.6: Oxides of 3d Metals
Oxidation Number Group
3 4 5 6 7 8 9 10 11 12
+7 Mn2O7
+6 CrO3
+5 V2O5
+4 TiO2 V2O4 CrO2 MnO2
+3 Sc2O3 Ti2O3 V2O3 Cr2O3 Mn2O3 Fe2O3
Mn3O4* Fe3O4* Co3O4*
+2 TiO VO (CrO) MnO FeO CoO NiO CuO ZnO
Cu2O

Example 8.5
How would you account for the increasing oxidising power in the series VO2+ < Cr2O72– < MnO4 ?
Solution
This is due to the increasing stability of the lower species to which they are reduced.

Intext Question
8.5 How would you account for the irregular variation of ionisation enthalpies (first and second) in the first series of the transition elements?

8.3.8 Chemical Reactivity and EΘ Values

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

The metals of the first series with the exception of copper are relatively more reactive and are oxidised by 1M H+ , though the actual rate at which these metals react with oxidising agents like hydrogen ion (H+) is sometimes slow. For example, titanium and vanadium, in practice, are passive to dilute non oxidising acids at room temperature. The EΘ values for M2+ /M (Table 8.2) indicate a decreasing tendency to form divalent cations across the series. This general trend towards less negative E&Theta values is related to the increase in the sum of the first and second ionisation enthalpies. It is interesting to note that the EΘ values for Mn, Ni and Zn are more negative than expected from the general trend. Whereas the stabilities of half-filled d subshell (d5) in Mn2+ and completely filled d subshell (d10 ) in zinc are related to their Ee values; for nickel, EΘ value is related to the highest negative enthalpy of hydration.

An examination of the EΘ values for the redox couple M3+/M2+ (Table 8.2) shows that Mn3+ and Co3+ ions are the strongest oxidising agents in aqueous solutions. The ions Ti2+ , V2+ and Cr2+ are strong reducing agents and will liberate hydrogen from a dilute acid, e.g.,
2 Cr2+(aq) + 2 H+(aq) → 2 Cr3+(aq) + H2(g)

Example 8.6 For the first row transition metals the EΘ values are:
EΘ     V     Cr     Mn      Fe     Co     Ni     Cu
(M2+/M) –1.18       – 0.91      –1.18       – 0.44       – 0.28     – 0.25     +0.34
Explain the irregularity in the above values.
Solution
The EΘ (M2+/M) values are not regular which can be explained from the irregular variation of ionisation enthalpies ( ΔiH1 + ΔiH2 ) and also the sublimation enthalpies which are relatively much less for manganese and vanadium.

Example 8.7
Why is the EΘ value for the Mn3+/Mn2+ couple much more positive than that for Cr3+ /Cr2+ or Fe3+ /Fe2+ ? Explain.
Solution
Much larger third ionisation energy of Mn (where the required change is d5 to d4 ) is mainly responsible for this. This also explains why the +3 state of Mn is of little importance.

Intext Questions
8.6 Why is the highest oxidation state of a metal exhibited in its oxide or fluoride only?
8.7 Which is a stronger reducing agent Cr2+ or Fe2+ and why ?

8.3.9 Magnetic Properties

When a magnetic field is applied to substances, mainly two types of magnetic behaviour are observed: diamagnetism and paramagnetism (Unit 1). Diamagnetic substances are repelled by the applied field while the paramagnetic substances are attracted. Substances which are attracted very strongly are said to be ferromagnetic. In fact, ferromagnetism is an extreme form of paramagnetism. Many of the transition metal ions are paramagnetic.

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

where n is the number of unpaired electrons and μ is the magnetic moment in units of Bohr magneton (BM). A single unpaired electron has a magnetic moment of 1.73 Bohr magnetons (BM).
The magnetic moment increases with the increasing number of unpaired electrons. Thus, the observed magnetic moment gives a useful indication about the number of unpaired electrons present in the atom, molecule or ion. The magnetic moments calculated from the ‘spin-only’ formula and those derived experimentally for some ions of the first row transition elements are given in Table 8.7. The experimental data are mainly for hydrated ions in solution or in the solid state.

Table 8.7: Calculated and Observed Magnetic Moments (BM)
Ion configuration Unpaired electron(s) Group
Calculated Observed
Sc3+ 3d0 0 0 0
Ti3+ 3d1 1 1.73 1.75
TL2+ 3d2 2 2.84 2.76
V+2 3d3 3 3.87 3.86
Cr2+ 3d4 4 4.90 4.80
Mn2+ 3d5 5 5.92 5.96
Fe2+ 3d6 4 4.90 5.3 – 5.5
Co+2 3d7 3 3.87 4.4 – 5.2
Ni2+ 3d8 2 2.84 2.9 – 3,4
Cu2+ 3d9 1 1.73
Zn2+ 3d10 O O

Example 8.8
Calculate the magnetic moment of a divalent ion in aqueous solution if its atomic number is 25.
Solution
With atomic number 25, the divalent ion in aqueous solution will have d5 configuration (five unpaired electrons). The magnetic moment, μ is

Intext Question
8.8 Calculate the ‘spin only’ magnetic moment of M2+ (aq) ion (Z = 27).

8.3.10 Formation of Coloured Ions

When an electron from a lower energy d orbital is excited to a higher energy d orbital, the energy of excitation corresponds to the frequency of light absorbed (Unit 9). This frequency generally lies in the visible region. The colour observed corresponds to the complementary colour of the light absorbed. The frequency of the light absorbed is determined by the nature of the ligand. In aqueous solutions where water molecules are the ligands, the colours of the ions observed are listed in Table 8.8. A few coloured solutions of d–block elements are illustrated in Fig. 8.5.

3d0

Table 8.8: Colours of Some of the First Row (aquated) Transition Metal Ions
Configuration Example Colour
3d0 Sc3+ Colourless
Ti4+ Colourless
3d1 Ti3+ purple
3d1 V4+ blue
3d2 V3+ green
3d3 V2+ violet
3d3 Cr3+ Violet
3d4 Mn3+ Violet
3d4 Cr2+ blue
3d5 Mn2+ pink
3d5 Fe3+ Yellow
3d6 Fe2+ green
3d63d7 Co3+Co2+ bluepink
3d8 Ni2+ green
3d9 Cu2+ blue
3d10 Zn2+ Colorless

8.3.11 Formation of Complex Compounds

Complex compounds are those in which the metal ions bind a number of anions or neutral molecules giving complex species with characteristic properties. A few examples are: [Fe(CN)6]3–, [Fe(CN)6]4–, [Cu(NH3)4]2+ and [PtCl4]2+ . (The chemistry of complex compounds is dealt with in detail in Unit 9). The transition metals form a large number of complex compounds. This is due to the comparatively smaller sizes of the metal ions, their high ionic charges and the availability of d orbitals for bond formation.

8.3.12 Catalytic Properties

The transition metals and their compounds are known for their catalytic activity. This activity is ascribed to their ability to adopt multiple oxidation states and to form complexes. Vanadium(V) oxide (in Contact Process), finely divided iron (in Haber’s Process), and nickel (in Catalytic Hydrogenation) are some of the examples. Catalysts at a solid surface involve the formation of bonds between reactant molecules and atoms of the surface of the catalyst (first row transition metals utilise 3d and 4s electrons for bonding). This has the effect of increasing the concentration of the reactants at the catalyst surface and also weakening of the bonds in the reacting molecules (the activation energy is lowering ). Also because the transition metal ions can change their oxidation states, they become more effective as catalysts. For example, iron(III) catalyses the reaction between iodide and persulphate ions.
2 I + S2O82− → I2 + 2 SO42−
An explanation of this catalytic action can be given as:
2 Fe3+ + 2 I → 2 Fe2+ + I2
2 Fe2+ + S2O82− → 2 Fe3+ + 2SO42−

8.3.13 Formation of Interstitial Compounds

Interstitial compounds are thosewhich are formed when small atoms like H, C or N are trapped inside the crystal lattices of metals. They are usually non stoichiometric and are neither typically ionic nor covalent, for example, TiC, Mn4N, Fe3H, VH0.56 and TiH1.7, etc. The formulas quoted do not, of course, correspond to any normal oxidation state of the metal. Because of the nature of their composition, these compounds are referred to as interstitial compounds. The principal physical and chemical characteristics of these compounds are as follows:
(i) They have high melting points, higher than than those of pure metals.
(ii) They are very hard, some borides approach diamond in hardness.
(iii) They retain metallic conductivity.
(iv) They are chemically inert.

8.3.14 Alloy Formation

An alloy is a blend of metals prepared by mixing the components. Alloys may be homogeneous solid solutions in which the atoms of one metal are distributed randomly among the atoms of the other. Such alloys are formed by atoms with metallic radii that are within about 15 percent of each other. Because of similar radii and other characteristics of transition metals, alloys are readily formed by these metals. The alloys so formed are hard and have often high melting points. The best known are ferrous alloys: chromium, vanadium, tungsten, molybdenum and manganese are used for the production of a variety of steels and stainless steel. Alloys of transition metals with non transition metals such as brass (copper-zinc) and bronze (copper-tin), are also of considerable industrial importance.

Example 8.9
What is meant by ‘disproportionation’ of an oxidation state? Give an example.
Solution
When a particular oxidation state becomes less stable relative to other oxidation states, one lower, one higher, it is said to undergo disproportionation. For example, manganese (VI) becomes unstable relative to manganese(VII) and manganese (IV) in acidic solution.
3 MnVIO42– + 4 H+ → 2 MnVIIO–4 + MnIVO2 + 2H2O

Intext Question
8.9 Explain why Cu+ ion is not stable in aqueous solutions?

8.4 Some Important Compounds of Transition Elements

8.4.1 Oxides and Oxoanions of Metals

These oxides are generally formed by the reaction of metals with oxygen at high temperatures. All the metals except scandium form MO oxides which are ionic. The highest oxidation number in the oxides, coincides with the group number and is attained in Sc2O3 to Mn2O7. Beyond group 7, no higher oxides of iron above Fe2O3 are known. Besides the oxides, the oxocations stabilise Vv as VO2+ , VIV as VO2+ and TiIV as TiO2+ .
As the oxidation number of a metal increases ionic character decreases. In the case of Mn, Mn2O7 is a covalent green oil. Even CrO3 and V2O5 have low melting points. In these higher oxides, the acidic character is predominant.
Thus, Mn2O7 gives HMnO4 and CrO3 gives H2CrO4 and H2Cr2O7. V2O5 is, however, amphoteric though mainly acidic and it gives VO43− as well as VO2+ salts. In vanadium there is gradual change from the basic V2O3 to less basic V2O4 and to amphoteric V2O5. V2O4 dissolves in acids to give VO2+ salts. Similarly, V2O5 reacts with alkalies as well as acids to give VO3 −4 and VO4+ respectively. The well characterised CrO is basic but Cr2O3 is amphoteric.

Potassium dichromate K2Cr2O7

Potassium dichromate is a very important chemical used in leather industry and as an oxidant for preparation of many azo compounds. Dichromates are generally prepared from chromate, which in turn are obtained by the fusion of chromite ore (FeCr2O4) with sodium or potassium carbonate in free access of air. The reaction with sodium carbonate occurs as follows:
4FeCr2O4 + 8Na2CO3 + 7O2 → 8Na2CrO4 + 2Fe2O3 + 8CO2
The yellow solution of sodium chromate is filtered and acidified with sulphuric acid to give a solution from which orange sodium dichromate, Na2Cr2O7. 2H2O can be crystallised.
2Na2CrO4 + 2H → Na2Cr2O7 + 2Na + H2O
Sodium dichromate is more soluble than potassium dichromate. The latter is therefore, prepared by treating the solution of sodium dichromate with potassium chloride.
Na2Cr2O>sub>7 + 2KCl → K2Cr2O7 + 2NaCl
Orange crystals of potassium dichromate crystallise out. The chromates and dichromates are interconvertible in aqueous solution depending upon pH of the solution. The oxidation state of chromium in chromate and dichromate is the same.
2CrO4 + 2H → Cr2O7 + H2O
Cr2O7 + 2OH → 2CrO4 + H2O
The structures of chromate ion, CrO42− and the dichromate ion, Cr2O72− are shown below. The chromate ion is tetrahedral whereas the dichromate ion consists of two tetrahedra sharing one corner with Cr–O–Cr bond angle of 126°. Sodium and potassium dichromates are strong oxidising agents; the sodium salt has a greater solubility in water and is extensively used as an oxidising agent in organic chemistry. Potassium dichromate is used as a primary standard in volumetric analysis. In acidic solution, its oxidising action can be represented as follows:
Cr2O72− + 14H+ + 6e → 2Cr3+ + 7H2O (Eθ = 1.33V)
Thus, acidified potassium dichromate will oxidise iodides to iodine, sulphides to sulphur, tin(II) to tin(IV) and iron(II) salts to iron(III). The half-reactions are noted below:
6 I → 3I2 + 6e ;      3Sn2+ → 3Sn4+ + 6 e
3 H2S → 6H+ + 3S + 6e ;      6 Fe2+ → 6Fe3+ + 6e
The full ionic equation may be obtained by adding the half-reaction for potassium dichromate to the half-reaction for the reducing agent, for e.g.,
Cr2O72− + 14H+ + 6Fe2+ → 2Cr3+ + 6Fe3+ + 7H2O

Potassium permanganate KMnO4

Potassium permanganate is prepared by fusion of MnO2 with an alkali metal hydroxide and an oxidising agent like KNO3. This produces the dark green K2MnO4 which disproportionates in a neutral or acidic solution to give permanganate.
2MnO2 + 4KOH + O2 → 2K2MnO4 + 2H2O
3MnO42– + 4H+ → 2MnO4 + MnO2 + 2H2O

Commercially it is prepared by the alkaline oxidative fusion of MnO2 followed by the electrolytic oxidation of manganate (Vl).

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

2Mn2+ + 5S2O82− + 8H2O → 2MnO4 + 10SO42− + 16H+

Potassium permanganate forms dark purple (almost black) crystals which are isostructural with those of KClO4. The salt is not very soluble in water (6.4 g/100 g of water at 293 K), but when heated it decomposes at 513 K.

2KMnO4 → K2MnO4 + MnO2 + O2

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

The manganate and permanganate ions are tetrahedral; the green manganate is paramagnetic with one unpaired electron but the permanganate is diamagnetic.

The π-bonding takes place by overlap of p orbitals of oxygen with d orbitals of manganese.

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

The full reaction can be written by adding the half-reaction for KMnO4 to the half-reaction of the reducing agent, balancing wherever necessary.

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

MnO4 + e → MnO42–      (EΘ = + 0.56 V)
MnO4 + 4H+ + 3e → MnO2 + 2H2O
(EΘ = + 1.52 V)
MnO4 + 8H+ + 5e → Mn2+ + 4H2O

We can very well see that the hydrogen ion concentration of the solution plays an important part in influencing the reaction. Although many reactions can be understood by consideration of redox potential, kinetics of the reaction is also an important factor. permanganate at [H+ ] = 1 should oxidise water but in practice the reaction is extremely slow unless either manganese(ll) ions are present or the temperature is raised.

A few important oxidising reactions of KMnO4 are given below:

1. In acid solutions:
(a) Iodine is liberated from potassium iodide :
10I + 2MnO4 + 16H+ ——> 2Mn2+ + 8H2O + 5I2
(b) Fe2+ ion (green) is converted to Fe3+ (yellow):
5Fe2+ + MnO4 + 8H+ ——> Mn2+ + 4H2O + 5Fe3+
(c) Oxalate ion or oxalic acid is oxidised at 333 K:
5C2O42−+ 2MnO4minus; + 16H+ ——> 2Mn2+ + 8H2O + 10CO2
(d) Hydrogen sulphide is oxidised, sulphur being precipitated:
H2S —> 2H+ + S2−
5S2− + 2MnO4 + 16H+ ——> 2Mn2+ + 8H2O + 5S
(e) Sulphurous acid or sulphite is oxidised to a sulphate or sulphuric acid:
5SO32− + 2MnO4 + 6H+ ——> 2Mn2+ + 3H2O + 5SO42−
(f) Nitrite is oxidised to nitrate:
5NO2 + 2MnO4 + 6H+ ——> 2Mn2+ + 5NO3 + 3H2O

2. In neutral or faintly alkaline solutions:
(a) A notable reaction is the oxidation of iodide to iodate:
2MnO4 + H2O + I ——> 2MnO2 + 2OH + IO3
(b) Thiosulphate is oxidised almost quantitatively to sulphate:
8MnO4 + 3S2O32– + H2O ——> 8MnO2 + 6SO42– + 2OH
(c) Manganous salt is oxidised to MnO2; the presence of zinc sulphate or zinc oxide catalyses the oxidation:
2MnO4 + 3Mn2+ + 2H2O ——> 5MnO2 + 4H+
Note: Permanganate titrations in presence of hydrochloric acid are unsatisfactory since hydrochloric acid is oxidised to chlorine.

Uses:
Besides its use in analytical chemistry, potassium permanganate is used as a favourite oxidant in preparative organic chemistry. Its uses for the bleaching of wool, cotton, silk and other textile fibres and for the decolourisation of oils are also dependent on its strong oxidising power.

THE INNER TRANSITION ELEMENTS ( f-BLOCK)

The f-block consists of the two series, lanthanoids (the fourteen elements following lanthanum) and actinoids (the fourteen elements following actinium). Because lanthanum closely resembles the lanthanoids, it is usually included in any discussion of the lanthanoids for which the general symbol Ln is often used. Similarly, a discussion of the actinoids includes actinium besides the fourteen elements constituting the series. The lanthanoids resemble one another more closely than do the members of ordinary transition elements in any series. They have only one stable oxidation state and their chemistry provides an excellent opportunity to examine the effect of small changes in size and nuclear charge along a series of otherwise similar elements. The chemistry of the actinoids is, on the other hand, much more complicated. The complication arises partly owing to the occurrence of a wide range of oxidatoxidation states in these elements and partly because their radioactivity creates special problems in their study; the two series will be considered separately here.

8.5 The Lanthanoids

The names, symbols, electronic configurations of atomic and some ionic states and atomic and ionic radii of lanthanum and lanthanoids (for which the general symbol Ln is used) are given in Table 8.9.

8.5.1 Electronic Configurations

It may be noted that atoms of these elements have electronic configuration with 6s2 common but with variable occupancy of 4f level (Table 8.9). However, the electronic configurations of all the tripositive ions (the most stable oxidation state of all the lanthanoids) are of the form 4fn (n = 1 to 14 with increasing atomic number).

8.5.2 Atomic and Ionic Sizes

The overall decrease in atomic and ionic radii from lanthanum to lutetium (the lanthanoid contraction) is a unique feature in the chemistry of the lanthanoids. It has far reaching consequences in the chemistry of the third transition series of the elements. The decrease in atomic radii (derived from the structures of metals) is not quite regular as it is regular in M3+ ions (Fig. 8.6). This contraction is, of course, similar to that observed in an ordinary transition series and is attribute d to the same cause, the imperfect shielding of one electron by another in the same sub-shell. However, the shielding of one 4 f electron by another is less than one d electron by another with the increase in nuclear charge along the series. There is fairly regular decrease in the sizes with increasing atomic number.
The cumulative effect of the contraction of the lanthanoid series, known as lanthanoid contraction, causes the radii of the members of the third transition series to be very similar to those of the corresponding members of the second series. The almost identical radii of (160 pm) and Hf (159 pm), a consequence of the lanthanoid contraction, account for their Atomic number occurrence together in nature and for the difficulty faced in their separation.

8.5.3 Oxidation States

In the lanthanoids, La(III) and Ln(III) compounds are predominant species. However, occasionally +2 and +4 ions in solution or in solid compounds are also obtained. This irregularity (as in ionisation enthalpies) arises mainly from the extra stability of empty, half-filled or filled f subshell. Thus, the formation of CeIV is favoured by its noble gas configuration, but it is a strong oxidant reverting to the common +3 state. The EΘ value for Ce4+/ Ce3+ is + 1.74 V which suggests that it can oxidise water. However, the reaction rate is very slow and hence Ce(IV) is a good analytical reagent. Pr, Nd, Tb and Dy also exhibit +4 state but only in oxides, MO2. Eu2+ is formed by losing the two s electrons and its f7 configuration accounts for the formation of this ion. However, Eu2+ is a strong reducing agent changing to the common +3 state. Similarly Yb2+ which has f 14 configuration is a reductant. TbIV has half-filled f-orbitals and is an oxidant. The behaviour of samarium is very much like europium, exhibiting both +2 and +3 oxidation states.

Table 8.9: Electronic Configurations and Radii of Lanthanum and Lanthanoids
Atomic Number Name symbol Electronic configurations* Radii/pm
Ln Ln2+ Ln3+ Ln4+ Ln Ln3+
57 Lanthanum La 5d16s2 5d1 4f0 187 106
58 Cerium Ce 4f15d16s2 4f2 4f1 4f0 183 103
59 Praseodymium Pr 4f36s2 4f3 4f2 4f1 182 101
60 Neodymium Nd 4f46s2 4f4 4f3 4f2 181 99
61 Promethium Pm 4f56s2 4f5 4f4 181 98
62 Samarium Sm 4f66s2 4f6 4f5 180 90
63 Europium Eu 4f76f2 4f7 4f6 199 95
64 Gadolinium Gd 4f75d16s2 4f75d1 4f7 180 94
65 Terbium Tb 4f96f2 4f9 4f8 4f7 178 92
66 Dysprosium Dy 4f106s2 4f10 4f9 4f8 177 91
67 Holmium Ho 4f116s2 4f11 4f10 176 89
68 Erbium Er 4f126s2 4f12 4f11 175 87
69 Thulium Tm 4f136s2 4f13 4f12 174 87
70 Ytterbium Yb 4f146s2 4f14 4f13 173 86
71 Lutetium Lu 4f145d16s2 4f145d1 4f14 - - -

8.5.4 General Characteristics

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

Many trivalent lanthanoid ions are coloured both in the solid state and in aqueous solutions. Colour of these ions may be attributed to the presence of f electrons. Neither La3+M nor Lu3+ ion shows any colour but the rest do so. However, absorption bands are narrow, probably because of the excitation within f level. The lanthanoid ions other than the f° type (La3+ and Ce4+ ) and the f14 type (Yb2+ and Lu3+ ) are all paramagnetic. The paramagnetism rises to maximum in neodymium.

The first ionisation enthalpies of the lanthanoids are around 600 kJ mol–1, the second about 1200 kJ mol–1 comparable with those of calcium. A detailed discussion of the variation of the third ionisation enthalpies indicates that the exchange enthalpy considerations (as in 3d orbitals of the first transition series), appear to impart a certain degree of stability to empty, half-filled and completely filled orbitals f level. This is indicated from the abnormally low value of the third ionisation enthalpy of lanthanum, gadolinium and lutetium.
In their chemical behaviour, in general, the earlier members of the series are quite reactive similar to calcium but, with increasing atomic number, they behave more like aluminium. Values for EΘ for the half-reaction:
Ln3+ (aq) + 3e → Ln(s)

are in the range of –2.2 to –2.4 V except for Eu for which the value is – 2.0 V. This is, of course, a small variation. The metals combine with hydrogen when gently heated in the gas. The carbides, Ln3C, Ln2C3 and LnC2 are formed when the metals are heated with with carbon. They liberate hydrogen from dilute acids and burn in halogens to form halides. They form oxides M2O3 and hydroxides M(OH)3. The hydroxides are definite compounds, not just hydrated oxides. They are basic like alkaline earth metal oxides and hydroxides. Their general reactions are
depicted in Fig. 8.7.

The best single use of the lanthanoids is for the production of alloy steels for plates and pipes. A well known alloy is mischmetall which consists of a lanthanoid metal (~ 95%) and iron (~ 5%) and traces of S, C, Ca and Al. A good deal of mischmetall is used in Mg-based alloy to produce bullets, shell and lighter flint. Mixed oxides of lanthanoids are employed as catalysts in petroleum cracking. Some individual Ln oxides are used as phosphors in television screens and similar fluorescing surfaces.

8.6 The Actinoids

The actinoids include the fourteen elements from Th to Lr. The names, symbols and some properties of these elements are given in Table 8.10.

4+

Table 8.9: Electronic Configurations and Radii of Lanthanum and Lanthanoids
Atomic Number Name symbol Electronic configurations* Radii/pm
M M3+ M4+ M3+ M
89 Actinium Ac 6d17s2 5f0 111
90 Thorium Th 6d27s2 5f1 5f0 99
91 Proctactiniium Pa 5f26d17s2 5f2 5f1 96
92 Uranium U 5f36d17s2 5f3 5f2 103 93
93 Neptunium Np 5f46d17s2 5f4 5f3 101 92
94 Plutonium Pu 5f67s2 5f5 5f4 100 90
95 Americium Am 5f77s2 5f6 5f5 99 89
96 Curium Cm 5f76d17s2 5f7 5f7 99 98
97 Berkelium Bk 5f97s2 5f85f79886
98 Californium Cf 5f107s2 5f9 5f8 98 86
99 Einstenium Es 5f117s2 5f10 5f9 - -
100 Fermium Fm 5f127s2 5f11 5f10 - -
101 Mendelevium Md 5f137s2 5f12 5f11 - -
102 Nobelium No 5f147s2 5f13 5f12 - -
103 Lawrencium Lr 5f146s17s2 5f14 5f13 - -

The actinoids are radioactive elements and the earlier members have relatively long half-lives, the latter ones have half-life values ranging from a day to 3 minutes for lawrencium (Z =103). The latter members could be prepared only in nanogram quantities. These facts render their study more difficult.

8.6.1 Electronic Configurations

All the actinoids are believed to have the electronic configuration of 7s2 and variable occupancy of the 5f and 6d subshells. The fourteen electrons are formally added to 5f, though not in thorium (Z = 90) but from Pa onwards the 5f orbitals are complete at element 103. The irregularities in the electronic configurations of the actinoids, like those in the lanthanoids are related to the stabilities of the f° , f7 and f14 occupancies of the 5f orbitals. Thus, the configurations of Am and Cm are [Rn] 5f77s2 and [Rn] 5f7 6d1 7s2 . Although the 5f orbitals resemble the 4f orbitals in their angular part of the wave-function, they are not as buried as 4f orbitals and hence 5f electrons can participate in bonding to a far greater extent.8.6.2 Ionic Sizes
The general trend in lanthanoids is observable in the actinoids as well. There is a gradual decrease in the size of atoms or M3+ ions across the series. This may be referred to as the actinoid contraction (like lanthanoid contraction). The contraction is, however, greater from element to element in this series resulting from poor shielding by 5f electrons.

8.6.3 Oxidation States

There is a greater range of oxidation states, which is in part attributed to the fact that the 5f, 6d and 7s levels are of comparable energies. The known oxidation states of actinoids are listed in Table 8.11.

Table 8.11: Oxidation States of Actinium and Actinoids
Ac Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr
3 3 3 3 3 3 3 3 3 3 3 3 3 3 3
4 4 4 4 4 4 4 4
5 5 5 5 5
6 6 6 6
7 7

The actinoids show in general +3 oxidation state. The elements, in the first half of the series frequently exhibit higher oxidation states. For example, the maximum oxidation state increases from +4 in Th to +5, +6 and +7 respectively in Pa, U and Np but decreases in succeeding elements (Table 8.11). The actinoids resemble the lanthanoids in having more compounds in +3 state than in the +4 state. However, +3 and +4 ions tend to hydrolyse. Because the distribution of oxidation states among the actinoids is so uneven and so different for the earlier and latter elements, it is unsatisfactory to review their chemistry in terms of oxidation states.

8.6.4 General Characteristics and Comparison with Lanthanoids

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

The actinoids are highly reactive metals, especially when finely divided. The action of boiling water on them, for example, gives a mixture of oxide and hydride and combination with most non metals takes place at moderate temperatures. Hydrochloric acid attacks all metals but most are slightly affected by nitric acid owing to the formation of protective oxide layers; alkalies have no action.

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

It is evident from the behaviour of the actinoids that the ionisation enthalpies of the early actinoids, though not accurately known, but are lower than for the early lanthanoids. This is quite reasonable since it is to be expected that when 5f orbitals are beginning to be occupied, they willpenetrate less into the inner core of electrons. The 5f electrons, will therefore, be more effectively shielded from the nuclear charge than the 4f electrons of the corresponding lanthanoids. Because the outer electrons are less firmly held, they are available for bonding in the actinoids.

A comparison of the actinoids with the lanthanoids, with respect to different characteristics as discussed above, reveals that behaviour similar to that of the lanthanoids is not evident until the second half of the actinoid series. However, even the early actinoids resemble the lanthanoids in showing close similarities with each other and in gradual variation in properties which do not entail change in oxidation state. The lanthanoid and actinoid contractions, have extended effects on the sizes, and therefore, the properties of the elements succeeding them in their respective periods. The lanthanoid contraction is more important because the chemistry of elements succeeding the actinoids are much less known at the present time.

Example 8.10 Name a member of the lanthanoid series which is well known to exhibit +4 oxidation state.
Solution
Cerium (Z = 58)

Intext Question
8.10 Actinoid contraction is greater from element to element than lanthanoid contraction. Why?

8.7 Some Applications of d- and f-Block Elements

Iron and steels are the most important construction materials. Their production is based on the reduction of iron oxides, the removal of impurities and the addition of carbon and alloying metals such as Cr, Mn and Ni. Some compounds are manufactured for special purposes such as TiO for the pigment industry and MnO2 for use in dry battery cells. The battery industry also requires Zn and Ni/Cd. The elements of Group 11 are still worthy of being called the coinage metals, although Ag and Au are restricted to collection items and the contemporary UK ‘copper’ coins are copper-coated steel. The ‘silver’ UK coins are a Cu/Ni alloy. Many of the metals and/or their compounds are essential catalysts in the chemical industry. V2O5 catalyses the oxidation of SO2 in the manufacture of sulphuric acid. TiCl4 with Al(CH3)3 forms the basis of the Ziegler catalysts used to manufacture polyethylene (polythene). Iron catalysts are used in the Haber process for the production of ammonia from N2/H2 mixtures. Nickel catalysts enable the hydrogenation of fats to proceed. In the Wacker process the oxidation of ethyne to ethanal is catalysed by PdCl2. Nickel complexes are useful in the polymerisation of alkynes and other organic compounds such as benzene. The photographic industry relies on the special light-sensitive properties of AgBr.

Summary

The d-block consisting of Groups 3-12 occupies the large middle section of the periodic table. In these elements the inner d orbitals are progressively filled. The f-block is placed outside at the bottom of the periodic table and in the elements of this block, 4f and 5f orbitals are progressively filled.

Corresponding to the filling of 3d, 4d and 5d orbitals, three series of transition elements are well recognised. All the transition elements exhibit typical metallic properties such as –high tensile strength, ductility, malleability, thermal and electrical conductivity and metallic character. Their melting and boiling points are high which are attributed to the involvement of (n –1) d electrons resulting into strong interatomic bonding. In many of these properties, the maxima occur at about the middle of each series which indicates that one unpaired electron per d orbital is particularly a favourable configuration for strong interatomic interaction.
Successive ionisation enthalpies do not increase as steeply as in the main group elements with increasing atomic number. Hence, the loss of variable number of electrons from (n –1) d orbitals is not energetically unfavourable. The involvement of (n –1) d electrons in the behaviour of transition elements impart certain distinct characteristics to these elements. Thus, in addition to variable oxidation states, they exhibit paramagnetic behaviour, catalytic properties and tendency for the formation of coloured ions, interstitial compounds and complexes.

The transition elements vary widely in their chemical behaviour. Many of them are sufficiently electropositive to dissolve in mineral acids, although a few are ‘noble’. Of the first series, with the exception of copper, all the metals are relatively reactive.

The transition metals react with a number of non-metals like oxygen, nitrogen, sulphur and halogens to form binary compounds. The first series transition metal oxides are generally formed from the reaction of metals with oxygen at high temperatures. These oxides dissolve in acids and bases to form oxometallic salts. Potassium dichromate and potassium permanganate are common examples. Potassium dichromate is prepared from the chromite ore by fusion with alkali in presence of air and acidifying the extract. Pyrolusite ore (MnO2) is used for the preparation of potassium permanganate. Both the dichromate and the permanganate ions are strong oxidising agents.

The two series of inner transition elements, lanthanoids and actinoids constitute the f-block of the periodic table. With the successive filling of the inner orbitals, 4f, there is a gradual decrease in the atomic and ionic sizes of these metals along the series (lanthanoid contraction). This has far reacting consequences in the chemistry of the elements succeeding them. Lanthanum and all the lanthanoids are rather soft white metals. They react easily with water to give solutions giving +3 ions. The principal oxidation state is +3, although +4 and +2 oxidation states are also exhibited by some occasionally. The chemistry of the actinoids is more complex in view of their ability to exist in different oxidation states. Furthermore, many of the actinoid elements are radioactive which make the study of these elements rather difficult.

There are many useful applications of the d- and f-block elements and their compounds, notable among them being in varieties of steels, catalysts, complexes, organic syntheses, etc.

Exercises
8.1 Write down the electronic configuration of:
(i) Cr3+
(ii) Pm3+
(iii) Cu+
(v) Co2+
(iv) Ce4+
(vi) Lu2+
(vii) Mn2+
(viii) Th4+

8.2 Why are Mn2+ compounds more stable than Fe2+ towards oxidation to their +3 states?
8.3 Explain briefly how +2 state becomes more and more stable in the first half of the first row transition elements with increasing atomic number?
8.4 To what extent do the electronic configurations decide the stability of oxidation states in the first series of the transition elements? Illustrate your answer with examples.
8.5 What may be the stable oxidation state of the transition element with the following d electron configurations in the ground state of their atoms : 3d3 , 3d5 , 3d8 and 3d4 ?
8.6 Name the oxometal anions of the first series of the transition metals in which the metal exhibits the oxidation state equal to its group number.
8.7 What is lanthanoid contraction? What are the consequences of lanthanoid contraction?
8.8 What are the characteristics of the transition elements and why are they called transition elements? Which of the d-block elements may not be regarded as the transition elements?
8.9 In what way is the electronic configuration of the transition elements different from that of the non transition elements?
8.10 What are the different oxidation states exhibited by the lanthanoids?
8.11 Explain giving reasons:
(i) Transition metals and many of their compounds show paramagnetic behaviour.
(ii) The enthalpies of atomisation of the transition metals are high.
(iii) The transition metals generally form coloured compounds.
(iv) Transition metals and their many compounds act as good catalyst.
8.12 What are interstitial compounds? Why are such compounds well known for transition metals?
8.13 How is the variability in oxidation states of transition metals different from that of the non transition metals? Illustrate with examples.
8.14 Describe the preparation of potassium dichromate from iron chromite ore. What is the effect of increasing pH on a solution of potassium dichromate?
8.15 Describe the oxidising action of potassium dichromate and write the ionic
equations for its reaction with:
(i) iodide
(ii) iron(II) solution and
(iii) H2S
8.16 Describe the preparation of potassium permanganate. How does the acidified permanganate solution react with (i) iron(II) ions (ii) SO2 and (iii) oxalic acid? Write the ionic equations for the reactions.
8.17 For M2+/M and M3+/M2+ systems the EΘ values for some metals are as follows:
Cr2+/Cr    -0.9V    Cr3/Cr2+    -0.4V
Mn2+/Mn    -1.2V    Mn3+/Mn2+    +1.5V
Fe2+/Fe    -0.4V    Fe3+/Fe2+    +0.8V
Use this data to comment upon:
(i) the stability of Fe3+ in acid solution as compared to that of Cr3+ or Mn3+ and
(ii) the ease with which iron can be oxidised as compared to a similar process for either chromium or manganese metal.
8.18 Predict which of the following will be coloured in aqueous solution? Ti3+, V3+, Cu+, Sc3+, Mn2+, Fe3+ and Co2+. Give reasons for each.
8.19 Compare the stability of +2 oxidation state for the elements of the first transition series.
8.20 Compare the chemistry of actinoids with that of the lanthanoids with special reference to:
(i) electronic configuration
(ii) atomic and ionic sizes and
(iii) oxidation state
(iv) chemical reactivity.
8.21 How would you account for the following:
(i) Of the d4 species, Cr2+ is strongly reducing while manganese(III) is strongly oxidising.
(ii) Cobalt(II) is stable in aqueous solution but in the presence of complexing reagents it is easily oxidised.
(iii) The d1 configuration is very unstable in ions.
8.22 What is meant by ‘disproportionation’? Give two examples of disproportionation reaction in aqueous solution.
8.23 Which metal in the first series of transition metals exhibits +1 oxidation state most frequently and why?
8.24 Calculate the number of unpaired electrons in the following gaseous ions: Mn3+, Cr3+ , V3+ and Ti3+ . Which one of these is the most stable in aqueous solution?
8.25 Give examples and suggest reasons for the following features of the transition metal chemistry:
(i) The lowest oxide of transition metal is basic, the highest is amphoteric/acidic.
(ii) A transition metal exhibits highest oxidation state in oxides and fluorides.
(iii) The highest oxidation state is exhibited in oxoanions of a metal.
8.26 Indicate the steps in the preparation of:
(i) K2Cr2O7 from chromite ore.
(ii) KMnO4 from pyrolusite ore.
8.27 What are alloys? Name an important alloy which contains some of the lanthanoid metals. Mention its uses.
8.28 What are inner transition elements? Decide which of the following atomic numbers are the atomic numbers of the inner transition elements : 29, 59, 74, 95, 102, 104.
8.29 The chemistry of the actinoid elements is not so smooth as that of the lanthanoids. Justify this statement by giving some examples from the oxidation state of these elements.
8.30 Which is the last element in the series of the actinoids? Write the electronic configuration of this element. Comment on the possible oxidation state of this element.
8.31 Use Hund’s rule to derive the electronic configuration of Ce3+ ion, and calculate its magnetic moment on the basis of ‘spin-only’ formula.
8.32 Name the members of the lanthanoid series which exhibit +4 oxidation states and those which exhibit +2 oxidation states. Try to correlate this type of behaviour with the electronic configurations of these elements.
8.33 Compare the chemistry of the actinoids with that of lanthanoids with reference to:
(i) electronic configuration
(ii) oxidation states and
(iii) chemical reactivity.
8.34 Write the electronic configurations of the elements with the atomic numbers 61, 91, 101, and 109.
8.35 Compare the general characteristics of the first series of the transition metals with those of the second and third series metals in the respective vertical columns. Give special emphasis on the following points:
(i) electronic configurations
(ii) oxidation states
(iii) ionisation enthalpies
and (iv) atomic sizes.
8.36 Write down the number of 3d electrons in each of the following ions: Ti2+, V2+,Cr3+, Mn2+, Fe2+, Fe3+, Co2+, Ni2+ and Cu2+. Indicate how would you expect the five 3d orbitals to be occupied for these hydrated ions (octahedral).
8.37 Comment on the statement that elements of the first transition series possess many properties different from those of heavier transition elements.
8.38 What can be inferred from the magnetic moment values of the following complex species ?
Example               Magnetic Moment (BM)
K4[Mn(CN)6)               2.2
[Fe(H2O)6]               5.3
K2[MnCl4]               5.9

Answers to Some Intext Questions
8.1 Silver (Z = 47) can exhibit +2 oxidation state wherein it will have incompletely filled d-orbitals (4d), hence a transition element.
8.2 In the formation of metallic bonds, no eletrons from 3d-orbitals are involved in case of zinc, while in all other metals of the 3d series, electrons from the d-orbitals are always involved in the formation of metallic bonds.
8.3 Manganese (Z = 25), as its atom has the maximum number of unpaired electrons.
8.5 Irregular variation of ionisation enthalpies is mainly attributed to varying degree of stability of different 3d-configurations (e.g., d0 , d5 , d10 are exceptionally stable).
8.6 Because of small size and high electronegativity oxygen or fluorine can oxidise the metal to its highest oxidation state.
8.7 Cr2+ is stronger reducing agent than Fe2+
Reason: d4 → d3 occurs in case of Cr2+ to Cr3+
But d6→ d5 occurs in case of Fe2+ to Fe3+
In a medium (like water) d3 is more stable as compared to d5 (see CFSE)
8.9 Cu+ in aqueous solution underoes disproportionation, i.e.,
2Cu+(aq) → Cu2+(aq) + Cu(s)
The E0 value for this is favourable.
8.10 The 5f electrons are more effectively shielded from nuclear charge. In other words the 5f electrons themselves provide poor shielding from element to element in the series.

I. Multiple Choice Questions (Type-I)

1. Electronic configuration of a transition element X in +3 oxidation state is [Ar]3d5. What is its atomic number?

(i) 25
(ii) 26
(iii) 27
(iv) 24

2. The electronic configuration of Cu(II) is 3d9 whereas that of Cu(I) is 3d10. Which of the following is correct?

(i) Cu(II) is more stable
(ii) Cu(II) is less stable
(iii) Cu(I) and Cu(II) are equally stable
(iv) Stability of Cu(I) and Cu(II) depends on nature of copper salts

3.Metallic radii of some transition elements are given below. Which of these elements will have highest density?

NiCu

Element Fe Co
Metallic radii/pm 126 125 125 128

(i) Fe
(ii) Ni
(iii) Co
(iv) Cu

4. Generally transition elements form coloured salts due to the presence of unpaired electrons. Which of the following compounds will be coloured in solid state?

(i) Ag2SO4
(ii) CuF2
(iii) ZnF2
(iv) Cu2Cl2

5. On addition of small amount of KMnO4 to concentrated H2SO4, a green oily compound is obtained which is highly explosive in nature. Identify the compound from the following.

(i) Mn2O7
(ii) MnO2
(iii) MnSO4
(iv) Mn2O3

6. The magnetic nature of elements depends on the presence of unpaired electrons. Identify the configuration of transition element, which shows highest magnetic moment.

(i) 3d7
(ii) 3d5
(iii) 3d8
(iv) 3d2

7. Which of the following oxidation state is common for all lanthanoids?

(i) +2
(ii) +3
(iii) +4
(iv) +5

8. Which of the following reactions are disproportionation reactions?

(a) Cu+ → Cu2+ + Cu
(b) 3MnO4 + 4H+ → 2MnO4 + MnO2 + 2H2O
(c) 2KMnO4 → K2MnO4 + MnO2 + O2
(d) 2MnO4 + 3Mn2+ + 2H2O → 5MnO2 + 4H+

(i) a, b
(ii) a, b, c
(iii) b, c, d
(iv) a, d

9. When KMnO4 solution is added to oxalic acid solution, the decolourisation is slow in the beginning but becomes instantaneous after some time because

(i) CO2 is formed as the product.
(ii) Reaction is exothermic.
(iii) MnO4 catalyses the reaction.
(iv) Mn2+ acts as autocatalyst.

10. There are 14 elements in actinoid series. Which of the following elements does not belong to this series?

(i) U
(ii) Np
(iii) Tm
(iv) Fm

11. KMnO4 acts as an oxidising agent in acidic medium. The number of moles of KMnO4 that will be needed to react with one mole of sulphide ions in acidic solution is

(i) 2/5
(ii) 3/5
(iii) 4/5
(iv) 1/5

12. Which of the following is amphoteric oxide?

Mn2O7, CrO3, Cr2O3, CrO, V2O5, V2O4

(i) V2O5, Cr2O3
(ii) Mn2O7, CrO3
(iii) CrO, V2O5
(iv) V2O5, V2O4

13. Gadolinium belongs to 4f series. It’s atomic number is 64. Which of the following is the correct electronic configuration of gadolinium?

(i) [Xe] 4f75d16s2
(ii) [Xe] 4f65d26s2
(iii) [Xe] 4f86d2
(iv) [Xe] 4f95s1

14. Interstitial compounds are formed when small atoms are trapped inside the crystal lattice of metals. Which of the following is not the characteristic property of interstitial compounds?

(i) They have high melting points in comparison to pure metals.
(ii) They are very hard.
(iii) They retain metallic conductivity.
(iv) They are chemically very reactive.

15. The magnetic moment is associated with its spin angular momentum and orbital angular momentum. Spin only magnetic moment value of Cr3+ ion is ___________.

(i) 2.87 B.M.
(ii) 3.87 B.M.
(iii) 3.47 B.M.
(iv) 3.57 B.M.

16. KMnO4 acts as an oxidising agent in alkaline medium. When alkaline KMnO4 is treated with KI, iodide ion is oxidised to ____________.

(i) I2
(ii) IO
(iii) IO3
(iv) IO4

17. Which of the following statements is not correct?

(i) Copper liberates hydrogen from acids.
(ii) In its higher oxidation states, manganese forms stable compounds with oxygen and fluorine.
(iii) Mn3+ and Co3+ are oxidising agents in aqueous solution.
(iv) Ti2+ and Cr2+ are reducing agents in aqueous solution.

18. When acidified K2Cr2O7 solution is added to Sn2+ salts then Sn2+ changes to

(i) Sn
(ii) Sn3+
(iii) Sn4+
(iv) Sn+

19. Highest oxidation state of manganese in fluoride is +4 (MnF4) but highest oxidation state in oxides is +7 (Mn2O7) because ____________.

(i) fluorine is more electronegative than oxygen.
(ii) fluorine does not possess d-orbitals.
(iii) fluorine stabilises lower oxidation state.
(iv) in covalent compounds fluorine can form single bond only while oxygen forms double bond.

20. Although Zirconium belongs to 4d transition series and Hafnium to 5d transition series even then they show similar physical and chemical properties because___________.

(i) both belong to d-block.
(ii) both have same number of electrons.
(iii) both have similar atomic radius.
(iv) both belong to the same group of the periodic table.

21. Why is HCl not used to make the medium acidic in oxidation reactions of KMnO4 in acidic medium?

(i) Both HCl and KMnO4 act as oxidising agents.
(ii) KMnO4 oxidises HCl into Cl2 which is also an oxidising agent.
(iii) KMnO4 is a weaker oxidising agent than HCl.
(iv) KMnO4 acts as a reducing agent in the presence of HCl.

II. Multiple Choice Questions (Type-II)

Note : In the following questions two or more options may be correct.

22. Generally transition elements and their salts are coloured due to the presence of unpaired electrons in metal ions. Which of the following compounds are coloured?

(i) KMnO4
(ii) Ce(SO4)2
(iii) TiCl4
(iv) Cu2Cl2

23. Transition elements show magnetic moment due to spin and orbital motion of electrons. Which of the following metallic ions have almost same spin only magnetic moment?

(i) Co2+
(ii) Cr2+
(iii) Mn2+
(iv) Cr3+

24. In the form of dichromate, Cr (VI) is a strong oxidising agent in acidic medium but Mo (VI) in MoO3 and W (VI) in WO3 are not because ___________.

(i) Cr (VI) is more stable than Mo(VI) and W(VI).
(ii) Mo(VI) and W(VI) are more stable than Cr(VI).
(iii) Higher oxidation states of heavier members of group-6 of transition series are more stable.
(iv) Lower oxidation states of heavier members of group-6 of transition series are more stable.

25. Which of the following actinoids show oxidation states upto +7?

(i) Am
(ii) Pu
(iii) U
(iv) Np

26. General electronic configuration of actionoids is (n–2)f1–14 (n-1)d0–2ns2.Which of the following actinoids have one electron in 6d orbital?

(i) U (Atomic no. 92)
(ii) Np (Atomic no.93)
(iii) Pu (Atomic no. 94)
(iv) Am (Atomic no. 95)

27. Which of the following lanthanoids show +2 oxidation state besides the characteristic oxidation state +3 of lanthanoids?

(i) Ce
(ii) Eu
(iii) Yb
(iv) Ho

28. Which of the following ions show higher spin only magnetic moment value?

(i) Ti3+
(ii) Mn2+
(iii) Fe2+
(iv) Co3+

29. Transition elements form binary compounds with halogens. Which of the following elements will form MF3 type compounds?

(i) Cr
(ii) Co
(iii) Cu
(iv) Ni

30. Which of the following will not act as oxidising agents?

(i) CrO3
(ii) MoO3
(iii) WO3
(iv) CrO42–

31. Although +3 is the characteristic oxidation state for lanthanoids but cerium also shows +4 oxidation state because ___________.

(i) it has variable ionisation enthalpy
(ii) it has a tendency to attain noble gas configuration
(iii) it has a tendency to attain f 0 configuration
(iv) it resembles Pb4+

III. Short Answer Type

32. Why does copper not replace hydrogen from acids?
33. Why EΘ values for Mn, Ni and Zn are more negative than expected?
34. Why first ionisation enthalpy of Cr is lower than that of Zn ?
35. Transition elements show high melting points. Why?
36. When Cu2+ ion is treated with KI, a white precipitate is formed. Explain the reaction with the help of chemical equation.

37. Out of Cu2Cl2 and CuCl2, which is more stable and why?
38. When a brown compound of manganese (A) is treated with HCl it gives a gas
(B). The gas taken in excess, reacts with NH3 to give an explosive compound
(C). Identify compounds A, B and C.

39. Although fluorine is more electronegative than oxygen, but the ability of oxygen to stabilise higher oxidation states exceeds that of fluorine. Why?

40. Although Cr3+ and Co2+ ions have same number of unpaired electrons but the magnetic moment of Cr3+ is 3.87 B.M. and that of Co2+ is 4.87 B.M. Why?

41. Ionisation enthalpies of Ce, Pr and Nd are higher than Th, Pa and U. Why?
42. Although Zr belongs to 4d and Hf belongs to 5d transition series but it is quite difficult to separate them. Why?
43. Although +3 oxidation states is the characteristic oxidation state of lanthanoids but cerium shows +4 oxidation state also. Why?
44. Explain why does colour of KMnO4 disappear when oxalic acid is added to its solution in acidic medium.
45. When orange solution containing Cr2O72– ion is treated with an alkali, a yellow solution is formed and when H+ ions are added to yellow solution, an orange solution is obtained. Explain why does this happen?

46. A solution of KMnO4 on reduction yields either a colourless solution or a brown precipitate or a green solution depending on pH of the solution. What different stages of the reduction do these represent and how are they carried out?

47. The second and third rows of transition elements resemble each other much more than they resemble the first row. Explain why?

48. EΘ of Cu is + 0.34V while that of Zn is – 0.76V. Explain.
49. The halides of transition elements become more covalent with increasing oxidation state of the metal. Why?
50. While filling up of electrons in the atomic orbitals, the 4s orbital is filled before the 3d orbital but reverse happens during the ionisation of the atom. Explain why?

51. Reactivity of transition elements decreases almost regularly from Sc to Cu. Explain.

IV. Matching Type

Note : Match the items of Column I and Column II in the following questions.

52. Match the catalysts given in Column I with the processes given in Column II.

Column I (Catalyst) Column II (Process)
(i) Ni in the presence of hydrogen (a) Zieglar Natta catalyst
(ii) Cu2Cl2 (b) Contact process
(iii) V2O5 (c) Vegetable oil to ghee
(iv) Finely divided iron (d) Sandmeyer reaction
(v) TiCl4 + Al(CH3)3 (e) Haber’s Process
(f) Decomposition of KClO3

53. Match the compounds/elements given in Column I with uses given in Column II.

Column I(Compound/element) Column II(Use)
(i) Lanthanoid oxide (a) Production of iron alloy
(ii) Lanthanoid (b) Television screen
(iii) Misch metal (c) Petroleum cracking
(iv) Magnesium based alloy is constituent of (d) Lanthanoid metal + iron
(v) Mixed oxides of lanthanoids are employed (e) Bullets
(f) In X-ray screen

54. Match the properties given in Column I with the metals given in Column II.

Column I(Property) Column II(Metal)
(i) An element which can show +8 oxidation state (a) Mn
(ii) 3d block element that can show upto +7 oxidation state (b) Cr
(iii) 3d block element with highest melting point (c) Os
(d) Fe

55. Match the statements given in Column I with the oxidation states given in Column II.

Column I Column II
(i) Oxidation state of Mn in MnO2 is (a) + 2
(ii) Most stable oxidation state of Mn is (b) + 3
(iii) Most stable oxidation state of Mn in oxides is (c) + 4
(iv) Characteristic oxidation state of lanthanoids is (d) + 5
(e) + 7

56. Match the solutions given in Column I and the colours given in Column II.

Column I(Aqueous solution of salt) Column II(Colour)
(i) FeSO4.7H2O (a) Green
(ii) NiCl2.4H2O (b) Light pink
(iii) MnCl2.4H2O (c) Blue
(iv) CoCl2.6H2O (d) Pale green
(v) Cu2Cl2 (e) Pink
(f) Colourless

57. Match the property given in Column I with the element given in Column II.

Column I(Property) Column II(Element)
(i) Lanthanoid which shows +4 oxidation state (a) Pm
(ii) Lanthanoid which can show +2 oxidation state (b) Ce
(iii) Radioactive lanthanoid (c) Lu
(iv) Lanthanoid which has 4f7 electronic configuration in +3 oxidation state (d) Eu
(v) Lanthanoid which has 4f14 electronic configuration in +3 oxidation state (e) Gd
(f) Dy

58. Match the properties given in Column I with the metals given in Column II.

Column I(Property) Column II(Metal)
(i) Element with highest second ionisation enthalpy (a) Co
(ii) Element with highest third ionisation enthalpy (b) Cr
(iii) M in M (CO)6 is (c) Cu
(iv) Element with highest heat of atomisation (d) Zn
(e) Ni

V. Assertion and Reason Type

Note : In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.

(i) Both assertion and reason are true, and reason is the correct explanation of the assertion.
(ii) Both assertion and reason are true but reason is not the correct explanation of assertion.
(iii) Assertion is not true but reason is true.
(iv) Both assertion and reason are false.

59. Assertion : Cu2+ iodide is not known.
Reason : Cu2+ oxidises I to iodine.

60. Assertion : Separation of Zr and Hf is difficult.
Reason : Because Zr and Hf lie in the same group of the periodic table.

61. Assertion : Actinoids form relatively less stable complexes as compared to lanthanoids.
Reason : Actinoids can utilise their 5f orbitals along with 6d orbitals in bonding but lanthanoids do not use their 4f orbital for bonding.

62. Assertion : Cu cannot liberate hydrogen from acids.
Reason : Because it has positive electrode potential.

63. Assertion : The highest oxidation state of osmium is +8.
Reason : Osmium is a 5d-block element.

VI. Long Answer Type

64. Identify A to E and also explain the reactions involved.

65. When a chromite ore (A) is fused with sodium carbonate in free excess of air and the product is dissolved in water, a yellow solution of compound (B) is obtained. After treatment of this yellow solution with sulphuric acid,
compound (C) can be crystallised from the solution. When compound (C) is treated with KCl, orange crystals of compound (D) crystallise out. Identify A to D and also explain the reactions.

66. When an oxide of manganese (A) is fused with KOH in the presence of an oxidising agent and dissolved in water, it gives a dark green solution of compound (B). Compound (B) disproportionates in neutral or acidic solution to give purple compound (C). An alkaline solution of compound (C) oxidises potassium iodide solution to a compound (D) and compound (A) is also formed. Identify compounds A to D and also explain the reactions involved.

67. On the basis of Lanthanoid contraction, explain the following :

(i) Nature of bonding in La2O3 and Lu2O3.
(ii) Trends in the stability of oxo salts of lanthanoids from La to Lu.
(iii) Stability of the complexes of lanthanoids.
(iv) Radii of 4d and 5d block elements.
(v) Trends in acidic character of lanthanoid oxides.

68. (a) Answer the following questions :

(i) Which element of the first transition series has highest second ionisation enthalpy?
(ii) Which element of the first transition series has highest third ionisation enthalpy?
(iii) Which element of the first transition series has lowest enthalpy of atomisation?

(b) Identify the metal and justify your answer.
(i) Carbonyl M(CO)5
(ii) MO3F

69. Mention the type of compounds formed when small atoms like H, C and N get trapped inside the crystal lattice of transition metals. Also give physical and chemical characteristics of these compounds.

70. (a) Transition metals can act as catalysts because these can change their oxidation state. How does Fe(III) catalyse the reaction between iodide and persulphate ions?

(b) Mention any three processes where transition metals act as catalysts.

71. A violet compound of manganese (A) decomposes on heating to liberate oxygen and compounds (B) and (C) of manganese are formed. Compound (C) reacts with KOH in the presence of potassium nitrate to give compound (B). On heating compound (C) with conc. H2SO4 and NaCl, chlorine gas is liberated and a compound (D) of manganese along with other products is formed. Identify compounds A to D and also explain the reactions involved.

ANSWERS

I. Multiple Choice Questions (Type-I)

1. (ii)     2. (i)     3. (iv)     4. (ii)     5. (i)     6. (ii)     7. (ii)     8. (i)     9. (iv)     10. (iii)     11. (i)     12. (i)     13. (i)     14. (iv)     15. (ii)     16. (iii)     17. (i)     18. (iii)     19. (iv)     20. (iii)     21. (ii)

II. Multiple Choice Questions (Type-II)

22. (i), (ii)     23. (i), (iv)     24. (ii), (iii)     25. (ii), (iv)     26. (i), (ii)     27. (ii), (iii)     28. (ii), (iii)     29. (i), (ii)     30. (ii), (iii)     31. (ii), (iii)

III. Short Answer Type

32. Cu shows positive EΘ value.
33. Hint : Negative EΘ values for Mn2+ and Zn2+ are related to stabilities of half filled and fully filled configuration respectively. But for Ni2+ , EΘ value is related to the highest negative enthalpy of hydration.

34. Ionisation enthalpy of Cr is lower due to stability of d5 and the value for Zn is higher because its electron comes out from 4s orbital.

35. The high melting points of transition metals are attributed to the involvement of greater number of electrons in the interatomic metallic bonding from (n-1) d-orbitals in addition to ns electrons

36. Hint : Cu2+ gets reduced to Cu+
2Cu2+ + 4I → Cu2I2 + I2
(white precipitate)

37. Hint : CuCl2 is more stable than Cu2Cl2. The stability of Cu2+ (aq.) rather than Cu+(aq.) is due to the much more negative ΔhydHV of Cu2+ (aq.) than
Cu+(aq.).

38. A = MnO2 B = Cl2 C = NCl3
MnO2 + 4HCl → MnCl2 + Cl2 + 2H2O
(A)                                      (B)

NH3 + 3Cl2 → NCl3 + 3HCl
(excess)    (C)

39. Hint : It is due to the ability of oxygen to form multiple bonds to metals.
40. Hint : Due to symmetrical electronic configuration there is no orbital
contribution in Cr3+ ion. However appreciable orbital contribution takes place in Co2+ ion.

41. Hint : It is because in the beginning, when 5f orbitals begin to be occupied, they will penetrate less into the inner core of electrons. The 5f electrons will therefore, be more effectively shielded from the nuclear charge than 4f electrons of the corresponding lanthanoids. Therefore outer electrons are less firmly held and they are available for bonding in the actinoids.

42. Hint : Due to lanthanoid contraction, they have almost same size (Zr, 160 pm) and (Hf, 159 pm).

43. It is because after losing one more electron Ce acquires stable 4f 0 electronic configuration.

44. KMnO4 acts as oxidising agent. It oxidises oxalic acid to CO2 and itself changes to Mn2+ ion which is colourless.

5C2O42– + 2MnO4 + 16H+ → 2Mn2+ + 8H2O + 10CO2
(Coloured)                    (Colourless)

46. Oxidising behaviour of KMnO4 depends on pH of the solution. In acidic medium (pH < 7)

MnO4 + 8H+ + 5e → Mn2+ + 4H2O
(Colourless)

In alkaline medium (pH>7)

MnO4 + e → MnO42–
(Green)

In neutral medium(pH=7)

MnO4 + 2H2O + 3e → MnO2 + 4OH
(Brown precipitate)

47. Due to lanthanoid contraction, the atomic radii of the second and third row transition elements is almost same. So they resemble each other much more as compared to first row elements.

48. Hint : High ionisation enthalpy to transform Cu(s) to Cu2+ (aq) is not balanced by its hydration enthalpy. However, in case of Zn after removal of electrons from 4s-orbital, stable 3d10 configuration is acquired.

49. As the oxidation state increases, size of the ion of transition element decreases. As per Fajan’s rule, as the size of metal ion decreases, covalent character of the bond formed increases.

50. n + l rule : For 3d = n + l = 5
4s = n + l = 4
So electron will enter in 4s orbital.
Ionisation enthalpy is responsible for the ionisation of atom. 4s electrons are loosely held by the nucleus. So electrons are removed from 4s orbital prior to 3d.

51. Hint : It is due to regular increase in ionisation enthalpy.

IV. Matching Type

52. (i) → (c) (ii) → (d) (iii) → (b) (iv) → (e) (v) → (a)
53. (i) → (b) (ii) → (a) (iii) → (d) (iv) → (e) (v) → (c)
54. (i) → (c) (ii) → (a) (iii) → (b)
55. (i) → (c) (ii) → (a) (iii) → (e) (iv) → (b)
56. (i) → (d) (ii) → (a) (iii) → (b) (iv) → (e) (v) → (f)
57. (i) → (b) (ii) → (d) (iii) → (a) (iv) → (e) (v) → (c)
58. (i) → (c) (ii) → (d) (iii) → (b) (iv) → (a)

V. Assertion and Reason Type

59. (i) 60. (ii) 61. (iii) 62. (i) 63. (ii)

VI. Long Answer Type

64. A = Cu B = Cu(NO3)2 C = [Cu(NH3)4] D = CO2
E = CaCO3 F = Cu2[Fe(CN)6] G = Ca(HCO3)2
CuCO3 → CuO + CO2

CaCO3 + H2O + CO2 → Ca(HCO3)2

65. A = FeCr2O4 B = Na2CrO4 C = Na2Cr2O7.2H2O D = K2Cr2O7

66. A = MnO2 (B) K2MnO4 (C) KMnO4 (D) KIO3

67. Hint : (i) As the size decreases covalent character increases. Therefore
La2O3 is more ionic and Lu2O3 is more covalent.

(ii) As the size decreases from La to Lu, stability of oxosalts also decreases.

(iii) Stability of complexes increases as the size of lanthanoids decreases.
(iv) Radii of 4d and 5d block elements will be almost same.
(v) Acidic character of oxides increases from La to Lu.

68. (a) (i) Cu, because the electronic configuration of Cu is 3d104s1. So second electron needs to be removed from completely filled d-orbital.

(ii) Zn [Hint : As above]
(iii) Zn [Hint : No unpaired electron for metallic bonding]
(b) (i) Fe(CO)5 [Hint : EAN rule]
(ii) MnO3F [Hint : Mn shows +7 oxidation state; d-electrons are not involved in bonding.]

69. Interstitial compounds.
Characteristic properties :
(i) High melting points, higher than those of pure metals.
(ii) Very hard.
(iii) Retain metallic conductivity.
(iv) Chemically inert.

70. (a) Reaction between iodide and persulphate ions is :


Role of Fe (III) ions :
2Fe3+ + 2I → 2Fe2+ + I2
2Fe2+ + S2O82– → 2Fe3+ + 2SO42–

(b) (i) Vanadium (V) oxide in contact process for oxidation of SO2 to SO3.
(ii) Finely divided iron in Haber’s process in conversion of N2 and H2 to NH3.
(iii) MnO2 in preparation of oxygen from KClO3.

71. A = KMnO4 B = K2MnO4 C = MnO2 D = MnCl2


MnO2 + KOH + O2 → 2K2MnO2 + 2H2O
MnO2 + 4NaCl + 4H2SO4 → MnCl2 + 2NaHSO4 + 2H2O + Cl2
(D)

41 thoughts on “8. The d- and f- block Elements

  1. This is a very well-organized website with high value for education of students in general chemistry. May I ask if you happen to know which of the low-lying orbitals are used in Iodine atom in the IF7 molecule, i.e., what is the electronic configuration of the central atom ? Would the vacant 6s orbital be
    used ? Or would the 5d and 4f vacant orbitals be used first ?
    I am guessing that probably the 6s orbital would be filled first, then when the next pair of electrons coming from fluorine they would go into the 5d orbital first, which would lower the energy level of 5d to below that of 6s. With more pairs of electrons being added we’d fill up the 5d orbitals with one more pair
    of electrons left to go. If this pair would go into the 4f orbital we’d have:
    sd5f configuration.
    But since the Lanthanide series have 6s5dand 4f orbitals filled first, would the following be more reasonable: sdf5 configuration, filling the 6s, one 5d, and
    5 4f orbitals making up the IF7 case ?
    Please reply. Thank you very much.

  2. Sir U r doing a great job.
    its v easy to down load the content
    if u could do the same for physics-class 12,it would be great help
    Thanx a lot

    • sir, u have done a great job…great notes…i have gone through…sir would you provide me notes for VSEPR model for Fluorides of xenon and oxides of xenon to M.Sc level…..pls sir….i expecting very shortly….i persuing my masters in org chemistry in central college…pls sir.

  3. This material is very useful. Because it is short, clear and precise to understand. I appreciate it.
    I need further explanation why La is grouped with other f block elements specifically with lanthanide series.
    thank you.

  4. some more info about the metallic lattices of oxides and chlorides could have been added…
    otherwise the info is very helpful

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