3.10 In a reaction between A and B, the initial rate of reaction (r0) was measured for different initial concentrations of A and B as given below: A/ mol L–1 0.20 0.20 0.40 B/ mol L–1 0.30 0.10 0.05 r0/mol L–1s–1 5.07 × 10–5 5.07 × 10–5 1.43 × 10–4 What is the order of the reaction with respect to A and B? 3.11 The following results have been obtained during the kinetic studies of the reaction: 2A + B ® C + D Experiment [A]/mol L–1 [B]/mol L–1 Initial rate of formation of D/mol L–1 min–1 I 0.1 0.1 6.0 × 10–3 II 0.3 0.2 7.2 × 10–2 III 0.3 0.4 2.88 × 10–1 IV 0.4 0.1 2.40 × 10–2 Determine the rate law and the rate constant for the reaction. 3.12 The reaction between A and B is first order with respect to A and zero order with respect to B. Fill in the blanks in the following table: Experiment [A]/ mol L–1 [B]/ mol L–1 Initial rate/ mol L–1 min–1 I 0.1 0.1 2.0 × 10–2 II – 0.2 4.0 × 10–2 III 0.4 0.4 – IV – 0.2 2.0 × 10–2 3.13 Calculate the half-life of a first order reaction from their rate constants given below: (i) 200 s–1 (ii) 2 min–1 (iii) 4 years–1 3.14 The half-life for radioactive decay of 14C is 5730 years. An archaeological artifact containing wood had only 80% of the 14C found in a living tree. Estimate the age of the sample. 3.15 The experimental data for decomposition of N2O5 [2N2O5 ® 4NO2 + O2] in gas phase at 318K are given below: t/s 0 400 800 1200 1600 2000 2400 2800 3200 102 × [N2O5]/ 1.63 1.36 1.14 0.93 0.78 0.64 0.53 0.43 0.35 mol L–1 (i) Plot [N2O5] against t. (ii) Find the half-life period for the reaction. (iii) Draw a graph between log[N2O5] and t. (iv) What is the rate law ? Chemistry 86 Reprint 2025-26 (v) Calculate the rate constant. (vi) Calculate the half-life period from k and compare it with (ii).

3.27 The rate constant for the first order decomposition of H2O2 is given by the following equation: log k = 14.34 – 1.25 × 104K/T Calculate Ea for this reaction and at what temperature will its half-period be 256 minutes? 3.28 The decomposition of A into product has value of k as 4.5 × 103 s–1 at 10°C and energy of activation 60 kJ mol–1. At what temperature would k be 1.5 × 104s–1? 3.29 The time required for 10% completion of a first order reaction at 298K is equal to that required for its 25% completion at 308K. If the value of A is 4 × 1010s–1. Calculate k at 318K and Ea. 3.30 The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction assuming that it does not change with temperature. Answers to Some Intext Questions 3.1 rav = 6.66 × 10–6 Ms–1 3.2 Rate of reaction = rate of diappearance of A = 0.005 mol litre–1min–1 3.3 Order of the reaction is 2.5 3.4 X ® Y Rate = k[X]2 The rate will increase 9 times 3.5 t = 444 s 3.6 1.925 × 10–4 s–1 3.8 Ea = 52.897 kJ mol–1 3.9 1.471 × 10–19 Chemistry 88 Reprint 2025-26 UnitUnitUnitUnit Unit44 TheThe dd-- andand f-f-Objectives After studying this Unit, you will beable to BlockBlock ElementsElements • learn the positions of the d– and f-block elements in the periodic table; Iron, copper, silver and gold are among the transition elements that • know the electronic configurations have played important roles in the development of human civilisation. of the transition (d-block) and the The inner transition elements such as Th, Pa and U are proving inner transition (f-block) elements; excellent sources of nuclear energy in modern times. • appreciate the relative stability of various oxidation states in terms of electrode potential values; The d-block of the periodic table contains the elements of the groups 3-12 in which the d orbitals are• describe the preparation, progressively filled in each of the four long periods. properties, structures and uses of some important compounds The f-block consists of elements in which 4 f and 5 f such as K2Cr2O7 and KMnO4; orbitals are progressively filled. They are placed in a • understand the general separate panel at the bottom of the periodic table. The characteristics of the d– and names transition metals and inner transition metals f–block elements and the general are often used to refer to the elements of d-and horizontal and group trends in f-blocks respectively. them; There are mainly four series of the transition metals, • describe the properties of the 3d series (Sc to Zn), 4d series (Y to Cd), 5d series (La f-block elements and give a and Hf to Hg) and 6d series which has Ac and elements comparative account of the from Rf to Cn. The two series of the inner transition lanthanoids and actinoids with metals; 4f (Ce to Lu) and 5f (Th to Lr) are known as respect to their electronic lanthanoids and actinoids respectively. configurations, oxidation states Originally the name transition metals was derived and chemical behaviour. from the fact that their chemical properties were transitional between those of s and p-block elements. Now according to IUPAC, transition metals are defined as metals which have incomplete d subshell either in neutral atom or in their ions. 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 3d, 4d and 5d transition series, respectively, their chemistry is studied along with the chemistry of the transition metals. The presence of partly filled d or f orbitals in their atoms makes transition elements different from that of Reprint 2025-26 the non-transition elements. Hence, transition elements and their compounds are studied separately. However, the usual theory of valence as applicable to the non- transition elements can be applied successfully to the transition elements also. Various precious metals such as silver, gold and platinum and industrially important metals like iron, copper and titanium belong to the transition metals series. In this Unit, we shall first deal with the electronic configuration, occurrence and general characteristics of transition elements with special emphasis on the trends in the properties of the first row (3d) transition metals along with 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) 4.14.14.14.14.1 PositionPositionPositionPositionPosition ininininin thethethethethe The d–block occupies the large middle section of the periodic table PeriodicPeriodicPeriodicPeriodicPeriodic TableTableTableTableTable flanked between s– and p– blocks in the periodic table. The d–orbitals of the penultimate energy level of atoms receive electrons giving rise to four rows of the transition metals, i.e., 3d, 4d, 5d and 6d. All these series of transition elements are shown in Table 4.1. 4.24.24.24.24.2 ElectronicElectronicElectronicElectronicElectronic In general1– the electronic configuration of outer orbitals of these elements is (n-1)d 10ns1–2except for Pd where its electronic configuration is 4d105s0. ConfigurationsConfigurationsConfigurationsConfigurationsConfigurations The (n–1) stands for the inner d orbitals which may have one to ten ofofofofof thethethethethe d-Blockd-Blockd-Blockd-Blockd-Block electrons and the outermost ns orbital may have one or two electrons. ElementsElementsElementsElementsElements 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. For example, consider the case of Cr, which has 3d 5 4s 1 configuration instead of 3d44s 2; 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 3d104s 1 and not 3d 94s2. The ground state electronic configurations of the outer orbitals of transition elements are given in Table 4.1. Table 4.1: Electronic Configurations of outer orbitals of the Transition Elements (ground state) 1st Series Sc Ti V Cr Mn Fe Co Ni Cu Zn Z 21 22 23 24 25 26 27 28 29 30 4s 2 2 2 1 2 2 2 2 1 2 3d 1 2 3 5 5 6 7 8 10 10 Chemistry 90 Reprint 2025-26 2nd Series Y Zr Nb Mo Tc Ru Rh Pd Ag Cd Z 39 40 41 42 43 44 45 46 47 48 5s 2 2 1 1 1 1 1 0 1 2 4d 1 2 4 5 6 7 8 10 10 10 3rd Series La Hf Ta W Re Os Ir Pt Au Hg Z 57 72 73 74 75 76 77 78 79 80 6s 2 2 2 2 2 2 2 1 1 2 5d 1 2 3 4 5 6 7 9 10 10 4th Series Ac Rf Db Sg Bh Hs Mt Ds Rg Cn Z 89 104 105 106 107 108 109 110 111 112 7s 2 2 2 2 2 2 2 2 1 2 6d 1 2 3 4 5 6 7 8 10 10 The electronic configurations of outer orbitals of Zn, Cd, Hg and Cn are represented by the general formula (n-1)d 10ns2. 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 protrude 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 affect 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 similarities in the properties of the transition elements of a horizontal row in contrast to the non-transition 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. On what ground can you say that scandium (Z = 21) is a transition ExampleExampleExampleExampleExample 4.14.14.14.14.1 element but zinc (Z = 30) is not? On the basis of incompletely filled 3d orbitals in case of scandium atom SolutionSolutionSolutionSolutionSolution 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. 91 The d- and f- Block Elements Reprint 2025-26 IntextIntextIntextIntextIntext QuestionQuestionQuestionQuestionQuestion 4.1 Silver atom has completely filled d orbitals (4d10) in its ground state. How can you say that it is a transition element? We will discuss the properties of elements of first transition series only in the following sections. 4.34.34.34.34.3 GeneralGeneralGeneralGeneralGeneral 4.3.1 Physical Properties PropertiesPropertiesPropertiesPropertiesProperties ofofofofof Nearly all the transition elements display typical metallic properties thethethethethe TransitionTransitionTransitionTransitionTransition such as high tensile strength, ductility, malleability, high thermal and electrical conductivity and metallic lustre. With the exceptions of Zn, ElementsElementsElementsElementsElements Cd, Hg and Mn, they have one or more typical metallic structures at (d-Block)(d-Block)(d-Block)(d-Block)(d-Block) normal temperatures. Lattice Structures of Transition Metals Sc Ti V Cr Mn Fe Co Ni Cu Zn hcp hcp bcc bcc X bcc ccp ccp ccp X (bcc) (bcc) (bcc, ccp) (hcp) (hcp) (hcp) Y Zr Nb Mo Tc Ru Rh Pd Ag Cd hcp hcp bcc bcc hcp hcp ccp ccp ccp X (bcc) (bcc) (hcp) La Hf Ta W Re Os Ir Pt Au Hg hcp hcp bcc bcc hcp hcp ccp ccp ccp X (ccp,bcc) (bcc) 4 (bcc = body centred cubic; hcp = hexagonal close packed; ccp = cubic close packed; X = a typical metal structure). W The transition metals (with the exception Re Ta of Zn, Cd and Hg) are very hard and have low volatility. Their melting and boiling points are 3 Mo Os high. Fig. 4.1 depicts the melting points of Nb Ru transition metals belonging to 3d, 4d and 5d Ir series. The high melting points of these metals Hf Tc K are attributed to the involvement of greater 3 Cr Rh number of electrons from (n-1)d in addition to Zr V Pt 2 the ns electrons in the interatomic metallic bonding. In any row the melting points of these M.p./10 Ti Fe Co Pd 5 metals rise to a maximum at d except for Ni anomalous values of Mn and Tc and fall Mn Cu regularly as the atomic number increases. Au Ag They have high enthalpies of atomisation which 1 are shown in Fig. 4.2. The maxima at about Atomic number the middle of each series indicate that one Fig. 4.1: Trends in melting points of unpaired electron per d orbital is particularly transition elements Chemistry 92 Reprint 2025-26 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. 4.2 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. –1 mol V/kJ DaH Fig. 4.2 Trends in enthalpies of atomisation of transition elements 4.3.2 Variation in In general, ions of the same charge in a given series show progressive Atomic and decrease in radius with increasing atomic number. This is because the Ionic Sizes new electron enters a d orbital each time the nuclear charge increases of by unity. It may be recalled that the shielding effect of a d electron is Transition not that effective, hence the net electrostatic attraction between the Metals 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. 4.3 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 93 The d- and f- Block Elements Reprint 2025-26 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. 19 The factor responsible for the lanthanoid 18 contraction is somewhat similar to that observed in an ordinary transition series and is attributed 17 to similar cause, i.e., the imperfect shielding of 16 one electron by another in the same set of orbitals. However, the shielding of one 4f electron by 15 Radius/nm another is less than that of one d electron by 14 another, and as the nuclear charge increases 13 along the series, there is fairly regular decrease in the size of the entire 4f n orbitals. 12 Sc Ti V Cr Mn Fe Co Ni Cu Zn The decrease in metallic radius coupled with Y Zr Nb Mo Tc Ru Rh Pd Ag Cd increase in atomic mass results in a general increase in the density of these elements. Thus, La Hf Ta W Re Os Ir Pt Au Hg from titanium (Z = 22) to copper (Z = 29) the Fig. 4.3: Trends in atomic radii of significant increase in the density may be noted transition elements (Table 4.2). Table 4.2: Electronic Configurations and some other Properties of the First Series of Transition Elements Element Sc Ti V Cr Mn Fe Co Ni Cu Zn Atomic number 21 22 23 24 25 26 27 28 29 30 Electronic configuration M 3d 14s 2 3d 24s 2 3d 34s 2 3d 54s 1 3d 54s 2 3d 64s 2 3d 74s 2 3d 84s 2 3d 104s 1 3d 104s 2 M + 3d 14s 1 3d 24s 1 3d 34s 1 3d 5 3d 54s 1 3d 64s 1 3d 74s 1 3d 84s 1 3d 10 3d 104s 1 M 2+ 3d 1 3d 2 3d 3 3d 4 3d 5 3d 6 3d 7 3d 8 3d 9 3d 10 M 3+ [Ar] 3d 1 3d 2 3d 3 3d 4 3d 5 3d 6 3d 7 – – Enthalpy of atomisation, DaH o/kJ mol–1 326 473 515 397 281 416 425 430 339 126 Ionisation enthalpy/DiH o/kJ mol –1 DiHo I 631 656 650 653 717 762 758 736 745 906 DiHo II 1235 1309 1414 1592 1509 1561 1644 1752 1958 1734 DiHo III 2393 2657 2833 2990 3260 2962 3243 3402 3556 3837 Metallic/ionic M 164 147 135 129 137 126 125 125 128 137 radii/pm M 2+ – – 79 82 82 77 74 70 73 75 M 3+ 73 67 64 62 65 65 61 60 – – Standard electrode M 2+/M – –1.63 –1.18 –0.90 –1.18 –0.44 –0.28 –0.25 +0.34 -0.76 potential Eo/V M 3+/M 2+ – –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 Chemistry 94 Reprint 2025-26 Why do the transition elements exhibit higher enthalpies of ExampleExampleExampleExampleExample 4.24.24.24.24.2 atomisation? Because of large number of unpaired electrons in their atoms they SolutionSolutionSolutionSolutionSolution have stronger interatomic interaction and hence stronger bonding between atoms resulting in higher enthalpies of atomisation. IntextIntextIntextIntextIntext QuestionQuestionQuestionQuestionQuestion 4.2 In the series Sc (Z = 21) to Zn (Z = 30), the enthalpy of atomisation of zinc is the lowest, i.e., 126 kJ mol–1. Why? 4.3.3 Ionisation There is an increase in ionisation enthalpy along each series of the Enthalpies transition elements from left to right due to an increase in nuclear charge which accompanies the filling of the inner d orbitals. Table 4.2 gives the values of the first three ionisation enthalpies of the first series of transition elements. These values show that the successive enthalpies of these elements do not increase as steeply as in the case of non-transition elements. The variation in ionisation enthalpy along a series of transition elements is much less in comparison to the variation along a period of non-transition elements. The first ionisation enthalpy, in general, increases, but the magnitude of the increase in the second and third ionisation enthalpies for the successive elements, is much higher along a series. The irregular trend in the first ionisation enthalpy of the metals of 3d series, 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. You have learnt that when d-block elements form ions, ns electrons are lost before (n – 1) d electrons. As we move along the period in 3d series, we see that nuclear charge increases from scandium to zinc but electrons are added to the orbital of inner subshell, i.e., 3d orbitals. These 3d electrons shield the 4s electrons from the increasing nuclear charge somewhat more effectively than the outer shell electrons can shield one another. Therefore, the atomic radii decrease less rapidly. Thus, ionization energies increase only slightly along the 3d series. The doubly or more highly charged ions have dn configurations with no 4s electrons. A general trend of increasing values of second ionisation enthalpy is expected as the effective nuclear charge increases because one d electron does not shield another electron from the influence of nuclear charge because d-orbitals differ in direction. However, the trend of steady increase in second and third ionisation enthalpy breaks for the formation of Mn2+ and Fe3+ respectively. In both the cases, ions have d5 configuration. Similar breaks occur at corresponding elements in the later transition series. The interpretation of variation in ionisation enthalpy for an electronic configuration dn is as follows: The three terms responsible for the value of ionisation enthalpy are attraction of each electron towards nucleus, repulsion between the 95 The d- and f- Block Elements Reprint 2025-26 electrons and the exchange energy. Exchange energy is responsible for the stabilisation of energy state. Exchange energy is approximately proportional to the total number of possible pairs of parallel spins in the degenerate orbitals. When several electrons occupy a set of degenerate orbitals, the lowest energy state corresponds to the maximum possible extent of single occupation of orbital and parallel spins (Hunds rule). The loss of exchange energy increases the stability. As the stability increases, the ionisation becomes more difficult. There is no loss of exchange energy at d6 configuration. Mn+ has 3d54s1 configuration and configuration of Cr+ is d5, therefore, ionisation enthalpy of Mn+ is lower than Cr+. In the same way, Fe2+ has d6 configuration and Mn2+ has 3d5 configuration. Hence, ionisation enthalpy of Fe2+ is lower than the Mn2+. In other words, we can say that the third ionisation enthalpy of Fe is lower than that of Mn. The lowest common oxidation state of these metals is +2. To form the M 2+ ions from the gaseous atoms, the sum of the first and second ionisation enthalpy is required in addition to the enthalpy of atomisation. The dominant term is the second ionisation enthalpy which shows unusually high values for Cr and Cu where M + ions have the d 5 and d 10 configurations respectively. The value for Zn is correspondingly low as the ionisation causes the removal of one 4s electron which results in the formation of stable d 10 configuration. The trend in the third ionisation enthalpies is not complicated by the 4s orbital factor and shows the greater difficulty of removing an electron from the d 5 (Mn 2+) and d 10 (Zn 2+) ions. In general, the third ionisation enthalpies are quite high. Also the high values for third ionisation enthalpies of 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 amenable to ready generalisation. 4.3.4 Oxidation One of the notable features of a transition elements is the great variety States of oxidation states these may show in their compounds. Table 4.3 lists the common oxidation states of the first row transition elements. Table 4.3: Oxidation States of the first row Transition Metal (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 Chemistry 96 Reprint 2025-26 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 (Ti IVO2, VVO2 +, Cr V1O42–, MnVIIO4–) followed by a rather abrupt decrease in stability of higher oxidation states, so that the typical species to follow are FeII,III, Co II,III, NiII, CuI,II, Zn II. 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., V II, V III, 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 p-acceptor character in addition to the s-bonding. For example, in Ni(CO)4 and Fe(CO)5, the oxidation state of nickel and iron is zero. Name a transition element which does not exhibit variable ExampleExampleExampleExampleExample 4.34.34.34.34.3 oxidation states. Scandium (Z = 21) does not exhibit variable oxidation states. SolutionSolutionSolutionSolutionSolution IntextIntextIntextIntextIntext QuestionQuestionQuestionQuestionQuestion 4.3 Which of the 3d series of the transition metals exhibits the largest number of oxidation states and why? 97 The d- and f- Block Elements Reprint 2025-26 4.3.5 Trends in the Table 4.4 contains the thermochemical parameters related to the M2+/M transformation of the solid metal atoms to M2+ ions in solution and their V Standard standard electrode potentials. The observed values of E and those Electrode calculated using the data of Table 4.4 are compared in Fig. 4.4. Potentials The unique behaviour of Cu, having a positive EV, 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 V enthalpy. The general trend towards less negative E values across the Fig. 4.4: Observed and calculated values for the standard electrode potentials (M2+ ® M°) of the elements Ti to Zn series is related to the general increase in the sum of the first and second V 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. Why is Cr2+ reducing and Mn3+ oxidising when both have d4 configuration? ExampleExampleExampleExampleExample 4.44.44.44.44.4 Cr 2+ is reducing as its configuration changes from d 4 to d 3, the latter SolutionSolutionSolutionSolutionSolution having a half-filled t2g level (see Unit 5). On the other hand, the change from Mn3+ to Mn2+ results in the half-filled (d5) configuration which has extra stability. IntextIntextIntextIntextIntext QuestionQuestionQuestionQuestionQuestion 4.4 The E o(M2+/M) value for copper is positive (+0.34V). What is possible reason for this? (Hint: consider its high DaH o and low DhydH o) Chemistry 98 Reprint 2025-26 Table 4.4: Thermochemical data (kJ mol-1) for the first row Transition Elements and the Standard Electrode Potentials for the Reduction of MII to M. Element (M) DaH o (M) DiH1o D1H2o DhydH o(M2+) Eo/V Ti 469 656 1309 -1866 -1.63 V 515 650 1414 -1895 -1.18 Cr 398 653 1592 -1925 -0.90 Mn 279 717 1509 -1862 -1.18 Fe 418 762 1561 -1998 -0.44 Co 427 758 1644 -2079 -0.28 Ni 431 736 1752 -2121 -0.25 Cu 339 745 1958 -2121 0.34 Zn 130 906 1734 -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 E o values, whereas E o for Ni is related to the highest negative DhydH o. 4.3.6 Trends in An examination of the E o(M3+/M2+) values (Table 4.2) shows the varying the M3+/M2+ trends. The low value for Sc reflects the stability of Sc3+ which has a Standard noble gas configuration. The highest value for Zn is due to the removal Electrode of an electron from the stable d 10 configuration of Zn 2+. The Potentials comparatively high value for Mn shows that Mn 2+(d5) is particularly stable, whereas comparatively low value for Fe shows the extra stability of Fe 3+ (d5). The comparatively low value for V is related to the stability of V 2+ (half-filled t2g level, Unit 5). 4.3.7 Trends in Table 4.5 shows the stable halides of the 3d series of transition metals. Stability of The highest oxidation numbers are achieved in TiX4 (tetrahalides), VF5 Higher and CrF6. The +7 state for Mn is not represented in simple halides but Oxidation MnO3F is known, and beyond Mn no metal has a trihalide except FeX3 States 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 V +5 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) Table 4.5: Formulas of Halides of 3d Metals Oxidation Number + 6 CrF6 + 5 VF5 CrF5 + 4 TiX4 VXI4 CrX4 MnF4 + 3 TiX3 VX3 CrX3 MnF3 FeXI3 CoF3 + 2 TiX2III VX2 CrX2 MnX2 FeX2 CoX2 NiX2 CuX2II ZnX2 + 1 CuXIII Key: X = F ® I; XI = F ® Br; XII = F, CI; XIII = CI ® I 99 The d- and f- Block Elements Reprint 2025-26 and the same applies to CuX. On the other hand, all Cu II halides are known except the iodide. In this case, Cu 2+ oxidises I – to I2: 2Cu 2   4I   Cu2 I2 s  I2 However, many copper (I) compounds are unstable in aqueous solution and undergo disproportionation. 2Cu + ® Cu 2+ + Cu The stability of Cu 2+ (aq) rather than Cu+(aq) is due to the much more negative DhydH o of Cu 2+ (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 4.6) 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 V v as VO2 +, V IV as VO2+ and Ti IV as TiO 2+. 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 V V, Cr Vl, Mn V, Mn Vl and Mn VII. Table 4.6: Oxides of 3d Metals Oxidation Groups Number 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 + 1 Cu2O * mixed oxides How would you account for the increasing oxidising power in the ExampleExampleExampleExampleExample 4.54.54.54.54.5 series VO2+ < Cr2O7 2– < MnO4 – ? This is due to the increasing stability of the lower species to which they SolutionSolutionSolutionSolutionSolution are reduced. IntextIntextIntextIntextIntext QuestionQuestionQuestionQuestionQuestion 4.5 How would you account for the irregular variation of ionisation enthalpies (first and second) in the first series of the transition elements? Chemistry 100 Reprint 2025-26 4.3.8 Chemical Transition metals vary widely in their chemical reactivity. Many of Reactivity them are sufficiently electropositive to dissolve in mineral acids, although and Eo a few are ‘noble’—that is, they are unaffected by single acids. Values 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 E o values for M2+/M (Table 4.2) indicate a decreasing tendency to form divalent cations across the series. This general trend towards less negative E o values is related to the increase in the sum of the first and second ionisation enthalpies. It is interesting to note that the E o 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 E e values; for nickel, Eo value is related to the highest negative enthalpy of hydration. An examination of the E o values for the redox couple M 3+/M2+ (Table 4.2) shows that Mn 3+ and Co 3+ ions are the strongest oxidising agents in aqueous solutions. The ions Ti 2+, V 2+ and Cr2+ are strong reducing agents and will liberate hydrogen from a dilute acid, e.g., 2 Cr 2+(aq) + 2 H+(aq) ® 2 Cr 3+(aq) + H2(g) ExampleExampleExampleExampleExample 4.64.64.64.64.6 For the first row transition metals the Eo values are: E o 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. SolutionSolutionSolutionSolutionSolution The E o (M2+/M) values are not regular which can be explained from the irregular variation of ionisation enthalpies (  i H1  i H 2 ) and also the sublimation enthalpies which are relatively much less for manganese and vanadium. ExampleExampleExampleExampleExample 4.74.74.74.74.7 Why is the E o value for the Mn3+/Mn 2+ couple much more positive than that for Cr 3+/Cr2+ or Fe 3+/Fe 2+? Explain. SolutionSolutionSolutionSolutionSolution 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. IntextIntextIntextIntextIntext QuestionsQuestionsQuestionsQuestionsQuestions 4.6 Why is the highest oxidation state of a metal exhibited in its oxide or fluoride only? 4.7 Which is a stronger reducing agent Cr2+ or Fe2+ and why ? 4.3.9 Magnetic When a magnetic field is applied to substances, mainly two types of Properties magnetic behaviour are observed: diamagnetism and paramagnetism. Diamagnetic substances are repelled by the applied field while the paramagnetic substances are attracted. Substances which are 101 The d- and f- Block Elements Reprint 2025-26 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.,  n  n  2  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 4.7. The experimental data are mainly for hydrated ions in solution or in the solid state. Table 4.7: Calculated and Observed Magnetic Moments (BM) Ion Configuration Unpaired Magnetic moment electron(s) Calculated Observed Sc3+ 3d0 0 0 0 Ti 3+ 3d1 1 1.73 1.75 Tl2+ 3d2 2 2.84 2.76 V2+ 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 Co2+ 3d7 3 3.87 4.4 – 5.2 Ni2+ 3d8 2 2.84 2.9 – 3, 4 Cu 2+ 3d9 1 1.73 1.8 – 2.2 Zn2+ 3d10 0 0 Calculate the magnetic moment of a divalent ion in aqueous solution ExampleExampleExampleExampleExample 4.84.84.84.84.8 if its atomic number is 25. With atomic number 25, the divalent ion in aqueous solution will have SolutionSolutionSolutionSolutionSolution d5 configuration (five unpaired electrons). The magnetic moment, µ is  5  5  2   5.92BM Chemistry 102 Reprint 2025-26 IntextIntextIntextIntextIntext QuestionQuestionQuestionQuestionQuestion 4.8 Calculate the ‘spin only’ magnetic moment of M 2+ (aq) ion (Z = 27). 4.3.10 Formation When an electron from a lower energy d orbital is excited to a higher of Coloured energy d orbital, the energy of excitation corresponds to the frequency Ions of light absorbed (Unit 5). 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 4.8. A few coloured solutions of Fig. 4.5: Colours of some of the first row d–block elements are transition metal ions in aqueous solutions. From illustrated in Fig. 4.5. left to right: V4+,V3+,Mn2+,Fe3+,Co2+,Ni2+and Cu2+ . Table 4.8: Colours of Some of the First Row (aquated) Transition Metal Ions Configuration Example Colour 3d0 Sc3+ colourless 3d0 Ti 4+ colourless 3d1 Ti 3+ purple 3d1 V4+ blue 3d2 V3+ green 3d3 V2+ violet 3d3 Cr3+ violet 3d4 Mn 3+ violet 3d4 Cr2+ blue 3d5 Mn 2+ pink 3d5 Fe3+ yellow 3d6 Fe2+ green 3d63d7 Co3+Co2+ bluepink 3d8 Ni2+ green 3d9 Cu 2+ blue 3d10 Zn2+ colourless 4.3.11 Formation Complex compounds are those in which the metal ions bind a number of Complex of anions or neutral molecules giving complex species with Compounds 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 103 The d- and f- Block Elements Reprint 2025-26 dealt with in detail in Unit 5). 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. 4.3.12 Catalytic The transition metals and their compounds are known for their catalytic Properties 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– + S2O8 2– ® I2 + 2 SO4 2– An explanation of this catalytic action can be given as: 2 Fe 3+ + 2 I – ® 2 Fe 2+ + I2 2 Fe 2+ + S2O82– ® 2 Fe3+ + 2SO42– 4.3.13 Formation Interstitial compounds are those which are formed when small atoms of like H, C or N are trapped inside the crystal lattices of metals. They are Interstitial usually non stoichiometric and are neither typically ionic nor covalent, Compounds 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 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. 4.3.14 Alloy An alloy is a blend of metals prepared by mixing the components. Formation 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. Chemistry 104 Reprint 2025-26 ExampleExampleExampleExampleExample 4.94.94.94.94.9 What is meant by ‘disproportionation’ of an oxidation state? Give an example. SolutionSolutionSolutionSolutionSolution 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 Mn VIO4 2– + 4 H + ® 2 Mn VIIO –4 + Mn IVO2 + 2H2O IntextIntextIntextIntextIntext QuestionQuestionQuestionQuestionQuestion 4.9 Explain why Cu+ ion is not stable in aqueous solutions? 4.44.44.44.44.4 SomeSomeSomeSomeSome 4.4.1 Oxides and Oxoanions of Metals ImportantImportantImportantImportantImportant These oxides are generally formed by the reaction of metals with CompoundsCompoundsCompoundsCompoundsCompounds ofofofofof oxygen at high temperatures. All the metals except scandium form TransitionTransitionTransitionTransitionTransition MO oxides which are ionic. The highest oxidation number in the oxides, coincides with the group number and is attained in Sc2O3 to ElementsElementsElementsElementsElements Mn2O7. Beyond group 7, no higher oxides of iron above Fe2O3 are known. Besides the oxides, the oxocations stabilise V V as VO2 +, V IV as VO 2+ and Ti IV as TiO 2+. 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 VO4 3– 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 VO 2+ salts. Similarly, V2O5 reacts with alkalies as well as acids to give VO 34  and VO4 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: 4 FeCr2O4 + 8 Na2CO3 + 7 O2 ® 8 Na2CrO4 + 2 Fe2O3 + 8 CO2 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 + 2 H+ ® Na2Cr2O7 + 2 Na + + H2O 105 The d- and f- Block Elements Reprint 2025-26 Sodium dichromate is more soluble than potassium dichromate. The latter is therefore, prepared by treating the solution of sodium dichromate with potassium chloride. Na2Cr2O7 + 2 KCl ® K2Cr2O7 + 2 NaCl 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. 2 CrO4 2– + 2H + ® Cr2O7 2– + H2O Cr2O7 2– + 2 OH- ® 2 CrO4 2– + H2O The structures of chromate ion, CrO4 2– and the dichromate ion, Cr2O7 2– 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: Cr2O7 2– + 14H + + 6e – ® 2Cr 3+ + 7H2O (E o = 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 + 6 e – ; 3 Sn 2+ ® 3Sn 4+ + 6 e – 3 H2S ® 6H+ + 3S + 6e – ; 6 Fe 2+ ® 6Fe3+ + 6 e– The full ionic equation may be obtained by adding the half-reaction for potassium dichromate to the half-reaction for the reducing agent, for e.g., Cr2O7 2– + 14 H+ + 6 Fe2+ ® 2 Cr3+ + 6 Fe3+ + 7 H2O 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 3MnO4 2– + 4H+ ® 2MnO4 – + MnO2 + 2H2O Commercially it is prepared by the alkaline oxidative fusion of MnO2 followed by the electrolytic oxidation of manganate (Vl). F used with KOH, oxidised Electrolytic oxidation in MnO 2 →with air or KNO 3 MnO 24 − ; MnO 24  alkaline solution MnO 4 manganate ion manganate permanganate ion Chemistry 106 Reprint 2025-26 In the laboratory, a manganese (II) ion salt is oxidised by peroxodisulphate to permanganate. 2Mn2+ + 5S2O8 2– + 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: its intense colour and its diamagnetism along with temperature-dependent weak paramagnetism. These can be explained by the use of molecular orbital theory which is beyond the present scope. The manganate and permanganate ions are tetrahedral; the p- bonding takes place by overlap of p orbitals of oxygen with d orbitals of manganese. The green manganate is paramagnetic because of one unpaired electron but the permanganate is diamagnetic due to the absence of unpaired electron. 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: COO – 5 10CO2 + 10e – COO – 5 Fe2+ ® 5 Fe3+ + 5e– 5NO2 – + 5H2O ® 5NO3 – + 10H+ + l0e– 10I– ® 5I2 + 10e– 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– ® MnO4 2– (E o = + 0.56 V) MnO4 – + 4H+ + 3e– ® MnO2 + 2H2O (E o = + 1.69 V) MnO4 – + 8H+ + 5e– ® Mn2+ + 4H2O (E o = + 1.52 V) 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): 5Fe 2+ + MnO4 – + 8H+ ® Mn2+ + 4H2O + 5Fe 3+ 107 The d- and f- Block Elements Reprint 2025-26 (c) Oxalate ion or oxalic acid is oxidised at 333 K: 5C2O4 2– + 2MnO4 – + 16H + ——> 2Mn 2+ + 8H2O + 10CO2 (d) Hydrogen sulphide is oxidised, sulphur being precipitated: H2S —> 2H + + S2– 5S 2– + 2MnO – 4 + 16H + ——> 2Mn2+ + 8H2O + 5S (e) Sulphurous acid or sulphite is oxidised to a sulphate or sulphuric acid: 5SO3 2– + 2MnO4 – + 6H + ——> 2Mn 2+ + 3H2O + 5SO42– (f) Nitrite is oxidised to nitrate: 5NO2– + 2MnO4– + 6H + ——> 2Mn 2+ + 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 – + 3S2O3 2– + H2O ——> 8MnO2 + 6SO4 2– + 2OH – (c) Manganous salt is oxidised to MnO2; the presence of zinc sulphate or zinc oxide catalyses the oxidation: 2MnO4 – + 3Mn 2+ + 2H2O ——> 5MnO2 + 4H+ Note: Permanganate titrations in presence of hydrochloric acid are unsatisfactory since hydrochloric acid is oxidised to chlorine. UsesUsesUses:UsesUses 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 oxidation states in these elements and partly because their radioactivity creates special problems in their study; the two series will be considered separately here. 4.54.54.54.54.5 TheTheTheTheThe The names, symbols, electronic configurations of atomic and some LanthanoidsLanthanoidsLanthanoidsLanthanoidsLanthanoids ionic states and atomic and ionic radii of lanthanum and lanthanoids (for which the general symbol Ln is used) are given in Table 4.9. Chemistry 108 Reprint 2025-26 4.5.1 Electronic It may be noted that atoms of these elements have electronic Configurations configuration with 6s 2 common but with variable occupancy of 4f level (Table 4.9). However, the electronic configurations of all the tripositive ions (the most stable oxidation state of all the lanthanoids) are of the form 4f n (n = 1 to 14 with increasing atomic number). 4.5.2 Atomic and The overall decrease in atomic and ionic radii from lanthanum to Ionic Sizes lutetium (the lanthanoid contraction) is a unique feature in the chemistry of the lanthanoids. It has far reaching Sm 2+ consequences in the chemistry of the third 110 2+ transition series of the elements. The decrease Eu in atomic radii (derived from the structures of La3+ metals) is not quite regular as it is regular in 3+ M3+ ions (Fig. 4.6). This contraction is, of Ce course, similar to that observed in an ordinary Pr3+ transition series and is attributed to the same 100 Nd3+ cause, the imperfect shielding of one electron Pm 3+ by another in the same sub-shell. However, the Sm3+ shielding of one 4 f electron by another is less Eu3+ than one d electron by another with the increase Gd3+ Tm 2+radii/pm 2+ in nuclear charge along the series. There is Yb Ce 4+ Tb 3+ fairly regular decrease in the sizes with 3+ DyIonic Pr4+ 3+ increasing atomic number. 90 Ho Er 3+ The cumulative effect of the contraction of Tm3+ the lanthanoid series, known as lanthanoid Yb3+ 3+ contraction, causes the radii of the members 4+ Lu Tb of the third transition series to be very similar to those of the corresponding members of the second series. The almost identical radii of Zr 57 59 61 63 65 67 69 71 (160 pm) and Hf (159 pm), a consequence of the lanthanoid contraction, account for their Atomic number occurrence together in nature and for the Fig. 4.6: Trends in ionic radii of lanthanoids difficulty faced in their separation. 4.5.3 Oxidation In the lanthanoids, La(II) and Ln(III) compounds are predominant States 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 Ce IV is favoured by its noble gas configuration, but it is a strong oxidant reverting to the common +3 state. The E o value for Ce 4+/ Ce 3+ is + 1.74 V which suggests that it can oxidise water. However, the reaction rate is very slow and hence Ce(IV) is a good analytical reagent. 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 f 7 configuration accounts for the formation of this ion. However, Eu 2+ is a strong reducing agent changing to the common +3 state. Similarly Yb 2+ which has f 14 configuration is a reductant. Tb IV has half-filled f-orbitals and is an oxidant. The behaviour of samarium is very much like europium, exhibiting both +2 and +3 oxidation states. 109 The d- and f- Block Elements Reprint 2025-26 Table 4.9: Electronic Configurations and Radii of Lanthanum and Lanthanoids Electronic configurations* Radii/pm Atomic Name Symbol Ln Ln2+ Ln3+ Ln4+ Ln Ln3+ Number 57 Lanthanum La 5d16s2 5d1 4f 0 187 106 58 Cerium Ce 4f15d16s2 4f 2 4f 1 4f 0 183 103 59 Praseodymium Pr 4f 36s2 4f 3 4f 2 4f 1 182 101 60 Neodymium Nd 4f 46s2 4f 4 4f 3 4f 2 181 99 61 Promethium Pm 4f 56s2 4f 5 4f 4 181 98 62 Samarium Sm 4f 66s2 4f 6 4f 5 180 96 63 Europium Eu 4f 76s2 4f 7 4f 6 199 95 64 Gadolinium Gd 4f 75d16s2 4f 75d 1 4f 7 180 94 65 Terbium Tb 4f 96s2 4f 9 4f 8 4f 7 178 92 66 Dysprosium Dy 4f 106s2 4f 10 4f 9 4f 8 177 91 67 Holmium Ho 4f 116s2 4f 11 4f 10 176 89 68 Erbium Er 4f 126s2 4f 12 4f 11 175 88 69 Thulium Tm 4f 136s2 4f 13 4f 12 174 87 70 Ytterbium Yb 4f 146s2 4f 14 4f 13 173 86 71 Lutetium Lu 4f 145d16s2 4f 145d1 4f 14 – – – * Only electrons outside [Xe] core are indicated 4.5.4 General All the lanthanoids are silvery white soft metals and tarnish rapidly in air. Characteristics 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 La 3+ 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 0 type (La 3+ and Ce4+) and the f 14 type (Yb2+ and Lu3+) are all paramagnetic. 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 E o for the half-reaction: Ln 3+(aq) + 3e – ® Ln(s) Chemistry 110 Reprint 2025-26 Ln2 O 3 H2 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 acids variation. The metals combine with burns in with hydrogen when gently heated in the O2 gas. The carbides, Ln3C, Ln2C3 and LnC2 are formed when the metals are heated heated with S Ln with halogens with carbon. They liberate hydrogen Ln 2 S3 LnX 3 from dilute acids and burn in halogens N with with toandform hydroxideshalides. They formM(OH)3.oxides M2O3The C K H2 O hydroxides are definite compounds, not heated just hydrated oxides. They are basic with 2773 like alkaline earth metal oxides and Ln N LnC2 Ln(OH)3 + H2 hydroxides. Their general reactions are depicted in Fig. 4.7. Fig 4.7: Chemical reactions of the lanthanoids. The best single use of the lanthanoids is for the production of alloy steels for plates and pipes. A well known alloy is mischmetall which consists of a lanthanoid metal (~ 95%) and iron (~ 5%) and traces of S, C, Ca and Al. A good deal of mischmetall is used in Mg-based 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. 4.64.64.64.64.6 TheTheTheTheThe ActinoidsActinoidsActinoidsActinoidsActinoids The actinoids include the fourteen elements from Th to Lr. The names, symbols and some properties of these elements are given in Table 4.10. Table 4.10: Some Properties of Actinium and Actinoids Electronic conifigurations* Radii/pm Atomic Name Symbol M M3+ M4+ M3+ M4+ Number 89 Actinium Ac 6d 17s 2 5f 0 111 90 Thorium Th 6d 27s 2 5f 1 5f 0 99 91 Protactinium Pa 5f 26d 17s 2 5f 2 5f 1 96 92 Uranium U 5f 36d 17s 2 5f 3 5f 2 103 93 93 Neptunium Np 5f 46d 17s 2 5f 4 5f 3 101 92 94 Plutonium Pu 5f 67s 2 5f 5 5f 4 100 90 95 Americium Am 5f 77s 2 5f 6 5f 5 99 89 96 Curium Cm 5f 76d 17s 2 5f 7 5f 6 99 88 97 Berkelium Bk 5f 97s 2 5f 8 5f 7 98 87 98 Californium Cf 5f 107s 2 5f 9 5f 8 98 86 99 Einstenium Es 5f 117s 2 5f 10 5f 9 – – 100 Fermium Fm 5f 127s 2 5f 11 5f 10 – – 101 Mendelevium Md 5f 137s 2 5f 12 5f 11 – – 102 Nobelium No 5f 147s 2 5f 13 5f 12 – – 103 Lawrencium Lr 5f 146d 17s 2 5f 14 5f 13 – – 111 The d- and f- Block Elements Reprint 2025-26 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. 4.6.1 Electronic All the actinoids are believed to have the electronic configuration of 7s2 Configurations 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 0, f 7 and f 14 occupancies of the 5f orbitals. Thus, the configurations of Am and Cm are [Rn] 5f 77s2 and [Rn] 5f 76d17s2. 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. 4.6.2 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. 4.6.3 Oxidation There is a greater range of oxidation states, which is in part attributed to States the fact that the 5f, 6d and 7s levels are of comparable energies. The known oxidation states of actinoids are listed in Table 4.11. 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 4.11). 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 former and later elements, it is unsatisfactory to review their chemistry in terms of oxidation states. Table 4.11: Oxidation States of Actinium and Actinoids Ac Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr 3 3 3 3 3 3 3 3 3 3 3 3 3 3 4 4 4 4 4 4 4 4 5 5 5 5 5 6 6 6 6 7 7 4.6.4 General The actinoid metals are all silvery in appearance but display Characteristics a variety of structures. The structural variability is obtained and Comparison due to irregularities in metallic radii which are far greater with Lanthanoids than in lanthanoids. Chemistry 112 Reprint 2025-26 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 will penetrate less into the inner core of electrons. The 5f electrons, will therefore, be more effectively shielded from the nuclear charge than the 4f electrons of the corresponding lanthanoids. Because the outer electrons are less firmly held, they are 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. ExampleExampleExampleExampleExample 4.104.104.104.104.10 Name a member of the lanthanoid series which is well known to exhibit +4 oxidation state. SolutionSolutionSolutionSolutionSolution Cerium (Z = 58) IntextIntextIntextIntextIntext QuestionQuestionQuestionQuestionQuestion 4.10 Actinoid contraction is greater from element to element than lanthanoid contraction. Why? 4.74.74.74.74.7 SomeSomeSomeSomeSome Iron and steels are the most important construction materials. Their ApplicationsApplicationsApplicationsApplicationsApplications 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 ofofofofof d-d-d-d-d- andandandandand and Ni. Some compounds are manufactured for special purposes such as f-Blockf-Blockf-Blockf-Blockf-Block TiO for the pigment industry and MnO2 for use in dry battery cells. The ElementsElementsElementsElementsElements 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 113 The d- and f- Block Elements Reprint 2025-26 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 A1(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. SummarySummarySummarySummarySummary 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 reaching 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 Chemistry 114 Reprint 2025-26 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 4.1 Write down the electronic configuration of: (i) Cr3+ (iii) Cu+ (v) Co2+ (vii) Mn2+ (ii) Pm3+ (iv) Ce4+ (vi) Lu2+ (viii) Th4+ 4.2 Why are Mn2+ compounds more stable than Fe2+ towards oxidation to their +3 state? 4.3 Explain briefly how +2 state becomes more and more stable in the first half of the first row transition elements with increasing atomic number? 4.4 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. 4.5 What may be the stable oxidation state of the transition element with the following d electron configurations in the ground state of their atoms : 3d 3, 3d 5, 3d 8 and 3d 4? 4.6 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. 4.7 What is lanthanoid contraction? What are the consequences of lanthanoid contraction? 4.8 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? 4.9 In what way is the electronic configuration of the transition elements different from that of the non transition elements?

3.5 ELECTRONIC CONFIGURATIONS transition series of elements. This starts from OF ELEMENTS AND THE PERIODIC scandium (Z = 21) which has the electronic TABLE configuration 3d14s2. The 3d orbitals are filled In the preceding unit we have learnt that an at zinc (Z=30) with electronic configuration electron in an atom is characterised by a set 3d104s2. The fourth period ends at krypton of four quantum numbers, and the principal with the filling up of the 4p orbitals. Altogether quantum number (n ) defines the main energy we have 18 elements in this fourth period. The level known as shell. We have also studied fifth period (n = 5) beginning with rubidium about the filling of electrons into different is similar to the fourth period and contains subshells, also referred to as orbitals (s, p, the 4d transition series starting at yttrium (Z = 39). This period ends at xenon with thed, f ) in an atom. The distribution of electrons filling up of the 5p orbitals. The sixth periodinto orbitals of an atom is called its electronic configuration. An element’s location in the (n = 6) contains 32 elements and successive electrons enter 6s, 4f, 5d and 6p orbitals, inPeriodic Table reflects the quantum numbers the order — filling up of the 4f orbitals beginsof the last orbital filled. In this section we with cerium (Z = 58) and ends at lutetiumwill observe a direct connection between the (Z = 71) to give the 4f-inner transition serieselectronic configurations of the elements and which is called the lanthanoid series. Thethe long form of the Periodic Table. seventh period (n = 7) is similar to the sixth (a) Electronic Configurations in Periods period with the successive filling up of the 7s, 5f, 6d and 7p orbitals and includes mostThe period indicates the value of n for the of the man-made radioactive elements. Thisoutermost or valence shell. In other words, period will end at the element with atomicsuccessive period in the Periodic Table is number 118 which would belong to the nobleassociated with the filling of the next higher gas family. Filling up of the 5f orbitals afterprincipal energy level (n = 1, n = 2, etc.). It can 82 chemistry actinium (Z = 89) gives the 5f-inner transition a theoretical foundation for the periodic series known as the actinoid series. The 4f- classification. The elements in a vertical column and 5f-inner transition series of elements of the Periodic Table constitute a group or are placed separately in the Periodic Table family and exhibit similar chemical behaviour. to maintain its structure and to preserve the This similarity arises because these elements principle of classification by keeping elements have the same number and same distribution with similar properties in a single column. of electrons in their outermost orbitals. We can classify the elements into four blocks viz., Problem 3.2 s-block, p-block, d-block and f-block How would you justify the presence depending on the type of atomic orbitals that of 18 elements in the 5th period of the are being filled with electrons. This is illustrated Periodic Table? in Fig. 3.3. We notice two exceptions to this Solution categorisation. Strictly, helium belongs to the s-block but its positioning in the p-block When n = 5, l = 0, 1, 2, 3. The order along with other group 18 elements is in which the energy of the available justified because it has a completely filled orbitals 4d, 5s and 5p increases is 5s < 4d < 5p. The total number of orbitals valence shell (1s2) and as a result, exhibits available are 9. The maximum number properties characteristic of other noble gases. of electrons that can be accommodated The other exception is hydrogen. It has only is 18; and therefore 18 elements are one s-electron and hence can be placed in there in the 5th period. group 1 (alkali metals). It can also gain an electron to achieve a noble gas (b) Groupwise Electronic Configurations arrangement and hence it can behave similar to a group 17 (halogen family)Elements in the same vertical column or elements. Because it is a special case, wegroup have similar valence shell electronic shall place hydrogen separately at the top ofconfigurations, the same number of electrons the Periodic Table as shown in Fig. 3.2 andin the outer orbitals, and similar properties. Fig. 3.3. We will briefly discuss the salientFor example, the Group 1 elements (alkali metals) all have ns1 valence shell electronic features of the four types of elements marked in configuration as shown below. the Periodic Table. More about these elements Atomic number Symbol Electronic configuration 3 Li 1s22s1 (or) [He]2s1 11 Na 1s22s22p63s1 (or) [Ne]3s1 19 K 1s22s22p63s23p64s1 (or) [Ar]4s1 37 Rb 1s22s22p63s23p63d104s24p65s1 (or) [Kr]5s1 55 Cs 1s22s22p63s23p63d104s24p64d105s25p66s1 (or) [Xe]6s1 87 Fr [Rn]7s1 Thus it can be seen that the properties of will be discussed later. During the description an element have periodic dependence upon of their features certain terminology has been its atomic number and not on relative atomic used which has been classified in section 3.7. mass. 3.6.1 The s-Block Elements3.6 ELECTRONIC CONFIGURATIONS A N D T Y P E S O F E L E M E N T S : The elements of Group 1 (alkali metals) and s-, p-, d-, f- BLOCKS Group 2 (alkaline earth metals) which have The aufbau (build up) principle and the ns1 and ns2 outermost electronic configuration electronic configuration of atoms provide belong to the s-Block Elements. They are all Classification of Elements and Periodicity in Properties 83 Og Ts Mc that Nh ). METALS ( orbitalsinto the on elements of METALLOIDS based and ) Tabledivision broad ( Periodicthe theis in shown NON-METALS Also), elements of filled. types being Theare( 3.3 Fig. 84 chemistry reactive metals with low ionization enthalpies. valence (oxidation states), paramagnetism and They lose the outermost electron(s) readily to oftenly used as catalysts. However, Zn, Cd and form 1+ ion (in the case of alkali metals) or 2+ Hg which have the electronic configuration, ion (in the case of alkaline earth metals). The (n-1) d10ns2 do not show most of the properties metallic character and the reactivity increase of transition elements. In a way, transition as we go down the group. Because of high metals form a bridge between the chemically reactivity they are never found pure in nature. active metals of s-block elements and the The compounds of the s-block elements, with less active elements of Groups 13 and 14 and thus take their familiar name “Transitionthe exception of those of lithium and beryllium Elements”.are predominantly ionic. 3.6.4 The f-Block Elements3.6.2 The p-Block Elements (Inner-Transition Elements) The p-Block Elements comprise those The two rows of elements at the bottom ofbelonging to Group 13 to 18 and these the Periodic Table, called the Lanthanoids,together with the s-Block Elements are Ce(Z = 58) – Lu(Z = 71) and Actinoids,called the Representative Elements or Main Th(Z = 90) – Lr (Z = 103) are characterised by Group Elements. The outermost electronic the outer electronic configuration (n-2)f1-14 configuration varies from ns2np1 to ns2np6 (n-1)d0–1ns2. The last electron added to each in each period. At the end of each period is element is filled in f- orbital. These two series a noble gas element with a closed valence of elements are hence called the Inner- shell ns2np6 configuration. All the orbitals Transition Elements (f-Block Elements). in the valence shell of the noble gases are They are all metals. Within each series, the completely filled by electrons and it is very properties of the elements are quite similar. difficult to alter this stable arrangement by The chemistry of the early actinoids is the addition or removal of electrons. The more complicated than the corresponding lanthanoids, due to the large number ofnoble gases thus exhibit very low chemical oxidation states possible for these actinoidreactivity. Preceding the noble gas family elements. Actinoid elements are radioactive.are two chemically important groups of non- Many of the actinoid elements have been mademetals. They are the halogens (Group 17) and only in nanogram quantities or even less bythe chalcogens (Group 16). These two groups nuclear reactions and their chemistry is not of elements have highly negative electron fully studied. The elements after uranium are gain enthalpies and readily add one or two called Transuranium Elements. electrons respectively to attain the stable noble gas configuration. The non-metallic Problem 3.3 character increases as we move from left to The elements Z = 117 and 120 have not yetright across a period and metallic character been discovered. In which family/group increases as we go down the group. would you place these elements and also give the electronic configuration in3.6.3 The d-Block Elements (Transition each case. Elements) SolutionThese are the elements of Group 3 to 12 in the centre of the Periodic Table. These are We see from Fig. 3.2, that element characterised by the filling of inner d orbitals with Z = 117, would belong to the halogen family (Group 17) and theby electrons and are therefore referred to as electronic configuration would be [Rn]d-Block Elements. These elements have 5f146d107s27p5. The element with Z = 120,the general outer electronic configuration will be placed in Group 2 (alkaline earth (n-1)d1-10ns0-2 except for Pd where its electronic metals), and will have the electronic configuration is 4d105s0.. They are all metals. configuration [Uuo]8s2. They mostly form coloured ions, exhibit variable Classification of Elements and Periodicity in Properties 85 3.6.5 Metals, Non-metals and Metalloids SolutionIn addition to displaying the classification Metallic character increases down aof elements into s-, p-, d-, and f-blocks, group and decreases along a period asFig. 3.3 shows another broad classification we move from left to right. Hence the of elements based on their properties. The order of increasing metallic character elements can be divided into Metals and is: P < Si < Be < Mg < Na. Non-Metals. Metals comprise more than 78% of all known elements and appear on 3.7 PERIODIC TRENDS IN PROPERTIES the left side of the Periodic Table. Metals are OF ELEMENTS usually solids at room temperature [mercury There are many observable patterns in theis an exception; gallium and caesium also physical and chemical properties of elements have very low melting points (303K and as we descend in a group or move across a 302K, respectively)]. Metals usually have high period in the Periodic Table. For example, melting and boiling points. They are good within a period, chemical reactivity tends to conductors of heat and electricity. They are be high in Group 1 metals, lower in elements malleable (can be flattened into thin sheets by towards the middle of the table, and increases hammering) and ductile (can be drawn into to a maximum in the Group 17 non-metals. wires). In contrast, non-metals are located at Likewise within a group of representative the top right hand side of the Periodic Table. metals (say alkali metals) reactivity increases In fact, in a horizontal row, the property of on moving down the group, whereas within a elements change from metallic on the left to group of non-metals (say halogens), reactivity non-metallic on the right. Non-metals are decreases down the group. But why do the usually solids or gases at room temperature properties of elements follow these trends? with low melting and boiling points (boron And how can we explain periodicity? To and carbon are exceptions). They are poor answer these questions, we must look into the conductors of heat and electricity. Most non- theories of atomic structure and properties metallic solids are brittle and are neither of the atom. In this section we shall discuss malleable nor ductile. The elements become the periodic trends in certain physical and more metallic as we go down a group; the chemical properties and try to explain them non-metallic character increases as one goes in terms of number of electrons and energy from left to right across the Periodic Table. levels. The change from metallic to non-metallic 3.7.1 Trends in Physical Propertiescharacter is not abrupt as shown by the thick There are numerous physical properties ofzig-zag line in Fig. 3.3. The elements (e.g., elements such as melting and boiling points,silicon, germanium, arsenic, antimony and heats of fusion and vaporization, energytellurium) bordering this line and running of atomization, etc. which show periodicdiagonally across the Periodic Table show variations. However, we shall discuss theproperties that are characteristic of both periodic trends with respect to atomic andmetals and non-metals. These elements are ionic radii, ionization enthalpy, electron gaincalled Semi-metals or Metalloids. enthalpy and electronegativity. Problem 3.4 (a) Atomic Radius Considering the atomic number and You can very well imagine that finding the position in the periodic table, arrange size of an atom is a lot more complicated than the following elements in the increasing measuring the radius of a ball. Do you know order of metallic character : Si, Be, Mg, why? Firstly, because the size of an atom Na, P. (~ 1.2 Å i.e., 1.2 × 10–10 m in radius) is very 86 chemistry small. Secondly, since the electron cloud The atomic radii of a few elements are listed surrounding the atom does not have a sharp in Table 3.6. Two trends are obvious. We can boundary, the determination of the atomic explain these trends in terms of nuclear charge size cannot be precise. In other words, there and energy level. The atomic size generally is no practical way by which the size of an decreases across a period as illustrated in individual atom can be measured. However, Fig. 3.4(a) for the elements of the second an estimate of the atomic size can be made by period. It is because within the period the knowing the distance between the atoms in outer electrons are in the same valence shell the combined state. One practical approach to and the effective nuclear charge increases estimate the size of an atom of a non-metallic as the atomic number increases resulting in element is to measure the distance between the increased attraction of electrons to the two atoms when they are bound together nucleus. Within a family or vertical column by a single bond in a covalent molecule and of the periodic table, the atomic radius from this value, the “Covalent Radius” of the increases regularly with atomic number as element can be calculated. For example, the illustrated in Fig. 3.4(b). For alkali metals bond distance in the chlorine molecule (Cl2) and halogens, as we descend the groups, is 198 pm and half this distance (99 pm), is the principal quantum number (n) increases taken as the atomic radius of chlorine. For and the valence electrons are farther frommetals, we define the term “Metallic Radius” the nucleus. This happens because the innerwhich is taken as half the internuclear energy levels are filled with electrons, whichdistance separating the metal cores in the serve to shield the outer electrons from themetallic crystal. For example, the distance pull of the nucleus. Consequently the size ofbetween two adjacent copper atoms in solid the atom increases as reflected in the atomiccopper is 256 pm; hence the metallic radius radii.of copper is assigned a value of 128 pm. For simplicity, in this book, we use the term Note that the atomic radii of noble gases Atomic Radius to refer to both covalent or are not considered here. Being monoatomic, metallic radius depending on whether the their (non-bonded radii) values are very element is a non-metal or a metal. Atomic large. In fact radii of noble gases should be radii can be measured by X-ray or other compared not with the covalent radii but with spectroscopic methods. the van der Waals radii of other elements. Table 3.6(a) Atomic Radii/pm Across the Periods Atom (Period II) Li Be B C N O F Atomic radius 152 111 88 77 74 66 64 Atom (Period III) Na Mg Al Si P S Cl Atomic radius 186 160 143 117 110 104 99 Table 3.6(b) Atomic Radii/pm Down a Family Atom Atomic Atom Atomic (Group I) Radius (Group 17) Radius Li 152 F 64 Na 186 Cl 99 K 231 Br 114 Rb 244 I 133 Cs 262 At 140 Classification of Elements and Periodicity in Properties 87 Fig. 3.4 (a) Variation of atomic radius with atomic Fig. 3.4 (b) Variation of atomic radius with number across the second period atomic number for alkali metals and halogens (b) Ionic Radius cation with the greater positive charge will have a smaller radius because of the greaterThe removal of an electron from an atom attraction of the electrons to the nucleus.results in the formation of a cation, whereas Anion with the greater negative charge willgain of an electron leads to an anion. The have the larger radius. In this case, the netionic radii can be estimated by measuring repulsion of the electrons will outweigh thethe distances between cations and anions nuclear charge and the ion will expand in size.in ionic crystals. In general, the ionic radii of elements exhibit the same trend as the Problem 3.5atomic radii. A cation is smaller than its parent atom because it has fewer electrons Which of the following species will have while its nuclear charge remains the same. the largest and the smallest size? Mg, Mg2+, Al, Al3+.The size of an anion will be larger than that of the parent atom because the addition of one Solution or more electrons would result in increased Atomic radii decrease across a period. repulsion among the electrons and a decrease Cations are smaller than their parent in effective nuclear charge. For example, the atoms. Among isoelectronic species, ionic radius of fluoride ion (F–) is 136 pm the one with the larger positive nuclear whereas the atomic radius of fluorine is only charge will have a smaller radius. 64 pm. On the other hand, the atomic radius Hence the largest species is Mg; the of sodium is 186 pm compared to the ionic smallest one is Al3+. radius of 95 pm for Na+. When we find some atoms and ions which (c) Ionization Enthalpy contain the same number of electrons, we call A quantitative measure of the tendency of them isoelectronic species*. For example, an element to lose electron is given by its O2–, F–, Na+ and Mg2+ have the same number Ionization Enthalpy. It represents the of electrons (10). Their radii would be different energy required to remove an electron from an because of their different nuclear charges. The isolated gaseous atom (X) in its ground state. * Two or more species with same number of atoms, same number of valence electrons and same structure, regardless of the nature of elements involved. 88 chemistry In other words, the first ionization enthalpy for an element X is the enthalpy change (∆i H) for the reaction depicted in equation 3.1. X(g) → X+(g) + e– (3.1) The ionization enthalpy is expressed in units of kJ mol–1. We can define the second ionization enthalpy as the energy required to remove the second most loosely bound electron; it is the energy required to carry out the reaction shown in equation 3.2. X+(g) → X2+(g) + e– (3.2) Energy is always required to remove Fig. 3.5 Variation of first ionization enthalpieselectrons from an atom and hence ionization (∆iH) with atomic number for elementsenthalpies are always positive. The second with Z = 1 to 60ionization enthalpy will be higher than the first ionization enthalpy because it is more can be correlated with their high reactivity. difficult to remove an electron from a positively In addition, you will notice two trends the charged ion than from a neutral atom. In the first ionization enthalpy generally increases same way the third ionization enthalpy will be as we go across a period and decreases higher than the second and so on. The term as we descend in a group. These trends “ionization enthalpy”, if not qualified, is taken are illustrated in Figs. 3.6(a) and 3.6(b) as the first ionization enthalpy. respectively for the elements of the second The first ionization enthalpies of elements period and the first group of the periodic having atomic numbers up to 60 are plotted table. You will appreciate that the ionization in Fig. 3.5. The periodicity of the graph is enthalpy and atomic radius are closely related quite striking. You will find maxima at the properties. To understand these trends, we noble gases which have closed electron shells have to consider two factors : (i) the attraction and very stable electron configurations. On of electrons towards the nucleus, and (ii) the the other hand, minima occur at the alkali repulsion of electrons from each other. The metals and their low ionization enthalpies effective nuclear charge experienced by a 3.6 (a) 3.6 (b) Fig. 3.6(a) First ionization enthalpies (∆iH) of elements of the second period as a function of atomic number (Z) and Fig. 3.6(b) ∆iH of alkali metals as a function of Z. Classification of Elements and Periodicity in Properties 89 valence electron in an atom will be less than the 2s electrons of beryllium. Therefore, it is the actual charge on the nucleus because of easier to remove the 2p-electron from boron “shielding” or “screening” of the valence compared to the removal of a 2s- electron from electron from the nucleus by the intervening beryllium. Thus, boron has a smaller first core electrons. For example, the 2s electron ionization enthalpy than beryllium. Another in lithium is shielded from the nucleus by “anomaly” is the smaller first ionization the inner core of 1s electrons. As a result, the enthalpy of oxygen compared to nitrogen. This valence electron experiences a net positive arises because in the nitrogen atom, three charge which is less than the actual charge 2p-electrons reside in different atomic orbitals of +3. In general, shielding is effective when (Hund’s rule) whereas in the oxygen atom, the orbitals in the inner shells are completely two of the four 2p-electrons must occupy the filled. This situation occurs in the case of same 2p-orbital resulting in an increased alkali metals which have single outermost electron-electron repulsion. Consequently, ns-electron preceded by a noble gas electronic it is easier to remove the fourth 2p-electron configuration. from oxygen than it is, to remove one of the When we move from lithium to fluorine three 2p-electrons from nitrogen. across the second period, successive electrons are added to orbitals in the same principal Problem 3.6 quantum level and the shielding of the nuclear The first ionization enthalpy (∆i H ) values charge by the inner core of electrons does of the third period elements, Na, Mg and not increase very much to compensate for Si are respectively 496, 737 and 786 kJ the increased attraction of the electron to the mol–1. Predict whether the first ∆i H value nucleus. Thus, across a period, increasing for Al will be more close to 575 or 760 kJ nuclear charge outweighs the shielding. mol–1 ? Justify your answer. Consequently, the outermost electrons are Solution held more and more tightly and the ionization It will be more close to 575 kJ mol–1.enthalpy increases across a period. As we go The value for Al should be lower thandown a group, the outermost electron being that of Mg because of effective shielding increasingly farther from the nucleus, there is of 3p electrons from the nucleus by an increased shielding of the nuclear charge 3s-electrons. by the electrons in the inner levels. In this case, increase in shielding outweighs the (d) Electron Gain Enthalpy increasing nuclear charge and the removal of When an electron is added to a neutralthe outermost electron requires less energy gaseous atom (X) to convert it into a negativedown a group. ion, the enthalpy change accompanying the From Fig. 3.6(a), you will also notice that process is defined as the Electron Gain the first ionization enthalpy of boron (Z = 5) Enthalpy (∆egH). Electron gain enthalpyis slightly less than that of beryllium (Z = 4) provides a measure of the ease with which even though the former has a greater nuclear an atom adds an electron to form anion as charge. When we consider the same principal represented by equation 3.3. quantum level, an s-electron is attracted to the X(g) + e– → X –(g) (3.3)nucleus more than a p-electron. In beryllium, the electron removed during the ionization is Depending on the element, the process an s-electron whereas the electron removed of adding an electron to the atom can be during ionization of boron is a p-electron. The either endothermic or exothermic. For many penetration of a 2s-electron to the nucleus is elements energy is released when an electron more than that of a 2p-electron; hence the 2p is added to the atom and the electron gain electron of boron is more shielded from the enthalpy is negative. For example, group nucleus by the inner core of electrons than 17 elements (the halogens) have very high 90 chemistry Table 3.7 Electron Gain Enthalpies* / (kJ mol–1) of Some Main Group Elements Group 1 ∆egH Group 16 ∆egH Group 17 ∆egH Group 0 ∆egH H – 73 He + 48 Li – 60 O – 141 F – 328 Ne + 116 Na – 53 S – 200 Cl – 349 Ar + 96 K – 48 Se – 195 Br – 325 Kr + 96 Rb – 47 Te – 190 I – 295 Xe + 77 Cs – 46 Po – 174 At – 270 Rn + 68 negative electron gain enthalpies because they can attain stable noble gas electronic Problem 3.7 configurations by picking up an electron. Which of the following will have the most On the other hand, noble gases have large negative electron gain enthalpy and positive electron gain enthalpies because the which the least negative? electron has to enter the next higher principal P, S, Cl, F. quantum level leading to a very unstable Explain your answer. electronic configuration. It may be noted that Solution electron gain enthalpies have large negative Electron gain enthalpy generallyvalues toward the upper right of the periodic becomes more negative across atable preceding the noble gases. period as we move from left to right. The variation in electron gain enthalpies of Within a group, electron gain enthalpy elements is less systematic than for ionization becomes less negative down a group. enthalpies. As a general rule, electron gain However, adding an electron to the enthalpy becomes more negative with increase 2p-orbital leads to greater repulsion in the atomic number across a period. The than adding an electron to the larger effective nuclear charge increases from left to 3p-orbital. Hence the element with right across a period and consequently it will most negative electron gain enthalpy is be easier to add an electron to a smaller atom chlorine; the one with the least negative since the added electron on an average would electron gain enthalpy is phosphorus. be closer to the positively charged nucleus. We should also expect electron gain enthalpy to (e) Electronegativity become less negative as we go down a group A qualitative measure of the ability of an atombecause the size of the atom increases and in a chemical compound to attract sharedthe added electron would be farther from the electrons to itself is called electronegativity.nucleus. This is generally the case (Table Unlike ionization enthalpy and electron gain3.7). However, electron gain enthalpy of O or enthalpy, it is not a measureable quantity.F is less negative than that of the succeeding However, a number of numerical scales ofelement. This is because when an electron is added to O or F, the added electron goes to electronegativity of elements viz., Pauling the smaller n = 2 quantum level and suffers scale, Mulliken-Jaffe scale, Allred-Rochow significant repulsion from the other electrons scale have been developed. The one which present in this level. For the n = 3 quantum is the most widely used is the Pauling scale. level (S or Cl), the added electron occupies Linus Pauling, an American scientist, in 1922 a larger region of space and the electron- assigned arbitrarily a value of 4.0 to fluorine, electron repulsion is much less. the element considered to have the greatest * In many books, the negative of the enthalpy change for the process depicted in equation 3.3 is defined as the ELECTRON AFFINITY (Ae ) of the atom under consideration. If energy is released when an electron is added to an atom, the electron affinity is taken as positive, contrary to thermodynamic convention. If energy has to be supplied to add an electron to an atom, then the electron affinity of the atom is assigned a negative sign. However, electron affinity is defined as absolute zero and, therefore at any other temperature (T) heat capacities of the reactants and the products have to be taken into account in ∆egH = –Ae – 5/2 RT. Classification of Elements and Periodicity in Properties 91 ability to attract electrons. Approximate On the same account electronegativity values values for the electronegativity of a few decrease with the increase in atomic radii elements are given in Table 3.8(a) down a group. The trend is similar to that of ionization enthalpy. The electronegativity of any given element Knowing the relationship betweenis not constant; it varies depending on the electronegativity and atomic radius, canelement to which it is bound. Though it is you now visualise the relationship betweennot a measurable quantity, it does provide a electronegativity and non-metallic properties?means of predicting the nature of force that Non-metallic elements have strong tendencyholds a pair of atoms together – a relationship that you will explore later. Electronegativity generally increases across a period from left to right (say from lithium to fluorine) and decrease down a group (say from fluorine to astatine) in the periodic table. How can these trends be explained? Can the electronegativity be related to atomic radii, which tend to decrease across each period from left to right, but increase down each group ? The attraction between the outer (or valence) electrons and the nucleus increases as the atomic radius decreases in a period. The electronegativity also increases. Fig. 3.7 The periodic trends of elements in the periodic table Table 3.8(a) Electronegativity Values (on Pauling scale) Across the Periods Atom (Period II) Li Be B C N O F Electronegativity 1.0 1.5 2.0 2.5 3.0 3.5 4.0 Atom (Period III) Na Mg Al Si P S Cl Electronegativity 0.9 1.2 1.5 1.8 2.1 2.5 3.0 Table 3.8(b) Electronegativity Values (on Pauling scale) Down a Family Atom Electronegativity Atom Electronegativity (Group I) Value (Group 17) Value Li 1.0 F 4.0 Na 0.9 Cl 3.0 K 0.8 Br 2.8 Rb 0.8 I 2.5 Cs 0.7 At 2.2 92 chemistry to gain electrons. Therefore, electronegativity is with outer electronic configuration 2s22p5, directly related to that non-metallic properties shares one electron with oxygen in the OF2 of elements. It can be further extended to say molecule. Being highest electronegative that the electronegativity is inversely related element, fluorine is given oxidation state to the metallic properties of elements. Thus, –1. Since there are two fluorine atoms in the increase in electronegativities across this molecule, oxygen with outer electronic a period is accompanied by an increase configuration 2s22p4 shares two electrons in non-metallic properties (or decrease in with fluorine atoms and thereby exhibits metallic properties) of elements. Similarly, the oxidation state +2. In Na2O, oxygen being decrease in electronegativity down a group is more electronegative accepts two electrons, accompanied by a decrease in non-metallic one from each of the two sodium atoms and, properties (or increase in metallic properties) thus, shows oxidation state –2. On the other of elements. hand sodium with electronic configuration All these periodic trends are summarised 3s1 loses one electron to oxygen and is given in Figure 3.7. oxidation state +1. Thus, the oxidation state of an element in a particular compound can 3.7.2 Periodic Trends in Chemical be defined as the charge acquired by its atom Properties on the basis of electronegative consideration Most of the trends in chemical properties of from other atoms in the molecule. elements, such as diagonal relationships, inert pair effect, effects of lanthanoid contraction Problem 3.8 etc. will be dealt with along the discussion Using the Periodic Table, predict the of each group in later units. In this section formulas of compounds which might we shall study the periodicity of the valence be formed by the following pairs of state shown by elements and the anomalous elements; (a) silicon and bromine properties of the second period elements (from (b) aluminium and sulphur. lithium to fluorine). Solution (a) Periodicity of Valence or Oxidation (a) Silicon is group 14 element with States a valence of 4; bromine belongs to the halogen family with a valenceThe valence is the most characteristic property of 1. Hence the formula of theof the elements and can be understood in compound formed would be SiBr4.terms of their electronic configurations. The (b) Aluminium belongs to groupvalence of representative elements is usually 13 with a valence of 3; sulphur(though not necessarily) equal to the number belongs to group 16 elements withof electrons in the outermost orbitals and/or a valence of 2. Hence, the formulaequal to eight minus the number of outermost of the compound formed would be electrons as shown below. Al2S3. Nowadays the term oxidation state is Some periodic trends observed in thefrequently used for valence. Consider the valence of elements (hydrides and oxides)two oxygen containing compounds: OF2 and are shown in Table 3.9. Other such periodicNa2O. The order of electronegativity of the trends which occur in the chemical behaviourthree elements involved in these compounds of the elements are discussed elsewhere inis F > O > Na. Each of the atoms of fluorine, Group 1 2 13 14 15 16 17 18 Number of valence 1 2 3 4 5 6 7 8 electron alence 1 2 3 4 3,5 2,6 1,7 0,8 Classification of Elements and Periodicity in Properties 93 Table 3.9 Periodic Trends in Valence of Elements as shown by the Formulas of Their Compounds Group 1 2 13 14 15 16 17 Formula of LiH CaH2 B2H6 CH4 NH3 H2O HF hydride NaH AlH3 SiH4 PH3 H2S HCl KH GeH4 AsH3 H2Se HBr SnH4 H2Te HI Formula Li2O MgO B2O3 CO2 N2O3, N2O5 – of oxide Na2O CaO Al2O3 SiO2 P4O6, P4O10 SO3 Cl2 O7 SrO K2O Ga2O3 GeO2 As2O3, As2O5 SeO3 – BaO In2O3 SnO2 Sb2O3, Sb2O5 TeO3 – PbO2 Bi2O3 – – this book. There are many elements which the second element of the following group exhibit variable valence. This is particularly i.e., magnesium and aluminium, respectively. characteristic of transition elements and This sort of similarity is commonly referred actinoids, which we shall study later. to as diagonal relationship in the periodic properties. (b) Anomalous Properties of Second Period Elements What are the reasons for the different chemical behaviour of the first member ofThe first element of each of the groups 1 a group of elements in the s- and p-blocks(lithium) and 2 (beryllium) and groups 13-17 compared to that of the subsequent members(boron to fluorine) differs in many respects in the same group? The anomalous behaviourfrom the other members of their respective is attributed to their small size, large charge/group. For example, lithium unlike other radius ratio and high electronegativity of the alkali metals, and beryllium unlike other elements. In addition, the first member of alkaline earth metals, form compounds with group has only four valence orbitals (2s and pronounced covalent character; the other 2p) available for bonding, whereas the second members of these groups predominantly member of the groups have nine valence form ionic compounds. In fact the behaviour orbitals (3s, 3p, 3d). As a consequence of of lithium and beryllium is more similar with this, the maximum covalency of the first member of each group is 4 (e.g., boron Property Element can only form  BF4 , whereas the other members of the groups can expand their Metallic radius M/pm Li Be B valence shell to accommodate more than 152 111 88 four pairs of electrons e.g., aluminium forms). Furthermore, the first Na Mg Al  AlF6 3  186 160 143 member of p-block elements displays greater ability to form pπ – pπ multiple Ionic radius M+/pm Li Be bonds to itself (e.g., C = C, C ≡ C, 76 31 N = N, N ≡ Ν) and to other second period Na Mg elements (e.g., C = O, C = N, C ≡ N, 102 72 N = O) compared to subsequent members of the same group. 94 chemistry here it can be directly related to the metallic Problem 3.9 and non-metallic character of elements. Thus, Are the oxidation state and covalency of the metallic character of an element, which Al in [AlCl(H2O)5]2+ same ? is highest at the extremely left decreases and Solution the non-metallic character increases while moving from left to right across the period. No. The oxidation state of Al is +3 and the covalency is 6. The chemical reactivity of an element can be best shown by its reactions with oxygen and 3.7.3 Periodic Trends and Chemical halogens. Here, we shall consider the reaction Reactivity of the elements with oxygen only. Elements on two extremes of a period easily combineWe have observed the periodic trends in with oxygen to form oxides. The normal oxidecertain fundamental properties such as formed by the element on extreme left is theatomic and ionic radii, ionization enthalpy, most basic (e.g., Na2O), whereas that formedelectron gain enthalpy and valence. We know by now that the periodicity is related to by the element on extreme right is the most electronic configuration. That is, all chemical acidic (e.g., Cl2O7). Oxides of elements in the and physical properties are a manifestation of centre are amphoteric (e.g., Al2O3, As2O3) or the electronic configuration of elements. We neutral (e.g., CO, NO, N2O). Amphoteric oxides shall now try to explore relationships between behave as acidic with bases and as basic with these fundamental properties of elements with acids, whereas neutral oxides have no acidic their chemical reactivity. or basic properties. The atomic and ionic radii, as we know, Problem 3.10generally decrease in a period from left to right. As a consequence, the ionization enthalpies Show by a chemical reaction with water that Na2O is a basic oxide and Cl2O7 isgenerally increase (with some exceptions as an acidic oxide.outlined in section 3.7.1(a)) and electron gain enthalpies become more negative across a Solution period. In other words, the ionization enthalpy Na2O with water forms a strong base of the extreme left element in a period is the whereas Cl2O7 forms strong acid. least and the electron gain enthalpy of the Na2O + H2O → 2NaOH element on the extreme right is the highest Cl2O7 + H2O → 2HClO4 negative (note : noble gases having completely Their basic or acidic nature can befilled shells have rather positive electron qualitatively tested with litmus paper.gain enthalpy values). This results into high chemical reactivity at the two extremes and the lowest in the centre. Thus, the maximum Among transition metals (3d series), the chemical reactivity at the extreme left (among change in atomic radii is much smaller as alkali metals) is exhibited by the loss of an compared to those of representative elements electron leading to the formation of a cation across the period. The change in atomic radii and at the extreme right (among halogens) is still smaller among inner-transition metals shown by the gain of an electron forming (4f series). The ionization enthalpies are an anion. This property can be related with intermediate between those of s- and p-blocks. the reducing and oxidizing behaviour of the As a consequence, they are less electropositive elements which you will learn later. However, than group 1 and 2 metals. Classification of Elements and Periodicity in Properties 95 In a group, the increase in atomic and increases down the group and non-metallic ionic radii with increase in atomic number character decreases. This trend can be related generally results in a gradual decrease in with their reducing and oxidizing property ionization enthalpies and a regular decrease which you will learn later. In the case of (with exception in some third period elements transition elements, however, a reverse trend as shown in section 3.7.1(d)) in electron is observed. This can be explained in terms of gain enthalpies in the case of main group atomic size and ionization enthalpy. elements. Thus, the metallic character SUMMARY In this Unit, you have studied the development of the Periodic Law and the Periodic Table. Mendeleev’s Periodic Table was based on atomic masses. Modern Periodic Table arranges the elements in the order of their atomic numbers in seven horizontal rows (periods) and eighteen vertical columns (groups or families). Atomic numbers in a period are consecutive, whereas in a group they increase in a pattern. Elements of the same group have similar valence shell electronic configuration and, therefore, exhibit similar chemical properties. However, the elements of the same period have incrementally increasing number of electrons from left to right, and, therefore, have different valencies. Four types of elements can be recognized in the periodic table on the basis of their electronic configurations. These are s-block, p-block, d-block and f-block elements. Hydrogen with one electron in the 1s orbital occupies a unique position in the periodic table. Metals comprise more than seventy eight per cent of the known elements. Non-metals, which are located at the top of the periodic table, are less than twenty in number. Elements which lie at the border line between metals and non-metals (e.g., Si, Ge, As) are called metalloids or semi-metals. Metallic character increases with increasing atomic number in a group whereas decreases from left to right in a period. The physical and chemical properties of elements vary periodically with their atomic numbers. Periodic trends are observed in atomic sizes, ionization enthalpies, electron gain enthalpies, electronegativity and valence. The atomic radii decrease while going from left to right in a period and increase with atomic number in a group. Ionization enthalpies generally increase across a period and decrease down a group. Electronegativity also shows a similar trend. Electron gain enthalpies, in general, become more negative across a period and less negative down a group. There is some periodicity in valence, for example, among representative elements, the valence is either equal to the number of electrons in the outermost orbitals or eight minus this number. Chemical reactivity is highest at the two extremes of a period and is lowest in the centre. The reactivity on the left extreme of a period is because of the ease of electron loss (or low ionization enthalpy). Highly reactive elements do not occur in nature in free state; they usually occur in the combined form. Oxides formed of the elements on the left are basic and of the elements on the right are acidic in nature. Oxides of elements in the centre are amphoteric or neutral. 96 chemistry Exercises 3.1 What is the basic theme of organisation in the periodic table? 3.2 Which important property did Mendeleev use to classify the elements in his periodic table and did he stick to that? 3.3 What is the basic difference in approach between the Mendeleev’s Periodic Law and the Modern Periodic Law? 3.4 On the basis of quantum numbers, justify that the sixth period of the periodic table should have 32 elements. 3.5 In terms of period and group where would you locate the element with Z =114? 3.6 Write the atomic number of the element present in the third period and seventeenth group of the periodic table. 3.7 Which element do you think would have been named by (i) Lawrence Berkeley Laboratory (ii) Seaborg’s group? 3.8 Why do elements in the same group have similar physical and chemical properties? 3.9 What does atomic radius and ionic radius really mean to you? 3.10 How do atomic radius vary in a period and in a group? How do you explain the variation? 3.11 What do you understand by isoelectronic species? Name a species that will be isoelectronic with each of the following atoms or ions. (i) F– (ii) Ar (iii) Mg2+ (iv) Rb+ 3.12 Consider the following species : N3–, O2–, F–, Na+, Mg2+ and Al3+ (a) What is common in them? (b) Arrange them in the order of increasing ionic radii. 3.13 Explain why cation are smaller and anions larger in radii than their parent atoms? 3.14 What is the significance of the terms — ‘isolated gaseous atom’ and ‘ground state’ while defining the ionization enthalpy and electron gain enthalpy? Hint : Requirements for comparison purposes. 3.15 Energy of an electron in the ground state of the hydrogen atom is –2.18×10–18J. Calculate the ionization enthalpy of atomic hydrogen in terms of J mol–1. Hint: Apply the idea of mole concept to derive the answer. 3.16 Among the second period elements the actual ionization enthalpies are in the order Li < B < Be < C < O < N < F < Ne. Explain why (i) Be has higher ∆i H than B (ii) O has lower ∆i H than N and F? Classification of Elements and Periodicity in Properties 97 3.17 How would you explain the fact that the first ionization enthalpy of sodium is lower than that of magnesium but its second ionization enthalpy is higher than that of magnesium? 3.18 What are the various factors due to which the ionization enthalpy of the main group elements tends to decrease down a group? 3.19 The first ionization enthalpy values (in kJ mol–1) of group 13 elements are : B Al Ga In Tl 801 577 579 558 589 How would you explain this deviation from the general trend ? 3.20 Which of the following pairs of elements would have a more negative electron gain enthalpy? (i) O or F (ii) F or Cl 3.21 Would you expect the second electron gain enthalpy of O as positive, more negative or less negative than the first? Justify your answer. 3.22 What is the basic difference between the terms electron gain enthalpy and electronegativity? 3.23 How would you react to the statement that the electronegativity of N on Pauling scale is 3.0 in all the nitrogen compounds? 3.24 Describe the theory associated with the radius of an atom as it (a) gains an electron (b) loses an electron 3.25 Would you expect the first ionization enthalpies for two isotopes of the same element to be the same or different? Justify your answer. 3.26 What are the major differences between metals and non-metals? 3.27 Use the periodic table to answer the following questions. (a) Identify an element with five electrons in the outer subshell. (b) Identify an element that would tend to lose two electrons. (c) Identify an element that would tend to gain two electrons. (d) Identify the group having metal, non-metal, liquid as well as gas at the room temperature. 3.28 The increasing order of reactivity among group 1 elements is Li < Na < K < Rb <Cs whereas that among group 17 elements is F > CI > Br > I. Explain. 3.29 Write the general outer electronic configuration of s-, p-, d- and f- block elements. 3.30 Assign the position of the element having outer electronic configuration (i) ns2np4 for n=3 (ii) (n-1)d2ns2 for n=4, and (iii) (n-2) f 7 (n-1)d1ns2 for n=6, in the periodic table. 98 chemistry 3.31 The first (∆iH1) and the second (∆iH2) ionization enthalpies (in kJ mol–1) and the (∆egH) electron gain enthalpy (in kJ mol–1) of a few elements are given below: Elements ∆H1 ∆H2 ∆egH I 520 7300 –60 II 419 3051 –48 III 1681 3374 –328 IV 1008 1846 –295 V 2372 5251 +48 VI 738 1451 –40 Which of the above elements is likely to be : (a) the least reactive element. (b) the most reactive metal. (c) the most reactive non-metal. (d) the least reactive non-metal. (e) the metal which can form a stable binary halide of the formula MX2(X=halogen). (f) the metal which can form a predominantly stable covalent halide of the formula MX (X=halogen)? 3.32 Predict the formulas of the stable binary compounds that would be formed by the combination of the following pairs of elements. (a) Lithium and oxygen (b) Magnesium and nitrogen (c) Aluminium and iodine (d) Silicon and oxygen (e) Phosphorus and fluorine (f) Element 71 and fluorine 3.33 In the modern periodic table, the period indicates the value of : (a) atomic number (b) atomic mass (c) principal quantum number (d) azimuthal quantum number. 3.34 Which of the following statements related to the modern periodic table is incorrect? (a) The p-block has 6 columns, because a maximum of 6 electrons can occupy all the orbitals in a p-shell. (b) The d-block has 8 columns, because a maximum of 8 electrons can occupy all the orbitals in a d-subshell. (c) Each block contains a number of columns equal to the number of electrons that can occupy that subshell. (d) The block indicates value of azimuthal quantum number (l) for the last subshell that received electrons in building up the electronic configuration. Classification of Elements and Periodicity in Properties 99 3.35 Anything that influences the valence electrons will affect the chemistry of the element. Which one of the following factors does not affect the valence shell? (a) Valence principal quantum number (n) (b) Nuclear charge (Z ) (c) Nuclear mass (d) Number of core electrons. 3.36 The size of isoelectronic species — F–, Ne and Na+ is affected by (a) nuclear charge (Z ) (b) valence principal quantum number (n) (c) electron-electron interaction in the outer orbitals (d) none of the factors because their size is the same. 3.37 Which one of the following statements is incorrect in relation to ionization enthalpy? (a) Ionization enthalpy increases for each successive electron. (b) The greatest increase in ionization enthalpy is experienced on removal of electron from core noble gas configuration. (c) End of valence electrons is marked by a big jump in ionization enthalpy. (d) Removal of electron from orbitals bearing lower n value is easier than from orbital having higher n value. 3.38 Considering the elements B, Al, Mg, and K, the correct order of their metallic character is : (a) B > Al > Mg > K (b) Al > Mg > B > K (c) Mg > Al > K > B (d) K > Mg > Al > B 3.39 Considering the elements B, C, N, F, and Si, the correct order of their non-metallic character is : (a) B > C > Si > N > F (b) Si > C > B > N > F (c) F > N > C > B > Si (d) F > N > C > Si > B 3.40 Considering the elements F, Cl, O and N, the correct order of their chemical reactivity in terms of oxidizing property is : (a) F > Cl > O > N (b) F > O > Cl > N (c) Cl > F > O > N (d) O > F > N > Cl Unit 4 CHEMICAL BONDING AND MOLECULAR STRUCTURE Scientists are constantly discovering new compounds, orderly arranging the facts about them, trying to explain with the existing knowledge, organising to modify the earlier views or evolve theories for explaining the newly After studying this Unit, you will be observed facts. able to • understand Kössel-Lewis approach to chemical bonding; • explain the octet rule and its Matter is made up of one or different type of elements. limitations, draw Lewis structures Under normal conditions no other element exists as an of simple molecules; independent atom in nature, except noble gases. However, • explain the formation of different a group of atoms is found to exist together as one species types of bonds; having characteristic properties. Such a group of atoms is called a molecule. Obviously there must be some force • describe the VSEPR theory and which holds these constituent atoms together in the predict the geometry of simple molecules. The attractive force which holds various molecules; constituents (atoms, ions, etc.) together in different • explain the valence bond chemical species is called a chemical bond. Since the approach for the formation of formation of chemical compounds takes place as a result of covalent bonds; combination of atoms of various elements in different ways, • predict the directional properties it raises many questions. Why do atoms combine? Why are of covalent bonds; only certain combinations possible? Why do some atoms combine while certain others do not? Why do molecules• explain the different types of hybridisation involving s, p and possess definite shapes? To answer such questions different d orbitals and draw shapes of theories and concepts have been put forward from time simple covalent molecules; to time. These are Kössel-Lewis approach, Valence Shell Electron Pair Repulsion (VSEPR) Theory, Valence Bond (VB)• describe the molecular orbital theory of homonuclear diatomic Theory and Molecular Orbital (MO) Theory. The evolution molecules; of various theories of valence and the interpretation of the nature of chemical bonds have closely been related to • explain the concept of hydrogen the developments in the understanding of the structure bond. of atom, the electronic configuration of elements and the periodic table. Every system tends to be more stable and bonding is nature’s way of lowering the energy of the system to attain stability. Reprint 2025-26 Chemical Bonding And Molecular Structure 101