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?

2.10 The conductivity of sodium chloride at 298 K has been determined at different concentrations and the results are given below: Concentration/M 0.001 0.010 0.020 0.050 0.100 102 × k/S m–1 1.237 11.85 23.15 55.53 106.74 Calculate Λm for all concentrations and draw a plot between Λm and c½. Find the value of 0m . 2.11 Conductivity of 0.00241 M acetic acid is 7.896 × 10–5 S cm–1. Calculate its molar conductivity. If 0m for acetic acid is 390.5 S cm2 mol–1, what is its dissociation constant? 2.12 How much charge is required for the following reductions: (i) 1 mol of Al3+ to Al? (ii) 1 mol of Cu2+ to Cu? (iii) 1 mol of MnO4– to Mn2+? 2.13 How much electricity in terms of Faraday is required to produce (i) 20.0 g of Ca from molten CaCl2? (ii) 40.0 g of Al from molten Al2O3? 2.14 How much electricity is required in coulomb for the oxidation of (i) 1 mol of H2O to O2? (ii) 1 mol of FeO to Fe2O3? 2.15 A solution of Ni(NO3)2 is electrolysed between platinum electrodes using a current of 5 amperes for 20 minutes. What mass of Ni is deposited at the cathode? 2.16 Three electrolytic cells A,B,C containing solutions of ZnSO4, AgNO3 and CuSO4, respectively are connected in series. A steady current of 1.5 amperes was passed through them until 1.45 g of silver deposited at the cathode of cell B. How long did the current flow? What mass of copper and zinc were deposited? 2.17 Using the standard electrode potentials given in Table 3.1, predict if the reaction between the following is feasible: (i) Fe3+(aq) and I–(aq) (ii) Ag+ (aq) and Cu(s) (iii) Fe3+ (aq) and Br– (aq) (iv) Ag(s) and Fe 3+ (aq) (v) Br2 (aq) and Fe2+ (aq). 2.18 Predict the products of electrolysis in each of the following: (i) An aqueous solution of AgNO3 with silver electrodes. (ii) An aqueous solution of AgNO3 with platinum electrodes. (iii) A dilute solution of H2SO4 with platinum electrodes. (iv) An aqueous solution of CuCl2 with platinum electrodes. Answers to Some Intext Questions 2.5 E(cell) = 0.91 V 2.6 ∆ rG o = −45.54 kJ mol −1 , Kc = 9.62 ×107 2.9 0.114, 3.67 × 10–4 mol L–1 Chemistry 60 Reprint 2025-26 UnitUnitUnit Unit33Unit Objectives ChemicalChemical KineticsKinetics After studying this Unit, you will be able to · define the average and Chemical Kinetics helps us to understand how chemical reactions instantaneous rate of a reaction; occur. · express the rate of a reaction in terms of change in concentration Chemistry, by its very nature, is concerned with change. of either of the reactants or Substances with well defined properties are converted products with time; by chemical reactions into other substances with · distinguish between elementary different properties. For any chemical reaction, chemists and complex reactions; try to find out · differentiate between the (a) the feasibility of a chemical reaction which can be molecularity and order of a reaction; predicted by thermodynamics ( as you know that a · define rate constant; reaction with DG < 0, at constant temperature and pressure is feasible);· discuss the dependence of rate of reactions on concentration, (b) extent to which a reaction will proceed can be temperature and catalyst; determined from chemical equilibrium; · derive integrated rate equations (c) speed of a reaction i.e. time taken by a reaction to for the zero and first order reach equilibrium. reactions; Along with feasibility and extent, it is equally · determine the rate constants for important to know the rate and the factors controlling zeroth and first order reactions; the rate of a chemical reaction for its complete · describe collision theory. understanding. For example, which parameters determine as to how rapidly food gets spoiled? How to design a rapidly setting material for dental filling? Or what controls the rate at which fuel burns in an auto engine? All these questions can be answered by the branch of chemistry, which deals with the study of reaction rates and their mechanisms, called chemical kinetics. The word kinetics is derived from the Greek word ‘kinesis’ meaning movement. Thermodynamics tells only about the feasibility of a reaction whereas chemical kinetics tells about the rate of a reaction. For example, thermodynamic data indicate that diamond shall convert to graphite but in reality the conversion rate is so slow that the change is not perceptible at all. Therefore, most people think Reprint 2025-26 that diamond is forever. Kinetic studies not only help us to determine the speed or rate of a chemical reaction but also describe the conditions by which the reaction rates can be altered. The factors such as concentration, temperature, pressure and catalyst affect the rate of a reaction. At the macroscopic level, we are interested in amounts reacted or formed and the rates of their consumption or formation. At the molecular level, the reaction mechanisms involving orientation and energy of molecules undergoing collisions, are discussed. In this Unit, we shall be dealing with average and instantaneous rate of reaction and the factors affecting these. Some elementary ideas about the collision theory of reaction rates are also given. However, in order to understand all these, let us first learn about the reaction rate. 3.13.13.13.13.1 RateRateRateRateRate ofofofofof aaaaa Some reactions such as ionic reactions occur very fast, for example, ChemicalChemicalChemicalChemicalChemical precipitation of silver chloride occurs instantaneously by mixing of aqueous solutions of silver nitrate and sodium chloride. On the other ReactionReactionReactionReactionReaction hand, some reactions are very slow, for example, rusting of iron in the presence of air and moisture. Also there are reactions like inversion of cane sugar and hydrolysis of starch, which proceed with a moderate speed. Can you think of more examples from each category? You must be knowing that speed of an automobile is expressed in terms of change in the position or distance covered by it in a certain period of time. Similarly, the speed of a reaction or the rate of a reaction can be defined as the change in concentration of a reactant or product in unit time. To be more specific, it can be expressed in terms of: (i) the rate of decrease in concentration of any one of the reactants, or (ii) the rate of increase in concentration of any one of the products. Consider a hypothetical reaction, assuming that the volume of the system remains constant. R ® P One mole of the reactant R produces one mole of the product P. If [R]1 and [P]1 are the concentrations of R and P respectively at time t1 and [R]2 and [P]2 are their concentrations at time t2 then, Dt = t2 – t1 D[R] = [R]2 – [R]1 D [P] = [P]2 – [P]1 The square brackets in the above expressions are used to express molar concentration. Rate of disappearance of R Decrease in concentration of R ∆ [ R ] = = − (3.1) Time taken ∆ t Chemistry 62 Reprint 2025-26 Rate of appearance of P Increase in concentration of P ∆ [ P ] = = + (3.2) Time taken ∆t Since, D[R] is a negative quantity (as concentration of reactants is decreasing), it is multiplied with –1 to make the rate of the reaction a positive quantity. Equations (3.1) and (3.2) given above represent the average rate of a reaction, rav. Average rate depends upon the change in concentration of reactants or products and the time taken for that change to occur (Fig. 3.1). { } Fig. 3.1: Instantaneous and average rate of a reaction Units of rate of a reaction From equations (3.1) and (3.2), it is clear that units of rate are concentration time–1. For example, if concentration is in mol L–1 and time is in seconds then the units will be mol L-1s–1. However, in gaseous reactions, when the concentration of gases is expressed in terms of their partial pressures, then the units of the rate equation will be atm s–1. From the concentrations of C4H9Cl (butyl chloride) at different times given ExampleExampleExampleExampleExample 3.13.13.13.13.1 below, calculate the average rate of the reaction: C4H9Cl + H2O ® C4H9OH + HCl during different intervals of time. t/s 0 50 100 150 200 300 400 700 800 [C4H9Cl]/mol L–1 0.100 0.0905 0.0820 0.0741 0.0671 0.0549 0.0439 0.0210 0.017 We can determine the difference in concentration over different intervals SolutionSolutionSolutionSolutionSolution of time and thus determine the average rate by dividing D[R] by Dt (Table 3.1). 63 Chemical Kinetics Reprint 2025-26 Table 3.1: Average rates of hydrolysis of butyl chloride [C4H9CI]t1 / [C4H9CI]t2 / t1/s t2/s rav × 104/mol L–1s–1 × 10 4 mol L–1 mol L–1 = – / ( t 2 − t1 ) C 4 H 9 Cl ]t 2 – [ C 4 H 9 Cl ]t1 {[ } 0.100 0.0905 0 50 1.90 0.0905 0.0820 50 100 1.70 0.0820 0.0741 100 150 1.58 0.0741 0.0671 150 200 1.40 0.0671 0.0549 200 300 1.22 0.0549 0.0439 300 400 1.10 0.0439 0.0335 400 500 1.04 0.0210 0.017 700 800 0.4 It can be seen (Table 3.1) that the average rate falls from 1.90 × 0-4 mol L-1s-1 to 0.4 × 10-4 mol L-1s-1. However, average rate cannot be used to predict the rate of a reaction at a particular instant as it would be constant for the time interval for which it is calculated. So, to express the rate at a particular moment of time we determine the instantaneous rate. It is obtained when we consider the average rate at the smallest time interval say dt ( i.e. when Dt approaches zero). Hence, mathematically for an infinitesimally small dt instantaneous rate is given by −∆ [ R ] ∆ [ P ] rav = = (3.3) ∆t ∆ t  d  R  d P As Dt ® 0 or rinst   d t d t Fig 3.2 Instantaneous rate of hydrolysis of butyl chloride(C4H9Cl) Chemistry 64 Reprint 2025-26 It can be determined graphically by drawing a tangent at time t on either of the curves for concentration of R and P vs time t and calculating its slope (Fig. 3.1). So in problem 3.1, rinst at 600s for example, can be calculated by plotting concentration of butyl chloride as a function of time. A tangent is drawn that touches the curve at t = 600 s (Fig. 3.2). The slope of this tangent gives the instantaneous rate. So, rinst at 600 s = – mol L–1 = 5.12 × 10–5 mol L–1s–1 At t = 250 s rinst = 1.22 × 10–4 mol L–1s–1 t = 350 s rinst = 1.0 × 10–4 mol L–1s–1 t = 450 s rinst = 6.4 ×× 10–5 mol L–1s–1 Now consider a reaction Hg(l) + Cl2 (g) ® HgCl2(s) Where stoichiometric coefficients of the reactants and products are same, then rate of the reaction is given as ∆ [ Hg ] ∆ [ Cl 2 ] ∆ [ HgCl 2 ] Rate of reaction = – = – = ∆t ∆t ∆ t i.e., rate of disappearance of any of the reactants is same as the rate of appearance of the products. But in the following reaction, two moles of HI decompose to produce one mole each of H2 and I2, 2HI(g) ® H2(g) + I2(g) For expressing the rate of such a reaction where stoichiometric coefficients of reactants or products are not equal to one, rate of disappearance of any of the reactants or the rate of appearance of products is divided by their respective stoichiometric coefficients. Since rate of consumption of HI is twice the rate of formation of H2 or I2, to make them equal, the term D[HI] is divided by 2. The rate of this reaction is given by 1 ∆ [ HI ] ∆ [ H 2 ] ∆ [ I 2 ] Rate of reaction = − = = 2 ∆ t ∆ t ∆ t Similarly, for the reaction 5 Br- (aq) + BrO3– (aq) + 6 H+ (aq) ® 3 Br2 (aq) + 3 H2O (l) − − + 1 ∆ [ Br BrO 3 1 ∆ [ H ] 1 ∆ [ Br2 ] 1 ∆ [ H 2 O ] ] ∆  Rate = − = − = − = = 5 ∆ t ∆ t 6 ∆t 3 ∆ t 3 ∆t For a gaseous reaction at constant temperature, concentration is directly proportional to the partial pressure of a species and hence, rate can also be expressed as rate of change in partial pressure of the reactant or the product. 65 Chemical Kinetics Reprint 2025-26 ExampleExampleExampleExampleExample 3.23.23.23.23.2 The decomposition of N2O5 in CCl4 at 318K has been studied by monitoring the concentration of N2O5 in the solution. Initially the concentration of N2O5 is 2.33 mol L–1 and after 184 minutes, it is reduced to 2.08 mol L–1. The reaction takes place according to the equation 2 N2O5 (g) ® 4 NO2 (g) + O2 (g) Calculate the average rate of this reaction in terms of hours, minutes and seconds. What is the rate of production of NO2 during this period? 1  ( 2.08 − 2 .33 ) mol L−1  1 ∆ [ N 2 O5 ] SolutionSolutionSolutionSolutionSolution Average Rate = − = − 184 min 2  ∆t  2   = 6.79 × 10–4 mol L–1/min = (6.79 × 10–4 mol L–1 min–1) × (60 min/1h) = 4.07 × 10–2 mol L–1/h = 6.79 × 10–4 mol L–1 × 1min/60s = 1.13 × 10–5 mol L–1s–1 It may be remembered that 1 ∆ [ NO 2 ] Rate = 4  ∆t  ∆ [ NO 2 ] = 6.79 × 10–4 × 4 mol L–1 min–1 = 2.72 × 10–3 mol L–1min–1 ∆t IntextIntextIntextIntextIntext QuestionsQuestionsQuestionsQuestionsQuestions 3.1 For the reaction R ® P, the concentration of a reactant changes from 0.03M to 0.02M in 25 minutes. Calculate the average rate of reaction using units of time both in minutes and seconds. 3.2 In a reaction, 2A ® Products, the concentration of A decreases from 0.5 mol L–1 to 0.4 mol L–1 in 10 minutes. Calculate the rate during this interval? 3.23.23.23.23.2 FactorsFactorsFactorsFactorsFactors InfluencingInfluencingInfluencingInfluencingInfluencing Rate of reaction depends upon the experimental conditions such RateRateRateRateRate ofofofofof aaaaa ReactionReactionReactionReactionReaction as concentration of reactants (pressure in case of gases), temperature and catalyst. 3.2.1 Dependence The rate of a chemical reaction at a given temperature may depend on of Rate on the concentration of one or more reactants and products. The Concentration representation of rate of reaction in terms of concentration of the reactants is known as rate law. It is also called as rate equation or rate expression. 3.2.2 Rate The results in Table 3.1 clearly show that rate of a reaction decreases with Expression the passage of time as the concentration of reactants decrease. Conversely, and Rate rates generally increase when reactant concentrations increase. So, rate of Constant a reaction depends upon the concentration of reactants. Chemistry 66 Reprint 2025-26 Consider a general reaction aA + bB ® cC + dD where a, b, c and d are the stoichiometric coefficients of reactants and products. The rate expression for this reaction is Rate µ [A] x [B] y (3.4) where exponents x and y may or may not be equal to the stoichiometric coefficients (a and b) of the reactants. Above equation can also be written as Rate = k [A] x [B] y (3.4a) d [ R ] x y − = k [ A ] [ B ] (3.4b) d t This form of equation (3.4 b) is known as differential rate equation, where k is a proportionality constant called rate constant. The equation like (3.4), which relates the rate of a reaction to concentration of reactants is called rate law or rate expression. Thus, rate law is the expression in which reaction rate is given in terms of molar concentration of reactants with each term raised to some power, which may or may not be same as the stoichiometric coefficient of the reacting species in a balanced chemical equation. For example: 2NO(g) + O2(g) ® 2NO2 (g) We can measure the rate of this reaction as a function of initial concentrations either by keeping the concentration of one of the reactants constant and changing the concentration of the other reactant or by changing the concentration of both the reactants. The following results are obtained (Table 3.2). Table 3.2: Initial rate of formation of NO2 Experiment Initial [NO]/ mol L-1 Initial [O2]/ mol L-1 Initial rate of formation of NO2/ mol L-1s-1 1. 0.30 0.30 0.096 2. 0.60 0.30 0.384 3. 0.30 0.60 0.192 4. 0.60 0.60 0.768 It is obvious, after looking at the results, that when the concentration of NO is doubled and that of O2 is kept constant then the initial rate increases by a factor of four from 0.096 to 0.384 mol L–1s–1. This indicates that the rate depends upon the square of the concentration of NO. When concentration of NO is kept constant and concentration of O2 is doubled the rate also gets doubled indicating that rate depends on concentration of O2 to the first power. Hence, the rate equation for this reaction will be Rate = k [NO] 2[O2] 67 Chemical Kinetics Reprint 2025-26 The differential form of this rate expression is given as d [ R ] 2 − = k [ NO ] [ O 2 ] d t Now, we observe that for this reaction in the rate equation derived from the experimental data, the exponents of the concentration terms are the same as their stoichiometric coefficients in the balanced chemical equation. Some other examples are given below: Reaction Experimental rate expression 1. CHCl3 + Cl2 ® CCl4 + HCl Rate = k [CHCl3 ] [Cl2]1/2 2. CH3COOC2H5 + H2O ® CH3COOH + C2H5OH Rate = k [CH3COOC2H5]1 [H2O]0 In these reactions, the exponents of the concentration terms are not the same as their stoichiometric coefficients. Thus, we can say that: Rate law for any reaction cannot be predicted by merely looking at the balanced chemical equation, i.e., theoretically but must be determined experimentally. 3.2.3 Order of a In the rate equation (3.4) Reaction Rate = k [A]x [B]y x and y indicate how sensitive the rate is to the change in concentration of A and B. Sum of these exponents, i.e., x + y in (3.4) gives the overall order of a reaction whereas x and y represent the order with respect to the reactants A and B respectively. Hence, the sum of powers of the concentration of the reactants in the rate law expression is called the order of that chemical reaction. Order of a reaction can be 0, 1, 2, 3 and even a fraction. A zero order reaction means that the rate of reaction is independent of the concentration of reactants. ExampleExampleExampleExampleExample 3.33.33.33.33.3 Calculate the overall order of a reaction which has the rate expression (a) Rate = k [A]1/2 [B]3/2 (b) Rate = k [A]3/2 [B]–1 SolutionSolutionSolutionSolutionSolution (a) Rate = k [A]x [B]y order = x + y So order = 1/2 + 3/2 = 2, i.e., second order (b) order = 3/2 + (–1) = 1/2, i.e., half order. A balanced chemical equation never gives us a true picture of how a reaction takes place since rarely a reaction gets completed in one step. The reactions taking place in one step are called elementary reactions. When a sequence of elementary reactions (called mechanism) gives us the products, the reactions are called complex reactions. Chemistry 68 Reprint 2025-26 These may be consecutive reactions (e.g., oxidation of ethane to CO2 and H2O passes through a series of intermediate steps in which alcohol, aldehyde and acid are formed), reverse reactions and side reactions (e.g., nitration of phenol yields o-nitrophenol and p-nitrophenol). Units of rate constant For a general reaction aA + bB ® cC + dD Rate = k [A]x [B]y Where x + y = n = order of the reaction Rate k = x [A] [B]y concentration 1 = × n ( where [A] = [B]) time ( concentration ) Taking SI units of concentration, mol L –1 and time, s, the units of k for different reaction order are listed in Table 3.3 Table 3.3: Units of rate constant Reaction Order Units of rate constant mol L−1 1 −1 − 1 × 0 = mol L s −1 Zero order reaction 0 s ( mol L ) −1 mol L 1 −1 × = s − 1 1 First order reaction 1 s ( mol L ) − 1 mol L 1 − 1 −1 × = mol L s −1 2 Second order reaction 2 s ( mol L ) Identify the reaction order from each of the following rate constants. ExampleExampleExampleExampleExample 3.43.43.43.43.4 (i) k = 2.3 × 10–5 L mol–1 s–1 (ii) k = 3 × 10–4 s–1 (i) The unit of second order rate constant is L mol–1 s–1, therefore SolutionSolutionSolutionSolutionSolution k = 2.3 × 10–5 L mol–1 s–1 represents a second order reaction. (ii) The unit of a first order rate constant is s–1 therefore k = 3 × 10–4 s–1 represents a first order reaction. 3.2.4 Molecularity Another property of a reaction called molecularity helps in of a understanding its mechanism. The number of reacting species Reaction (atoms, ions or molecules) taking part in an elementary reaction, which must collide simultaneously in order to bring about a chemical reaction is called molecularity of a reaction. The reaction can be unimolecular when one reacting species is involved, for example, decomposition of ammonium nitrite. 69 Chemical Kinetics Reprint 2025-26 NH4NO2 ® N2 + 2H2O Bimolecular reactions involve simultaneous collision between two species, for example, dissociation of hydrogen iodide. 2HI ® H2 + I2 Trimolecular or termolecular reactions involve simultaneous collision between three reacting species, for example, 2NO + O2 ® 2NO2 The probability that more than three molecules can collide and react simultaneously is very small. Hence, reactions with the molecularity three are very rare and slow to proceed. It is, therefore, evident that complex reactions involving more than three molecules in the stoichiometric equation must take place in more than one step. KClO3 + 6FeSO4 + 3H2SO4 ® KCl + 3Fe2(SO4)3 + 3H2O This reaction which apparently seems to be of tenth order is actually a second order reaction. This shows that this reaction takes place in several steps. Which step controls the rate of the overall reaction? The question can be answered if we go through the mechanism of reaction, for example, chances to win the relay race competition by a team depend upon the slowest person in the team. Similarly, the overall rate of the reaction is controlled by the slowest step in a reaction called the rate determining step. Consider the decomposition of hydrogen peroxide which is catalysed by iodide ion in an alkaline medium. -I 2H2O2  2H2O + O2 Alkaline medium The rate equation for this reaction is found to be d  H 2 O 2   Rate  k  H2 O 2 I dt This reaction is first order with respect to both H2O2 and I–. Evidences suggest that this reaction takes place in two steps (1) H2O2 + I– ® H2O + IO– (2) H2O2 + IO– ® H2O + I– + O2 Both the steps are bimolecular elementary reactions. Species IO- is called as an intermediate since it is formed during the course of the reaction but not in the overall balanced equation. The first step, being slow, is the rate determining step. Thus, the rate of formation of intermediate will determine the rate of this reaction. Thus, from the discussion, till now, we conclude the following: (i) Order of a reaction is an experimental quantity. It can be zero and even a fraction but molecularity cannot be zero or a non integer. (ii) Order is applicable to elementary as well as complex reactions whereas molecularity is applicable only for elementary reactions. For complex reaction molecularity has no meaning. Chemistry 70 Reprint 2025-26 (iii) For complex reaction, order is given by the slowest step and molecularity of the slowest step is same as the order of the overall reaction. IntextIntextIntextIntextIntext QuestionsQuestionsQuestionsQuestionsQuestions 3.3 For a reaction, A + B ® Product; the rate law is given by, r = k [ A]1/2 [B]2. What is the order of the reaction? 3.4 The conversion of molecules X to Y follows second order kinetics. If concentration of X is increased to three times how will it affect the rate of formation of Y ? 3.33.33.33.33.3 IntegratedIntegratedIntegratedIntegratedIntegrated We have already noted that the concentration dependence of rate is RateRateRateRateRate called differential rate equation. It is not always convenient to determine the instantaneous rate, as it is measured by determination EquationsEquationsEquationsEquationsEquations of slope of the tangent at point ‘t’ in concentration vs time plot (Fig. 3.1). This makes it difficult to determine the rate law and hence the order of the reaction. In order to avoid this difficulty, we can integrate the differential rate equation to give a relation between directly measured experimental data, i.e., concentrations at different times and rate constant. The integrated rate equations are different for the reactions of different reaction orders. We shall determine these equations only for zero and first order chemical reactions. 3.3.1 Zero Order Zero order reaction means that the rate of the reaction is proportional Reactions to zero power of the concentration of reactants. Consider the reaction, R ® P d  R  Rate =   k  R 0 d t As any quantity raised to power zero is unity d  R  Rate =   k × 1 d t d[R] = – k dt Integrating both sides [R] = – k t + I (3.5) where, I is the constant of integration. At t = 0, the concentration of the reactant R = [R]0, where [R]0 is initial concentration of the reactant. Substituting in equation (3.5) [R]0 = –k × 0 + I [R]0 = I Substituting the value of I in the equation (3.5) [R] = -kt + [R]0 (3.6) 71 Chemical Kinetics Reprint 2025-26 Comparing (3.6) with equation of a straight line, y = mx + c, if we plot [R] against t, we get a straight [R0 ] line (Fig. 3.3) with slope = –k and intercept equal to [R]0. Further simplifying equation (3.6), we get the rateR k = -slope constant, k as of [ R ]0 − [ R ] k = (3.7) t Zero order reactions are relatively uncommon but they occur under special conditions. Some enzyme Concentration catalysed reactions and reactions which occur on metal surfaces are a few examples of zero order 0 Time reactions. The decomposition of gaseous ammonia on a hot platinum surface is a zero order reaction at Fig. 3.3: Variation in the concentration high pressure. vs time plot for a zero order 1130K reaction 2NH 3 ( g ) →Pt catalyst N 2 ( g ) +3H 2 ( g ) Rate = k [NH3]0 = k In this reaction, platinum metal acts as a catalyst. At high pressure, the metal surface gets saturated with gas molecules. So, a further change in reaction conditions is unable to alter the amount of ammonia on the surface of the catalyst making rate of the reaction independent of its concentration. The thermal decomposition of HI on gold surface is another example of zero order reaction. 3.3.2 First Order In this class of reactions, the rate of the reaction is proportional to the Reactions first power of the concentration of the reactant R. For example, R ® P d [ R ] Rate = − = k [ R ] d t d [ R ] or = – kdt [ R ] Integrating this equation, we get ln [R] = – kt + I (3.8) Again, I is the constant of integration and its value can be determined easily. When t = 0, R = [R]0, where [R]0 is the initial concentration of the reactant. Therefore, equation (3.8) can be written as ln [R]0 = –k × 0 + I ln [R]0 = I Substituting the value of I in equation (3.8) ln[R] = –kt + ln[R]0 (3.9) Chemistry 72 Reprint 2025-26 Rearranging this equation [ R ] ln = −kt [ R ]0 1 R 0 or k  ln (3.10) t R At time t1 from equation (3.8) *ln[R]1 = – kt1 + *ln[R]0 (3.11) At time t2 ln[R]2 = – kt2 + ln[R]0 (3.12) where, [R]1 and [R]2 are the concentrations of the reactants at time t1 and t2 respectively. Subtracting (3.12) from (3.11) ln[R]1– ln[R]2 = – kt1 – (–kt2) [ R ]1 ln = k (t 2 − t 1 ) [ R ]2 1 [ R ]1 k = ln (t 2 − t 1 ) [ R ]2 (3.13) Equation (3.9) can also be written as [ R ] ln = −kt [ R ]0 Taking antilog of both sides [R] = [R]0 e–kt (3.14) Comparing equation (3.9) with y = mx + c, if we plot ln [R] against t (Fig. 3.4) we get a straight line with slope = –k and intercept equal to ln [R]0 The first order rate equation (3.10) can also be written in the form 2.303 [ R ]0 k = log (3.15) t [ R ] [ R ]0 kt * log = [ R ] 2.303 If we plot a graph between log [R]0/[R] vs t, (Fig. 3.5), the slope = k/2.303 Hydrogenation of ethene is an example of first order reaction. C2H4(g) + H2 (g) ® C2H6(g) Rate = k [C2H4] All natural and artificial radioactive decay of unstable nuclei take place by first order kinetics. * Refer to Appendix-IV for ln and log (logarithms). 73 Chemical Kinetics Reprint 2025-26 /[R]) 0] Slope = k /2.303 ([R log 0 Time Fig. 3.4: A plot between ln[R] and t Fig. 3.5: Plot of log [R]0/[R] vs time for a for a first order reaction first order reaction 226 88 Ra  24 He  22286 Rn Rate = k [Ra] Decomposition of N2O5 and N2O are some more examples of first order reactions. ExampleExampleExampleExampleExample 3.53.53.53.53.5 The initial concentration of N2O5 in the following first order reaction N2O5(g) ® 2 NO2(g) + 1/2O2 (g) was 1.24 × 10–2 mol L–1 at 318 K. The concentration of N2O5 after 60 minutes was 0.20 × 10–2 mol L–1. Calculate the rate constant of the reaction at 318 K. SolutionSolutionSolutionSolutionSolution For a first order reaction  R 1 k t 2  t 1  log =  R 2 2.303 2.303  R 1 log k =  t 2  t 1   R 2 2.303 1.24  10  2 mol L1 log  2  1 =  60 min  0 min  0.20  10 mol L 2.303  1 = log 6.2 min 60 k = 0.0304 min-1 Let us consider a typical first order gas phase reaction A(g) ® B(g) + C(g) Let pi be the initial pressure of A and pt the total pressure at time ‘t’. Integrated rate equation for such a reaction can be derived as Total pressure pt = pA + pB + pC (pressure units) Chemistry 74 Reprint 2025-26 pA, pB and pC are the partial pressures of A, B and C, respectively. If x atm be the decrease in pressure of A at time t and one mole each of B and C is being formed, the increase in pressure of B and C will also be x atm each. A(g) ® B(g) + C(g) At t = 0 pi atm 0 atm 0 atm At time t (pi–x) atm x atm x atm where, pi is the initial pressure at time t = 0. pt = (pi – x) + x + x = pi + x x = (pt - pi) where, pA = pi – x = pi – (pt – pi) = 2pi – pt  2.303   p i  k =   log  (3.16)  t  p A  2.303 p i log = t  2 p i  p t  The following data were obtained during the first order thermal ExampleExampleExampleExampleExample 3.63.63.63.63.6 decomposition of N2O5 (g) at constant volume: 2N 2 O5 ( g ) → 2N 2 O 4 ( g ) + O 2 ( g ) S.No. Time/s Total Pressure/(atm) 1. 0 0.5 2. 100 0.512 Calculate the rate constant. Let the pressure of N2O5(g) decrease by 2x atm. As two moles of SolutionSolutionSolutionSolutionSolution N2O5 decompose to give two moles of N2O4(g) and one mole of O2 (g), the pressure of N2O4 (g) increases by 2x atm and that of O2 (g) increases by x atm. 2N 2 O5 ( g ) → 2N 2 O 4 ( g ) + O 2 ( g ) Start t = 0 0.5 atm 0 atm 0 atm At time t (0.5 – 2x) atm 2x atm x atm pt = p N 2 O 5  p N 2 O 4  p O 2 = (0.5 – 2x) + 2x + x = 0.5 + x x = tp − 0.5 p N 2 O5 = 0.5 – 2x = 0.5 – 2 (pt – 0.5) = 1.5 – 2pt At t = 100 s; pt = 0.512 atm 75 Chemical Kinetics Reprint 2025-26 p N 2 O 5 = 1.5 – 2 × 0.512 = 0.476 atm Using equation (3.16) 2.303 p i 2.303 0.5 atm k  log  log t p A 100s 0.476 atm 2.303  4 1   0.0216  4.98  10 s 100s 3.3.3 Half-Life of The half-life of a reaction is the time in which the concentration of a a Reaction reactant is reduced to one half of its initial concentration. It is represented as t1/2. For a zero order reaction, rate constant is given by equation 3.7. [ R ]0 − [ R ] k = t 1 [ R ]0 At t = t 1/2 , [ R ] = 2 The rate constant at t1/2 becomes [ R ]0 − 1/2 [ R ]0 k = t 1/2 [ R ]0 t 1/2 = 2k It is clear that t1/2 for a zero order reaction is directly proportional to the initial concentration of the reactants and inversely proportional to the rate constant. For the first order reaction, 2.303 [ R ]0 k = log (3.15) t [ R ] [ R ]0 at t1/2 [ R ] = (3.16) 2 So, the above equation becomes 2.303 [ R ]0 k = log t 1/2 [ R ] /2 2.303 or t1/2  log 2 k 2.303 t 1/2 = × 0.301 k 0.693 t 1/2 = (3.17) k Chemistry 76 Reprint 2025-26 It can be seen that for a first order reaction, half-life period is constant, i.e., it is independent of initial concentration of the reacting species. The half-life of a first order equation is readily calculated from the rate constant and vice versa. For zero order reaction t1/2 µ [R]0. For first order reaction t1/2 is independent of [R]0. A first order reaction is found to have a rate constant, k = 5.5 × 10-14 s-1. ExampleExampleExampleExampleExample 3.73.73.73.73.7 Find the half-life of the reaction. Half-life for a first order reaction is SolutionSolutionSolutionSolutionSolution 0.693 t 1/2 = k 0.693 t 1/2 = –14 –1 = 1.26 × 1013s 5.5×10 s Show that in a first order reaction, time required for completion of 99.9% is 10 times of half-life (t1/2) of the reaction. When reaction is completed 99.9%, [R]n = [R]0 – 0.999[R]0 ExampleExampleExampleExampleExample 3.83.83.83.83.8 2.303  R 0 log k = SolutionSolutionSolutionSolutionSolution t  R  2.303  R 0 2.303 3 log = = log10 t  R 0  0.999  R 0 t t = 6.909/k For half-life of the reaction t1/2 = 0.693/k t 6.909 k =   10 t1/2 k 0.693 Table 3.4 summarises the mathematical features of integrated laws of zero and first order reactions. Table 3.4: Integrated Rate Laws for the Reactions of Zero and First Order Order Reaction Differential Integrated Straight Half- Units of k type rate law rate law line plot life 0 R® P d[R]/dt = -k kt = [R]0-[R] [R] vs t [R]0/2k conc time-1 or mol L–1s–1 1 R® P d[R]/dt = -k[R] [R] = [R]0e-kt ln[R] vs t ln 2/k time-1 or s–1 or kt = ln{[R]0/[R]} 77 Chemical Kinetics Reprint 2025-26 The order of a reaction is sometimes altered by conditions. There are many reactions which obey first order rate law although they are higher order reactions. Consider the hydrolysis of ethyl acetate which is a chemical reaction between ethyl acetate and water. In reality, it is a second order reaction and concentration of both ethyl acetate and water affect the rate of the reaction. But water is taken in large excess for hydrolysis, therefore, concentration of water is not altered much during the reaction. Thus, the rate of reaction is affected by concentration of ethyl acetate only. For example, during the hydrolysis of 0.01 mol of ethyl acetate with 10 mol of water, amounts of the reactants and products at the beginning (t = 0) and completion (t) of the reaction are given as under. H CH3COOH + C2H5OH CH3COOC2H5 + H2O  t = 0 0.01 mol 10 mol 0 mol 0 mol t 0 mol 9.99 mol 0.01 mol 0.01 mol The concentration of water does not get altered much during the course of the reaction. So, the reaction behaves as first order reaction. Such reactions are called pseudo first order reactions. Inversion of cane sugar is another pseudo first order reaction. C12H22O11 + H2O →H+ C6H12O6 + C6H12O6 Cane sugar Glucose Fructose Rate = k [C12H22O11] IntextIntextIntextIntextIntext QuestionsQuestionsQuestionsQuestionsQuestions 3.5 A first order reaction has a rate constant 1.15 × 10-3 s-1. How long will 5 g of this reactant take to reduce to 3 g? 3.6 Time required to decompose SO2Cl2 to half of its initial amount is 60 minutes. If the decomposition is a first order reaction, calculate the rate constant of the reaction. 3.43.43.43.43.4 TemperatureTemperatureTemperatureTemperatureTemperature Most of the chemical reactions are accelerated by increase in temperature. For example, in decomposition of N2O5, the time taken for half of the DependenceDependenceDependenceDependenceDependence ofofofofof original amount of material to decompose is 12 min at 50oC, 5 h at thethethethethe RateRateRateRateRate ofofofofof aaaaa 25oC and 10 days at 0oC. You also know that in a mixture of potassium ReactionReactionReactionReactionReaction permanganate (KMnO4) and oxalic acid (H2C2O4), potassium permanganate gets decolourised faster at a higher temperature than that at a lower temperature. It has been found that for a chemical reaction with rise in temperature by 10°, the rate constant is nearly doubled. The temperature dependence of the rate of a chemical reaction can be accurately explained by Arrhenius equation (3.18). It was first proposed by Dutch chemist, J.H. van’t Hoff but Swedish chemist, Arrhenius provided its physical justification and interpretation. Chemistry 78 Reprint 2025-26 k = A e -Ea /RT (3.18) where A is the Arrhenius factor or the frequency factor. It is also called pre-exponential factor. It is a constant specific to a particular reaction. R is gas constant and Ea is activation energy measured in joules/mole (J mol –1). It can be understood clearly using the following simple reaction H 2 g  I 2 g  2HI g According to Arrhenius, this reaction can take place only when a molecule of hydrogen and a molecule of iodine Intermediate collide to form an unstable intermediate (Fig. 3.6). It exists for a very short time and then breaks up to form two Fig. 3.6: Formation of HI through molecules of hydrogen iodide. the intermediate The energy required to form this intermediate, called activated complex (C), is known as activation energy (Ea). Fig. 3.7 is obtained by plotting potential energy vs reaction coordinate. Reaction coordinate represents the profile of energy change when reactants change into products. Some energy is released when the complex decomposes to form products. So, the final enthalpy of the reaction depends upon the nature of reactants and products. All the molecules in the reacting species do not have the same kinetic Fig. 3.7: Diagram showing plot of potential energy. Since it is difficult to predict the energy vs reaction coordinate behaviour of any one molecule with precision, Ludwig Boltzmann and James Clark Maxwell used statistics to predict the behaviour of large number of molecules. According to them, the distribution of kinetic energy may be described by plotting the fraction of molecules (NE/NT) with a given kinetic energy (E) vs kinetic energy (Fig. 3.8). Here, NE is the number of molecules with energy E and NT is total number of molecules. The peak of the curve corresponds to the most probable kinetic energy, i.e., kinetic energy of maximum fraction of molecules. There are decreasing number Fig. 3.8: Distribution curve showing energies of molecules with energies higher or among gaseous molecules lower than this value. When the 79 Chemical Kinetics Reprint 2025-26 temperature is raised, the maximum of the curve moves to the higher energy value (Fig. 3.9) and the curve broadens out, i.e., spreads to the right such that there is a greater proportion of molecules with much higher energies. The area under the curve must be constant since total probability must be one at all times. We can mark the position of Ea on Fig. 3.9: Distribution curve showing temperature Maxwell Boltzmann distribution curve dependence of rate of a reaction (Fig. 3.9). Increasing the temperature of the substance increases the fraction of molecules, which collide with energies greater than Ea. It is clear from the diagram that in the curve at (t + 10), the area showing the fraction of molecules having energy equal to or greater than activation energy gets doubled leading to doubling the rate of a reaction. In the Arrhenius equation (3.18) the factor e -Ea /RT corresponds to the fraction of molecules that have kinetic energy greater than Ea. Taking natural logarithm of both sides of equation (3.18) E a ln k = – + ln A (3.19) RT The plot of ln k vs 1/T gives a straight line according to the equation (3.19) as shown in Fig. 3.10. Thus, it has been found from Arrhenius equation (3.18) that increasing the temperature or decreasing the activation energy will result in an increase in the rate of the reaction and an exponential increase in the rate constant. E a In Fig. 3.10, slope = – and intercept = ln R A. So we can calculate Ea and A using these values. At temperature T1, equation (3.19) is E a ln k1 = – RT1 + ln A (3.20) At temperature T2, equation (3.19) is E a ln k2 = – RT2 + ln A (3.21) (since A is constant for a given reaction) k1 and k2 are the values of rate constants at temperatures T1 and T2 respectively. Fig. 3.10: A plot between ln k and 1/T Chemistry 80 Reprint 2025-26 Subtracting equation (3.20) from (3.21), we obtain E a E a ln k2 – ln k1 = RT1 – RT2 k 2 E a  1 1  ln = − k1 R  T1 T2  k 2 E a  1 1  log = − (3.22) k1 2.303 R  T1 T2  k 2 E a  T2 − T1  log = k1 2. 303R  T1T2  ExampleExampleExampleExampleExample 3.93.93.93.93.9 The rate constants of a reaction at 500K and 700K are 0.02s–1 and 0.07s–1 respectively. Calculate the values of Ea and A. k 2 E a  T2  T1 SolutionSolutionSolutionSolutionSolution log =   k1 2.303 R  T1T2  0.07  E a   700  500  log =      700  500 0.02  2.303  8.314 JK  1 mol  1  0.544 = Ea × 5.714 × 10-4/19.15 Ea = 0.544 × 19.15/5.714 × 10–4 = 18230.8 J Since k = Ae-Ea/RT × 500 0.02 = Ae-18230.8/8.314 A = 0.02/0.012 = 1.61 ExampleExampleExampleExampleExample 3.103.103.103.103.10 The first order rate constant for the decomposition of ethyl iodide by the reaction C2H5I(g) ® C2H4 (g) + HI(g) at 600K is 1.60 × 10–5 s–1. Its energy of activation is 209 kJ/mol. Calculate the rate constant of the reaction at 700K. SolutionSolutionSolutionSolutionSolution We know that E a  1 1     log k2 – log k1 = 2.303R  T1 T2  81 Chemical Kinetics Reprint 2025-26 E a  1 1  log k2 = log k1     2.303R  T1 T2  209000 J mol L 1  1 1  5 = log 1.60  10     1  1    2.303  8.314 J mol L K  600 K 700K  log k2 = – 4.796 + 2.599 = – 2.197 k2 = 6.36 × 10–3 s–1 3.4.1 Effect of A catalyst is a substance which increases the rate of a reaction without Catalyst itself undergoing any permanent chemical change. For example, MnO2 catalyses the following reaction so as to increase its rate considerably. 2KClO3 MnO2 2 KCl + 3O2 The word catalyst should not be used when the added substance reduces the rate of raction. The substance is then called inhibitor. The action of the catalyst can be explained by intermediate complex theory. According to this theory, a catalyst participates in a chemical reaction by forming temporary bonds with the reactants resulting in an intermediate complex. This has a transitory existence and decomposes to yield products and the catalyst. It is believed that the catalyst provides an alternate pathway or reaction mechanism by reducing the activation energy between reactants and products and hence lowering the potential energy barrier as shown in Fig. 3.11. It is clear from Arrhenius equation (3.18) that lower the value of activation energy faster will be the rate of a reaction. A small amount of the catalyst can catalyse a large amount of reactants. A catalyst does Fig. 3.11: Effect of catalyst on activation energy not alter Gibbs energy, DG of a reaction. It catalyses the spontaneous reactions but does not catalyse non-spontaneous reactions. It is also found that a catalyst does not change the equilibrium constant of a reaction rather, it helps in attaining the equilibrium faster, that is, it catalyses the forward as well as the backward reactions to the same extent so that the equilibrium state remains same but is reached earlier. 3.53.53.53.53.5 CollisionCollisionCollisionCollisionCollision Though Arrhenius equation is applicable under a wide range of TheoryTheoryTheoryTheoryTheory ofofofofof circumstances, collision theory, which was developed by Max Trautz ChemicalChemicalChemicalChemicalChemical and William Lewis in 1916 -18, provides a greater insight into the energetic and mechanistic aspects of reactions. It is based on kinetic ReactionsReactionsReactionsReactionsReactions theory of gases. According to this theory, the reactant molecules are Chemistry 82 Reprint 2025-26 assumed to be hard spheres and reaction is postulated to occur when molecules collide with each other. The number of collisions per second per unit volume of the reaction mixture is known as collision frequency (Z). Another factor which affects the rate of chemical reactions is activation energy (as we have already studied). For a bimolecular elementary reaction A + B ® Products rate of reaction can be expressed as a / RT (3.23) Rate = Z AB e − E where ZAB represents the collision frequency of reactants, A and B and e-Ea /RT represents the fraction of molecules with energies equal to or greater than Ea. Comparing (3.23) with Arrhenius equation, we can say that A is related to collision frequency. Equation (3.23) predicts the value of rate constants fairly accurately for the reactions that involve atomic species or simple molecules but for complex molecules significant deviations are observed. The reason could be that all collisions do not lead to the formation of products. The collisions in which molecules collide with sufficient kinetic energy (called threshold energy*) and proper orientation, so as to facilitate breaking of bonds between reacting species and formation of new bonds to form products are called as effective collisions. For example, formation of methanol from bromoethane depends upon the orientation of reactant molecules as shown in Fig. 3.12. The proper orientation of reactant molecules lead to bond formation whereas improper orientation makes them simply bounce back and no products are formed. Fig. 3.12: Diagram showing molecules having proper and To account for effective collisions, improper orientation another factor P, called the probability or steric factor is introduced. It takes into account the fact that in a collision, molecules must be properly oriented i.e., − E a / RT Rate = PZ AB e Thus, in collision theory activation energy and proper orientation of the molecules together determine the criteria for an effective collision and hence the rate of a chemical reaction. Collision theory also has certain drawbacks as it considers atoms/ molecules to be hard spheres and ignores their structural aspect. You will study details about this theory and more on other theories in your higher classes. * Threshold energy = Activation Energy + energy possessed by reacting species. 83 Chemical Kinetics Reprint 2025-26 IntextIntextIntextIntextIntext QuestionsQuestionsQuestionsQuestionsQuestions 3.7 What will be the effect of temperature on rate constant ? 3.8 The rate of the chemical reaction doubles for an increase of 10K in absolute temperature from 298K. Calculate Ea. 3.9 The activation energy for the reaction 2 HI(g) ® H2 + I2 (g) is 209.5 kJ mol–1 at 581K.Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy? SummarySummarySummarySummarySummary Chemical kinetics is the study of chemical reactions with respect to reaction rates, effect of various variables, rearrangement of atoms and formation of intermediates. The rate of a reaction is concerned with decrease in concentration of reactants or increase in the concentration of products per unit time. It can be expressed as instantaneous rate at a particular instant of time and average rate over a large interval of time. A number of factors such as temperature, concentration of reactants, catalyst, affect the rate of a reaction. Mathematical representation of rate of a reaction is given by rate law. It has to be determined experimentally and cannot be predicted. Order of a reaction with respect to a reactant is the power of its concentration which appears in the rate law equation. The order of a reaction is the sum of all such powers of concentration of terms for different reactants. Rate constant is the proportionality factor in the rate law. Rate constant and order of a reaction can be determined from rate law or its integrated rate equation. Molecularity is defined only for an elementary reaction. Its values are limited from 1 to 3 whereas order can be 0, 1, 2, 3 or even a fraction. Molecularity and order of an elementary reaction are same. Temperature dependence of rate constants is described by Arrhenius equation (k = Ae–Ea/RT). Ea corresponds to the activation energy and is given by the energy difference between activated complex and the reactant molecules, and A (Arrhenius factor or pre-exponential factor) corresponds to the collision frequency. The equation clearly shows that increase of temperature or lowering of Ea will lead to an increase in the rate of reaction and presence of a catalyst lowers the activation energy by providing an alternate path for the reaction. According to collision theory, another factor P called steric factor which refers to the orientation of molecules which collide, is important and contributes to effective collisions, thus, modifying the Arrhenius equation to k  P Z AB e  E a / RT . Chemistry 84 Reprint 2025-26 ExercisesExercisesExercisesExercisesExercises 3.1 From the rate expression for the following reactions, determine their order of reaction and the dimensions of the rate constants. (i) 3NO(g) ® N2O (g) Rate = k[NO]2 (ii) H2O2 (aq) + 3I– (aq) + 2H+ ® 2H2O (l) + 3I Rate = k[H2O2][I-] (iii) CH3CHO (g) ® CH4 (g) + CO(g) Rate = k [CH3CHO]3/2 (iv) C2H5Cl (g) ® C2H4 (g) + HCl (g) Rate = k [C2H5Cl] 3.2 For the reaction: 2A + B ® A2B the rate = k[A][B]2 with k = 2.0 × 10–6 mol–2 L2 s–1. Calculate the initial rate of the reaction when [A] = 0.1 mol L–1, [B] = 0.2 mol L–1. Calculate the rate of reaction after [A] is reduced to 0.06 mol L–1. 3.3 The decomposition of NH3 on platinum surface is zero order reaction. What are the rates of production of N2 and H2 if k = 2.5 × 10–4 mol–1 L s–1? 3.4 The decomposition of dimethyl ether leads to the formation of CH4, H2 and CO and the reaction rate is given by Rate = k [CH3OCH3]3/2 The rate of reaction is followed by increase in pressure in a closed vessel, so the rate can also be expressed in terms of the partial pressure of dimethyl ether, i.e., 3/2 Rate = k ( p CH 3 OCH 3 ) If the pressure is measured in bar and time in minutes, then what are the units of rate and rate constants? 3.5 Mention the factors that affect the rate of a chemical reaction. 3.6 A reaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is (i) doubled (ii) reduced to half ? 3.7 What is the effect of temperature on the rate constant of a reaction? How can this effect of temperature on rate constant be represented quantitatively? 3.8 In a pseudo first order reaction in water, the following results were obtained: t/s 0 30 60 90 [A]/ mol L–1 0.55 0.31 0.17 0.085 Calculate the average rate of reaction between the time interval 30 to 60 seconds.