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

4.10 The 5f electrons are more effectively shielded from nuclear charge. In other words the 5f electrons themselves provide poor shielding from element to element in the series. 117 The d- and f- Block Elements Reprint 2025-26 UnitUnitUnitUnit Unit55 CoordinationCoordinationObjectives After studying this Unit, you will beable to CompoundsCompounds • appreciate the postulates of Werner’s theory of coordination compounds; Coordination Compounds are the backbone of modern inorganic • know the meaning of the terms: and bio–inorganic chemistry and chemical industry. coordination entity, central atom/ ion, ligand, coordination number, coordination sphere, coordination In the previous Unit we learnt that the transition metals polyhedron, oxidation number, form a large number of complex compounds in which homoleptic and heteroleptic; the metal atoms are bound to a number of anions or • learn the rules of nomenclature neutral molecules by sharing of electrons. In modern of coordination compounds; terminology such compounds are called coordination • write the formulas and names compounds. The chemistry of coordination compounds of mononuclear coordination is an important and challenging area of modern compounds; inorganic chemistry. New concepts of chemical bonding • define different types of isomerism and molecular structure have provided insights into in coordination compounds; the functioning of these compounds as vital components • understand the nature of bonding of biological systems. Chlorophyll, haemoglobin and in coordination compounds in vitamin B12 are coordination compounds of magnesium, terms of the Valence Bond and Crystal Field theories; iron and cobalt respectively. Variety of metallurgical processes, industrial catalysts and analytical reagents• appreciate the importance and applications of coordination involve the use of coordination compounds. compounds in our day to day life. Coordination compounds also find many applications in electroplating, textile dyeing and medicinal chemistry. 5.15.15.15.15.1 Werner’Werner’Werner’sWerner’Werner’ Alfred Werner (1866-1919), a Swiss chemist was the first to formulate his ideas about the structures of coordination compounds. He prepared TheoryTheoryTheoryTheoryTheory ofofofofof and characterised a large number of coordination compounds and CoordinationCoordinationCoordinationCoordinationCoordination studied their physical and chemical behaviour by simple experimental CompoundsCompoundsCompoundsCompoundsCompounds techniques. Werner proposed the concept of a primary valence and a secondary valence for a metal ion. Binary compounds such as CrCl3, CoCl2 or PdCl2 have primary valence of 3, 2 and 2 respectively. In a series of compounds of cobalt(III) chloride with ammonia, it was found that some of the chloride ions could be precipitated as AgCl on adding excess silver nitrate solution in cold but some remained in solution. Chemistry 118 Reprint 2025-26 1 mol CoCl3.6NH3 (Yellow) gave 3 mol AgCl 1 mol CoCl3.5NH3 (Purple) gave 2 mol AgCl 1 mol CoCl3.4NH3 (Green) gave 1 mol AgCl 1 mol CoCl3.4NH3 (Violet) gave 1 mol AgCl These observations, together with the results of conductivity measurements in solution can be explained if (i) six groups in all, either chloride ions or ammonia molecules or both, remain bonded to the cobalt ion during the reaction and (ii) the compounds are formulated as shown in Table 5.1, where the atoms within the square brackets form a single entity which does not dissociate under the reaction conditions. Werner proposed the term secondary valence for the number of groups bound directly to the metal ion; in each of these examples the secondary valences are six. Table 5.1: Formulation of Cobalt(III) Chloride-Ammonia Complexes Colour Formula Solution conductivity corresponds to Yellow [Co(NH3)6] 3+3Cl– 1:3 electrolyte Purple [CoCl(NH3)5] 2+2Cl – 1:2 electrolyte Green [CoCl2(NH3)4] +Cl – 1:1 electrolyte Violet [CoCl2(NH3)4] +Cl – 1:1 electrolyte Note that the last two compounds in Table 5.1 have identical empirical formula, CoCl3.4NH3, but distinct properties. Such compounds are termed as isomers. Werner in 1898, propounded his theory of coordination compounds. The main postulates are: 1. In coordination compounds metals show two types of linkages (valences)-primary and secondary. 2. The primary valences are normally ionisable and are satisfied by negative ions. 3. The secondary valences are non ionisable. These are satisfied by neutral molecules or negative ions. The secondary valence is equal to the coordination number and is fixed for a metal. 4. The ions/groups bound by the secondary linkages to the metal have characteristic spatial arrangements corresponding to different coordination numbers. In modern formulations, such spatial arrangements are called coordination polyhedra. The species within the square bracket are coordination entities or complexes and the ions outside the square bracket are called counter ions. He further postulated that octahedral, tetrahedral and square planar geometrical shapes are more common in coordination compounds of transition metals. Thus, [Co(NH3)6] 3+, [CoCl(NH3)5] 2+ and [CoCl2(NH3)4] + are octahedral entities, while [Ni(CO)4] and [PtCl4]2– are tetrahedral and square planar, respectively. 119 Coordination Compounds Reprint 2025-26 On the basis of the following observations made with aqueous solutions, ExampleExampleExampleExampleExample 5.15.15.15.15.1 assign secondary valences to metals in the following compounds: Formula Moles of AgCl precipitated per mole of the compounds with excess AgNO3 (i) PdCl2.4NH3 2 (ii) NiCl2.6H2O 2 (iii) PtCl4.2HCl 0 (iv) CoCl3.4NH3 1 (v) PtCl2.2NH3 0 (i) Secondary 4 (ii) Secondary 6 SolutionSolutionSolutionSolutionSolution (iii) Secondary 6 (iv) Secondary 6 (v) Secondary 4 Difference between a double salt and a complex Both double salts as well as complexes are formed by the combination of two or more stable compounds in stoichiometric ratio. However, they differ in the fact that double salts such as carnallite, KCl.MgCl2.6H2O, Mohr’s salt, FeSO4.(NH4)2SO4.6H2O, potash alum, KAl(SO4)2.12H2O, etc. dissociate into simple ions completely when dissolved in water. However, complex ions such as [Fe(CN)6] 4– of K4[Fe(CN)6] do not dissociate into Fe2+ and CN – ions. WernerWernerWernerWernerWerner was born on December 12, 1866, in Mülhouse, a small community in the French province of Alsace. His study of chemistry began in Karlsruhe (Germany) and continued in Zurich (Switzerland), where in his doctoral thesis in 1890, he explained the difference in properties of certain nitrogen containing organic (1866-1919)(1866-1919)(1866-1919)(1866-1919)(1866-1919) substances on the basis of isomerism. He extended vant Hoff’s theory of tetrahedral carbon atom and modified it for nitrogen. Werner showed optical and electrical differences between complex compounds based on physical measurements. In fact, Werner was the first to discover optical activity in certain coordination compounds. He, at the age of 29 years became a full professor at Technische Hochschule in Zurich in 1895. Alfred Werner was a chemist and educationist. His accomplishments included the development of the theory of coordination compounds. This theory, in which Werner proposed revolutionary ideas about how atoms and molecules are linked together, was formulated in a span of only three years, from 1890 to 1893. The remainder of his career was spent gathering the experimental support required to validate his new ideas. Werner became the first Swiss chemist to win the Nobel Prize in 1913 for his work on the linkage of atoms and the coordination theory. Chemistry 120 Reprint 2025-26 5.25.25.25.25.2 DefinitionsDefinitionsDefinitionsDefinitionsDefinitions ofofofofof ( a ) Coordination entity SomeSomeSomeSomeSome A coordination entity constitutes a central metal atom or ion bonded to a fixed number of ions or molecules. For example, [CoCl3(NH3)3] ImportantImportantImportantImportantImportant is a coordination entity in which the cobalt ion is surrounded by TermsTermsTermsTermsTerms three ammonia molecules and three chloride ions. Other examples are [Ni(CO)4], [PtCl2(NH3)2], [Fe(CN)6]4–, [Co(NH3)6]3+. PertainingPertainingPertainingPertainingPertaining tototototo CoordinationCoordinationCoordinationCoordinationCoordination (b) Central atom/ion In a coordination entity, the atom/ion to which a fixed number CompoundsCompoundsCompoundsCompoundsCompounds of ions/groups are bound in a definite geometrical arrangement around it, is called the central atom or ion. For example, the central atom/ion in the coordination entities: [NiCl2(H2O)4], [CoCl(NH3)5] 2+ and [Fe(CN)6] 3– are Ni 2+, Co 3+ and Fe3+, respectively. These central atoms/ions are also referred to as Lewis acids. ( c ) Ligands The ions or molecules bound to the central atom/ion in the coordination entity are called ligands. These may be simple ions such as Cl –, small molecules such as H2O or NH3, larger molecules such as H2NCH2CH2NH2 or N(CH2CH2NH2)3 or even macromolecules, such as proteins. When a ligand is bound to a metal ion through a single donor atom, as with Cl–, H2O or NH3, the ligand is said to be unidentate. When a ligand can bind through two donor atoms as in H2NCH2CH2NH2 (ethane-1,2-diamine) or C2O42– (oxalate), the ligand is said to be didentate and when several donor atoms are present in a single ligand as in N(CH2CH2NH2)3, the ligand is said to be polydentate. Ethylenediaminetetraacetate ion (EDTA4–) is an important hexadentate ligand. It can bind through two nitrogen and four oxygen atoms to a central metal ion. When a di- or polydentate ligand uses its two or more donor atoms simultaneously to bind a single metal ion, it is said to be a chelate ligand. The number of such ligating groups is called the denticity of the ligand. Such complexes, called chelate complexes tend to be more stable than similar complexes containing unidentate ligands. Ligand which has two different donor atoms and either of the two ligetes in the complex is called ambidentate ligand. Examples of such ligands are the NO2– and SCN– ions. NO2 – ion can coordinate either through nitrogen or through oxygen to a central metal atom/ion. Similarly, SCN– ion can coordinate through the sulphur or nitrogen atom. ( d ) Coordination number The coordination number (CN) of a metal ion in a complex can be defined as the number of ligand donor atoms to which the metal is directly bonded. For example, in the complex ions, [PtCl6]2– and [Ni(NH3)4] 2+, the coordination number of Pt and Ni are 6 and 4 respectively. Similarly, in the complex ions, [Fe(C2O4)3] 3– and [Co(en)3]3+, the coordination number of both, Fe and Co, is 6 because C2O4 2– and en (ethane-1,2-diamine) are didentate ligands. 121 Coordination Compounds Reprint 2025-26 It is important to note here that coordination number of the central atom/ion is determined only by the number of sigma bonds formed by the ligand with the central atom/ion. Pi bonds, if formed between the ligand and the central atom/ion, are not counted for this purpose. (e) Coordination sphere The central atom/ion and the ligands attached to it are enclosed in square bracket and is collectively termed as the coordination sphere. The ionisable groups are written outside the bracket and are called counter ions. For example, in the complex K4[Fe(CN)6], the coordination sphere is [Fe(CN)6]4– and the counter ion is K +. (f) Coordination polyhedron The spatial arrangement of the ligand atoms which are directly attached to the central atom/ion defines a coordination polyhedron about the central atom. The most common coordination polyhedra are octahedral, square planar and tetrahedral. For example, [Co(NH3)6] 3+ is octahedral, [Ni(CO)4] is tetrahedral and [PtCl4] 2– is square planar. Fig. 5.1 shows the shapes of different coordination polyhedra. Fig. 5.1: Shapes of different coordination polyhedra. M represents the central atom/ion and L, a unidentate ligand. (g) Oxidation number of central atom The oxidation number of the central atom in a complex is defined as the charge it would carry if all the ligands are removed along with the electron pairs that are shared with the central atom. The oxidation number is represented by a Roman numeral in parenthesis following the name of the coordination entity. For example, oxidation number of copper in [Cu(CN)4]3– is +1 and it is written as Cu(I). ( h ) Homoleptic and heteroleptic complexes Complexes in which a metal is bound to only one kind of donor groups, e.g., [Co(NH3)6]3+, are known as homoleptic. Complexes in which a metal is bound to more than one kind of donor groups, e.g., [Co(NH3)4Cl2] +, are known as heteroleptic. 5.35.35.35.35.3 NomenclatureNomenclatureNomenclatureNomenclatureNomenclature Nomenclature is important in Coordination Chemistry because of the ofofofofof need to have an unambiguous method of describing formulas and writing systematic names, particularly when dealing with isomers. The CoordinationCoordinationCoordinationCoordinationCoordination formulas and names adopted for coordination entities are based on the CompoundsCompoundsCompoundsCompoundsCompounds recommendations of the International Union of Pure and Applied Chemistry (IUPAC). Chemistry 122 Reprint 2025-26 5.3.1 Formulas of The formula of a compound is a shorthand tool used to provide basic Mononuclear information about the constitution of the compound in a concise and Coordination convenient manner. Mononuclear coordination entities contain a single Entities central metal atom. The following rules are applied while writing the formulas: (i) The central atom is listed first. (ii) The ligands are then listed in alphabetical order. The placement of a ligand in the list does not depend on its charge. (iii) Polydentate ligands are also listed alphabetically. In case of abbreviated ligand, the first letter of the abbreviation is used to determine the position of the ligand in the alphabetical order. (iv) The formula for the entire coordination entity, whether charged or not, is enclosed in square brackets. When ligands are polyatomic, their formulas are enclosed in parentheses. Ligand abbreviations are also enclosed in parentheses. (v) There should be no space between the ligands and the metal within a coordination sphere. (vi) When the formula of a charged coordination entity is to be written Note: The 2004 IUPAC without that of the counter ion, the charge is indicated outside the draft recommends that square brackets as a right superscript with the number before theligands will be sorted alphabetically, sign. For example, [Co(CN)6]3–, [Cr(H2O)6]3+, etc. irrespective of charge. (vii) The charge of the cation(s) is balanced by the charge of the anion(s). 5.3.2 Naming of The names of coordination compounds are derived by following the Mononuclear principles of additive nomenclature. Thus, the groups that surround the Coordination central atom must be identified in the name. They are listed as prefixes Compounds to the name of the central atom along with any appropriate multipliers. The following rules are used when naming coordination compounds: (i) The cation is named first in both positively and negatively charged coordination entities. (ii) The ligands are named in an alphabetical order before the name of the central atom/ion. (This procedure is reversed from writing formula). (iii) Names of the anionic ligands end in –o, those of neutral and cationic ligands are the same except aqua for H2O, ammine for NH3, carbonyl for CO and nitrosyl for NO. While writing the formula of coordination entity, these are enclosed in brackets ( ). (iv) Prefixes mono, di, tri, etc., are used to indicate the number of the individual ligands in the coordination entity. When the names of the ligands include a numerical prefix, then the terms, bis, tris, tetrakis are used, the ligand to which they refer being placed in parentheses. For example, [NiCl2(PPh3)2] is named as dichloridobis(triphenylphosphine)nickel(II). (v) Oxidation state of the metal in cation, anion or neutral coordination entity is indicated by Roman numeral in parenthesis. (vi) If the complex ion is a cation, the metal is named same as the element. For example, Co in a complex cation is called cobalt and Note: The 2004 IUPAC draft Pt is called platinum. If the complex ion is an anion, the name of recommends that the metal ends with the suffix – ate. For example, Co in a complex 2  anionic ligands will anion,   Co  SCN  4  is called cobaltate. For some metals, the Latinend with–ido so that names are used in the complex anions, e.g., ferrate for Fe.chloro would become chlorido, etc. 123 Coordination Compounds Reprint 2025-26 (vii) The neutral complex molecule is named similar to that of the complex cation. The following examples illustrate the nomenclature for coordination compounds. 1. [Cr(NH3)3(H2O)3]Cl3 is named as: triamminetriaquachromium(III) chloride Explanation: The complex ion is inside the square bracket, which is a cation. The amine ligands are named before the aqua ligands according to alphabetical order. Since there are three chloride ions in the compound, the charge on the complex ion must be +3 (since the compound is electrically neutral). From the charge on the complex ion and the charge on the ligands, we can calculate the oxidation number of the metal. In this example, all the ligands are neutral molecules. Therefore, the oxidation number of chromium must be the same as the charge of the complex ion, +3. 2. [Co(H2NCH2CH2NH2)3]2(SO4)3 is named as: tris(ethane-1,2–diamine)cobalt(III) sulphate Explanation: The sulphate is the counter anion in this molecule. Since it takes 3 sulphates to bond with two complex cations, the charge on each complex cation must be +3. Further, ethane-1,2– Notice how the name diamine is a neutral molecule, so the oxidation number of cobalt of the metal differs in in the complex ion must be +3. Remember that you never have to cation and anion even though they contain the indicate the number of cations and anions in the name of an same metal ions. ionic compound. 3. [Ag(NH3)2][Ag(CN)2] is named as: diamminesilver(I)dicyanidoargentate(I) ExampleExampleExampleExampleExample 5.25.25.25.25.2 Write the formulas for the following coordination compounds: (a) tetraammineaquachloridocobalt(III) chloride (b) potassium tetrahydroxidozincate(II) (c) potassium trioxalatoaluminate(III) (d) dichloridobis(ethane-1,2-diamine)cobalt(III) (e) tetracarbonylnickel(0) SolutionSolutionSolutionSolutionSolution (a) [Co(NH3)4(H2O)Cl]Cl2 (b) K2[Zn(OH)4] (c) K3[Al(C2O4)3] (d) [CoCl2(en)2]+ (e) [Ni(CO)4] ExampleExampleExampleExampleExample 5.35.35.35.35.3 Write the IUPAC names of the following coordination compounds: (a) [Pt(NH3)2Cl(NO2)] (b) K3[Cr(C2O4)3] (c) [CoCl2(en)2]Cl (d) [Co(NH3)5(CO3)]Cl (e) Hg[Co(SCN)4] SolutionSolutionSolutionSolutionSolution (a) diamminechloridonitrito-N-platinum(II) (b) potassium trioxalatochromate(III) (c) dichloridobis(ethane-1,2-diamine)cobalt(III) chloride (d) pentaamminecarbonatocobalt(III) chloride (e) mercury (I) tetrathiocyanato-S-cobaltate(III) Chemistry 124 Reprint 2025-26 IntextIntextIntextIntextIntext QuestionsQuestionsQuestionsQuestionsQuestions 5.1 Write the formulas for the following coordination compounds: (i) tetraamminediaquacobalt(III) chloride (ii) potassium tetracyanidonickelate(II) (iii) tris(ethane–1,2–diamine) chromium(III) chloride (iv) amminebromidochloridonitrito-N-platinate(II) (v) dichloridobis(ethane–1,2–diamine)platinum(IV) nitrate (vi) iron(III) hexacyanidoferrate(II) 5.2 Write the IUPAC names of the following coordination compounds: (i) [Co(NH3)6]Cl3 (ii) [Co(NH3)5Cl]Cl2 (iii) K3[Fe(CN)6] (iv) K3[Fe(C2O4)3] (v) K2[PdCl4] (vi) [Pt(NH3)2Cl(NH2CH3)]Cl 5.45.45.45.45.4 IsomerismIsomerismIsomerismIsomerismIsomerism ininininin Isomers are two or more compounds that have the same chemical CoordinationCoordinationCoordinationCoordinationCoordination formula but a different arrangement of atoms. Because of the different arrangement of atoms, they differ in one or more physical or chemical CompoundsCompoundsCompoundsCompoundsCompounds properties. Two principal types of isomerism are known among coordination compounds. Each of which can be further subdivided. (a) Stereoisomerism (i) Geometrical isomerism (ii) Optical isomerism (b) Structural isomerism (i) Linkage isomerism (ii) Coordination isomerism (iii) Ionisation isomerism (iv) Solvate isomerism Stereoisomers have the same chemical formula and chemical bonds but they have different spatial arrangement. Structural isomers have different bonds. A detailed account of these isomers are given below. 5.4.1 Geometric Isomerism This type of isomerism arises in heteroleptic complexes due to different possible geometric arrangements of the ligands. Important examples of this behaviour are found with coordination numbers 4 and 6. In a square planar complex of formula [MX2L2] (X and L are unidentate), the two ligands X may be arranged adjacent to each Fig. 5.2: Geometrical isomers (cis and trans) other in a cis isomer, or opposite to each other of Pt [NH3)2Cl2] in a trans isomer as depicted in Fig. 5.2. + + Other square planar complex of the type Cl Cl MABXL (where A, B, X, L are unidentates) N H3 Cl N H3 N H3 shows three isomers-two cis and one trans. Co Co You may attempt to draw these structures. N H3 N H3 N H3 N H3 Such isomerism is not possible for a tetrahedral N H3 Cl geometry but similar behaviour is possible in cis trans octahedral complexes of formula [MX2L4] in which the two ligands X may be oriented cis or Fig. 5.3: Geometrical isomers (cis and trans) trans to each other (Fig. 5.3). of [Co(NH3)4Cl2]+ 125 Coordination Compounds Reprint 2025-26 This type of isomerism also arises when didentate ligands L – L [e.g., NH2 CH2 CH2 NH2 (en)] are present in complexes of formula [MX2(L – L)2] (Fig. 5.4). Another type of geometrical isomerism occurs in octahedral Fig. 5.4: Geometrical isomers (cis and trans) coordination entities of the type of [CoCl2(en)2] [Ma3b3] like [Co(NH3)3(NO2)3]. If three donor atoms of the same ligands occupy adjacent positions at the corners of an octahedral Fig. 5.5 face, we have the facial (fac) The facial (fac) and isomer. When the positions are meridional (mer) around the meridian of the isomers of octahedron, we get the meridional [Co(NH3)3(NO2)3] (mer) isomer (Fig. 5.5). Why is geometrical isomerism not possible in tetrahedral complexes ExampleExampleExampleExampleExample 5.45.45.45.45.4 having two different types of unidentate ligands coordinated with the central metal ion ? Tetrahedral complexes do not show geometrical isomerism because SolutionSolutionSolutionSolutionSolution the relative positions of the unidentate ligands attached to the central metal atom are the same with respect to each other. 5.4.2 Optical Isomerism Optical isomers are mirror images that cannot be superimposed on one another. These are called as enantiomers. The molecules or ions that cannot be superimposed are called chiral. The two forms are called dextro (d) and laevo (l) depending upon the direction they rotate the plane of polarised light in a polarimeter (d rotates to the right, l to the left). Optical isomerism is common in octahedral complexes involving Fig.5.6: Optical isomers (d and l) of [Co(en)3] 3+ didentate ligands (Fig. 5.6). In a coordination entity of the type [PtCl2(en)2] 2+, only the cis-isomer shows optical activity (Fig. 5.7). Fig.5.7 Optical isomers (d and l) of cis- [PtCl2(en)2]2+ Chemistry 126 Reprint 2025-26 ExampleExampleExampleExampleExample 5.55.55.55.55.5 Draw structures of geometrical isomers of [Fe(NH3)2(CN)4] – SolutionSolutionSolutionSolutionSolution ExampleExampleExampleExampleExample 5.65.65.65.65.6 Out of the following two coordination entities which is chiral (optically active)? (a) cis-[CrCl2(ox)2]3– (b) trans-[CrCl2(ox)2]3– SolutionSolutionSolutionSolutionSolution The two entities are represented as Out of the two, (a) cis - [CrCl2(ox)2]3- is chiral (optically active). 5.4.3 Linkage Linkage isomerism arises in a coordination compound containing Isomerism ambidentate ligand. A simple example is provided by complexes containing the thiocyanate ligand, NCS–, which may bind through the nitrogen to give M–NCS or through sulphur to give M–SCN. Jørgensen discovered such behaviour in the complex [Co(NH3)5(NO2)]Cl2, which is obtained as the red form, in which the nitrite ligand is bound through oxygen (–ONO), and as the yellow form, in which the nitrite ligand is bound through nitrogen (–NO2). 5.4.4 Coordination This type of isomerism arises from the interchange of ligands between Isomerism cationic and anionic entities of different metal ions present in a complex. An example is provided by [Co(NH3)6][Cr(CN)6], in which the NH3 ligands are bound to Co 3+ and the CN – ligands to Cr 3+. In its coordination isomer [Cr(NH3)6][Co(CN)6], the NH3 ligands are bound to Cr3+ and the CN– ligands to Co 3+. 5.4.5 Ionisation This form of isomerism arises when the counter ion in a complex salt Isomerism is itself a potential ligand and can displace a ligand which can then become the counter ion. An example is provided by the ionisation isomers [Co(NH3)5(SO4)]Br and [Co(NH3)5Br]SO4. 127 Coordination Compounds Reprint 2025-26 5.4.6 Solvate This form of isomerism is known as ‘hydrate isomerism’ in case where Isomerism water is involved as a solvent. This is similar to ionisation isomerism. Solvate isomers differ by whether or not a solvent molecule is directly bonded to the metal ion or merely present as free solvent moleculesin the crystal lattice. An example is provided by the aqua complex [Cr(H2O)6]Cl3 (violet) and its solvate isomer [Cr(H2O)5Cl]Cl2.H2O (grey-green). IntextIntextIntextIntextIntext QuestionsQuestionsQuestionsQuestionsQuestions 5.3 Indicate the types of isomerism exhibited by the following complexes and draw the structures for these isomers: (i) K[Cr(H2O)2(C2O4)2 (ii) [Co(en)3]Cl3 (iii) [Co(NH3)5(NO2)](NO3)2 (iv) [Pt(NH3)(H2O)Cl2] 5.4 Give evidence that [Co(NH3)5Cl]SO4 and [Co(NH3)5(SO4)]Cl are ionisation isomers. 5.55.55.55.55.5 BondingBondingBondingBondingBonding ininininin Werner was the first to describe the bonding features in coordination CoordinationCoordinationCoordinationCoordinationCoordination compounds. But his theory could not answer basic questions like: CompoundsCompoundsCompoundsCompoundsCompounds (i) Why only certain elements possess the remarkable property of forming coordination compounds? (ii) Why the bonds in coordination compounds have directional properties? (iii) Why coordination compounds have characteristic magnetic and optical properties? Many approaches have been put forth to explain the nature of bonding in coordination compounds viz. Valence Bond Theory (VBT), Crystal Field Theory (CFT), Ligand Field Theory (LFT) and Molecular Orbital Theory (MOT). We shall focus our attention on elementary treatment of the application of VBT and CFT to coordination compounds. 5.5.1 Valence According to this theory, the metal atom or ion under the influence of Bond ligands can use its (n-1)d, ns, np or ns, np, nd orbitals for hybridisation Theory to yield a set of equivalent orbitals of definite geometry such as octahedral, tetrahedral, square planar and so on (Table 5.2). These hybridised orbitals are allowed to overlap with ligand orbitals that can donate electron pairs for bonding. This is illustrated by the following examples. Table 5.2: Number of Orbitals and Types of Hybridisations Coordination Type of Distribution of hybrid number hybridisation orbitals in space 4 sp3 Tetrahedral 4 dsp2 Square planar 5 sp3d Trigonal bipyramidal 6 sp3d2 Octahedral 6 d2sp3 Octahedral Chemistry 128 Reprint 2025-26 It is usually possible to predict the geometry of a complex from the knowledge of its magnetic behaviour on the basis of the valence bond theory. In the diamagnetic octahedral complex, [Co(NH3)6]3+, the cobalt ion is in +3 oxidation state and has the electronic configuration 3d 6. The hybridisation scheme is as shown in diagram. Six pairs of electrons, one from each NH3 molecule, occupy the six hybrid orbitals. Thus, the complex has octahedral geometry and is diamagnetic because of the absence of unpaired electron. In the formation of this complex, since the inner d orbital (3d) is used in hybridisation, the complex, [Co(NH3)6]3+ is called an inner orbital or low spin or spin paired complex. The paramagnetic octahedral complex, [CoF6]3– uses outer orbital (4d ) in hybridisation (sp 3d2). It is thus called outer orbital or high spin or spin free complex. Thus: Orbitals of Co3+ ion 3d 4s 4p 4d sp3d2 hybridised orbitals of Co3+ 3d sp3d3 hybrid [CoF6]3– (outer orbital or high spin complex) 3d Six pairs of electrons from six F – ions In tetrahedral complexes one s and three p orbitals are hybridised to form four equivalent orbitals oriented tetrahedrally. This is ill- ustrated below for [NiCl4]2-. Here nickel is in +2 oxidation state and the ion has the electronic configuration 3d 8. The hybridisation scheme is as shown in diagram. Each Cl – ion donates a pair of electrons. The compound is paramagnetic since it contains two unpaired electrons. Similarly, [Ni(CO)4] has tetrahedral geometry but is diamagnetic since nickel is in zero oxidation state and contains no unpaired electron. 129 Coordination Compounds Reprint 2025-26 In the square planar complexes, the hybridisation involved is dsp2. An example is [Ni(CN)4]2–. Here nickel is in +2 oxidation state and has the electronic configuration 3d8. The hybridisation scheme is as shown in diagram: Orbitals of Ni2+ ion 3d 4s 4p dsp2 hybridised orbitals of Ni2+ 3d dsp2 hydrid 4p [Ni(CN)4 ]2– (low spin complex) 3d Four pairs of electrons 4p from 4 CN– groups Each of the hybridised orbitals receives a pair of electrons from a cyanide ion. The compound is diamagnetic as evident from the absence of unpaired electron. It is important to note that the hybrid orbitals do not actually exist. In fact, hybridisation is a mathematical manipulation of wave equation for the atomic orbitals involved. 5.5.2 Magnetic The magnetic moment of coordination compounds can be measured Properties by the magnetic susceptibility experiments. The results can be used to of obtain information about the number of unpaired electrons and hence Coordination structures adopted by metal complexes. Compounds A critical study of the magnetic data of coordination compounds of metals of the first transition series reveals some complications. For metal ions with upto three electrons in the d orbitals, like Ti3+ (d1); V 3+ (d2); Cr 3+ (d3); two vacant d orbitals are available for octahedral hybridisation with 4s and 4p orbitals. The magnetic behaviour of these free ions and their coordination entities is similar. When more than three 3d electrons are present, the required pair of 3d orbitals for octahedral hybridisation is not directly available (as a consequence of Hund’s rule). Thus, for d4 (Cr 2+, Mn 3+), d5 (Mn2+, Fe 3+), d6 (Fe 2+, Co 3+) cases, a vacant pair of d orbitals results only by pairing of 3d electrons which leaves two, one and zero unpaired electrons, respectively. The magnetic data agree with maximum spin pairing in many cases, especially with coordination compounds containing d6 ions. However, with species containing d4 and d5 ions there are complications. [Mn(CN)6]3– has magnetic moment of two unpaired electrons while [MnCl6]3– has a paramagnetic moment of four unpaired electrons. [Fe(CN)6]3– has magnetic moment of a single unpaired electron while [FeF6]3– has a paramagnetic moment of five unpaired electrons. [CoF6]3– is paramagnetic with four unpaired electrons while [Co(C2O4)3]3– is diamagnetic. This apparent anomaly is explained by valence bond theory in terms of formation of inner orbital and outer orbital coordination entities. [Mn(CN)6]3–, [Fe(CN)6]3– and [Co(C2O4)3]3– are inner orbital complexes involving d2sp3 hybridisation, the former two complexes are paramagnetic and the latter diamagnetic. On the other hand, [MnCl6]3–, [FeF6]3– and [CoF6-]3– are outer orbital complexes involving sp 3d 2 hybridisation and are paramagnetic corresponding to four, five and four unpaired electrons. Chemistry 130 Reprint 2025-26 ExampleExampleExampleExampleExample 5.75.75.75.75.7 The spin only magnetic moment of [MnBr4]2– is 5.9 BM. Predict the geometry of the complex ion ? SolutionSolutionSolutionSolutionSolution Since the coordination number of Mn2+ ion in the complex ion is 4, it will be either tetrahedral (sp 3 hybridisation) or square planar (dsp2 hybridisation). But the fact that the magnetic moment of the complex ion is 5.9 BM, it should be tetrahedral in shape rather than square planar because of the presence of five unpaired electrons in the d orbitals. 5.5.3 Limitations While the VB theory, to a larger extent, explains the formation, structures of Valence and magnetic behaviour of coordination compounds, it suffers from Bond the following shortcomings: Theory (i) It involves a number of assumptions. (ii) It does not give quantitative interpretation of magnetic data. (iii) It does not explain the colour exhibited by coordination compounds. (iv) It does not give a quantitative interpretation of the thermodynamic or kinetic stabilities of coordination compounds. (v) It does not make exact predictions regarding the tetrahedral and square planar structures of 4-coordinate complexes. (vi) It does not distinguish between weak and strong ligands. 5.5.4 Crystal The crystal field theory (CFT) is an electrostatic model which considers Field the metal-ligand bond to be ionic arising purely from electrostatic Theory interactions between the metal ion and the ligand. Ligands are treated as point charges in case of anions or point dipoles in case of neutral molecules. The five d orbitals in an isolated gaseous metal atom/ion have same energy, i.e., they are degenerate. This degeneracy is maintained if a spherically symmetrical field of negative charges surrounds the metal atom/ion. However, when this negative field is due to ligands (either anions or the negative ends of dipolar molecules like NH3 and H2O) in a complex, it becomes asymmetrical and the degeneracy of the d orbitals is lifted. It results in splitting of the d orbitals. The pattern of splitting depends upon the nature of the crystal field. Let us explain this splitting in different crystal fields. ( a ) Crystal field splitting in octahedral coordination entities In an octahedral coordination entity with six ligands surrounding the metal atom/ion, there will be repulsion between the electrons in metal d orbitals and the electrons (or negative charges) of the ligands. Such a repulsion is more when the metal d orbital is directed towards the ligand than when it is away from the ligand. Thus, the d x 2  y 2 2 orbitals which point towards the axes along the direction of and d z the ligand will experience more repulsion and will be raised in energy; and the dxy, dyz and dxz orbitals which are directed between the axes will be lowered in energy relative to the average energy in the spherical crystal field. Thus, the degeneracy of the d orbitals has been removed due to ligand electron-metal electron repulsions in the octahedral complex to yield three orbitals of lower energy, t2g set and two orbitals of higher energy, eg set. This splitting of the 131 Coordination Compounds Reprint 2025-26 degenerate levels due to the presence of ligands in a definite geometry is termed as crystal field splitting and the energy separation is denoted by Do (the subscript o is for octahedral) (Fig.5.8). Thus, the energy of the two eg orbitals will increase by (3/5) Do and that of the three t2g will decrease by (2/5)Do. The crystal field splitting, Do, depends upon the field produced by the ligand and charge on the metal ion. Some ligands are able to produce strong fields in which case, the splitting will be large whereas others produce weak fields Fig.5.8: d orbital splitting in an octahedral crystal field and consequently result in small splitting of d orbitals. In general, ligands can be arranged in a series in the order of increasing field strength as given below: I – < Br– < SCN – < Cl – < S2– < F – < OH – < C2O4 2– < H2O < NCS – < edta 4– < NH3 < en < CN – < CO Such a series is termed as spectrochemical series. It is an experimentally determined series based on the absorption of light by complexes with different ligands. Let us assign electrons in the d orbitals of metal ion in octahedral coordination entities. Obviously, the single d electron occupies one of the lower energy t2g orbitals. In d 2 and d3 coordination entities, the d electrons occupy the t2g orbitals singly in accordance with the Hund’s rule. For d 4 ions, two possible patterns of electron distribution arise: (i) the fourth electron could either enter the t2g level and pair with an existing electron, or (ii) it could avoid paying the price of the pairing energy by occupying the eg level. Which of these possibilities occurs, depends on the relative magnitude of the crystal field splitting, Do and the pairing energy, P (P represents the energy required for electron pairing in a single orbital). The two options are: (i) If Do < P, the fourth electron enters one of the eg orbitals giving the configuration t 2g3 e 1g . Ligands for which Do < P are known as weak field ligands and form high spin complexes. (ii) If Do > P, it becomes more energetically favourable for the fourth electron to occupy a t2g orbital with configuration t2g 4eg 0. Ligands which produce this effect are known as strong field ligands and form low spin complexes. Calculations show that d4 to d7 coordination entities are more stable for strong field as compared to weak field cases. Chemistry 132 Reprint 2025-26 ( b ) Crystal field splitting in tetrahedral coordination entities In tetrahedral coordination entity formation, the d orbital splitting (Fig. 5.9) is inverted and is smaller as compared to the octahedral field splitting. For the same metal, the same ligands and metal-ligand distances, it can be shown that Dt = (4/9) D0. Consequently, the orbital splitting energies are not sufficiently large for forcing pairing and, therefore, low spin configurations are rarely observed. The ‘g’ subscript is used for the octahedral and square planar complexes which have centre of symmetry. Since Fig.5.9: d orbital splitting in a tetrahedral tetrahedral complexes lack symmetry, ‘g’ crystal field. subscript is not used with energy levels. 5.5.5 Colour in In the previous Unit, we learnt that one of the most distinctive Coordination properties of transition metal complexes is their wide range of colours. Compounds This means that some of the visible spectrum is being removed from white light as it passes through the sample, so the light that emerges is no longer white. The colour of the complex is complementary to that which is absorbed. The complementary colour is the colour generated from the wavelength left over; if green light is absorbed by the complex, it appears red. Table 5.3 gives the relationship of the different wavelength absorbed and the colour observed. Table 5.3: Relationship between the Wavelength of Light absorbed and the Colour observed in some Coordination Entities Coordinaton Wavelength of light Colour of light Colour of coordination entity absorbed (nm) absorbed entity [CoCl(NH3)5] 2+ 535 Yellow Violet [Co(NH3)5(H2O)]3+ 500 Blue Green Red [Co(NH3)6]3+ 475 Blue Yellow Orange 3– Not in visible [Co(CN)6] 310 Ultraviolet region Pale Yellow [Cu(H2O)4] 2+ 600 Red Blue [Ti(H2O)6]3+ 498 Blue Green Violet The colour in the coordination compounds can be readily explained in terms of the crystal field theory. Consider, for example, the complex [Ti(H2O)6] 3+, which is violet in colour. This is an octahedral complex where the single electron (Ti3+ is a 3d1 system) in the metal d orbital is in the t2g level in the ground state of the complex. The next higher state available for the electron is the empty eg level. If light corresponding to the energy of blue-green region is absorbed by the complex, it would excite the electron from t2g level to the eg level (t2g1eg 0 ® t2g0eg 1). Consequently, the complex appears violet in colour (Fig. 5.10). The crystal field theory attributes the colour of the coordination compounds to d-d transition of the electron. 133 Coordination Compounds Reprint 2025-26 It is important to note that in the absence of ligand, crystal field splitting does not occur and hence the substance is colourless. For example, removal of water from [Ti(H2O)6]Cl3 on heating renders it colourless. Similarly, anhydrous CuSO4 is white, but CuSO4.5H2O is Fig.5.10: Transition of an electron in blue in colour. The influence of the ligand on the colour of a complex may be illustrated by considering the [Ni(H2O)6]2+ complex, which forms when nickel(II) chloride is dissolved in water. If the didentate ligand, ethane-1,2-diamine(en) is progressively added in the molar ratios en:Ni, 1:1, 2:1, 3:1, the following series of reactions and their associated colour changes occur: [Ni(H2O)6]2+ (aq) + en (aq) = [Ni(H2O)4(en)]2+(aq) + 2H2O green pale blue [Ni(H2O)4 (en)] 2+(aq) + en (aq) = [Ni(H2O)2(en)2]2+(aq) + 2H2O blue/purple [Ni(H2O)2(en)2]2+(aq) + en (aq) = [Ni(en)3]2+(aq) + 2H2O violet This sequence is shown in Fig. 5.11. Fig.5.11 Aqueous solutions of complexes of nickel(II) with an 2+ 2+ [Ni(H 2O) 6] (aq) [Ni(en) 3] (aq)increasing number of ethane-1, 2-diamine ligands. [Ni(H 2O) 4en] 2+ (aq) [Ni(H 2O) 4en2] 2+ (aq) Colour of Some Gem Stones The colours produced by electronic transitions within the d orbitals of a transition metal ion occur frequently in everyday life. Ruby [Fig.5.12(a)] is aluminium oxide (Al2O3) containing about 0.5-1% Cr 3+ ions (d3), which are randomly distributed in positions normally occupied by Al 3+. We may view these chromium(III) species as octahedral chromium(III) complexes incorporated into the alumina lattice; d–d transitions at these centres give rise to the colour. Chemistry 134 Reprint 2025-26 In emerald [Fig.5.12(b)], Cr 3+ ions occupy octahedral sites in the mineral beryl (Be3Al2Si6O18). The absorption bands seen in the ruby shift (a) (b) to longer wavelength, namely yellow-red and blue, causing Fig.5.12: (a) Ruby: this gemstone was found in marble from Mogok, Myanmar; (b) emerald to transmit light in Emerald: this gemstone was found in the green region. Muzo, Columbia. 5.5.6 Limitations The crystal field model is successful in explaining the formation, of Crystal structures, colour and magnetic properties of coordination compounds Field to a large extent. However, from the assumptions that the ligands are Theory point charges, it follows that anionic ligands should exert the greatest splitting effect. The anionic ligands actually are found at the low end of the spectrochemical series. Further, it does not take into account the covalent character of bonding between the ligand and the central atom. These are some of the weaknesses of CFT, which are explained by ligand field theory (LFT) and molecular orbital theory which are beyond the scope of the present study. IntextIntextIntextIntextIntext QuestionsQuestionsQuestionsQuestionsQuestions 5.5 Explain on the basis of valence bond theory that [Ni(CN)4]2– ion with square planar structure is diamagnetic and the [NiCl4] 2– ion with tetrahedral geometry is paramagnetic. 5.6 [NiCl4] 2– is paramagnetic while [Ni(CO)4] is diamagnetic though both are tetrahedral. Why? 5.7 [Fe(H2O)6] 3+ is strongly paramagnetic whereas [Fe(CN)6] 3– is weakly paramagnetic. Explain. 5.8 Explain [Co(NH3)6] 3+ is an inner orbital complex whereas [Ni(NH3)6] 2+ is an outer orbital complex. 5.9 Predict the number of unpaired electrons in the square planar [Pt(CN)4]2– ion. 5.10 The hexaquo manganese(II) ion contains five unpaired electrons, while the hexacyanoion contains only one unpaired electron. Explain using Crystal Field Theory. 5.65.65.65.65.6 BondingBondingBondingBondingBonding ininininin The homoleptic carbonyls (compounds containing carbonyl ligands MetalMetalMetalMetalMetal only) are formed by most of the transition metals. These carbonyls have simple, well defined structures. Tetracarbonylnickel(0) is CarbonylsCarbonylsCarbonylsCarbonylsCarbonyls tetrahedral, pentacarbonyliron(0) is trigonalbipyramidal while hexacarbonyl chromium(0) is octahedral. Decacarbonyldimanganese(0) is made up of two square pyramidal Mn(CO)5 units joined by a Mn – Mn bond. Octacarbonyldicobalt(0) has a Co – Co bond bridged by two CO groups (Fig.5.13). 135 Coordination Compounds Reprint 2025-26 CO CO OC Ni Fe CO OC CO OC CO CO O Ni(CO) 4 Fe(CO)5 OC C CO Tetrahedral Trigonal bipyramidal OC Co Co CO OC CO C CO CO CO CO CO CO CO O Fig. 5.13 Cr OC Mn Mn CO [Co2 (CO) 8] Structures of some CO CO CO COrepresentative CO CO COhomoleptic metal carbonyls. Cr(CO)6 Octahedral [Mn 2 (CO)10 ] The metal-carbon bond in metal carbonyls possess both s and p character. The M–C s bond is formed by the donation of lone pair of electrons on the carbonyl carbon into a vacant orbital of the metal. The M–C p bond is formed by the donation of a pair of electrons from a filled d orbital of metal into the vacant antibonding p* orbital of carbon monoxide. The metal to ligand bonding Fig. 5.14: Example of synergic bonding creates a synergic effect which strengthens the interactions in a carbonyl bond between CO and the metal (Fig.5.14). complex. 5.75.75.75.75.7 ImportanceImportanceImportanceImportanceImportance The coordination compounds are of great importance. These compounds andandandandand are widely present in the mineral, plant and animal worlds and are known to play many important functions in the area of analytical ApplicationsApplicationsApplicationsApplicationsApplications chemistry, metallurgy, biological systems, industry and medicine. These ofofofofof are described below: CoordinationCoordinationCoordinationCoordinationCoordination • Coordination compounds find use in many qualitative and quantitative chemical analysis. The familiar colour reactions given CompoundsCompoundsCompoundsCompoundsCompounds by metal ions with a number of ligands (especially chelating ligands), as a result of formation of coordination entities, form the basis for their detection and estimation by classical and instrumental methods of analysis. Examples of such reagents include EDTA, DMG (dimethylglyoxime), a–nitroso–b–naphthol, cupron, etc. • Hardness of water is estimated by simple titration with Na2EDTA. The Ca2+ and Mg2+ ions form stable complexes with EDTA. The selective estimation of these ions can be done due to difference in the stability constants of calcium and magnesium complexes. • Some important extraction processes of metals, like those of silver and gold, make use of complex formation. Gold, for example, combines with cyanide in the presence of oxygen and water to form the coordination entity [Au(CN)2]– in aqueous solution. Gold can be separated in metallic form from this solution by the addition of zinc. • Similarly, purification of metals can be achieved through formation and subsequent decomposition of their coordination compounds. Chemistry 136 Reprint 2025-26 For example, impure nickel is converted to [Ni(CO)4], which is decomposed to yield pure nickel. • Coordination compounds are of great importance in biological systems. The pigment responsible for photosynthesis, chlorophyll, is a coordination compound of magnesium. Haemoglobin, the red pigment of blood which acts as oxygen carrier is a coordination compound of iron. Vitamin B12, cyanocobalamine, the anti– pernicious anaemia factor, is a coordination compound of cobalt. Among the other compounds of biological importance with coordinated metal ions are the enzymes like, carboxypeptidase A and carbonic anhydrase (catalysts of biological systems). • Coordination compounds are used as catalysts for many industrial processes. Examples include rhodium complex, [(Ph3P)3RhCl], a Wilkinson catalyst, is used for the hydrogenation of alkenes. • Articles can be electroplated with silver and gold much more smoothly and evenly from solutions of the complexes, [Ag(CN)2]– and [Au(CN)2] – than from a solution of simple metal ions. • In black and white photography, the developed film is fixed by washing with hypo solution which dissolves the undecomposed AgBr to form a complex ion, [Ag(S2O3)2]3–. • There is growing interest in the use of chelate therapy in medicinal chemistry. An example is the treatment of problems caused by the presence of metals in toxic proportions in plant/animal systems. Thus, excess of copper and iron are removed by the chelating ligands D–penicillamine and desferrioxime B via the formation of coordination compounds. EDTA is used in the treatment of lead poisoning. Some coordination compounds of platinum effectively inhibit the growth of tumours. Examples are: cis–platin and related compounds. SummarySummarySummarySummarySummary The chemistry of coordination compounds is an important and challenging area of modern inorganic chemistry. During the last fifty years, advances in this area, have provided development of new concepts and models of bonding and molecular structure, novel breakthroughs in chemical industry and vital insights into the functioning of critical components of biological systems. The first systematic attempt at explaining the formation, reactions, structure and bonding of a coordination compound was made by A. Werner. His theory postulated the use of two types of linkages (primary and secondary) by a metal atom/ion in a coordination compound. In the modern language of chemistry these linkages are recognised as the ionisable (ionic) and non-ionisable (covalent) bonds, respectively. Using the property of isomerism, Werner predicted the geometrical shapes of a large number of coordination entities. The Valence Bond Theory (VBT) explains with reasonable success, the formation, magnetic behaviour and geometrical shapes of coordination compounds. It, however, fails to provide a quantitative interpretation of magnetic behaviour and has nothing to say about the optical properties of these compounds. The Crystal Field Theory (CFT) to coordination compounds is based on the effect of different crystal fields (provided by the ligands taken as point charges), 137 Coordination Compounds Reprint 2025-26 on the degeneracy of d orbital energies of the central metal atom/ion. The splitting of the d orbitals provides different electronic arrangements in strong and weak crystal fields. The treatment provides for quantitative estimations of orbital separation energies, magnetic moments and spectral and stability parameters. However, the assumption that ligands consititute point charges creates many theoretical difficulties. The metal–carbon bond in metal carbonyls possesses both s and p character. The ligand to metal is s bond and metal to ligand is p bond. This unique synergic bonding provides stability to metal carbonyls. Coordination compounds are of great importance. These compounds provide critical insights into the functioning and structures of vital components of biological systems. Coordination compounds also find extensive applications in metallurgical processes, analytical and medicinal chemistry. Exercises 5.1 Explain the bonding in coordination compounds in terms of Werner’s postulates. 5.2 FeSO4 solution mixed with (NH4)2SO4 solution in 1:1 molar ratio gives the test of Fe 2+ ion but CuSO4 solution mixed with aqueous ammonia in 1:4 molar ratio does not give the test of Cu2+ ion. Explain why? 5.3 Explain with two examples each of the following: coordination entity, ligand, coordination number, coordination polyhedron, homoleptic and heteroleptic. 5.4 What is meant by unidentate, didentate and ambidentate ligands? Give two examples for each. 5.5 Specify the oxidation numbers of the metals in the following coordination entities: (i) [Co(H2O)(CN)(en)2] 2+ (iii) [PtCl4] 2– (v) [Cr(NH3)3Cl3] (ii) [CoBr2(en)2] + (iv) K3[Fe(CN)6] 5.6 Using IUPAC norms write the formulas for the following: (i) Tetrahydroxidozincate(II) (vi) Hexaamminecobalt(III) sulphate (ii) Potassium tetrachloridopalladate(II) (vii) Potassium tri(oxalato)chromate(III) (iii) Diamminedichloridoplatinum(II) (viii) Hexaammineplatinum(IV) (iv) Potassium tetracyanidonickelate(II) (ix) Tetrabromidocuprate(II) (v) Pentaamminenitrito-O-cobalt(III) (x) Pentaamminenitrito-N-cobalt(III) 5.7 Using IUPAC norms write the systematic names of the following: (i) [Co(NH3)6]Cl3 (iv) [Co(NH3)4Cl(NO2)]Cl (vii) [Ni(NH3)6]Cl2 (ii) [Pt(NH3)2Cl(NH2CH3)]Cl (v) [Mn(H2O)6]2+ (viii) [Co(en)3] 3+ (iii) [Ti(H2O)6] 3+ (vi) [NiCl4] 2– (ix) [Ni(CO)4] 5.8 List various types of isomerism possible for coordination compounds, giving an example of each. 5.9 How many geometrical isomers are possible in the following coordination entities? (i) [Cr(C2O4)3] 3– (ii) [Co(NH3)3Cl3] 5.10 Draw the structures of optical isomers of: (i) [Cr(C2O4)3] 3– (ii) [PtCl2(en)2] 2+ (iii) [Cr(NH3)2Cl2(en)] + Chemistry 138 Reprint 2025-26

4.16 Describe the preparation of potassium permanganate. How does the acidified permanganate solution react with (i) iron(II) ions (ii) SO2 and (iii) oxalic acid? Write the ionic equations for the reactions. 4.17 For M2+/M and M3+/M 2+ systems the E o values for some metals are as follows: Cr2+/Cr -0.9V Cr3/Cr2+ -0.4 V Mn 2+/Mn -1.2V Mn3+/Mn2+ +1.5 V Fe2+/Fe -0.4V Fe3+/Fe2+ +0.8 V Use this data to comment upon: (i) the stability of Fe3+ in acid solution as compared to that of Cr3+ or Mn3+ and (ii) the ease with which iron can be oxidised as compared to a similar process for either chromium or manganese metal. 4.18 Predict which of the following will be coloured in aqueous solution? Ti 3+, V3+, Cu+, Sc3+, Mn 2+, Fe3+ and Co 2+. Give reasons for each. 4.19 Compare the stability of +2 oxidation state for the elements of the first transition series. 4.20 Compare the chemistry of actinoids with that of the lanthanoids with special reference to: (i) electronic configuration (iii) oxidation state (ii) atomic and ionic sizes and (iv) chemical reactivity. 4.21 How would you account for the following: (i) Of the d4 species, Cr2+ is strongly reducing while manganese(III) is strongly oxidising. (ii) Cobalt(II) is stable in aqueous solution but in the presence of complexing reagents it is easily oxidised. (iii) The d1 configuration is very unstable in ions. 4.22 What is meant by ‘disproportionation’? Give two examples of disproportionation reaction in aqueous solution. 4.23 Which metal in the first series of transition metals exhibits +1 oxidation state most frequently and why? 4.24 Calculate the number of unpaired electrons in the following gaseous ions: Mn3+, Cr3+, V3+ and Ti3+. Which one of these is the most stable in aqueous solution? 4.25 Give examples and suggest reasons for the following features of the transition metal chemistry: (i) The lowest oxide of transition metal is basic, the highest is amphoteric/acidic. (ii) A transition metal exhibits highest oxidation state in oxides and fluorides. (iii) The highest oxidation state is exhibited in oxoanions of a metal. 4.26 Indicate the steps in the preparation of: (i) K2Cr2O7 from chromite ore. (ii) KMnO4 from pyrolusite ore. 4.27 What are alloys? Name an important alloy which contains some of the lanthanoid metals. Mention its uses. 4.28 What are inner transition elements? Decide which of the following atomic numbers are the atomic numbers of the inner transition elements : 29, 59, 74, 95, 102, 104. 4.29 The chemistry of the actinoid elements is not so smooth as that of the lanthanoids. Justify this statement by giving some examples from the oxidation state of these elements. 4.30 Which is the last element in the series of the actinoids? Write the electronic configuration of this element. Comment on the possible oxidation state of this element. Chemistry 116 Reprint 2025-26 4.31 Use Hund’s rule to derive the electronic configuration of Ce3+ ion, and calculate its magnetic moment on the basis of ‘spin-only’ formula. 4.32 Name the members of the lanthanoid series which exhibit +4 oxidation states and those which exhibit +2 oxidation states. Try to correlate this type of behaviour with the electronic configurations of these elements. 4.33 Compare the chemistry of the actinoids with that of lanthanoids with reference to: (i) electronic configuration (ii) oxidation states and (iii) chemical reactivity. 4.34 Write the electronic configurations of the elements with the atomic numbers 61, 91, 101, and 109. 4.35 Compare the general characteristics of the first series of the transition metals with those of the second and third series metals in the respective vertical columns. Give special emphasis on the following points: (i) electronic configurations (ii) oxidation states (iii) ionisation enthalpies and (iv) atomic sizes. 4.36 Write down the number of 3d electrons in each of the following ions: Ti 2+, V 2+, Cr3+, Mn 2+, Fe2+, Fe3+, Co2+, Ni2+ and Cu2+. Indicate how would you expect the five 3d orbitals to be occupied for these hydrated ions (octahedral). 4.37 Comment on the statement that elements of the first transition series possess many properties different from those of heavier transition elements. 4.38 What can be inferred from the magnetic moment values of the following complex species ? Example Magnetic Moment (BM) K4[Mn(CN)6) 2.2 [Fe(H2O)6]2+ 5.3 K2[MnCl4] 5.9 Answers to Some Intext Questions 4.1 Silver (Z = 47) can exhibit +2 oxidation state wherein it will have incompletely filled d-orbitals (4d), hence a transition element. 4.2 In the formation of metallic bonds, no eletrons from 3d-orbitals are involved in case of zinc, while in all other metals of the 3d series, electrons from the d-orbitals are always involved in the formation of metallic bonds. 4.3 Manganese (Z = 25), as its atom has the maximum number of unpaired electrons. 4.5 Irregular variation of ionisation enthalpies is mainly attributed to varying degree of stability of different 3d-configurations (e.g., d 0, d 5, d 10 are exceptionally stable). 4.6 Because of small size and high electronegativity oxygen or fluorine can oxidise the metal to its highest oxidation state. 4.7 Cr 2+ is stronger reducing agent than Fe 2+ Reason: d 4  d 3 occurs in case of Cr 2+ to Cr 3+ But d 6  d 5 occurs in case of Fe2+ to Fe 3+ In a medium (like water) d 3 is more stable as compared to d 5 (see CFSE) 4.9 Cu + in aqueous solution underoes disproportionation, i.e., 2Cu +(aq) ® Cu 2+(aq) + Cu(s) The E0 value for this is favourable.