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

1.8 Mole concept and Molar Masses molecules Atoms and molecules are extremely small 1 mol of sodium chloride = 6.022 ×1023 formula in size and their numbers in even a small units of sodium chloride amount of any substance is really very large. Having defined the mole, it is easier toTo handle such large numbers, a unit of convenient magnitude is required. know the mass of one mole of a substance Just as we denote one dozen for 12 items, or the constituent entities. The mass of one score for 20 items, gross for 144 items, we mole of a substance in grams is called its use the idea of mole to count entities at the molar mass. The molar mass in grams is microscopic level (i.e., atoms, molecules, numerically equal to atomic/molecular/ particles, electrons, ions, etc). formula mass in u. In SI system, mole (symbol, mol) was Molar mass of water = 18.02 g mol–1introduced as seventh base quantity for the amount of a substance. Molar mass of sodium chloride = 58.5 g mol–1 The mole, symbol mol, is the SI unit of 1.9 Percentage Compositionamount of substance. One mole contains exactly 6.02214076 × 1023 elementary entities. So far, we were dealing with the number of This number is the fixed numerical value of entities present in a given sample. But many the Avogadro constant, NA, when expressed a time, information regarding the percentage in the unit mol–1 and is called the Avogadro of a particular element present in a compound number. The amount of substance, symbol is required. Suppose, an unknown or new n, of a system is a measure of the number of compound is given to you, the first questionspecified elementary entities. An elementary entity may be an atom, a molecule, an ion, an electron, any other particle or specified group of particles. It may be emphasised that the mole of a substance always contains the same number of entities, no matter what the substance may be. In order to determine this number precisely, the mass of a carbon–12 atom was determined by a mass spectrometer and found to be equal to 1.992648 × 10–23 g. Knowing that one mole of carbon weighs 12 g, the number of atoms in it is equal to: 12 g / mol 12 C 1 .992648  10 23 g /12 Catom  6 .0221367  1023 atoms/mol Fig. 1.11 One mole of various substances Reprint 2025-26 Some Basic Concepts of Chemistry 19 you would ask is: what is its formula or what 1.9.1 Empirical Formula for Molecular are its constituents and in what ratio are they Formula present in the given compound? For known An empirical formula represents the simplest compounds also, such information provides a whole number ratio of various atoms present check whether the given sample contains the in a compound, whereas, the molecular same percentage of elements as present in a formula shows the exact number of different pure sample. In other words, one can check types of atoms present in a molecule of a the purity of a given sample by analysing this compound. data. If the mass per cent of various elements Let us understand it by taking the example present in a compound is known, its empirical of water (H2O). Since water contains hydrogen formula can be determined. Molecular formula and oxygen, the percentage composition of can further be obtained if the molar mass is both these elements can be calculated as known. The following example illustrates follows: this sequence. Mass % of an element = mass of that element in the compound × 100 molar mass of the compound Problem 1.2 A compound contains 4.07% hydrogen,Molar mass of water = 18.02 g 24.27% carbon and 71.65% chlorine. Its molar mass is 98.96 g. What are itsMass % of hydrogen = empirical and molecular formulas? = 11.18 Solution 16.00Mass % of oxygen = × 100 Step 1. Conversion of mass per cent 18.02 to grams = 88.79 Since we are having mass per cent, it is Let us take one more example. What is the convenient to use 100 g of the compound percentage of carbon, hydrogen and oxygen as the starting material. Thus, in the in ethanol? 100 g sample of the above compound, 4.07g hydrogen, 24.27g carbon andMolecular formula of ethanol is: C2H5OH 71.65g chlorine are present.Molar mass of ethanol is: Step 2. Convert into number moles of(2×12.01 + 6×1.008 + 16.00) g = 46.068 g each element Mass per cent of carbon Divide the masses obtained above by 24.02g = × 100 = 52.14% respective atomic masses of various 46.068g elements. This gives the number of Mass per cent of hydrogen moles of constituent elements in the compound 6.048g = × 100 = 13.13% 46.068g 4.07 g Moles of hydrogen = = 4.04 1.008gMass per cent of oxygen 16.00g = × 100 = 34.73% 24.27 g Moles of carbon = = 2 .021 46.068g 12 .01 g After understanding the calculation of 71.65g per cent of mass, let us now see what Moles of chlorine = = 2 .021 35 .453 ginformation can be obtained from the per cent composition data. Reprint 2025-26 20 chemistry equation of a given reaction. Let us consider Step 3. Divide each of the mole values the combustion of methane. A balanced obtained above by the smallest number equation for this reaction is as given below: amongst them CH4 (g) + 2O2 (g) → CO2 (g) + 2 H2O (g) Since 2.021 is smallest value, division Here, methane and dioxygen are called by it gives a ratio of 2:1:1 for H:C:Cl. reactants and carbon dioxide and water are In case the ratios are not whole numbers, called products. Note that all the reactants then they may be converted into whole and the products are gases in the above number by multiplying by the suitable reaction and this has been indicated by coefficient. letter (g) in the brackets next to its formula. Step 4. Write down the empirical Similarly, in case of solids and liquids, (s) and formula by mentioning the numbers (l) are written respectively. after writing the symbols of respective The coefficients 2 for O2 and H2O are elements called stoichiometric coefficients. Similarly CH2Cl is, thus, the empirical formula the coefficient for CH4 and CO2 is one in each of the above compound. case. They represent the number of molecules (and moles as well) taking part in the reaction Step 5. Writing molecular formula or formed in the reaction. (a) Determine empirical formula mass by Thus, according to the above chemical adding the atomic masses of various reaction, atoms present in the empirical formula. For CH2Cl, empirical formula mass is • One mole of CH4(g) reacts with two moles 12.01 + (2 ×1.008) + 35.453 of O2(g) to give one mole of CO2(g) and = 49.48 g two moles of H2O(g) (b) Divide Molar mass by empirical • One molecule of CH4(g) reacts with formula mass 2 molecules of O2(g) to give one molecule of CO2(g) and 2 molecules of H2O(g) • 22.7 L of CH4(g) reacts with 45.4 L of O2 = 2 = (n) (g) to give 22.7 L of CO2 (g) and 45.4 L of (c) Multiply empirical formula by n H2O(g) obtained above to get the molecular • 16 g of CH4 (g) reacts with 2×32 g of O2 formula (g) to give 44 g of CO2 (g) and 2×18 g of Empirical formula = CH2Cl, n = 2. Hence H2O (g). molecular formula is C2H4Cl2. From these relationships, the given data can be interconverted as follows: 1.10 Stoichiometry and mass Stoichiometric Calculations The word ‘stoichiometry’ is derived from two Greek words — stoicheion (meaning, Mass = Density element) and metron (meaning, measure). Volume Stoichiometry, thus, deals with the calculation of masses (sometimes volumes also) of the 1.10.1 Limiting Reagent reactants and the products involved in a Many a time, reactions are carried out with chemical reaction. Before understanding how the amounts of reactants that are different to calculate the amounts of reactants required than the amounts as required by a balanced or the products produced in a chemical chemical reaction. In such situations, one reaction, let us study what information reactant is in more amount than the amount is available from the balanced chemical required by balanced chemical reaction. The Reprint 2025-26 Some Basic Concepts of Chemistry 21 reactant which is present in the least amount important to understand as how the amount gets consumed after sometime and after that of substance is expressed when it is present in further reaction does not take place whatever the solution. The concentration of a solution be the amount of the other reactant. Hence, or the amount of substance present in its the reactant, which gets consumed first, given volume can be expressed in any of the limits the amount of product formed and is, following ways. therefore, called the limiting reagent. 1. Mass per cent or weight per cent (w/w %) In performing stoichiometric calculations, 2. Mole fraction this aspect is also to be kept in mind. 3. Molarity 1.10.2 Reactions in Solutions 4. Molality A majority of reactions in the laboratories Let us now study each one of them in are carried out in solutions. Therefore, it is detail. Balancing a chemical equation According to the law of conservation of mass, a balanced chemical equation has the same number of atoms of each element on both sides of the equation. Many chemical equations can be balanced by trial and error. Let us take the reactions of a few metals and non-metals with oxygen to give oxides 4 Fe(s) + 3O2(g) → 2Fe2O3(s) (a) balanced equation 2 Mg(s) + O2(g) → 2MgO(s) (b) balanced equation P4(s) + O2 (g) → P4O10(s) (c) unbalanced equation Equations (a) and (b) are balanced, since there are same number of metal and oxygen atoms on each side of the equations. However equation (c) is not balanced. In this equation, phosphorus atoms are balanced but not the oxygen atoms. To balance it, we must place the coefficient 5 on the left of oxygen on the left side of the equation to balance the oxygen atoms appearing on the right side of the equation. P4(s) + 5O2(g) → P4O10(s) balanced equation Now, let us take combustion of propane, C3H8. This equation can be balanced in steps. Step 1 Write down the correct formulas of reactants and products. Here, propane and oxygen are reactants, and carbon dioxide and water are products. C3H8(g) + O2(g) → CO2 (g) + H2O(l) unbalanced equation Step 2 Balance the number of C atoms: Since 3 carbon atoms are in the reactant, therefore, three CO2 molecules are required on the right side. C3H8 (g) + O2 (g) → 3CO2 (g) + H2O (l) Step 3 Balance the number of H atoms: on the left there are 8 hydrogen atoms in the reactants however, each molecule of water has two hydrogen atoms, so four molecules of water will be required for eight hydrogen atoms on the right side. C3H8 (g) +O2 (g) → 3CO2 (g)+4H2O (l) Step 4 Balance the number of O atoms: There are 10 oxygen atoms on the right side (3 × 2 = 6 in CO2 and 4×1= 4 in water). Therefore, five O2 molecules are needed to supply the required 10 CO2 and 4×1= 4 in water). Therefore, five O2 molecules are needed to supply the required 10 oxygen atoms. C3H8 (g) +5O2 (g) → 3CO2 (g) + 4H2O (l) Step 5 Verify that the number of atoms of each element is balanced in the final equation. The equation shows three carbon atoms, eight hydrogen atoms, and 10 oxygen atoms on each side. All equations that have correct formulas for all reactants and products can be balanced. Always remember that subscripts in formulas of reactants and products cannot be changed to balance an equation. Reprint 2025-26 22 chemistry Problem 1.3 the limiting reagent in the production of NH3 in this situation. Calculate the amount of water (g) produced by the combustion of 16 g Solution of methane. A balanced equation for the above Solution reaction is written as follows : The balanced equation for the combustion of methane is : CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (g) Calculation of moles : (i) 16 g of CH4 corresponds to one mole. Number of moles of N2 (ii) From the above equation, 1 mol of 1000 g N 2 1 mol N 2 × 2 × CH4 (g) gives 2 mol of H2O (g). = 50 .0 kg N 1 kg N 2 28 .0 g N 2 2 mol of water (H2O) = 2×(2+16) = 17.86×102 mol = 2×18 = 36 g Number of moles of H2 H 2 O ⇒18g 1000 g H 2 1 mol H 2 = 1 1 mol H2O = 18 g H2O × = 10 .00 kg H 2 × 1mol H 2 O 1 kg H 2 2 .016 g H 2 18g H 2 O = 4.96 × 103 mol Hence, 2 mol H2O× 1mol H 2 O According to the above equation, 1 mol N2 (g) requires 3 mol H2 (g), for the = 2×18 g H2O = 36 g H2O reaction. Hence, for 17.86×102 mol of Problem 1.4 N2, the moles of H2 (g) required would be How many moles of methane are 2 3 mol H 2 g 2  required to produce 22g CO2 (g) after 17 .86  10 mol N 1 mol N 2 g combustion? = 5.36 × 103 mol H2 Solution But we have only 4.96×103 mol H2. According to the chemical equation, Hence, dihydrogen is the limiting CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (g) reagent in this case. So, NH3(g) would be formed only from that amount of 44g CO2 (g) is obtained from 16 g CH4 (g). available dihydrogen i.e., 4.96 × 103 mol [∴1 mol CO2(g) is obtained from 1 mol Since 3 mol H2(g) gives 2 mol NH3(g) of CH4(g)] Number of moles of CO2 (g) 2 mol NH 3 g 2 g 4.96 ×103 mol H2 (g) × 3 mol H 2 g = 22 g CO2 (g) ×1 molCO 44 gCO 2 g = 3.30 × 103 mol NH3 (g) = 0.5 mol CO2 (g) 3.30 × 103 mol NH3 (g) is obtained. Hence, 0.5 mol CO2 (g) would be obtained from 0.5 mol CH4 (g) or 0.5 mol If they are to be converted to grams, it of CH4 (g) would be required to produce is done as follows : 22 g CO2 (g). 1 mol NH3 (g) = 17.0 g NH3 (g) Problem 1.5 .0 g NH 3 g 50.0 kg of N2 (g) and 10.0 kg of H2 (g) 3.30 ×103 mol NH3 (g) ×17 3 g are mixed to produce NH3 (g). Calculate 1 mol NH the amount of NH3 (g) formed. Identify Reprint 2025-26 Some Basic Concepts of Chemistry 23 3. Molarity = 3.30×103×17 g NH3 (g) It is the most widely used unit and is denoted = 56.1×103 g NH3 by M. It is defined as the number of moles of = 56.1 kg NH3 the solute in 1 litre of the solution. Thus, No. of moles of solute 1. Mass per cent Molarity (M) = Volume of solution in litres It is obtained by using the following relation: Suppose, we have 1 M solution of a substance, say NaOH, and we want to prepare a 0.2 M solution from it. 1 M NaOH means 1 mol of NaOH present in 1 litre of the solution. For 0.2 M solution, Problem 1.6 we require 0.2 moles of NaOH dissolved in 1 litre solution. A solution is prepared by adding 2 g of a substance A to 18 g of water. Calculate Hence, for making 0.2M solution from 1M the mass per cent of the solute. solution, we have to take that volume of 1M NaOH solution, which contains 0.2 mol of NaOH Solution and dilute the solution with water to 1 litre. Now, how much volume of concentrated (1M) NaOH solution be taken, which contains 0.2 moles of NaOH can be calculated as follows: If 1 mol is present in 1L or 1000 mL solution then, 0.2 mol is present in 1000 mL × 0 .2 mol solution 1 mol 2. Mole Fraction = 200 mL solutionIt is the ratio of number of moles of a particular component to the total number Thus, 200 mL of 1M NaOH are taken and enough water is added to dilute it to make it 1 litre.of moles of the solution. If a substance ‘A’ dissolves in substance ‘B’ and their number In fact for such calculations, a general of moles are nA and nB, respectively, then the formula, M1×V1 = M2 × V2 where M and V are molarity and volume, respectively, can be used.mole fractions of A and B are given as: In this case, M1 is equal to 0.2M; V1 = 1000 mL and, M2 = 1.0M; V2 is to be calculated. Substituting the values in the formula: 0.2 M × 1000 mL = 1.0 M × V2 Note that the number of moles of solute (NaOH) was 0.2 in 200 mL and it has remained the same, i.e., 0.2 even after dilution ( in 1000 mL) as we have changed just the amount of solvent (i.e., water) and have not done anything with respect to NaOH. But keep in mind the concentration. Reprint 2025-26 24 chemistry Problem 1.7 Problem 1.8 Calculate the molarity of NaOH in the The density of 3 M solution of NaCl is solution prepared by dissolving its 4 g 1.25 g mL–1. Calculate the molality of in enough water to form 250 mL of the the solution. solution. Solution Solution M = 3 mol L–1 Since molarity (M) Mass of NaCl in 1 L solution = 3 × 58.5 = 175.5 g Mass of 1L solution = 1000 × 1.25 = 1250 g (since density = 1.25 g mL–1) Mass of water in solution = 1250 –75.5 = 1074.5 g No. of moles of solute Molality = Mass of solvent in kg 3 mol = = 2.79 m Note that molarity of a solution depends 1 .0745 kg upon temperature because volume of a Often in a chemistry laboratory, a solution is temperature dependent. solution of a desired concentration is prepared by diluting a solution of known 4. Molality higher concentration. The solution of higher concentration is also known as It is defined as the number of moles of solute stock solution. Note that the molality present in 1 kg of solvent. It is denoted by m. of a solution does not change with temperature since mass remains No. of moles of solute Thus, Molality (m) = unaffected with temperature. Mass of solvent in kg Summary Chemistry, as we understand it today is not a very old discipline. People in ancient India, already had the knowledge of many scientific phenomenon much before the advent of modern science. They applied the knowledge in various walks of life. The study of chemistry is very important as its domain encompasses every sphere of life. Chemists study the properties and structure of substances and the changes undergone by them. All substances contain matter, which can exist in three states – solid, liquid or gas. The constituent particles are held in different ways in these states of matter and they exhibit their characteristic properties. Matter can also be classified into elements, compounds or mixtures. An element contains particles of only one type, which may be atoms or molecules. The compounds are formed where atoms of two or more elements combine in a fixed ratio to each other. Mixtures occur widely and many of the substances present around us are mixtures. When the properties of a substance are studied, measurement is inherent. The quantification of properties requires a system of measurement and units in which the quantities are to be expressed. Many systems of measurement exist, of which the English Reprint 2025-26 Some Basic Concepts of Chemistry 25 and the Metric Systems are widely used. The scientific community, however, has agreed to have a uniform and common system throughout the world, which is abbreviated as SI units (International System of Units). Since measurements involve recording of data, which are always associated with a certain amount of u is very important. The measurements of quantities in chemistry are spread over a wide range of 10–31 to 10+23. Hence, a convenient system of expressing the numbers in scientific notation is used. The u figures, in which the observations are reported. The dimensional analysis helps to express the measured quantities in different systems of units. Hence, it is possible to interconvert the results from one system of units to another. The combination of different atoms is governed by basic laws of chemical combination — these being the Law of Conservation of Mass, Law of Definite Proportions, Law of Multiple Proportions, Gay Lussac’s Law of Gaseous Volumes and Avogadro Law. All these laws led to the Dalton’s atomic theory, which states that atoms are building blocks of matter. The atomic mass of an element is expressed relative to 12C isotope of carbon, which has an exact value of 12u. Usually, the atomic mass used for an element is the average atomic mass obtained by taking into account the natural abundance of different isotopes of that element. The molecular mass of a molecule is obtained by taking sum of the atomic masses of different atoms present in a molecule. The molecular formula can be calculated by determining the mass per cent of different elements present in a compound and its molecular mass. The number of atoms, molecules or any other particles present in a given system are expressed in the terms of Avogadro constant (6.022 × 1023). This is known as 1 mol of the respective particles or entities. Chemical reactions represent the chemical changes undergone by different elements and compounds. A balanced chemical equation provides a lot of information. The coefficients indicate the molar ratios and the respective number of particles taking part in a particular reaction. The quantitative study of the reactants required or the products formed is called stoichiometry. Using stoichiometric calculations, the amount of one or more reactant(s) required to produce a particular amount of product can be determined and vice-versa. The amount of substance present in a given volume of a solution is expressed in number of ways, e.g., mass per cent, mole fraction, molarity and molality. exerciseS 1.1 Calculate the molar mass of the following: (i) H2O (ii) CO2 (iii) CH4 1.2 Calculate the mass per cent of different elements present in sodium sulphate (Na2SO4). 1.3 Determine the empirical formula of an oxide of iron, which has 69.9% iron and 30.1% dioxygen by mass. 1.4 Calculate the amount of carbon dioxide that could be produced when (i) 1 mole of carbon is burnt in air. (ii) 1 mole of carbon is burnt in 16 g of dioxygen. (iii) 2 moles of carbon are burnt in 16 g of dioxygen. 1.5 Calculate the mass of sodium acetate (CH3COONa) required to make 500 mL of 0.375 molar aqueous solution. Molar mass of sodium acetate is 82.0245 g mol–1. Reprint 2025-26 26 chemistry 1.6 Calculate the concentration of nitric acid in moles per litre in a sample which has a density, 1.41 g mL–1 and the mass per cent of nitric acid in it being 69%. 1.7 How much copper can be obtained from 100 g of copper sulphate (CuSO4)? 1.8 Determine the molecular formula of an oxide of iron, in which the mass per cent of iron and oxygen are 69.9 and 30.1, respectively. 1.9 Calculate the atomic mass (average) of chlorine using the following data: % Natural Abundance Molar Mass 35Cl 75.77 34.9689 37Cl 24.23 36.9659 1.10 In three moles of ethane (C2H6), calculate the following: (i) Number of moles of carbon atoms. (ii) Number of moles of hydrogen atoms. (iii) Number of molecules of ethane. 1.11 What is the concentration of sugar (C12H22O11) in mol L–1 if its 20 g are dissolved in enough water to make a final volume up to 2L? 1.12 If the density of methanol is 0.793 kg L–1, what is its volume needed for making 2.5 L of its 0.25 M solution? 1.13 Pressure is determined as force per unit area of the surface. The SI unit of pressure, pascal is as shown below: 1Pa = 1N m–2 If mass of air at sea level is 1034 g cm–2, calculate the pressure in pascal. 1.14 What is the SI unit of mass? How is it defined? 1.15 Match the following prefixes with their multiples: Prefixes Multiples (i) micro 106 (ii) deca 109 (iii) mega 10–6 (iv) giga 10–15 (v) femto 10 1.16 What do you mean by significant figures? 1.17 A sample of drinking water was found to be severely contaminated with chloroform, CHCl3, supposed to be carcinogenic in nature. The level of contamination was 15 ppm (by mass). (i) Express this in per cent by mass. (ii) Determine the molality of chloroform in the water sample. 1.18 Express the following in the scientific notation: (i) 0.0048 (ii) 234,000 (iii) 8008 (iv) 500.0 (v) 6.0012 1.19 How many significant figures are present in the following? (i) 0.0025 (ii) 208 (iii) 5005 Reprint 2025-26 Some Basic Concepts of Chemistry 27 (iv) 126,000 (v) 500.0 (vi) 2.0034 1.20 Round up the following upto three significant figures: (i) 34.216 (ii) 10.4107 (iii) 0.04597 (iv) 2808 1.21 The following data are obtained when dinitrogen and dioxygen react together to form different compounds: Mass of dinitrogen Mass of dioxygen (i) 14 g 16 g (ii) 14 g 32 g (iii) 28 g 32 g (iv) 28 g 80 g (a) Which law of chemical combination is obeyed by the above experimental data? Give its statement. (b) Fill in the blanks in the following conversions: (i) 1 km = ...................... mm = ...................... pm (ii) 1 mg = ...................... kg = ...................... ng (iii) 1 mL = ...................... L = ...................... dm3 1.22 If the speed of light is 3.0 × 108 m s–1, calculate the distance covered by light in 2.00 ns. 1.23 In a reaction A + B2  AB2 Identify the limiting reagent, if any, in the following reaction mixtures. (i) 300 atoms of A + 200 molecules of B (ii) 2 mol A + 3 mol B (iii) 100 atoms of A + 100 molecules of B (iv) 5 mol A + 2.5 mol B (v) 2.5 mol A + 5 mol B 1.24 Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical equation: N2 (g) + H2 (g)  2NH3 (g) (i) Calculate the mass of ammonia produced if 2.00 × 103 g dinitrogen reacts with 1.00 × 103 g of dihydrogen. (ii) Will any of the two reactants remain unreacted? (iii) If yes, which one and what would be its mass? 1.25 How are 0.50 mol Na2CO3 and 0.50 M Na2CO3 different? 1.26 If 10 volumes of dihydrogen gas reacts with five volumes of dioxygen gas, how many volumes of water vapour would be produced? 1.27 Convert the following into basic units: (i) 28.7 pm (ii) 15.15 pm (iii) 25365 mg Reprint 2025-26 28 chemistry 1.28 Which one of the following will have the largest number of atoms? (i) 1 g Au (s) (ii) 1 g Na (s) (iii) 1 g Li (s) (iv) 1 g of Cl2(g) 1.29 Calculate the molarity of a solution of ethanol in water, in which the mole fraction of ethanol is 0.040 (assume the density of water to be one). 1.30 What will be the mass of one 12C atom in g? 1.31 How many significant figures should be present in the answer of the following calculations? 0.02856 × 298.15 × 0.112 (i) (ii) 5 × 5.364 0 .5785 (iii) 0.0125 + 0.7864 + 0.0215 1.32 Use the data given in the following table to calculate the molar mass of naturally occuring argon isotopes: Isotope Isotopic molar mass Abundance 36Ar 35.96755 g mol–1 0.337% 38Ar 37.96272 g mol–1 0.063% 40Ar 39.9624 g mol–1 99.600% 1.33 Calculate the number of atoms in each of the following (i) 52 moles of Ar (ii) 52 u of He (iii) 52 g of He. 1.34 A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen gives 3.38 g carbon dioxide, 0.690 g of water and no other products. A volume of 10.0 L (measured at STP) of this welding gas is found to weigh 11.6 g. Calculate (i) empirical formula, (ii) molar mass of the gas, and (iii) molecular formula. 1.35 Calcium carbonate reacts with aqueous HCl to give CaCl2 and CO2 according to the reaction, CaCO3 (s) + 2 HCl (aq) → CaCl2 (aq) + CO2(g) + H2O(l) What mass of CaCO3 is required to react completely with 25 mL of 0.75 M HCl? 1.36 Chlorine is prepared in the laboratory by treating manganese dioxide (MnO2) with aqueous hydrochloric acid according to the reaction 4 HCl (aq) + MnO2(s) → 2H2O (l) + MnCl2(aq) + Cl2 (g) How many grams of HCl react with 5.0 g of manganese dioxide? Reprint 2025-26 Unit 2 structure of atom The rich diversity of chemical behaviour of different Objectives elementsstructure canof atomsbe tracedof theseto theelements.differences in the internal After studying this unit you will be able to • know about the discovery of The existence of atoms has been proposed since the time electron, proton and neutron and of early Indian and Greek philosophers (400 B.C.) who their characteristics; were of the view that atoms are the fundamental building • describe Thomson, Rutherford blocks of matter. According to them, the continued and Bohr atomic models; subdivisions of matter would ultimately yield atoms which would not be further divisible. The word ‘atom’ • understand the important features has been derived from the Greek word ‘a-tomio’ which of the quantum mechanical model means ‘uncut-able’ or ‘non-divisible’. These earlier ideas of atom; were mere speculations and there was no way to test • u n d e r s t a n d n a t u r e o f them experimentally. These ideas remained dormant for electromagnetic radiation and a very long time and were revived again by scientists in Planck’s quantum theory; the nineteenth century. • explain the photoelectric effect The atomic theory of matter was first proposed and describe features of atomic on a firm scientific basis by John Dalton, a British spectra; school teacher in 1808. His theory, called Dalton’s • state the de Broglie relation and atomic theory, regarded the atom as the ultimate Heisenberg u able to explain the law of conservation of mass, law of • define an atomic orbital in terms constant composition and law of multiple proportion of quantum numbers; very successfully. However, it failed to explain the results • state aufbau principle, Pauli of many experiments, for example, it was known that exclusion principle and Hund’s substances like glass or ebonite when rubbed with silk rule of maximum multiplicity; and or fur get electrically charged. • write the electronic configurations In this unit we start with the experimental observations of atoms. made by scientists towards the end of nineteenth and beginning of twentieth century. These established that atoms are made of sub-atomic particles, i.e., electrons, protons and neutrons — a concept very different from that of Dalton. Reprint 2025-26 30 chemistry

6.11 Dynamics of rotational the motion of extended bodies. motion about a fixed axis A large class of problems with extended bodies can be