75.4 and 24.6 per cent, respectively. Thus, the average mass of a chlorine atom is obtained by the weighted average of the masses of the two isotopes, which works out to be 75.4 × 34.98 + 24.6 × 36.98 = 100 = 35.47 u which agrees with the atomic mass of chlorine. Even the lightest element, hydrogen has three isotopes having masses 1.0078 u, 2.0141 u, and 3.0160 u. The nucleus of the lightest atom of hydrogen, which has a relative abundance of 99.985%, is called the proton. The mass of a proton is −27 m p = 1.00727 u = 1.67262 × 10 kg (13.2) This is equal to the mass of the hydrogen atom (= 1.00783u), minus the mass of a single electron (me = 0.00055 u). The other two isotopes of hydrogen are called deuterium and tritium. Tritium nuclei, being unstable, do not occur naturally and are produced artificially in laboratories. The positive charge in the nucleus is that of the protons. A proton carries one unit of fundamental charge and is stable. It was earlier thought that the nucleus may contain electrons, but this was ruled out later using arguments based on quantum theory. All the electrons of an atom are outside the nucleus. We know that the number of these electrons outside 307the nucleus of the atom is Z, the atomic number. The total charge of the Reprint 2025-26 Physics atomic electrons is thus (–Ze), and since the atom is neutral, the charge of the nucleus is (+Ze). The number of protons in the nucleus of the atom is, therefore, exactly Z, the atomic number. Discovery of Neutron Since the nuclei of deuterium and tritium are isotopes of hydrogen, they must contain only one proton each. But the masses of the nuclei of hydrogen, deuterium and tritium are in the ratio of 1:2:3. Therefore, the nuclei of deuterium and tritium must contain, in addition to a proton, some neutral matter. The amount of neutral matter present in the nuclei of these isotopes, expressed in units of mass of a proton, is approximately equal to one and two, respectively. This fact indicates that the nuclei of atoms contain, in addition to protons, neutral matter in multiples of a basic unit. This hypothesis was verified in 1932 by James Chadwick who observed emission of neutral radiation when beryllium nuclei were bombarded with alpha-particles (a-particles are helium nuclei, to be discussed in a later section). It was found that this neutral radiation could knock out protons from light nuclei such as those of helium, carbon and nitrogen. The only neutral radiation known at that time was photons (electromagnetic radiation). Application of the principles of conservation of energy and momentum showed that if the neutral radiation consisted of photons, the energy of photons would have to be much higher than is available from the bombardment of beryllium nuclei with a-particles. The clue to this puzzle, which Chadwick satisfactorily solved, was to assume that the neutral radiation consists of a new type of neutral particles called neutrons. From conservation of energy and momentum, he was able to determine the mass of new particle ‘as very nearly the same as mass of proton’. The mass of a neutron is now known to a high degree of accuracy. It is m n = 1.00866 u = 1.6749×10–27 kg (13.3) Chadwick was awarded the 1935 Nobel Prize in Physics for his discovery of the neutron. A free neutron, unlike a free proton, is unstable. It decays into a proton, an electron and a antineutrino (another elementary particle), and has a mean life of about 1000s. It is, however, stable inside the nucleus. The composition of a nucleus can now be described using the following terms and symbols: Z - atomic number = number of protons [13.4(a)] N - neutron number = number of neutrons [13.4(b)] A - mass number = Z + N = total number of protons and neutrons [13.4(c)] One also uses the term nucleon for a proton or a neutron. Thus the number of nucleons in an atom is its mass number A. Nuclear species or nuclides are shown by the notation ZA X where X is the chemical symbol of the species. For example, the nucleus of gold is denoted by 19779 Au . It contains 197 nucleons, of which 79 are protons 308 and the rest118 are neutrons. Reprint 2025-26 Nuclei The composition of isotopes of an element can now be readily explained. The nuclei of isotopes of a given element contain the same number of protons, but differ from each other in their number of neutrons. Deuterium, 12 H, which is an isotope of hydrogen, contains one proton and one neutron. Its other isotope tritium, 13 H, contains one proton and two neutrons. The element gold has 32 isotopes, ranging from A =173 to A = 204. We have already mentioned that chemical properties of elements depend on their electronic structure. As the atoms of isotopes have identical electronic structure they have identical chemical behaviour and are placed in the same location in the periodic table. All nuclides with same mass number A are called isobars. For example, the nuclides 13 H and 32He are isobars. Nuclides with same neutron number N but different atomic number Z, for example 19880 Hg and 19779 Au , are called isotones.

4.9 Hydrogen Bonding Hydrogen bond is represented by a dotted line (– – –) while a solid line represents theNitrogen, oxygen and fluorine are the highly covalent bond. Thus, hydrogen bond can beelectronegative elements. When they are attached to a hydrogen atom to form covalent defined as the attractive force which binds bond, the electrons of the covalent bond are hydrogen atom of one molecule with the shifted towards the more electronegative electronegative atom (F, O or N) of another atom. This partially positively charged molecule. hydrogen atom forms a bond with the other 4.9.1 Cause of Formation of Hydrogen more electronegative atom. This bond is Bond known as hydrogen bond and is weaker When hydrogen is bonded to stronglythan the covalent bond. For example, in HF electronegative element ‘X’, the electron pairmolecule, the hydrogen bond exists between shared between the two atoms moves farhydrogen atom of one molecule and fluorine away from hydrogen atom. As a result theatom of another molecule as depicted below : hydrogen atom becomes highly electropositive – – – Hδ+–Fδ– – – –Hδ+ – Fδ– – – – Hδ+ – Fδ– with respect to the other atom ‘X’. Since Here, hydrogen bond acts as a bridge between there is displacement of electrons towards two atoms which holds one atom by covalent X, the hydrogen acquires fractional positive bond and the other by hydrogen bond. charge (δ +) while ‘X’ attain fractional negative Reprint 2025-26 132 chemistry charge (δ–). This results in the formation of a H-bond in case of HF molecule, alcohol or polar molecule having electrostatic force of water molecules, etc. attraction which can be represented as: (2) Intramolecular hydrogen bond : It is formed when hydrogen atom is in between Hδ+ – Xδ– – – – Hδ+ – Xδ– – – – Hδ+ – Xδ– the two highly electronegative (F, O, N) The magnitude of H-bonding depends atoms present within the same molecule. For on the physical state of the compound. It is example, in o-nitrophenol the hydrogen is in maximum in the solid state and minimum in between the two oxygen atoms. the gaseous state. Thus, the hydrogen bonds have strong influence on the structure and properties of the compounds. 4.9.2 Types of H-Bonds There are two types of H-bonds (i) Intermolecular hydrogen bond (ii) Intramolecular hydrogen bond (1) Intermolecular hydrogen bond : It is formed between two different molecules of the Fig. 4.22 Intramolecular hydrogen bonding in same or different compounds. For example, o-nitrophenol molecule SUMMARY Kössel’s first insight into the mechanism of formation of electropositive and electronegative ions related the process to the attainment of noble gas configurations by the respective ions. Electrostatic attraction between ions is the cause for their stability. This gives the concept of electrovalency. The first description of covalent bonding was provided by Lewis in terms of the sharing of electron pairs between atoms and he related the process to the attainment of noble gas configurations by reacting atoms as a result of sharing of electrons. The Lewis dot symbols show the number of valence electrons of the atoms of a given element and Lewis dot structures show pictorial representations of bonding in molecules. An ionic compound is pictured as a three-dimensional aggregation of positive and negative ions in an ordered arrangement called the crystal lattice. In a crystalline solid there is a charge balance between the positive and negative ions. The crystal lattice is stabilized by the enthalpy of lattice formation. While a single covalent bond is formed by sharing of an electron pair between two atoms, multiple bonds result from the sharing of two or three electron pairs. Some bonded atoms have additional pairs of electrons not involved in bonding. These are called lone-pairs of electrons. A Lewis dot structure shows the arrangement of bonded pairs and lone pairs around each atom in a molecule. Important parameters, associated with chemical bonds, like: bond length, bond angle, bond enthalpy, bond order and bond polarity have significant effect on the properties of compounds. A number of molecules and polyatomic ions cannot be described accurately by a single Lewis structure and a number of descriptions (representations) based on the same skeletal structure are written and these taken together represent the molecule or ion. This is a very important and extremely useful concept called resonance. The contributing structures or canonical forms taken together constitute the resonance hybrid which represents the molecule or ion. Reprint 2025-26 Chemical Bonding And Molecular Structure 133 The VSEPR model used for predicting the geometrical shapes of molecules is based on the assumption that electron pairs repel each other and, therefore, tend to remain as far apart as possible. According to this model, molecular geometry is determined by repulsions between lone pairs and lone pairs; lone pairs and bonding pairs and bonding pairs and bonding pairs. The order of these repulsions being : lp-lp > lp-bp > bp-bp The valence bond (VB) approach to covalent bonding is basically concerned with the energetics of covalent bond formation about which the Lewis and VSEPR models are silent. Basically the VB theory discusses bond formation in terms of overlap of orbitals. For example the formation of the H2 molecule from two hydrogen atoms involves the overlap of the 1s orbitals of the two H atoms which are singly occupied. It is seen that the potential energy of the system gets lowered as the two H atoms come near to each other. At the equilibrium inter-nuclear distance (bond distance) the energy touches a minimum. Any attempt to bring the nuclei still closer results in a sudden increase in energy and consequent destabilization of the molecule. Because of orbital overlap the electron density between the nuclei increases which helps in bringing them closer. It is however seen that the actual bond enthalpy and bond length values are not obtained by overlap alone and other variables have to be taken into account. For explaining the characteristic shapes of polyatomic molecules Pauling introduced the concept of hybridisation of atomic orbitals. sp, sp2, sp3 hybridizations of atomic orbitals of Be, B, C, N and O are used to explain the formation and geometrical shapes of molecules like BeCl2, BCl3, CH4, NH3 and H2O. They also explain the formation of multiple bonds in molecules like C2H2 and C2H4. The molecular orbital (MO) theory describes bonding in terms of the combination and arrangment of atomic orbitals to form molecular orbitals that are associated with the molecule as a whole. The number of molecular orbitals are always equal to the number of atomic orbitals from which they are formed. Bonding molecular orbitals increase electron density between the nuclei and are lower in energy than the individual atomic orbitals. Antibonding molecular orbitals have a region of zero electron density between the nuclei and have more energy than the individual atomic orbitals. The electronic configuration of the molecules is written by filling electrons in the molecular orbitals in the order of increasing energy levels. As in the case of atoms, the Pauli exclusion principle and Hund’s rule are applicable for the filling of molecular orbitals. Molecules are said to be stable if the number of elctrons in bonding molecular orbitals is greater than that in antibonding molecular orbitals. Hydrogen bond is formed when a hydrogen atom finds itself between two highly electronegative atoms such as F, O and N. It may be intermolecular (existing between two or more molecules of the same or different substances) or intramolecular (present within the same molecule). Hydrogen bonds have a powerful effect on the structure and properties of many compounds. EXERCISES 4.1 Explain the formation of a chemical bond. 4.2 Write Lewis dot symbols for atoms of the following elements : Mg, Na, B, O, N, Br. 4.3 Write Lewis symbols for the following atoms and ions: S and S2–; Al and Al3+; H and H– 4.4 Draw the Lewis structures for the following molecules and ions : H2S, SiCl4, BeF2, CO32−, HCOOH 4.5 Define octet rule. Write its significance and limitations. Reprint 2025-26 134 chemistry 4.6 Write the favourable factors for the formation of ionic bond. 4.7 Discuss the shape of the following molecules using the VSEPR model: BeCl2, BCl3, SiCl4, AsF5, H2S, PH3 4.8 Although geometries of NH3 and H2O molecules are distorted tetrahedral, bond angle in water is less than that of ammonia. Discuss. 4.9 How do you express the bond strength in terms of bond order ? 4.10 Define the bond length. 4.11 Explain the important aspects of resonance with reference to the CO32− ion. 4.12 H3PO3 can be represented by structures 1 and 2 shown below. Can these two structures be taken as the canonical forms of the resonance hybrid representing H3PO3 ? If not, give reasons for the same. 4.13 Write the resonance structures for SO3, NO2 and NO3−. 4.14 Use Lewis symbols to show electron transfer between the following atoms to form cations and anions : (a) K and S (b) Ca and O (c) Al and N. 4.15 Although both CO2 and H2O are triatomic molecules, the shape of H2O molecule is bent while that of CO2 is linear. Explain this on the basis of dipole moment. 4.16 Write the significance/applications of dipole moment. 4.17 Define electronegativity. How does it differ from electron gain enthalpy ? 4.18 Explain with the help of suitable example polar covalent bond. 4.19 Arrange the bonds in order of increasing ionic character in the molecules: LiF, K2O, N2, SO2 and ClF3. 4.20 The skeletal structure of CH3COOH as shown below is correct, but some of the bonds are shown incorrectly. Write the correct Lewis structure for acetic acid. 4.21 Apart from tetrahedral geometry, another possible geometry for CH4 is square planar with the four H atoms at the corners of the square and the C atom at its centre. Explain why CH4 is not square planar ? 4.22 Explain why BeH2 molecule has a zero dipole moment although the Be–H bonds are polar. 4.23 Which out of NH3 and NF3 has higher dipole moment and why ? 4.24 What is meant by hybridisation of atomic orbitals? Describe the shapes of sp, sp2, sp3 hybrid orbitals. 4.25 Describe the change in hybridisation (if any) of the Al atom in the following reaction. AlCl 3  Cl   AlCl 4 Reprint 2025-26 Chemical Bonding And Molecular Structure 135 4.26 Is there any change in the hybridisation of B and N atoms as a result of the following reaction? 4.27 Draw diagrams showing the formation of a double bond and a triple bond between carbon atoms in C2H4 and C2H2 molecules. 4.28 What is the total number of sigma and pi bonds in the following molecules? (a) C2H2 (b) C2H4 4.29 Considering x-axis as the internuclear axis which out of the following will not form a sigma bond and why? (a) 1s and 1s (b) 1s and 2px; (c) 2py and 2py (d) 1s and 2s. 4.30 Which hybrid orbitals are used by carbon atoms in the following molecules? CH3–CH3; (b) CH3–CH=CH2; (c) CH3-CH2-OH; (d) CH3-CHO (e) CH3COOH 4.31 What do you understand by bond pairs and lone pairs of electrons? Illustrate by giving one exmaple of each type. 4.32 Distinguish between a sigma and a pi bond. 4.33 Explain the formation of H2 molecule on the basis of valence bond theory. 4.34 Write the important conditions required for the linear combination of atomic orbitals to form molecular orbitals. 4.35 Use molecular orbital theory to explain why the Be2 molecule does not exist. 4.36 Compare the relative stability of the following species and indicate their magnetic properties; (superoxide), O22− (peroxide) 4.37 Write the significance of a plus and a minus sign shown in representing the orbitals. 4.38 Describe the hybridisation in case of PCl5. Why are the axial bonds longer as compared to equatorial bonds? 4.39 Define hydrogen bond. Is it weaker or stronger than the van der Waals forces? 4.40 What is meant by the term bond order? Calculate the bond order of : N2, O2, O2+ and O2–. Reprint 2025-26 Unit 5 Thermodynamics It is the only physical theory of universal content concerning which I am convinced that, within the framework of the applicability of its basic concepts, it will never be overthrown. After studying this Unit, you will be Albert Einstein able to • explain the terms : system and surroundings; • discriminate between close, open and isolated systems; Chemical energy stored by molecules can be released as• explain internal energy, work and heat; heat during chemical reactions when a fuel like methane, • state first law of thermodynamics cooking gas or coal burns in air. The chemical energy may and express it mathematically; also be used to do mechanical work when a fuel burns • calculate energy changes as in an engine or to provide electrical energy through a work and heat contributions in galvanic cell like dry cell. Thus, various forms of energy chemical systems; are interrelated and under certain conditions, these may • explain state functions: U, H. be transformed from one form into another. The study • correlate ∆U and ∆H; of these energy transformations forms the subject matter • measure experimentally ∆U and of thermodynamics. The laws of thermodynamics deal ∆H; with energy changes of macroscopic systems involving• define standard states for ∆H; • calculate enthalpy changes for a large number of molecules rather than microscopic various types of reactions; systems containing a few molecules. Thermodynamics is • state and apply Hess’s law of not concerned about how and at what rate these energy constant heat summation; transformations are carried out, but is based on initial and • differentiate between extensive final states of a system undergoing the change. Laws of and intensive properties; thermodynamics apply only when a system is in equilibrium • define spontaneous and non- or moves from one equilibrium state to another equilibrium spontaneous processes; state. Macroscopic properties like pressure and temperature• e x p l a i n e n t r o p y a s a thermodynamic state function do not change with time for a system in equilibrium state. and apply it for spontaneity; In this unit, we would like to answer some of the important • explain Gibbs energy change (∆G); questions through thermodynamics, like: and How do we determine the energy changes involved in a • establish relationship between chemical reaction/process? Will it occur or not? ∆G and spontaneity, ∆G and equilibrium constant. What drives a chemical reaction/process? To what extent do the chemical reactions proceed? Reprint 2025-26 THERMODYNAMICS 137

1.6 Dalton’s Atomic Theory 1.7.1 Atomic Mass Although the origin of the idea that matter is The atomic mass or the mass of an atom is composed of small indivisible particles called actually very-very small because atoms are ‘a-tomio’ (meaning, indivisible), dates back extremely small. Today, we have sophisticated to the time of Democritus, techniques e.g., mass spectrometry for a Greek Philosopher (460– determining the atomic masses fairly 370 BC), it again started accurately. But in the nineteenth century, emerging as a result of several scientists could determine the mass of one experimental studies which atom relative to another by experimental led to the laws mentioned means, as has been mentioned earlier. above. Hydrogen, being the lightest atom was arbitrarily assigned a mass of 1 (without In 1808, Dalton published John Dalton any units) and other elements were assigned‘A New System of Chemical (1776–1884) masses relative to it. However, the presentPhilosophy’, in which he system of atomic masses is based onproposed the following : carbon-12 as the standard and has been1. Matter consists of indivisible atoms. agreed upon in 1961. Here, Carbon-12 is2. All atoms of a given element have identical one of the isotopes of carbon and can be properties, including identical mass. Atoms represented as 12C. In this system, 12C is of different elements differ in mass. assigned a mass of exactly 12 atomic mass 3. Compounds are formed when atoms of unit (amu) and masses of all other atoms are different elements combine in a fixed ratio. given relative to this standard. One atomic 4. Chemical reactions involve reorganisation mass unit is defined as a mass exactly equal of atoms. These are neither created nor to one-twelfth of the mass of one carbon – 12 destroyed in a chemical reaction. atom. Reprint 2025-26 Some Basic Concepts of Chemistry 17 And 1 amu = 1.66056×10–24 g 1.7.3 Molecular Mass Mass of an atom of hydrogen Molecular mass is the sum of atomic masses of the elements present in a molecule. It is = 1.6736×10–24 g obtained by multiplying the atomic massThus, in terms of amu, the mass of each element by the number of its atoms and adding them together. For example,of hydrogen atom = molecular mass of methane, which contains one carbon atom and four hydrogen atoms, = 1.0078 amu can be obtained as follows: = 1.0080 amu Molecular mass of methane, Similarly, the mass of oxygen - 16 (16O) (CH4) = (12.011 u) + 4 (1.008 u) atom would be 15.995 amu. = 16.043 u At present, ‘amu’ has been replaced by Similarly, molecular mass of water (H2O)‘u’, which is known as unified mass. = 2 × atomic mass of hydrogen + 1× atomic When we use atomic masses of elements mass of oxygen in calculations, we actually use average = 2 (1.008 u) + 16.00 uatomic masses of elements, which are explained below. = 18.02 u 1.7.2 Average Atomic Mass 1.7.4 Formula Mass Many naturally occurring elements exist Some substances, such as sodium chloride, as more than one isotope. When we take do not contain discrete molecules as their into account the existence of these isotopes constituent units. In such compounds, and their relative abundance (per cent positive (sodium ion) and negative (chloride ion) occurrence), the average atomic mass of entities are arranged in a three-dimensional that element can be computed. For example, structure, as shown in Fig. 1.10. carbon has the following three isotopes with relative abundances and masses as shown against each of them. Isotope Relative Atomic Mass Abundance (amu) (%) 12C 98.892 12 13C 1.108 13.00335 14C 2 ×10–10 14.00317 Fig. 1.10 Packing of Na+ and Cl– ions From the above data, the average atomic in sodium chloride mass of carbon will come out to be: (0.98892) (12 u) + (0.01108) (13.00335 u) + It may be noted that in sodium chloride, (2 × 10–12) (14.00317 u) = 12.011 u one Na+ ion is surrounded by six Cl– ion and Similarly, average atomic masses for vice-versa. other elements can be calculated. In the The formula, such as NaCl, is used to periodic table of elements, the atomic masses calculate the formula mass instead of mentioned for different elements actually molecular mass as in the solid state sodium represent their average atomic masses. chloride does not exist as a single entity. Reprint 2025-26 18 chemistry Thus, the formula mass of sodium chloride is This number of entities in 1 mol is so atomic mass of sodium + atomic mass of chlorine important that it is given a separate name and symbol. It is known as ‘Avogadro constant’, = 23.0 u + 35.5 u = 58.5 u or Avogadro number denoted by NA in honour Problem 1.1 of Amedeo Avogadro. To appreciate the Calculate the molecular mass of glucose largeness of this number, let us write it with (C6H12O6) molecule. all zeroes without using any powers of ten. Solution 602213670000000000000000 Hence, so many entities (atoms, molecules or Molecular mass of glucose (C6H12O6) any other particle) constitute one mole of a = 6 (12.011 u) + 12 (1.008 u) + 6 (16.00 u) particular substance. = (72.066 u) + (12.096 u) + We can, therefore, say that 1 mol of hydrogen (96.00 u) atoms = 6.022 × 1023 atoms = 180.162 u 1 mol of water molecules = 6.022 × 1023 water