chemistry-part-1-book

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. 2020-21

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 2020-21

100 CHEMISTRY CHEMICAL BONDING AND UNIT 4 MOLECULAR STRUCTURE After studying this Unit, you will be Scientists are constantly discovering new compounds, orderly able to arranging the facts about them, trying to explain with the existing knowledge, organising to modify the earlier views or • understand KÖssel-Lewis evolve theories for explaining the newly observed facts. approach to chemical bonding; Matter is made up of one or different type of elements. Under normal conditions no other element exists as an • explain the octet rule and its independent atom in nature, except noble gases. However, a group of atoms is found to exist together as one species limitations, draw Lewis having characteristic properties. Such a group of atoms is structures of simple molecules; called a molecule. Obviously there must be some force which holds these constituent atoms together in the • explain the formation of different molecules. The attractive force which holds various constituents (atoms, ions, etc.) together in different types of bonds; chemical species is called a chemical bond. Since the formation of chemical compounds takes place as a result • describe the VSEPR theory and of combination of atoms of various elements in different ways, it raises many questions. Why do atoms combine? predict the geometry of simple Why are only certain combinations possible? Why do some molecules; atoms combine while certain others do not? Why do molecules possess definite shapes? To answer such • explain the valence bond questions different theories and concepts have been put forward from time to time. These are Kössel-Lewis approach for the formation of approach, Valence Shell Electron Pair Repulsion (VSEPR) covalent bonds; Theory, Valence Bond (VB) Theory and Molecular Orbital (MO) Theory. The evolution of various theories of valence • predict the directional properties and the interpretation of the nature of chemical bonds have closely been related to the developments in the of covalent bonds; understanding of the structure of atom, the electronic configuration of elements and the periodic table. Every • explain the different types of system tends to be more stable and bonding is nature’s way of lowering the energy of the system to attain stability. hybridisation involving s, p and d orbitals and draw shapes of simple covalent molecules; • describe the molecular orbital theory of homonuclear diatomic molecules; • explain the concept of hydrogen bond. 2020-21

CHEMICAL BONDING AND MOLECULAR STRUCTURE 101 4.1 KÖSSEL-LEWIS APPROACH TO the number of valence electrons. This number CHEMICAL BONDING of valence electrons helps to calculate the common or group valence of the element. The In order to explain the formation of chemical group valence of the elements is generally bond in terms of electrons, a number of either equal to the number of dots in Lewis attempts were made, but it was only in 1916 symbols or 8 minus the number of dots or when Kössel and Lewis succeeded valence electrons. independently in giving a satisfactory explanation. They were the first to provide Kössel, in relation to chemical bonding, some logical explanation of valence which was drew attention to the following facts: based on the inertness of noble gases. • In the periodic table, the highly Lewis pictured the atom in terms of a positively charged ‘Kernel’ (the nucleus plus electronegative halogens and the highly the inner electrons) and the outer shell that electropositive alkali metals are separated could accommodate a maximum of eight by the noble gases; electrons. He, further assumed that these eight electrons occupy the corners of a cube • The formation of a negative ion from a which surround the ‘Kernel’. Thus the single outer shell electron of sodium would occupy halogen atom and a positive ion from an one corner of the cube, while in the case of a alkali metal atom is associated with the noble gas all the eight corners would be gain and loss of an electron by the occupied. This octet of electrons, represents respective atoms; a particularly stable electronic arrangement. Lewis postulated that atoms achieve the • The negative and positive ions thus stable octet when they are linked by chemical bonds. In the case of sodium and formed attain stable noble gas electronic chlorine, this can happen by the transfer of an electron from sodium to chlorine thereby configurations. The noble gases (with the giving the Na+ and Cl– ions. In the case of exception of helium which has a duplet other molecules like Cl2, H2, F2, etc., the bond is formed by the sharing of a pair of electrons of electrons) have a particularly stable between the atoms. In the process each atom outer shell configuration of eight (octet) attains a stable outer octet of electrons. electrons, ns2np6. Lewis Symbols: In the formation of a • The negative and positive ions are molecule, only the outer shell electrons take part in chemical combination and they are stabilized by electrostatic attraction. known as valence electrons. The inner shell electrons are well protected and are generally For example, the formation of NaCl from not involved in the combination process. sodium and chlorine, according to the above G.N. Lewis, an American chemist introduced scheme, can be explained as: simple notations to represent valence electrons in an atom. These notations are Na → Na+ + e– called Lewis symbols. For example, the Lewis symbols for the elements of second period are [Ne] 3s1 [Ne] as under: Cl + e– → Cl– Significance of Lewis Symbols : The [Ne] 3s2 3p5 [Ne] 3s2 3p6 or [Ar] number of dots around the symbol represents Na+ + Cl– → NaCl or Na+Cl– Similarly the formation of CaF2 may be shown as: Ca → Ca2+ + 2e– [Ar]4s2 [Ar] F + e– → F– [He] 2s2 2p5 [He] 2s2 2p6 or [Ne] Ca2+ + 2F– → CaF2 or Ca2+(F– )2 The bond formed, as a result of the electrostatic attraction between the positive and negative ions was termed as 2020-21

102 CHEMISTRY the electrovalent bond. The electrovalence chlorine atoms attain the outer shell octet of is thus equal to the number of unit the nearest noble gas (i.e., argon). charge(s) on the ion. Thus, calcium is assigned a positive electrovalence of two, The dots represent electrons. Such while chlorine a negative electrovalence of structures are referred to as Lewis dot one. structures. Kössel’s postulations provide the basis for The Lewis dot structures can be written the modern concepts regarding ion-formation for other molecules also, in which the by electron transfer and the formation of ionic combining atoms may be identical or crystalline compounds. His views have proved different. The important conditions being that: to be of great value in the understanding and • Each bond is formed as a result of sharing systematisation of the ionic compounds. At the same time he did recognise the fact that of an electron pair between the atoms. a large number of compounds did not fit into • Each combining atom contributes at least these concepts. one electron to the shared pair. 4.1.1 Octet Rule • The combining atoms attain the outer- Kössel and Lewis in 1916 developed an shell noble gas configurations as a result important theory of chemical combination of the sharing of electrons. between atoms known as electronic theory • Thus in water and carbon tetrachloride of chemical bonding. According to this, molecules, formation of covalent bonds atoms can combine either by transfer of can be represented as: valence electrons from one atom to another (gaining or losing) or by sharing of valence Thus, when two atoms share one electrons in order to have an octet in their electron pair they are said to be joined by valence shells. This is known as octet rule. a single covalent bond. In many compounds we have multiple bonds between atoms. The 4.1.2 Covalent Bond formation of multiple bonds envisages sharing of more than one electron pair Langmuir (1919) refined the Lewis between two atoms. If two atoms share two postulations by abandoning the idea of the pairs of electrons, the covalent bond stationary cubical arrangement of the octet, between them is called a double bond. For and by introducing the term covalent bond. example, in the carbon dioxide molecule, we The Lewis-Langmuir theory can be have two double bonds between the carbon understood by considering the formation of and oxygen atoms. Similarly in ethene the chlorine molecule,Cl2. The Cl atom with molecule the two carbon atoms are joined by electronic configuration, [Ne]3s2 3p5, is one a double bond. electron short of the argon configuration. The formation of the Cl2 molecule can be understood in terms of the sharing of a pair of electrons between the two chlorine atoms, each chlorine atom contributing one electron to the shared pair. In the process both or Cl – Cl Double bonds in CO2 molecule Covalent bond between two Cl atoms 2020-21

CHEMICAL BONDING AND MOLECULAR STRUCTURE 103 C2H4 molecule in subtraction of one electron from the total When combining atoms share three electron pairs as in the case of two number of valence electrons. For example, nitrogen atoms in the N2 molecule and the for the CO32– ion, the two negative charges two carbon atoms in the ethyne molecule, indicate that there are two additional a triple bond is formed. electrons than those provided by the N2 molecule + neutral atoms. For NH 4 ion, one positive C2H2 molecule 4.1.3 Lewis Representation of Simple charge indicates the loss of one electron Molecules (the Lewis Structures) from the group of neutral atoms. The Lewis dot structures provide a picture of bonding in molecules and ions in terms • Knowing the chemical symbols of the of the shared pairs of electrons and the octet rule. While such a picture may not combining atoms and having knowledge explain the bonding and behaviour of a molecule completely, it does help in of the skeletal structure of the compound understanding the formation and properties of a molecule to a large extent. Writing of (known or guessed intelligently), it is easy Lewis dot structures of molecules is, therefore, very useful. The Lewis dot to distribute the total number of electrons structures can be written by adopting the following steps: as bonding shared pairs between the • The total number of electrons required for atoms in proportion to the total bonds. writing the structures are obtained by adding the valence electrons of the • In general the least electronegative atom combining atoms. For example, in the CH4 molecule there are eight valence electrons occupies the central position in the available for bonding (4 from carbon and 4 from the four hydrogen atoms). molecule/ion. For example in the NF3 and • For anions, each negative charge would CO32–, nitrogen and carbon are the central mean addition of one electron. For cations, each positive charge would result atoms whereas fluorine and oxygen occupy the terminal positions. • After accounting for the shared pairs of electrons for single bonds, the remaining electron pairs are either utilized for multiple bonding or remain as the lone pairs. The basic requirement being that each bonded atom gets an octet of electrons. Lewis representations of a few molecules/ ions are given in Table 4.1. Table 4.1 The Lewis Representation of Some Molecules * Each H atom attains the configuration of helium (a duplet of electrons) 2020-21

104 CHEMISTRY Problem 4.1 each of the oxygen atoms completing the octets on oxygen atoms. This, however, Write the Lewis dot structure of CO does not complete the octet on nitrogen molecule. if the remaining two electrons constitute lone pair on it. Solution Hence we have to resort to multiple Step 1. Count the total number of bonding between nitrogen and one of the valence electrons of carbon and oxygen oxygen atoms (in this case a double atoms. The outer (valence) shell bond). This leads to the following Lewis configurations of carbon and oxygen dot structures. atoms are: 2s2 2p2 and 2s2 2p4, respectively. The valence electrons available are 4 + 6 =10. Step 2. The skeletal structure of CO is written as: C O Step 3. Draw a single bond (one shared electron pair) between C and O and complete the octet on O, the remaining two electrons are the lone pair on C. This does not complete the octet on 4.1.4 Formal Charge carbon and hence we have to resort to multiple bonding (in this case a triple Lewis dot structures, in general, do not bond) between C and O atoms. This represent the actual shapes of the molecules. satisfies the octet rule condition for both In case of polyatomic ions, the net charge is atoms. possessed by the ion as a whole and not by a particular atom. It is, however, feasible to Problem 4.2 assign a formal charge on each atom. The Write the Lewis structure of the nitrite formal charge of an atom in a polyatomic ion, NO2– . molecule or ion may be defined as the difference between the number of valence Solution electrons of that atom in an isolated or free Step 1. Count the total number of state and the number of electrons assigned valence electrons of the nitrogen atom, to that atom in the Lewis structure. It is the oxygen atoms and the additional one expressed as : negative charge (equal to one electron). Formal charge (F.C.) = N(2s2 2p3), O (2s2 2p4) on an atom in a Lewis 5 + (2 × 6) +1 = 18 electrons structure Step 2. The skeletal structure of NO2– is written as : O N O total number of valence total number of non Step 3. Draw a single bond (one shared electrons in the free — bonding (lone pair) electron pair) between the nitrogen and atom electrons total number of — (1/2) bonding(shared) electrons 2020-21

CHEMICAL BONDING AND MOLECULAR STRUCTURE 105 The counting is based on the assumption 4.1.5 Limitations of the Octet Rule that the atom in the molecule owns one The octet rule, though useful, is not universal. electron of each shared pair and both the It is quite useful for understanding the electrons of a lone pair. structures of most of the organic compounds and it applies mainly to the second period Let us consider the ozone molecule (O3). elements of the periodic table. There are three The Lewis structure of O3 may be drawn as : types of exceptions to the octet rule. The incomplete octet of the central atom The atoms have been numbered as 1, 2 In some compounds, the number of electrons and 3. The formal charge on: surrounding the central atom is less than eight. This is especially the case with elements • The central O atom marked 1 having less than four valence electrons. Examples are LiCl, BeH2 and BCl3. 1 = 6 – 2 – 2 (6) = +1 Li, Be and B have 1,2 and 3 valence electrons • The end O atom marked 2 only. Some other such compounds are AlCl3 and BF3. =6–4– 1 (4) = 0 Odd-electron molecules 2 In molecules with an odd number of electrons like nitric oxide, NO and nitrogen dioxide, • The end O atom marked 3 NO2, the octet rule is not satisfied for all the atoms 1 = 6 – 6 – 2 (2) = –1 The expanded octet Elements in and beyond the third period of Hence, we represent O3 along with the the periodic table have, apart from 3s and 3p formal charges as follows: orbitals, 3d orbitals also available for bonding. In a number of compounds of these elements We must understand that formal charges there are more than eight valence electrons do not indicate real charge separation within around the central atom. This is termed as the expanded octet. Obviously the octet rule the molecule. Indicating the charges on the does not apply in such cases. atoms in the Lewis structure only helps in Some of the examples of such compounds keeping track of the valence electrons in the are: PF5, SF6, H2SO4 and a number of molecule. Formal charges help in the coordination compounds. selection of the lowest energy structure from a number of possible Lewis structures for a given species. Generally the lowest energy structure is the one with the smallest formal charges on the atoms. The formal charge is a factor based on a pure covalent view of bonding in which electron pairs are shared equally by neighbouring atoms. 2020-21

106 CHEMISTRY Interestingly, sulphur also forms many Obviously ionic bonds will be formed compounds in which the octet rule is obeyed. more easily between elements with In sulphur dichloride, the S atom has an octet comparatively low ionization enthalpies of electrons around it. and elements with comparatively high negative value of electron gain enthalpy. Other drawbacks of the octet theory Most ionic compounds have cations • It is clear that octet rule is based upon derived from metallic elements and anions the chemical inertness of noble gases. from non-metallic elements. The However, some noble gases (for example ammonium ion, NH4+ (made up of two non- xenon and krypton) also combine with metallic elements) is an exception. It forms oxygen and fluorine to form a number of the cation of a number of ionic compounds. compounds like XeF2, KrF2, XeOF2 etc., Ionic compounds in the crystalline state • This theory does not account for the shape consist of orderly three-dimensional of molecules. arrangements of cations and anions held together by coulombic interaction energies. • It does not explain the relative stability of These compounds crystallise in different the molecules being totally silent about crystal structures determined by the size the energy of a molecule. of the ions, their packing arrangements and other factors. The crystal structure of 4.2 IONIC OR ELECTROVALENT BOND sodium chloride, NaCl (rock salt), for example is shown below. From the Kössel and Lewis treatment of the formation of an ionic bond, it follows that the Rock salt structure formation of ionic compounds would In ionic solids, the sum of the electron primarily depend upon: gain enthalpy and the ionization enthalpy may be positive but still the crystal • The ease of formation of the positive and structure gets stabilized due to the energy negative ions from the respective neutral released in the formation of the crystal atoms; lattice. For example: the ionization enthalpy for Na+(g) formation from Na(g) • The arrangement of the positive and is 495.8 kJ mol–1 ; while the electron gain negative ions in the solid, that is, the enthalpy for the change Cl(g) + e–→ lattice of the crystalline compound. Cl– (g) is, – 348.7 kJ mol–1 only. The sum of the two, 147.1 kJ mol-1 is more than The formation of a positive ion involves compensated for by the enthalpy of lattice formation of NaCl(s) (–788 kJ mol–1). ionization, i.e., removal of electron(s) from Therefore, the energy released in the the neutral atom and that of the negative ion involves the addition of electron(s) to the neutral atom. M(g) → M+(g) + e– ; Ionization enthalpy X(g) + e– → X – (g) ; Electron gain enthalpy M+(g) + X –(g) → MX(s) The electron gain enthalpy, ∆eg H, is the enthalpy change (Unit 3), when a gas phase atom in its ground state gains an electron. The electron gain process may be exothermic or endothermic. The ionization, on the other hand, is always endothermic. Electron affinity, is the negative of the energy change accompanying electron gain. 2020-21

CHEMICAL BONDING AND MOLECULAR STRUCTURE 107 processes is more than the energy absorbed. Fig. 4.1 The bond length in a covalent Thus a qualitative measure of the stability of an ionic compound is molecule AB. provided by its enthalpy of lattice R = rA + rB (R is the bond length and rA and rB formation and not simply by achieving octet of electrons around the ionic species are the covalent radii of atoms A and B in gaseous state. respectively) Since lattice enthalpy plays a key role in the formation of ionic compounds, it is covalent bond in the same molecule. The van important that we learn more about it. der Waals radius represents the overall size of the atom which includes its valence shell 4.2.1 Lattice Enthalpy in a nonbonded situation. Further, the van der Waals radius is half of the distance The Lattice Enthalpy of an ionic solid is between two similar atoms in separate defined as the energy required to molecules in a solid. Covalent and van der completely separate one mole of a solid Waals radii of chlorine are depicted in Fig.4.2 ionic compound into gaseous constituent ions. For example, the lattice enthalpy of NaCl rc = 99 pm 198 is 788 kJ mol–1. This means that 788 kJ of pm energy is required to separate one mole of solid NaCl into one mole of Na+ (g) and one pm mole of Cl– (g) to an infinite distance. = 180 r vdw This process involves both the attractive 360 pm forces between ions of opposite charges and the repulsive forces between ions of like Fig. 4.2 Covalent and van der Waals radii in a charge. The solid crystal being three- dimensional; it is not possible to calculate chlorine molecule. The inner circles lattice enthalpy directly from the interaction of forces of attraction and repulsion only. correspond to the size of the chlorine atom Factors associated with the crystal geometry have to be included. (rvdw and rc are van der Waals and covalent radii respectively). 4.3 BOND PARAMETERS 4.3.1 Bond Length Bond length is defined as the equilibrium distance between the nuclei of two bonded atoms in a molecule. Bond lengths are measured by spectroscopic, X-ray diffraction and electron-diffraction techniques about which you will learn in higher classes. Each atom of the bonded pair contributes to the bond length (Fig. 4.1). In the case of a covalent bond, the contribution from each atom is called the covalent radius of that atom. The covalent radius is measured approximately as the radius of an atom’s core which is in contact with the core of an adjacent atom in a bonded situation. The covalent radius is half of the distance between two similar atoms joined by a , 2020-21

108 CHEMISTRY Some typical average bond lengths for Table 4.2 Average Bond Lengths for Some single, double and triple bonds are shown in Single, Double and Triple Bonds Table 4.2. Bond lengths for some common molecules are given in Table 4.3. Bond Type Covalent Bond Length (pm) The covalent radii of some common O–H elements are listed in Table 4.4. C–H 96 N–O 107 4.3.2 Bond Angle C–O 136 C–N 143 It is defined as the angle between the orbitals C–C 143 containing bonding electron pairs around the C=O 154 central atom in a molecule/complex ion. Bond N=O 121 angle is expressed in degree which can be C=C 122 experimentally determined by spectroscopic C=N 133 methods. It gives some idea regarding the C≡N 138 distribution of orbitals around the central C≡C 116 atom in a molecule/complex ion and hence it 120 helps us in determining its shape. For example H–O–H bond angle in water can be Table 4.3 Bond Lengths in Some Common represented as under : Molecules 4.3.3 Bond Enthalpy Molecule Bond Length (pm) It is defined as the amount of energy required to break one mole of bonds of a particular H2 (H – H) 74 type between two atoms in a gaseous state. F2 (F – F) 144 The unit of bond enthalpy is kJ mol–1. For Cl2 (Cl – Cl) 199 example, the H – H bond enthalpy in hydrogen Br2 (Br – Br) 228 molecule is 435.8 kJ mol–1. I2 (I – I) 267 H2(g) → H(g) + H(g); ∆aHV = 435.8 kJ mol–1 N2 (N ≡ N) 109 O2 (O = O) 121 Similarly the bond enthalpy for molecules HF (H – F) 92 containing multiple bonds, for example O2 and 127 N2 will be as under : HCl (H – Cl) 141 O2 (O = O) (g) → O(g) + O(g); 160 HBr (H – Br) ∆aHV = 498 kJ mol–1 N2 (N ≡ N) (g) → N(g) + N(g); HI (H – I) ∆aHV = 946.0 kJ mol–1 Table 4.4 Covalent Radii, *rcov/(pm) It is important that larger the bond dissociation enthalpy, stronger will be the * The values cited are for single bonds, except where bond in the molecule. For a heteronuclear otherwise indicated in parenthesis. (See also Unit 3 for diatomic molecules like HCl, we have periodic trends). HCl (g) → H(g) + Cl (g); ∆aHV = 431.0 kJ mol–1 In case of polyatomic molecules, the measurement of bond strength is more complicated. For example in case of H2O molecule, the enthalpy needed to break the two O – H bonds is not the same. 2020-21

CHEMICAL BONDING AND MOLECULAR STRUCTURE 109 H2O(g) → H(g) + OH(g); ∆aH1V = 502 kJ mol–1 equally represented by the structures I and II shown below: OH(g) → H(g) + O(g); ∆aH2V = 427 kJ mol–1 Fig. 4.3 Resonance in the O3 molecule The difference in the ∆aHV value shows that (structures I and II represent the two canonical the second O – H bond undergoes some change forms while the structure III is the resonance because of changed chemical environment. hybrid) This is the reason for some difference in energy of the same O – H bond in different molecules In both structures we have a O–O single like C2H5OH (ethanol) and water. Therefore in bond and a O=O double bond. The normal polyatomic molecules the term mean or O–O and O=O bond lengths are 148 pm and average bond enthalpy is used. It is obtained 121 pm respectively. Experimentally by dividing total bond dissociation enthalpy determined oxygen-oxygen bond lengths in by the number of bonds broken as explained the O3 molecule are same (128 pm). Thus the below in case of water molecule, oxygen-oxygen bonds in the O3 molecule are intermediate between a double and a single Average bond enthalpy = 502 + 427 bond. Obviously, this cannot be represented 2 by either of the two Lewis structures shown above. = 464.5 kJ mol–1 The concept of resonance was introduced 4.3.4 Bond Order to deal with the type of difficulty experienced in the depiction of accurate structures of In the Lewis description of covalent bond, molecules like O3. According to the concept the Bond Order is given by the number of of resonance, whenever a single Lewis bonds between the two atoms in a structure cannot describe a molecule molecule. The bond order, for example in H2 accurately, a number of structures with (with a single shared electron pair), in O2 similar energy, positions of nuclei, bonding (with two shared electron pairs) and in N2 and non-bonding pairs of electrons are taken (with three shared electron pairs) is 1,2,3 as the canonical structures of the hybrid respectively. Similarly in CO (three shared which describes the molecule accurately. Thus for O3, the two structures shown above electron pairs between C and O) the bond constitute the canonical structures or order is 3. For N2, bond order is 3 and its resonance structures and their hybrid i.e., the ∆aHV is 946 kJ mol–1; being one of the III structure represents the structure of O3 highest for a diatomic molecule. more accurately. This is also called resonance hybrid. Resonance is represented by a double Isoelectronic molecules and ions have headed arrow. identical bond orders; for example, F2 and O22– have bond order 1. N2, CO and NO+ have bond order 3. A general correlation useful for understanding the stablities of molecules is that: with increase in bond order, bond enthalpy increases and bond length decreases. 4.3.5 Resonance Structures It is often observed that a single Lewis structure is inadequate for the representation of a molecule in conformity with its experimentally determined parameters. For example, the ozone, O3 molecule can be 2020-21

110 CHEMISTRY Some of the other examples of resonance Fig. 4.5 Resonance in CO2 molecule, I, II structures are provided by the carbonate ion and III represent the three and the carbon dioxide molecule. canonical forms. Problem 4.3 Explain the structure of CO32– ion in terms In general, it may be stated that of resonance. Solution • Resonance stabilizes the molecule as the The single Lewis structure based on the energy of the resonance hybrid is less presence of two single bonds and one than the energy of any single cannonical double bond between carbon and oxygen structure; and, atoms is inadequate to represent the molecule accurately as it represents • Resonance averages the bond unequal bonds. According to the characteristics as a whole. experimental findings, all carbon to oxygen bonds in CO32– are equivalent. Thus the energy of the O3 resonance Therefore the carbonate ion is best hybrid is lower than either of the two described as a resonance hybrid of the cannonical froms I and II (Fig 4.3). canonical forms I, II, and III shown below. Many misconceptions are associated Fig.4.4 Resonance in CO32–, I, II and with resonance and the same need to be III represent the three dispelled. You should remember that : canonical forms. • The cannonical forms have no real Problem 4.4 existence. Explain the structure of CO2 molecule. Solution • The molecule does not exist for a The experimentally determined carbon certain fraction of time in one to oxygen bond length in CO2 is cannonical form and for other 115 pm. The lengths of a normal fractions of time in other cannonical carbon to oxygen double bond (C=O) forms. and carbon to oxygen triple bond (C≡O) are 121 pm and 110 pm respectively. • There is no such equilibrium between The carbon-oxygen bond lengths in the cannonical forms as we have CO2 (115 pm) lie between the values between tautomeric forms (keto and for C=O and C≡O. Obviously, a single enol) in tautomerism. Lewis structure cannot depict this position and it becomes necessary to • The molecule as such has a single write more than one Lewis structures structure which is the resonance and to consider that the structure of hybrid of the cannonical forms and CO2 is best described as a hybrid of which cannot as such be depicted by the canonical or resonance forms I, II a single Lewis structure. and III. 4.3.6 Polarity of Bonds The existence of a hundred percent ionic or covalent bond represents an ideal situation. In reality no bond or a compound is either completely covalent or ionic. Even in case of covalent bond between two hydrogen atoms, there is some ionic character. When covalent bond is formed between two similar atoms, for example in H2, O2, Cl2, N2 or F2, the shared pair of electrons is equally 2020-21

CHEMICAL BONDING AND MOLECULAR STRUCTURE 111 attracted by the two atoms. As a result electron In case of polyatomic molecules the dipole pair is situated exactly between the two moment not only depend upon the individual identical nuclei. The bond so formed is called dipole moments of bonds known as bond nonpolar covalent bond. Contrary to this in dipoles but also on the spatial arrangement of case of a heteronuclear molecule like HF, the various bonds in the molecule. In such case, shared electron pair between the two atoms the dipole moment of a molecule is the vector gets displaced more towards fluorine since the sum of the dipole moments of various bonds. electronegativity of fluorine (Unit 3) is far For example in H2O molecule, which has a bent greater than that of hydrogen. The resultant structure, the two O–H bonds are oriented at covalent bond is a polar covalent bond. an angle of 104.50. Net dipole moment of 6.17 × 10–30 C m (1D = 3.33564 × 10–30 C m) is the As a result of polarisation, the molecule resultant of the dipole moments of two O–H possesses the dipole moment (depicted bonds. below) which can be defined as the product of the magnitude of the charge and the Net Dipole moment, µ = 1.85 D distance between the centres of positive and negative charge. It is usually designated by a = 1.85 × 3.33564 × 10–30 C m = 6.17 ×10–30 C m Greek letter ‘µ’. Mathematically, it is expressed as follows : The dipole moment in case of BeF2 is zero. This is because the two equal bond dipoles Dipole moment (µ) = charge (Q) × distance of point in opposite directions and cancel the separation (r) effect of each other. Dipole moment is usually expressed in In tetra-atomic molecule, for example in Debye units (D). The conversion factor is BF3, the dipole moment is zero although the 1 D = 3.33564 × 10–30 C m B–F bonds are oriented at an angle of 120o to where C is coulomb and m is meter. one another, the three bond moments give a Further dipole moment is a vector quantity net sum of zero as the resultant of any two is and by convention it is depicted by a small arrow with tail on the negative centre and head equal and opposite to the third. pointing towards the positive centre. But in chemistry presence of dipole moment is Let us study an interesting case of NH3 represented by the crossed arrow ( ) put and NF3 molecule. Both the molecules have on Lewis structure of the molecule. The cross pyramidal shape with a lone pair of electrons is on positive end and arrow head is on negative on nitrogen atom. Although fluorine is more end. For example the dipole moment of HF may electronegative than nitrogen, the resultant be represented as : HF This arrow symbolises the direction of the shift of electron density in the molecule. Note that the direction of crossed arrow is opposite to the conventional direction of dipole moment vector. Peter Debye, the Dutch chemist received Nobel prize in 1936 for his work on X-ray diffraction and dipole moments. The magnitude of the dipole moment is given in Debye units in order to honour him. 2020-21

112 CHEMISTRY dipole moment of NH3 ( 4.90 × 10–30 C m) is in terms of the following rules: greater than that of NF3 (0.8 × 10–30 C m). This is because, in case of NH3 the orbital dipole • The smaller the size of the cation and the due to lone pair is in the same direction as the larger the size of the anion, the greater the resultant dipole moment of the N – H bonds, covalent character of an ionic bond. whereas in NF3 the orbital dipole is in the direction opposite to the resultant dipole • The greater the charge on the cation, the moment of the three N–F bonds. The orbital greater the covalent character of the ionic bond. dipole because of lone pair decreases the effect of the resultant N – F bond moments, which • For cations of the same size and charge, results in the low dipole moment of NF3 as the one, with electronic configuration represented below : (n-1)dnnso, typical of transition metals, is more polarising than the one with a noble Dipole moments of some molecules are gas configuration, ns2 np6, typical of alkali shown in Table 4.5. and alkaline earth metal cations. The cation polarises the anion, pulling the Just as all the covalent bonds have electronic charge toward itself and thereby some partial ionic character, the ionic increasing the electronic charge between bonds also have partial covalent the two. This is precisely what happens in character. The partial covalent character a covalent bond, i.e., buildup of electron of ionic bonds was discussed by Fajans charge density between the nuclei. The polarising power of the cation, the polarisability of the anion and the extent of distortion (polarisation) of anion are the factors, which determine the per cent covalent character of the ionic bond. 4.4 THE VALENCE SHELL ELECTRON PAIR REPULSION (VSEPR) THEORY As already explained, Lewis concept is unable to explain the shapes of molecules. This theory provides a simple procedure to predict the shapes of covalent molecules. Sidgwick Table 4.5 Dipole Moments of Selected Molecules Type of Example Dipole Geometry Molecule Moment, µ(D) Molecule (AB) HF 1.78 linear HCl 1.07 linear Molecule (AB2) HBr 0.79 linear Molecule (AB3) HI 0.38 linear Molecule (AB4) H2 linear 0 H2O bent H2S 1.85 bent CO2 0.95 linear NH3 0 trigonal-pyramidal NF3 trigonal-pyramidal BF3 1.47 trigonal-planar 0.23 CH4 tetrahedral CHCl3 0 tetrahedral CCl4 tetrahedral 0 1.04 0 2020-21

CHEMICAL BONDING AND MOLECULAR STRUCTURE 113 and Powell in 1940, proposed a simple theory result in deviations from idealised shapes and based on the repulsive interactions of the alterations in bond angles in molecules. electron pairs in the valence shell of the atoms. It was further developed and redefined by For the prediction of geometrical shapes of Nyholm and Gillespie (1957). molecules with the help of VSEPR theory, it is convenient to divide molecules into two The main postulates of VSEPR theory are categories as (i) molecules in which the as follows: central atom has no lone pair and (ii) molecules in which the central atom has • The shape of a molecule depends upon one or more lone pairs. the number of valence shell electron pairs (bonded or nonbonded) around the central Table 4.6 (page114) shows the atom. arrangement of electron pairs about a central atom A (without any lone pairs) and • Pairs of electrons in the valence shell repel geometries of some molecules/ions of the type one another since their electron clouds are AB. Table 4.7 (page 115) shows shapes of some negatively charged. simple molecules and ions in which the central atom has one or more lone pairs. Table 4.8 • These pairs of electrons tend to occupy (page 116) explains the reasons for the such positions in space that minimise distortions in the geometry of the molecule. repulsion and thus maximise distance between them. As depicted in Table 4.6, in the compounds of AB2, AB3, AB4, AB5 and AB6, • The valence shell is taken as a sphere with the arrangement of electron pairs and the B the electron pairs localising on the atoms around the central atom A are : linear, spherical surface at maximum distance trigonal planar, tetrahedral, trigonal- from one another. bipyramidal and octahedral, respectively. Such arrangement can be seen in the • A multiple bond is treated as if it is a single molecules like BF3 (AB3), CH4 (AB4) and PCl5 electron pair and the two or three electron (AB5) as depicted below by their ball and pairs of a multiple bond are treated as a stick models. single super pair. Fig. 4.6 The shapes of molecules in which • Where two or more resonance structures central atom has no lone pair can represent a molecule, the VSEPR model is applicable to any such structure. The VSEPR Theory is able to predict geometry of a large number of molecules, The repulsive interaction of electron pairs especially the compounds of p-block elements decrease in the order: accurately. It is also quite successful in determining the geometry quite-accurately Lone pair (lp) – Lone pair (lp) > Lone pair (lp) even when the energy difference between – Bond pair (bp) > Bond pair (bp) – possible structures is very small. The Bond pair (bp) theoretical basis of the VSEPR theory regarding the effects of electron pair repulsions Nyholm and Gillespie (1957) refined the on molecular shapes is not clear and VSEPR model by explaining the important continues to be a subject of doubt and difference between the lone pairs and bonding discussion. pairs of electrons. While the lone pairs are localised on the central atom, each bonded pair is shared between two atoms. As a result, the lone pair electrons in a molecule occupy more space as compared to the bonding pairs of electrons. This results in greater repulsion between lone pairs of electrons as compared to the lone pair - bond pair and bond pair - bond pair repulsions. These repulsion effects 2020-21

114 CHEMISTRY Table 4.6 Geometry of Molecules in which the Central Atom has No Lone Pair of Electrons 2020-21

CHEMICAL BONDING AND MOLECULAR STRUCTURE 115 Table 4.7 Shape (geometry) of Some Simple Molecules/Ions with Central Ions having One or More Lone Pairs of Electrons(E). 2020-21

116 CHEMISTRY Table 4.8 Shapes of Molecules containing Bond Pair and Lone Pair Molecule No. of No. of Arrangement Shape Reason for the type bonding lone of electrons shape acquired pairs pairs AB2E 4 1 Bent Theoretically the shape should have been triangular planar but actually it is found to be bent or v-shaped. The reason being the lone pair- bond pair repulsion is much more as compared to the bond pair-bond pair repul- sion. So the angle is reduced to 119.5° from 120°. AB3E 31 Trigonal Had there been a bp in place pyramidal of lp the shape would have been tetrahedral but one lone pair is present and due to the repulsion between lp-bp (which is more than bp-bp repulsion) the angle between bond pairs is reduced to 107° from 109.5°. Bent The shape should have been tetrahedral if there were all bp AB2E2 2 2 but two lp are present so the shape is distorted tetrahedral or angular. The reason is lp-lp repulsion is more than lp-bp repulsion which is more than bp-bp repulsion. Thus, the angle is reduced to 104.5° from 109.5°. AB4E 41 See- In (a) the lp is present at axial saw position so there are three lp—bp repulsions at 90°. In(b) the lp is in an equatorial position, and there are two lp—bp repulsions. Hence, arrangement (b) is more stable. The shape shown in (b) is described as a distorted (More stable) tetrahedron, a folded square or a see-saw. 2020-21

CHEMICAL BONDING AND MOLECULAR STRUCTURE 117 Molecule No. of No. of Arrangement Shape Reason for the type bonding lone of electrons T -shape shape acquired pairs pairs AB3E2 3 2 In (a) the lp are at equatorial position so there are less lp-bp repulsions as compared to others in which the lp are at axial positions. So structure (a) is most stable. (T-shaped). 4.5 VALENCE BOND THEORY knowledge of atomic orbitals, electronic configurations of elements (Units 2), the As we know that Lewis approach helps in overlap criteria of atomic orbitals, the writing the structure of molecules but it fails hybridization of atomic orbitals and the to explain the formation of chemical bond. It principles of variation and superposition. A also does not give any reason for the difference rigorous treatment of the VB theory in terms in bond dissociation enthalpies and bond of these aspects is beyond the scope of this lengths in molecules like H2 (435.8 kJ mol-1, book. Therefore, for the sake of convenience, 74 pm) and F2 (155 kJ mol-1, 144 pm), valence bond theory has been discussed in although in both the cases a single covalent terms of qualitative and non-mathematical bond is formed by the sharing of an electron treatment only. To start with, let us consider pair between the respective atoms. It also gives the formation of hydrogen molecule which is no idea about the shapes of polyatomic the simplest of all molecules. molecules. Consider two hydrogen atoms A and B Similarly the VSEPR theory gives the approaching each other having nuclei NA and geometry of simple molecules but NB and electrons present in them are theoretically, it does not explain them and also represented by eA and eB. When the two atoms it has limited applications. To overcome these are at large distance from each other, there is limitations the two important theories based no interaction between them. As these two on quantum mechanical principles are atoms approach each other, new attractive and introduced. These are valence bond (VB) theory repulsive forces begin to operate. and molecular orbital (MO) theory. Attractive forces arise between: Valence bond theory was introduced by Heitler and London (1927) and developed (i) nucleus of one atom and its own electron further by Pauling and others. A discussion that is NA – eA and NB– eB. of the valence bond theory is based on the 2020-21

118 CHEMISTRY (ii) nucleus of one atom and electron of other hydrogen atoms are said to be bonded together atom i.e., NA– eB, NB– eA. to form a stable molecule having the bond length of 74 pm. Similarly repulsive forces arise between (i) electrons of two atoms like eA – eB, Since the energy gets released when the (ii) nuclei of two atoms NA – NB. bond is formed between two hydrogen atoms, the hydrogen molecule is more stable than that Attractive forces tend to bring the two of isolated hydrogen atoms. The energy so atoms close to each other whereas repulsive released is called as bond enthalpy, which is forces tend to push them apart (Fig. 4.7). corresponding to minimum in the curve depicted in Fig. 4.8. Conversely, 435.8 kJ of energy is required to dissociate one mole of H2 molecule. H2(g) + 435.8 kJ mol–1 → H(g) + H(g) Fig. 4.7 Forces of attraction and repulsion during Fig. 4.8 The potential energy curve for the the formation of H2 molecule. formation of H2 molecule as a function of internuclear distance of the H atoms. The Experimentally it has been found that the magnitude of new attractive force is more than minimum in the curve corresponds to the the new repulsive forces. As a result, two atoms approach each other and potential most stable state of H2. energy decreases. Ultimately a stage is reached where the net force of attraction 4.5.1 Orbital Overlap Concept balances the force of repulsion and system acquires minimum energy. At this stage two In the formation of hydrogen molecule, there is a minimum energy state when two hydrogen atoms are so near that their atomic orbitals undergo partial interpenetration. This partial merging of atomic orbitals is called overlapping of atomic orbitals which results in the pairing of electrons. The extent of overlap decides the strength of a covalent bond. In general, greater the overlap the stronger is the bond formed between two atoms. Therefore, according to orbital overlap concept, the formation of a covalent bond between two atoms results by pairing of electrons present in the valence shell having opposite spins. 2020-21

CHEMICAL BONDING AND MOLECULAR STRUCTURE 119 4.5.2 Directional Properties of Bonds Fig.4.9 Positive, negative and zero overlaps of As we have already seen, the covalent bond is s and p atomic orbitals formed by overlapping of atomic orbitals. The molecule of hydrogen is formed due to the hydrogen.The four atomic orbitals of carbon, overlap of 1s-orbitals of two H atoms. each with an unpaired electron can overlap with the 1s orbitals of the four H atoms which In case of polyatomic molecules like CH4, are also singly occupied. This will result in the NH3 and H2O, the geometry of the molecules is formation of four C-H bonds. It will, however, also important in addition to the bond be observed that while the three p orbitals of formation. For example why is it so that CH4 carbon are at 90° to one another, the HCH molecule has tetrahedral shape and HCH bond angle for these will also be 90°. That is three angles are 109.5°? Why is the shape of NH3 C-H bonds will be oriented at 90° to one molecule pyramidal ? another. The 2s orbital of carbon and the 1s orbital of H are spherically symmetrical and The valence bond theory explains the they can overlap in any direction. Therefore shape, the formation and directional properties the direction of the fourth C-H bond cannot of bonds in polyatomic molecules like CH4, NH3 be ascertained. This description does not fit and H2O, etc. in terms of overlap and in with the tetrahedral HCH angles of 109.5°. hybridisation of atomic orbitals. Clearly, it follows that simple atomic orbital overlap does not account for the directional 4.5.3 Overlapping of Atomic Orbitals characteristics of bonds in CH4. Using similar procedure and arguments, it can be seen that in the When orbitals of two atoms come close to form bond, their overlap may be positive, negative or zero depending upon the sign (phase) and direction of orientation of amplitude of orbital wave function in space (Fig. 4.9). Positive and negative sign on boundary surface diagrams in the Fig. 4.9 show the sign (phase) of orbital wave function and are not related to charge. Orbitals forming bond should have same sign (phase) and orientation in space. This is called positive overlap. Various overlaps of s and p orbitals are depicted in Fig. 4.9. The criterion of overlap, as the main factor for the formation of covalent bonds applies uniformly to the homonuclear/heteronuclear diatomic molecules and polyatomic molecules. We know that the shapes of CH4, NH3, and H2O molecules are tetrahedral, pyramidal and bent respectively. It would be therefore interesting to use VB theory to find out if these geometrical shapes can be explained in terms of the orbital overlaps. Let us first consider the CH4 (methane) molecule. The electronic configuration of carbon in its ground state is [He]2s2 2p2 which 2inptzh1.eTehxecietnedersgtyarteeqbueicroemd efosr[Hthei]s2esx1c2itpaxt1io2npiys1 compensated by the release of energy due to overlap between the orbitals of carbon and the 2020-21

120 CHEMISTRY case of NH3 aens dsHh2oOuml dolbeceul9e0s,° .thTehHi sNHi sani nd charged clouds above and below the plane HOH angl of the participating atoms. disagreement with the actual bond angles of 4.5.5 Strength of Sigma and pi Bonds Basically the strength of a bond depends upon 107° and 104.5° in the NH3 and H2O the extent of overlapping. In case of sigma bond, molecules respectively. the overlapping of orbitals takes place to a larger extent. Hence, it is stronger as compared 4.5.4 Types of Overlapping and Nature of to the pi bond where the extent of overlapping Covalent Bonds occurs to a smaller extent. Further, it is important to note that in the formation of The covalent bond may be classified into two multiple bonds between two atoms of a types depending upon the types of molecule, pi bond(s) is formed in addition to a overlapping: sigma bond. (i) Sigma(σ) bond, and (ii) pi(π) bond (i) Sigma(σ) bond : This type of covalent bond is formed by the end to end (head-on) overlap of bonding orbitals along the internuclear axis. This is called as head on overlap or axial overlap. This can be formed by any one of the following types of combinations of atomic orbitals. • s-s overlapping : In this case, there is overlap of two half filled s-orbitals along the internuclear axis as shown below : • s-p overlapping: This type of overlap 4.6 HYBRIDISATION occurs between half filled s-orbitals of one atom and half filled p-orbitals of another In order to explain the characteristic atom. geometrical shapes of polyatomic molecules like CH4, NH3 and H2O etc., Pauling introduced • p–p overlapping : This type of overlap the concept of hybridisation. According to him takes place between half filled p-orbitals the atomic orbitals combine to form new set of of the two approaching atoms. equivalent orbitals known as hybrid orbitals. Unlike pure orbitals, the hybrid orbitals are (ii) pi(π ) bond : In the formation of π bond used in bond formation. The phenomenon is the atomic orbitals overlap in such a way known as hybridisation which can be defined that their axes remain parallel to each other as the process of intermixing of the orbitals of and perpendicular to the internuclear axis. slightly different energies so as to redistribute The orbitals formed due to sidewise their energies, resulting in the formation of new overlapping consists of two saucer type set of orbitals of equivalent energies and shape. For example when one 2s and three 2p-orbitals of carbon hybridise, there is the formation of four new sp3 hybrid orbitals. Salient features of hybridisation: The main features of hybridisation are as under : 1. The number of hybrid orbitals is equal to the number of the atomic orbitals that get hybridised. 2. The hybridised orbitals are always equivalent in energy and shape. 2020-21

CHEMICAL BONDING AND MOLECULAR STRUCTURE 121 3. The hybrid orbitals are more effective in vacant 2p orbital to account for its bivalency. forming stable bonds than the pure atomic One 2s and one 2p-orbital gets hybridised to orbitals. form two sp hybridised orbitals. These two sp hybrid orbitals are oriented in opposite 4. These hybrid orbitals are directed in space direction forming an angle of 180°. Each of in some preferred direction to have the sp hybridised orbital overlaps with the minimum repulsion between electron 2p-orbital of chlorine axially and form two Be- pairs and thus a stable arrangement. Cl sigma bonds. This is shown in Fig. 4.10. Therefore, the type of hybridisation indicates the geometry of the molecules. Be Important conditions for hybridisation Fig.4.10 (a) Formation of sp hybrids from s and p orbitals; (b) Formation of the linear (i) The orbitals present in the valence shell BeCl2 molecule of the atom are hybridised. (II) sp2 hybridisation : In this hybridisation (ii) The orbitals undergoing hybridisation there is involvement of one s and two should have almost equal energy. p-orbitals in order to form three equivalent sp2 hybridised orbitals. For example, in BCl3 (iii) Promotion of electron is not essential molecule, the ground state electronic condition prior to hybridisation. configuration of central boron atom is 1s22s22p1. In the excited state, one of the 2s (iv) It is not necessary that only half filled electrons is promoted to vacant 2p orbital as orbitals participate in hybridisation. In some cases, even filled orbitals of valence Fig.4.11 Formation of sp2 hybrids and the BCl3 shell take part in hybridisation. molecule 4.6.1 Types of Hybridisation There are various types of hybridisation involving s, p and d orbitals. The different types of hybridisation are as under: (I) sp hybridisation: This type of hybridisation involves the mixing of one s and one p orbital resulting in the formation of two equivalent sp hybrid orbitals. The suitable orbitals for sp hybridisation are s and pz, if the hybrid orbitals are to lie along the z-axis. Each sp hybrid orbitals has 50% s-character and 50% p-character. Such a molecule in which the central atom is sp-hybridised and linked directly to two other central atoms possesses linear geometry. This type of hybridisation is also known as diagonal hybridisation. The two sp hybrids point in the opposite direction along the z-axis with projecting positive lobes and very small negative lobes, which provides more effective overlapping resulting in the formation of stronger bonds. Example of molecule having sp hybridisation BeCl2: The ground state electronic configuration of Be is 1s22s2. In the exited state one of the 2s-electrons is promoted to 2020-21

122 CHEMISTRY a result boron has three unpaired electrons. ground state is 2S2 2p1x 2p1y 2p1z having three These three orbitals (one 2s and two 2p) hybridise to form three sp2 hybrid orbitals. The unpaired electrons in the sp3 hybrid orbitals three hybrid orbitals so formed are oriented in and a lone pair of electrons is present in the a trigonal planar arrangement and overlap with fourth one. These three hybrid orbitals overlap 2p orbitals of chlorine to form three B-Cl with 1s orbitals of hydrogen atoms to form bonds. Therefore, in BCl3 (Fig. 4.11), the three N–H sigma bonds. We know that the force geometry is trigonal planar with ClBCl bond of repulsion between a lone pair and a bond angle of 120°. pair is more than the force of repulsion between two bond pairs of electrons. The (III) sp3 hybridisation: This type of molecule thus gets distorted and the bond hybridisation can be explained by taking the angle is reduced to 107° from 109.5°. The example of CH4 molecule in which there is geometry of such a molecule will be pyramidal mixing of one s-orbital and three p-orbitals of as shown in Fig. 4.13. the valence shell to form four sp3 hybrid orbital of equivalent energies and shape. There is 25% s-character and 75% p-character in each sp3 hybrid orbital. The four sp3 hybrid orbitals so formed are directed towards the four corners of the tetrahedron. The angle between sp3 hybrid orbital is 109.5° as shown in Fig. 4.12. σ Fig.4.13 Formation of NH3 molecule σσ In case of H2O molecule, the four oxygen σ orbitals (one 2s and three 2p) undergo sp3 hybridisation forming four sp3 hybrid orbitals out of which two contain one electron each and the other two contain a pair of electrons. These four sp3 hybrid orbitals acquire a tetrahedral geometry, with two corners occupied by hydrogen atoms while the other two by the lone pairs. The bond angle in this case is reduced to 104.5° from 109.5° (Fig. 4.14) and the molecule thus acquires a V-shape or angular geometry. Fig.4.12 For mation of sp3 hybrids by the Fig.4.14 Formation of H2O molecule combination of s , px , py and pz atomic orbitals of carbon and the formation of CH4 molecule The structure of NH3 and H2O molecules can also be explained with the help of sp3 hybridisation. In NH3, the valence shell (outer) electronic configuration of nitrogen in the 2020-21

CHEMICAL BONDING AND MOLECULAR STRUCTURE 123 4.6.2 Other Examples of sp3, sp2 and sp sp2 hybrid orbitals of each carbon atom are Hybridisation used for making sp2–s sigma bond with two sp3 Hybridisation in C2H6 molecule: In ethane molecule both the carbon atoms hydrogen aotnoemcsa.rTbhoenuanthomybroivdeisreladposrbsiidtaelw(2ispex assume sp3 hybrid state. One of the four sp3 or 2py) of hybrid orbitals of carbon atom overlaps axially with the similar orbital of the other carbon with similar orbitals of other atom to form sp3-sp3 sigma bond while the other three atom to form weak π bond, which consists of hybrid orbitals of each carbon atom are used in forming sp3–s sigma bonds with hydrogen two equal electron clouds distributed above atoms as discussed in section 4.6.1(iii). Therefore in ethane C–C bond length is 154 and below the plane of carbon and hydrogen pm and each C–H bond length is 109 pm. atoms. sp2 Hybridisation in C2H4: In the formation of ethene molecule, one of the sp2 hybrid Thus, in ethene molecule, the carbon- orbitals of carbon atom overlaps axially with carbon bond consists of one sp2–sp2 sigma sp2 hybridised orbital of another carbon atom bond and one pi (π ) bond between p orbitals to form C–C sigma bond. While the other two which are not used in the hybridisation and are perpendicular to the plane of molecule; the bond length 134 pm. The C–H bond is sp2–s sigma with bond length 108 pm. The H– C–H bond angle is 117.6° while the H–C–C angle is 121°. The formation of sigma and pi bonds in ethene is shown in Fig. 4.15. Fig. 4.15 Formation of sigma and pi bonds in ethene 2020-21

124 CHEMISTRY sp Hybridisation in C2H2 : In the formation 4.6.3 Hybridisation of Elements involving of ethyne molecule, both the carbon atoms d Orbitals undergo sp-hybridisation having two unhybridised orbital i.e., 2py and 2px. The elements present in the third period contain d orbitals in addition to s and p One sp hybrid orbital of one carbon atom orbitals. The energy of the 3d orbitals are overlaps axially with sp hybrid orbital of the comparable to the energy of the 3s and 3p other carbon atom to form C–C sigma bond, orbitals. The energy of 3d orbitals are also while the other hybridised orbital of each comparable to those of 4s and 4p orbitals. As carbon atom overlaps axially with the half a consequence the hybridisation involving filled s orbital of hydrogen atoms forming σ either 3s, 3p and 3d or 3d, 4s and 4p is bonds. Each of the two unhybridised p orbitals possible. However, since the difference in of both the carbon atoms overlaps sidewise to energies of 3p and 4s orbitals is significant, no form two π bonds between the carbon atoms. hybridisation involving 3p, 3d and 4s orbitals So the triple bond between the two carbon is possible. atoms is made up of one sigma and two pi bonds as shown in Fig. 4.16. The important hybridisation schemes involving s, p and d orbitals are summarised below: Shape of Hybridisation Atomic Examples molecules/ type orbitals [Ni(CN)4]2–, ions dsp2 d+s+p(2) [Pt(Cl)4]2– PF5, PCl5 Square s+p(3)+d BrF5 planar SF6, [CrF6]3– s+p(3)+d(2) [Co(NH3)6]3+ Trigonal sp3d bipyramidal s+p(3)+d(2) d(2)+s+p(3) Square sp3d2 pyramidal Octahedral sp3d2 d2sp3 (i) Formation of PCl5 (sp3d hybridisation): The ground state and the excited state outer electronic configurations of phosphorus (Z=15) are represented below. Fig.4.16 Formation of sigma and pi bonds in sp3d hybrid orbitals filled by electron pairs ethyne donated by five Cl atoms. 2020-21

CHEMICAL BONDING AND MOLECULAR STRUCTURE 125 Now the five orbitals (i.e., one s, three p and hybrid orbitals overlap with singly occupied one d orbitals) are available for hybridisation orbitals of fluorine atoms to form six S–F sigma to yield a set of five sp3d hybrid orbitals which bonds. Thus SF6 molecule has a regular are directed towards the five corners of a octahedral geometry as shown in Fig. 4.18. trigonal bipyramidal as depicted in the Fig. 4.17. sp3d2 hybridisation Fig. 4.17 Trigonal bipyramidal geometry of PCl5 Fig. 4.18 Octahedral geometry of SF6 molecule molecule 4.7 MOLECULAR ORBITAL THEORY It should be noted that all the bond angles in trigonal bipyramidal geometry are not Molecular orbital (MO) theory was developed equivalent. In PCl5 the five sp3d orbitals of by F. Hund and R.S. Mulliken in 1932. The phosphorus overlap with the singly occupied salient features of this theory are : p orbitals of chlorine atoms to form five P–Cl sigma bonds. Three P–Cl bond lie in one plane (i) The electrons in a molecule are present and make an angle of 120° with each other; in the various molecular orbitals as the these bonds are termed as equatorial bonds. electrons of atoms are present in the The remaining two P–Cl bonds–one lying various atomic orbitals. above and the other lying below the equatorial plane, make an angle of 90° with the plane. (ii) The atomic orbitals of comparable These bonds are called axial bonds. As the axial energies and proper symmetry combine bond pairs suffer more repulsive interaction to form molecular orbitals. from the equatorial bond pairs, therefore axial bonds have been found to be slightly longer (iii) While an electron in an atomic orbital is and hence slightly weaker than the equatorial influenced by one nucleus, in a bonds; which makes PCl5 molecule more molecular orbital it is influenced by two reactive. or more nuclei depending upon the number of atoms in the molecule. Thus, (ii) Formation of SF6 (sp3d2 hybridisation): In SF6 the central sulphur atom has the ground state outer electronic configuration 3s23p4. In the exited state the available six orbitals i.e., one s, three p and two d are singly occupied by electrons. These orbitals hybridise to form six new sp3d2 hybrid orbitals, which are projected towards the six corners of a regular octahedron in SF6. These six sp3d2 2020-21

126 CHEMISTRY an atomic orbital is monocentric while a Mathematically, the formation of molecular molecular orbital is polycentric. orbitals may be described by the linear combination of atomic orbitals that can take (iv) The number of molecular orbital formed place by addition and by subtraction of wave is equal to the number of combining functions of individual atomic orbitals as atomic orbitals. When two atomic shown below : orbitals combine, two molecular orbitals are formed. One is known as bonding ψMO = ψA + ψB molecular orbital while the other is called antibonding molecular orbital. Therefore, the two molecular orbitals σ and σ* are formed as : (v) The bonding molecular orbital has lower energy and hence greater stability than σ = ψA + ψB the corresponding antibonding σ* = ψA – ψB molecular orbital. The molecular orbital σ formed by the addition of atomic orbitals is called the (vi) Just as the electron probability bonding molecular orbital while the distribution around a nucleus in an molecular orbital σ* formed by the subtraction atom is given by an atomic orbital, the of atomic orbital is called antibonding electron probability distribution around molecular orbital as depicted in Fig. 4.19. a group of nuclei in a molecule is given by a molecular orbital. σ* = ψA – ψB ψB (vii) The molecular orbitals like atomic ψA orbitals are filled in accordance with the σ = ψA + ψB aufbau principle obeying the Pauli’s exclusion principle and the Hund’s rule. Fig.4.19 For mation of bonding (σ) and 4.7.1 Formation of Molecular Orbitals antibonding (σ*) molecular orbitals by the Linear Combination of Atomic Orbitals (LCAO) linear combination of atomic orbitals ψA and ψB centered on two atoms A and B According to wave mechanics, the atomic respectively. orbitals can be expressed by wave functions (ψ ’s) which represent the amplitude of the Qualitatively, the formation of molecular electron waves. These are obtained from the orbitals can be understood in terms of the solution of Schrödinger wave equation. constructive or destructive interference of the However, since it cannot be solved for any electron waves of the combining atoms. In the system containing more than one electron, formation of bonding molecular orbital, the molecular orbitals which are one electron wave two electron waves of the bonding atoms functions for molecules are difficult to obtain reinforce each other due to constructive directly from the solution of Schrödinger wave interference while in the formation of equation. To overcome this problem, an approximate method known as linear combination of atomic orbitals (LCAO) has been adopted. Let us apply this method to the homonuclear diatomic hydrogen molecule. Consider the hydrogen molecule consisting of two atoms A and B. Each hydrogen atom in the ground state has one electron in 1s orbital. The atomic orbitals of these atoms may be represented by the wave functions ψA and ψB. 2020-21

CHEMICAL BONDING AND MOLECULAR STRUCTURE 127 antibonding molecular orbital, the electron molecular axis. By convention z-axis is waves cancel each other due to destructive taken as the molecular axis. It is important interference. As a result, the electron density to note that atomic orbitals having same or in a bonding molecular orbital is located nearly the same energy will not combine if between the nuclei of the bonded atoms they do not have the same symmetry. For because of which the repulsion between the example, 2pz orbital of one atom can combine nuclei is very less while in case of an with 2pz orbital of the other atom but not antibonding molecular orbital, most of the with the 2px or 2py orbitals because of their electron density is located away from the space different symmetries. between the nuclei. Infact, there is a nodal plane (on which the electron density is zero) 3. The combining atomic orbitals must between the nuclei and hence the repulsion overlap to the maximum extent. Greater between the nuclei is high. Electrons placed in the extent of overlap, the greater will be the a bonding molecular orbital tend to hold the electron-density between the nuclei of a nuclei together and stabilise a molecule. molecular orbital. Therefore, a bonding molecular orbital always possesses lower energy than either of the 4.7.3 Types of Molecular Orbitals atomic orbitals that have combined to form it. In contrast, the electrons placed in the Molecular orbitals of diatomic molecules are antibonding molecular orbital destabilise the designated as σ (sigma), π (pi), δ (delta), etc. molecule. This is because the mutual repulsion of the electrons in this orbital is more than the In this nomenclature, the sigma (σ) attraction between the electrons and the nuclei, molecular orbitals are symmetrical around which causes a net increase in energy. the bond-axis while pi (π) molecular orbitals are not symmetrical. For example, the linear It may be noted that the energy of the combination of 1s orbitals centered on two antibonding orbital is raised above the energy nuclei produces two molecular orbitals which of the parent atomic orbitals that have are symmetrical around the bond-axis. Such combined and the energy of the bonding orbital molecular orbitals are of the σ type and are has been lowered than the parent orbitals. The designated as σ1s and σ*1s [Fig. 4.20(a),page total energy of two molecular orbitals, however, 124]. If internuclear axis is taken to be in remains the same as that of two original atomic the z-direction, it can be seen that a linear orbitals. combination of 2pz- orbitals of two atoms also produces two sigma molecular orbitals 4.7.2 Conditions for the Combination of designated as σ2pz and σ*2pz. [Fig. 4.20(b)] Atomic Orbitals Molecular orbitals obtained from 2px and The linear combination of atomic orbitals to 2py orbitals are not symmetrical around the form molecular orbitals takes place only if the bond axis because of the presence of positive following conditions are satisfied: lobes above and negative lobes below the molecular plane. Such molecular orbitals, are 1. The combining atomic orbitals must labelled as π and π * [Fig. 4.20(c)]. A π bonding have the same or nearly the same energy. MO has larger electron density above and This means that 1s orbital can combine with below the inter-nuclear axis. The π* another 1s orbital but not with 2s orbital antibonding MO has a node between the nuclei. because the energy of 2s orbital is appreciably higher than that of 1s orbital. This is not true 4.7.4 Energy Level Diagram for Molecular if the atoms are very different. Orbitals 2. The combining atomic orbitals must We have seen that 1s atomic orbitals on two have the same symmetry about the atoms form two molecular orbitals designated as σ1s and σ*1s. In the same manner, the 2s and 2p atomic orbitals (eight atomic orbitals 2020-21

128 CHEMISTRY Fig. 4.20 Contours and energies of bonding and antibonding molecular orbitals formed through combinations of (a) 1s atomic orbitals; (b) 2pz atomic orbitals and (c) 2px atomic orbitals. on two atoms) give rise to the following eight The energy levels of these molecular molecular orbitals: orbitals have been determined experimentally from spectroscopic data for homonuclear Antibonding MOs σ*2s σ*2pz π *2px π*2py diatomic molecules of second row elements of the periodic table. The increasing order of Bonding MOs σ2s σ2pz π2px π2py 2020-21

CHEMICAL BONDING AND MOLECULAR STRUCTURE 129 energies of various molecular orbitals for O2 The rules discussed above regarding the and F2 is given below: stability of the molecule can be restated in terms of bond order as follows: A positive bond σ1s < σ * 1s < σ2s < σ * 2s < σ2 pz < ( π 2 px = π 2 py ) order (i.e., Nb > Na) means a stable molecule while a negative (i.e., Nb<Na) or zero (i.e., < (π * 2 px = π * 2 py ) < σ * 2 pz Nb = Na) bond order means an unstable molecule. However, this sequence of energy levels of molecular orbitals is not correct for the Nature of the bond remaining molecules Li2, Be2, B2, C2, N2. For instance, it has been observed experimentally Integral bond order values of 1, 2 or 3 that for molecules such as B2, C2, N2, etc. the correspond to single, double or triple bonds increasing order of energies of various respectively as studied in the classical molecular orbitals is concept. σ1s < σ * 1s < σ2s < σ * 2s < ( π 2 px = π 2 py ) Bond-length < σ2 pz < (π * 2 px = π * 2 py ) < σ * 2 pz The bond order between two atoms in a molecule may be taken as an approximate The important characteristic feature of this measure of the bond length. The bond length order is that the energy of σ2pz decreases as bond order increases. molecular orbital is higher than that of π 2px and π 2py molecular orbitals. Magnetic nature 4.7.5 Electronic Configuration and If all the molecular orbitals in a molecule are Molecular Behaviour doubly occupied, the substance is diamagnetic (repelled by magnetic field). The distribution of electrons among various However if one or more molecular orbitals are molecular orbitals is called the electronic singly occupied it is paramagnetic (attracted configuration of the molecule. From the by magnetic field), e.g., O2 molecule. electronic configuration of the molecule, it is possible to get important information about 4.8 BONDING IN SOME HOMONUCLEAR the molecule as discussed below. DIATOMIC MOLECULES Stability of Molecules: If Nb is the number In this section we shall discuss bonding in of electrons occupying bonding orbitals and some homonuclear diatomic molecules. Na the number occupying the antibonding orbitals, then 1. Hydrogen molecule (H2 ): It is formed by the combination of two hydrogen atoms. Each (i) the molecule is stable if Nb is greater than hydrogen atom has one electron in 1s orbital. Na, and Therefore, in all there are two electrons in hydrogen molecule which are present in σ1s (ii) the molecule is unstable if Nb is less molecular orbital. So electronic configuration than Na. of hydrogen molecule is In (i) more bonding orbitals are occupied H2 : (σ1s)2 and so the bonding influence is stronger and a The bond order of H2 molecule can be stable molecule results. In (ii) the antibonding calculated as given below: influence is stronger and therefore the molecule is unstable. Bond order = Nb − Na = 2−0 =1 2 2 Bond order This means that the two hydrogen atoms Bond order (b.o.) is defined as one half the are bonded together by a single covalent bond. difference between the number of electrons The bond dissociation energy of hydrogen present in the bonding and the antibonding molecule has been found to be 438 kJ mol–1 orbitals i.e., Bond order (b.o.) = ½ (Nb–Na) 2020-21

130 CHEMISTRY and bond length equal to 74 pm. Since no molecules have indeed been detected in vapour unpaired electron is present in hydrogen phase. It is important to note that double bond molecule, therefore, it is diamagnetic. in C2 consists of both pi bonds because of the presence of four electrons in two pi molecular 2co.nHfigeuliruatmionmofohleelciuumlea(tHome2is):1Ts2h.eEaeclhechterloiunmic orbitals. In most of the other molecules a atom contains 2 electrons, therefore, in He2 double bond is made up of a sigma bond and molecule there would be 4 electrons. These a pi bond. In a similar fashion the bonding in electrons will be accommodated in σ1s and σ*1s N2 molecule can be discussed. molecular orbitals leading to electronic configuration: 5. Oxygen molecule (O2 ): The electronic configuration of oxygen atom is 1s2 2s2 2p4. He2 : (σ1s)2 (σ*1s)2 Each oxygen atom has 8 electrons, hence, in O2 molecule there are 16 electrons. The Bond order of He2 is ½(2 – 2) = 0 electronic configuration of O2 molecule, therefore, is He2 molecule is therefore unstable and does not exist. O2: ( σ1s)2( σ*1s)2( σ*2s)2( σ*2s)2( σ2pz )2 Similarly, it can be shown that Be2 ( ) ( )π2px2 ≡ π2py2 molecule (σ1s)2 (σ*1s)2 (σ2s)2 (σ*2s)2 also does π * 2p 1 ≡ π * 2py1 not exist. x  KK (σ2s)2(σ * 2s)2(σ2pz )2  3. Lithium molecule (iLsi12 s):2,T2hse1 electronic ( ) ( )O2 :  configuration of lithium . There are  π 2p 2 ≡ π 2p 2 , π * 2p1x ≡ π * 2p1y  x y six electrons in Li2. The electronic configuration of Li2 molecule, therefore, is From the electronic configuration of O2 molecule it is clear that ten electrons are present Li2 : (σ1s)2 (σ*1s)2 (σ2s)2 in bonding molecular orbitals and six electrons The above configuration is also written as are present in antibonding molecular orbitals. KK(σ2s)2 where KK represents the closed K shell Its bond order, therefore, is structure (σ1s)2 (σ*1s)2. 1 1 Bond order = 2 [ N − Na] = 2 [10 − 6] = 2 From the electronic configuration of Li2 b molecule it is clear that there are four electrons present in bonding molecular orbitals and two So in oxygen molecule, atoms are held electrons present in antibonding molecular orbitals. Its bond order, therefore, is ½ (4 – 2) by a double bond. Moreover, it may be noted = 1. It means that Li2 molecule is stable and since it has no unpaired electrons it should be that it contains two unpaired electrons in diamagnetic. Indeed diamagnetic Li2 molecules are known to exist in the vapour π * 2px and π * 2py molecular orbitals, therefore, O2 molecule should be phase. paramagnetic, a prediction that corresponds to experimental observation. In this way, the theory successfully explains the paramagnetic nature of oxygen. 4. Carbon molecule (C2 ): The electronic Similarly, the electronic configurations of configuration of carbon is 1s2 2s2 2p2. There other homonuclear diatomic molecules of the second row of the periodic table can be written. are twelve electrons in C2. The electronic In Fig.4.21 are given the molecular orbital configuration of C2 molecule, therefore, is occupancy and molecular properties for B2 through Ne2. The sequence of MOs and their ( )C2 : (σ1s)2 (σ *1s)2 (σ * 2s)2 electron population are shown. The bond π 2p 2 = π 2py2 energy, bond length, bond order, magnetic x properties and valence electron configuration appear below the orbital diagrams. or KK (σ2s)2 (σ * 2s)2 (π2p2 = π 2p2 ) xy The bond order of C2 is ½ (8 – 4) = 2 and C2 should be diamagnetic. Diamagnetic C2 2020-21

CHEMICAL BONDING AND MOLECULAR STRUCTURE 131 Fig. 4.21 MO occupancy and molecular properties for B2 through Ne2. 4.9 HYDROGEN BONDING Hydrogen bond is represented by a dotted line (– – –) while a solid line represents the covalent Nitrogen, oxygen and fluorine are the higly bond. Thus, hydrogen bond can be defined electronegative elements. When they are as the attractive force which binds attached to a hydrogen atom to form covalent hydrogen atom of one molecule with the bond, the electrons of the covalent bond are electronegative atom (F, O or N) of another shifted towards the more electronegative atom. molecule. This partially positively charged hydrogen atom forms a bond with the other more 4.9.1 Cause of Formation of Hydrogen electronegative atom. This bond is known as Bond hydrogen bond and is weaker than the covalent bond. For example, in HF molecule, When hydrogen is bonded to strongly the hydrogen bond exists between hydrogen electronegative element ‘X’, the electron pair atom of one molecule and fluorine atom of shared between the two atoms moves far away another molecule as depicted below : from hydrogen atom. As a result the hydrogen atom becomes highly electropositive with − − −Hδ+ – Fδ− − − − Hδ+ – Fδ− − − − Hδ+ – Fδ− respect to the other atom ‘X’. Since there is displacement of electrons towards X, the Here, hydrogen bond acts as a bridge between hydrogen acquires fractional positive charge two atoms which holds one atom by covalent (δ +) while ‘X’ attain fractional negative charge bond and the other by hydrogen bond. 2020-21

132 CHEMISTRY (δ–). This results in the formation of a polar bond in case of HF molecule, alcohol or water molecule having electrostatic force of attraction molecules, etc. which can be represented as : (2) Intramolecular hydrogen bond : It is formed when hydrogen atom is in between the Hδ+ − Xδ− − − − Hδ+ − X δ− − − − Hδ+ − Xδ− two highly electronegative (F, O, N) atoms present within the same molecule. For example, The magnitude of H-bonding depends on in o-nitrophenol the hydrogen is in between the physical state of the compound. It is the two oxygen atoms. maximum in the solid state and minimum in the gaseous state. Thus, the hydrogen bonds Fig. 4.22 Intramolecular hydrogen bonding in have strong influence on the structure and o-nitrophenol molecule 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 same or different compounds. For example, H- 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. 2020-21

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, CO23−, HCOOH 2020-21

134 CHEMISTRY 4.5 Define octet rule. Write its significance and limitations. 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 4.12 Explain the important aspects of resonance with reference to the CO 2 − ion. 3 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 4.15 cations and anions : (a) K and S (b) Ca and O (c) Al and N. 4.16 Although both CO2 and H2O are triatomic molecules, the shape of H2O molecule 4.17 is bent while that of CO2 is linear. Explain this on the basis of dipole moment. 4.18 Write the significance/applications of dipole moment. 4.19 Define electronegativity. How does it differ from electron gain enthalpy ? 4.20 Explain with the help of suitable example polar covalent bond. Arrange the bonds in order of increasing ionic character in the molecules: LiF, K2O, N2, SO2 and ClF3. 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 4.22 centre. Explain why CH4 is not square planar ? 4.23 Explain why BeH2 molecule has a zero dipole moment although the Be–H bonds 4.24 are polar. Which out of NH3 and NF3 has higher dipole moment and why ? What is meant by hybridisation of atomic orbitals? Describe the shapes of sp, sp2, sp3 hybrid orbitals. 2020-21

CHEMICAL BONDING AND MOLECULAR STRUCTURE 135 4.25 Describe the change in hybridisation (if any) of the Al atom in the following reaction. 4.26 AlCl3 + Cl− → AlCl4− 4.27 4.28 Is there any change in the hybridisation of B and N atoms as a result of the 4.29 following reaction? 4.30 BF3 + NH3 → F3B.NH3 4.31 4.32 Draw diagrams showing the formation of a double bond and a triple bond between 4.33 carbon atoms in C2H4 and C2H2 molecules. 4.34 What is the total number of sigma and pi bonds in the following molecules? 4.35 4.36 (a) C2H2 (b) C2H4 Considering x-axis as the internuclear axis which out of the following will not 4.37 form a sigma bond and why? (a) 1s and 1s (b) 1s and 2px ; (c) 2py and 2py 4.38 (d) 1s and 2s. 4.39 4.40 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 What do you understand by bond pairs and lone pairs of electrons? Illustrate by giving one exmaple of each type. Distinguish between a sigma and a pi bond. Explain the formation of H2 molecule on the basis of valence bond theory. Write the important conditions required for the linear combination of atomic orbitals to form molecular orbitals. Use molecular orbital theory to explain why the Be2 molecule does not exist. Compare the relative stability of the following species and indicate their magnetic properties; O2,O2+,O2− (superoxide), O22− (peroxide) Write the significance of a plus and a minus sign shown in representing the orbitals. Describe the hybridisation in case of PCl5. Why are the axial bonds longer as compared to equatorial bonds? Define hydrogen bond. Is it weaker or stronger than the van der Waals forces? What is meant by the term bond order? Calculate the bond order of : N2, O2, O2+ and O2–. 2020-21

136 CHEMISTRY STATES OF MATTER UNIT 5 After studying this unit you will be The snowflake falls, yet lays not long able to Its feath’ry grasp on Mother Earth Ere Sun returns it to the vapors Whence it came, • explain the existence of different Or to waters tumbling down the rocky slope. states of matter in terms of Rod O’ Connor balance between intermolecular forces and thermal energy of INTRODUCTION particles; In previous units we have learnt about the properties • explain the laws governing related to single particle of matter, such as atomic size, ionization enthalpy, electronic charge density, molecular behaviour of ideal gases; shape and polarity, etc. Most of the observable characteristics of chemical systems with which we are • apply gas laws in various real life familiar represent bulk properties of matter, i.e., the properties associated with a collection of a large number situations; of atoms, ions or molecules. For example, an individual molecule of a liquid does not boil but the bulk boils. • explain the behaviour of real Collection of water molecules have wetting properties; individual molecules do not wet. Water can exist as ice, gases; which is a solid; it can exist as liquid; or it can exist in the gaseous state as water vapour or steam. Physical • describe the conditions required properties of ice, water and steam are very different. In all the three states of water chemical composition of water for liquifaction of gases; remains the same i.e., H2O. Characteristics of the three states of water depend on the energies of molecules and • realise that there is continuity in on the manner in which water molecules aggregate. Same is true for other substances also. gaseous and liquid state; Chemical properties of a substance do not change with • differentiate between gaseous the change of its physical state; but rate of chemical reactions do depend upon the physical state. Many times state and vapours; and in calculations while dealing with data of experiments we require knowledge of the state of matter. Therefore, it • explain properties of liquids in becomes necessary for a chemist to know the physical terms of intermolecular attractions. 2020-21

STATES OF MATTER 137 laws which govern the behaviour of matter in so happen that momentarily electronic charge different states. In this unit, we will learn distribution in one of the atoms, say ‘A’, more about these three physical states of becomes unsymmetrical i.e., the charge cloud matter particularly liquid and gaseous states. is more on one side than the other (Fig. 5.1 b To begin with, it is necessary to understand and c). This results in the development of the nature of intermolecular forces, molecular instantaneous dipole on the atom ‘A’ for a very interactions and effect of thermal energy on short time. This instantaneous or transient the motion of particles because a balance dipole distorts the electron density of the between these determines the state of a other atom ‘B’, which is close to it and as a substance. consequence a dipole is induced in the atom ‘B’. 5.1 INTERMOLECULAR FORCES The temporary dipoles of atom ‘A’ and ‘B’ Intermolecular forces are the forces of attract each other. Similarly temporary dipoles attraction and repulsion between interacting are induced in molecules also. This force of particles (atoms and molecules). This term attraction was first proposed by the German does not include the electrostatic forces that physicist Fritz London, and for this reason exist between the two oppositely charged ions force of attraction between two temporary and the forces that hold atoms of a molecule together i.e., covalent bonds. Fig. 5.1 Dispersion forces or London forces between atoms. Attractive intermolecular forces are known as van der Waals forces, in honour of Dutch scientist Johannes van der Waals (1837- 1923), who explained the deviation of real gases from the ideal behaviour through these forces. We will learn about this later in this unit. van der Waals forces vary considerably in magnitude and include dispersion forces or London forces, dipole-dipole forces, and dipole-induced dipole forces. A particularly strong type of dipole-dipole interaction is hydrogen bonding. Only a few elements can participate in hydrogen bond formation, therefore it is treated as a separate category. We have already learnt about this interaction in Unit 4. At this point, it is important to note that attractive forces between an ion and a dipole are known as ion-dipole forces and these are not van der Waals forces. We will now learn about different types of van der Waals forces. 5.1.1 Dispersion Forces or London Forces Atoms and nonpolar molecules are electrically symmetrical and have no dipole moment because their electronic charge cloud is symmetrically distributed. But a dipole may develop momentarily even in such atoms and molecules. This can be understood as follows. Suppose we have two atoms ‘A’ and ‘B’ in the close vicinity of each other (Fig. 5.1a). It may 2020-21

138 CHEMISTRY dipoles is known as London force. Another proportional to 1/r 6, where r is the distance name for this force is dispersion force. These between polar molecules. Besides dipole- forces are always attractive and interaction dipole interaction, polar molecules can energy is inversely proportional to the sixth interact by London forces also. Thus power of the distance between two interacting cumulative effect is that the total of particles (i.e., 1/r 6 where r is the distance intermolecular forces in polar molecules between two particles). These forces are increase. important only at short distances (~500 pm) and their magnitude depends on the 5.1.3 Dipole–Induced Dipole Forces polarisability of the particle. This type of attractive forces operate between 5.1.2 Dipole - Dipole Forces the polar molecules having permanent dipole and the molecules lacking permanent dipole. Dipole-dipole forces act between the molecules Permanent dipole of the polar molecule possessing permanent dipole. Ends of the induces dipole on the electrically neutral dipoles possess “partial charges” and these molecule by deforming its electronic cloud charges are shown by Greek letter delta (δ). (Fig. 5.3). Thus an induced dipole is developed Partial charges are always less than the unit in the other molecule. In this case also electronic charge (1.6×10–19 C). The polar interaction energy is proportional to 1/r 6 molecules interact with neighbouring where r is the distance between two molecules. Fig 5.2 (a) shows electron cloud molecules. Induced dipole moment depends distribution in the dipole of hydrogen chloride upon the dipole moment present in the and Fig. 5.2 (b) shows dipole-dipole interaction permanent dipole and the polarisability of the between two HCl molecules. This interaction electrically neutral molecule. We have already is stronger than the London forces but is learnt in Unit 4 that molecules of larger size weaker than ion-ion interaction because only can be easily polarized. High polarisability partial charges are involved. The attractive increases the strength of attractive force decreases with the increase of distance interactions. between the dipoles. As in the above case here also, the interaction energy is inversely proportional to distance between polar molecules. Dipole-dipole interaction energy between stationary polar molecules (as in solids) is proportional to 1/r3 and that between rotating polar molecules is Fig. 5.2 (a) Distribution of electron cloud in HCl – Fig. 5.3 Dipole - induced dipole interaction a polar molecule, (b) Dipole-dipole between permanent dipole and induced interaction between two HCl molecules dipole In this case also cumulative effect of dispersion forces and dipole-induced dipole interactions exists. 5.1.4 Hydrogen bond As already mentioned in section (5.1); this is special case of dipole-dipole interaction. We have already learnt about this in Unit 4. This 2020-21

STATES OF MATTER 139 is found in the molecules in which highly polar thermal energy of the molecules tends to keep N–H, O–H or H–F bonds are present. Although them apart. Three states of matter are the result hydrogen bonding is regarded as being limited of balance between intermolecular forces and to N, O and F; but species such as Cl may the thermal energy of the molecules. also participate in hydrogen bonding. Energy of hydrogen bond varies between 10 to 100 When molecular interactions are very kJ mol–1. This is quite a significant amount of weak, molecules do not cling together to make energy; therefore, hydrogen bonds are liquid or solid unless thermal energy is powerful force in determining the structure and reduced by lowering the temperature. Gases properties of many compounds, for example do not liquify on compression only, although proteins and nucleic acids. Strength of the molecules come very close to each other and hydrogen bond is determined by the coulombic intermolecular forces operate to the maximum. interaction between the lone-pair electrons of However, when thermal energy of molecules the electronegative atom of one molecule and is reduced by lowering the temperature; the the hydrogen atom of other molecule. gases can be very easily liquified. Following diagram shows the formation of Predominance of thermal energy and the hydrogen bond. molecular interaction energy of a substance in three states is depicted as follows : δ+ δ− δ+ δ− We have already learnt the cause for the H− F⋅⋅⋅ H− F existence of the three states of matter. Now we will learn more about gaseous and liquid Intermolecular forces discussed so far are states and the laws which govern the all attractive. Molecules also exert repulsive behaviour of matter in these states. We shall forces on one another. When two molecules deal with the solid state in class XII. are brought into close contact with each other, the repulsion between the electron clouds and 5.4 THE GASEOUS STATE that between the nuclei of two molecules comes This is the simplest state of matter. into play. Magnitude of the repulsion rises very Throughout our life we remain immersed in rapidly as the distance separating the the ocean of air which is a mixture of gases. molecules decreases. This is the reason that We spend our life in the lowermost layer of liquids and solids are hard to compress. In the atmosphere called troposphere, which is these states molecules are already in close held to the surface of the earth by gravitational contact; therefore they resist further force. The thin layer of atmosphere is vital to compression; as that would result in the our life. It shields us from harmful radiations increase of repulsive interactions. and contains substances like dioxygen, dinitrogen, carbon dioxide, water vapour, etc. 5.2 THERMAL ENERGY Let us now focus our attention on the Thermal energy is the energy of a body arising behaviour of substances which exist in the from motion of its atoms or molecules. It is gaseous state under normal conditions of directly proportional to the temperature of the temperature and pressure. A look at the substance. It is the measure of average periodic table shows that only eleven elements kinetic energy of the particles of the matter and is thus responsible for movement of particles. This movement of particles is called thermal motion. 5.3 INTERMOLECULAR FORCES vs THERMAL INTERACTIONS We have already learnt that intermolecular forces tend to keep the molecules together but 2020-21

140 CHEMISTRY centuries on the physical properties of gases. The first reliable measurement on properties of gases was made by Anglo-Irish scientist Robert Boyle in 1662. The law which he formulated is known as Boyle’s Law. Later on attempts to fly in air with the help of hot air balloons motivated Jaccques Charles and Joseph Lewis Gay Lussac to discover additional gas laws. Contribution from Avogadro and others provided lot of information about gaseous state. Fig. 5.4 Eleven elements that exist as gases 5.5.1 Boyle’s Law (Pressure - Volume Relationship) exist as gases under normal conditions (Fig 5.4). On the basis of his experiments, Robert Boyle reached to the conclusion that at constant The gaseous state is characterized by the temperature, the pressure of a fixed following physical properties. amount (i.e., number of moles n) of gas • Gases are highly compressible. varies inversely with its volume. This is • Gases exert pressure equally in all known as Boyle’s law. Mathematically, it can be written as directions. • Gases have much lower density than the p∝ 1 ( at constant T and n) (5.1) V solids and liquids. • The volume and the shape of gases are ⇒ p = k1 1 (5.2) V not fixed. These assume volume and shape of the container. where k1 is the proportionality constant. The • Gases mix evenly and completely in all value of constant k1 depends upon the proportions without any mechanical aid. amount of the gas, temperature of the gas Simplicity of gases is due to the fact that and the units in which p and V are expressed. the forces of interaction between their molecules are negligible. Their behaviour is On rearranging equation (5.2) we obtain governed by same general laws, which were discovered as a result of their experimental pV = k1 (5.3) studies. These laws are relationships between measurable properties of gases. Some of these It means that at constant temperature, properties like pressure, volume, temperature and mass are very important because product of pressure and volume of a fixed relationships between these variables describe state of the gas. Interdependence of these amount of gas is constant. variables leads to the formulation of gas laws. In the next section we will learn about gas If a fixed amount of gas at constant laws. temperature T occupying volume V1 at 5.5 THE GAS LAWS pressure p1 undergoes expansion, so that volume becomes V2 and pressure becomes p2, The gas laws which we will study now are the then according to Boyle’s law : result of research carried on for several p V = p V = constant (5.4) 11 22 ⇒ p1 = V2 (5.5) p2 V1 2020-21

STATES OF MATTER 141 Figure 5.5 shows two conventional ways of graphically presenting Boyle’s law. Fig. 5.5 (a) is the graph of equation (5.3) at different temperatures. The value of k1 for each curve is different because for a given mass of gas, it varies only with temperature. Each curve corresponds to a different constant temperature and is known as an isotherm (constant temperature plot). Higher curves correspond to higher temperature. It should be noted that volume of the gas doubles if pressure is halved. Table 5.1 gives effect of pressure on volume of 0.09 mol of CO2 at 300 K. Fig 5.5 (b) represents the graph between p Fig. 5.5(a) Graph of pressure, p vs. Volume, V of 1 a gas at different temperatures. and V . It is a straight line passing through origin. However at high pressures, gases deviate from Boyle’s law and under such conditions a straight line is not obtained in the graph. Experiments of Boyle, in a quantitative manner prove that gases are highly compressible because when a given mass of a gas is compressed, the same number of molecules occupy a smaller space. This means that gases become denser at high pressure. A relationship can be obtained between density and pressure of a gas by using Boyle’s law: By definition, density ‘d’ is related to the mass ‘m’ and the volume ‘V’ by the relation d = m . If we put value of V in this equation V 1 from Boyle’s law equation, we obtain the Fig. 5.5 (b) Graph of pressure of a gas, p vs. V relationship. Table 5.1 Effect of Pressure on the Volume of 0.09 mol CO2 Gas at 300 K. Pressure/104 Pa Volume/10–3 m3 (1/V )/m–3 pV/102 Pa m3 2.0 112.0 8.90 22.40 2.5 89.2 11.2 22.30 3.5 64.2 15.6 22.47 4.0 56.3 17.7 22.50 6.0 37.4 26.7 22.44 8.0 28.1 35.6 22.48 10.0 22.4 44.6 22.40 2020-21

142 CHEMISTRY d = m  p = k′ p ⇒ Vt = V0  273.15 + t  (5.6)  k1  273.15 This shows that at a constant temperature, At this stage, we define a new scale of pressure is directly proportional to the density temperature such that t °C on new scale is given of a fixed mass of the gas. by T = 273.15 + t and 0 °C will be given by T0 = 273.15. This new temperature scale is Problem 5.1 called the Kelvin temperature scale or Absolute temperature scale. A balloon is filled with hydrogen at room temperature. It will burst if pressure Thus 0°C on the celsius scale is equal to exceeds 0.2 bar. If at 1 bar pressure the 273.15 K at the absolute scale. Note that gas occupies 2.27 L volume, upto what degree sign is not used while writing the volume can the balloon be expanded ? temperature in absolute temperature scale, i.e., Kelvin scale. Kelvin scale of temperature Solution is also called Thermodynamic scale of temperature and is used in all scientific According to Boyle’s Law p1V1 = p2V2 works. If p1 is 1 bar, V1 will be 2.27 L Thus we add 273 (more precisely 273.15) to the celsius temperature to obtain If p2 = 0.2 bar, then V2 = p1V1 temperature at Kelvin scale. p2 ⇒ V2 = 1bar × 2.27 L =11.35 L If we write Tt = 273.15 + t and T0 = 273.15 0.2 bar in the equation (5.6) we obtain the relationship Since balloon bursts at 0.2 bar pressure, Vt = V0  Tt  the volume of balloon should be less than  T0  11.35 L. 5.5.2 Charles’ Law (Temperature - Volume ⇒ Vt = Tt (5.7) Relationship) V0 T0 Charles and Gay Lussac performed several Thus we can write a general equation as experiments on gases independently to improve upon hot air balloon technology. follows. Their investigations showed that for a fixed mass of a gas at constant pressure, volume V2 = T2 (5.8) of a gas increases on increasing temperature V1 T1 and decreases on cooling. They found that for each degree rise in temperature, volume ⇒ V1 = V2 T1 T2 1 of a gas increases by 273.15 of the original ⇒V = constant = k2 (5.9) volume of the gas at 0 °C. Thus if volumes of T the gas at 0 °C and at t °C are V0 and Vt respectively, then Thus V = k2 T (5.10) The value of constant k2 is determined by the pressure of the gas, its amount and the Vt = V0 + t V0 units in which volume V is expressed. 273.15 Equation (5.10) is the mathematical ⇒ Vt = V0 1 + t  expression for Charles’ law, which states that 273.15 pressure remaining constant, the volume 2020-21

STATES OF MATTER 143 of a fixed mass of a gas is directly with 2 L air. What will be the volume of proportional to its absolute temperature. the balloon when the ship reaches Indian Charles found that for all gases, at any given ocean, where temperature is 26.1°C ? pressure, graph of volume vs temperature (in celsius) is a straight line and on extending to Solution zero volume, each line intercepts the temperature axis at – 273.15 °C. Slopes of V1 = 2 L T2 = 26.1 + 273 lines obtained at different pressure are T1 = (23.4 + 273) K = 299.1 K different but at zero volume all the lines meet the temperature axis at – 273.15 °C (Fig. 5.6). = 296.4 K From Charles law V1 = V2 T1 T2 ⇒ V2 = V1T2 T1 ⇒ V2 = 2 L × 299.1K 296.4 K = 2L × 1.009 = 2.018L 5.5.3 Gay Lussac’s Law (Pressure- Temperature Relationship) Fig. 5.6 Volume vs Temperature ( °C) graph Pressure in well inflated tyres of automobiles is almost constant, but on a hot summer day Each line of the volume vs temperature this increases considerably and tyre may graph is called isobar. burst if pressure is not adjusted properly. During winters, on a cold morning one may Observations of Charles can be interpreted find the pressure in the tyres of a vehicle if we put the value of t in equation (5.6) as decreased considerably. The mathematical – 273.15 °C. We can see that the volume of relationship between pressure and the gas at – 273.15 °C will be zero. This means temperature was given by Joseph Gay Lussac that gas will not exist. In fact all the gases get and is known as Gay Lussac’s law. It states liquified before this temperature is reached. that at constant volume, pressure of a fixed The lowest hypothetical or imaginary amount of a gas varies directly with the temperature at which gases are supposed to temperature. Mathematically, occupy zero volume is called Absolute zero. p ∝T All gases obey Charles’ law at very low pressures and high temperatures. ⇒ p = constant = k3 T Problem 5.2 This relationship can be derived from On a ship sailing in Pacific Ocean where Boyle’s law and Charles’ Law. Pressure vs temperature is 23.4°C, a balloon is filled temperature (Kelvin) graph at constant molar volume is shown in Fig. 5.7. Each line of this graph is called isochore. 2020-21

144 CHEMISTRY Fig. 5.7 Pressure vs temperature (K) graph we came across while discussing definition of (Isochores) of a gas. a ‘mole’ (Unit 1). Since volume of a gas is directly proportional to the number of moles; one mole of each gas at standard temperature and pressure (STP)* will have same volume. Standard temperature and pressure means 273.15 K (0°C) temperature and 1 bar (i.e., exactly 105 pascal) pressure. These values approximate freezing temperature of water and atmospheric pressure at sea level. At STP molar volume of an ideal gas or a combination of ideal gases is 22.71098 L mol–1. Molar volume of some gases is given in (Table 5.2). Table 5.2 Molar volume in litres per mole of some gases at 273.15 K and 1 bar (STP). 5.5.4 Avogadro Law (Volume - Amount Argon 22.37 Relationship) Carbon dioxide 22.54 Dinitrogen 22.69 In 1811 Italian scientist Amedeo Avogadro Dioxygen 22.69 tried to combine conclusions of Dalton’s atomic Dihydrogen 22.72 theory and Gay Lussac’s law of combining Ideal gas 22.71 volumes (Unit 1) which is now known as Avogadro law. It states that equal volumes Number of moles of a gas can be calculated of all gases under the same conditions of as follows temperature and pressure contain equal number of molecules. This means that as m (5.12) long as the temperature and pressure remain n= M constant, the volume depends upon number of molecules of the gas or in other words Where m = mass of the gas under amount of the gas. Mathematically we can write investigation and M = molar mass V ∝ n where n is the number of moles Thus, of the gas. V = k4 m (5.13) M ⇒ V = k4 n (5.11) Equation (5.13) can be rearranged as The number of molecules in one mole of a follows : gas has been determined to be 6.022 ×1023 and is known as Avogadro constant. You m (5.14) will find that this is the same number which M = k4 M = k4d * The previous standard is still often used, and applies to all chemistry data more than decade old. In this definition STP denotes the same temperature of 0°C (273.15 K), but a slightly higher pressure of 1 atm (101.325 kPa). One mole of any gas of a combination of gases occupies 22.413996 L of volume at STP. Standard ambient temperature and pressure (SATP), conditions are also used in some scientific works. SATP conditions means 298.15 K and 1 bar (i.e., exactly 105 Pa). At SATP (1 bar and 298.15 K), the molar volume of an ideal gas is 24.789 L mol–1. 2020-21

STATES OF MATTER 145 Here ‘d’ is the density of the gas. We can conclude that at constant temperature and pressure from equation (5.14) that the density of a gas is n moles of any gas will have the same volume directly proportional to its molar mass. because V = nRT and n,R,T and p are A gas that follows Boyle’s law, Charles’ law p and Avogadro law strictly is called an ideal gas. Such a gas is hypothetical. It is assumed constant. This equation will be applicable to that intermolecular forces are not present any gas, under those conditions when between the molecules of an ideal gas. Real behaviour of the gas approaches ideal gases follow these laws only under certain behaviour. Volume of one mole of an ideal specific conditions when forces of interaction gas under STP conditions (273.15 K and 1 are practically negligible. In all other situations bar pressure) is 22.710981 L mol–1. Value these deviate from ideal behaviour. You will of R for one mole of an ideal gas can be learn about the deviations later in this unit. calculated under these conditions as follows : 5.6 IDEAL GAS EQUATION = 8.314 Pa m3 K–1 mol–1 = 8.314 × 10–2 bar L K–1 mol–1 The three laws which we have learnt till now can be combined together in a single equation which is known as ideal gas equation. At constant T and n; V ∝ 1 Boyle’s Law = 8.314 J K–1 mol–1 p At constant p and n; V ∝ T Charles’ Law At STP conditions used earlier At constant p and T ; V ∝ n Avogadro Law (0 °C and 1 atm pressure), value of R is 8.20578 × 10–2 L atm K–1 mol–1. Thus, Ideal gas equation is a relation between V ∝ nT (5.15) four variables and it describes the state of p any gas, therefore, it is also called equation of state. ⇒ V = R nT p (5.16) Let us now go back to the ideal gas equation. This is the relationship for the where R is proportionality constant. On simultaneous variation of the variables. If rearranging the equation (5.16) we obtain temperature, volume and pressure of a fixed amount of gas vary from T1, V1 and p1 to T2, pV = n RT (5.17) V2 and p2 then we can write ⇒ R = pV (5.18) p1V1 = nR and p2V2 = nR nT T1 T2 R is called gas constant. It is same for all gases. ⇒ p1V1 = p2V2 (5.19) Therefore it is also called Universal Gas T1 T2 Constant. Equation (5.17) is called ideal gas equation. Equation (5.19) is a very useful equation. Equation (5.18) shows that the value of If out of six, values of five variables are known, R depends upon units in which p, V and T are measured. If three variables in this the value of unknown variable can be equation are known, fourth can be calculated. From this equation we can see calculated from the equation (5.19). This equation is also known as Combined gas law. 2020-21

146 CHEMISTRY Problem 5.3 by the mixture of non-reactive gases is At 25°C and 760 mm of Hg pressure a equal to the sum of the partial pressures gas occupies 600 mL volume. What will of individual gases i.e., the pressures which be its pressure at a height where these gases would exert if they were enclosed temperature is 10°C and volume of the separately in the same volume and under the gas is 640 mL. same conditions of temperature. In a mixture of gases, the pressure exerted by the individual Solution gas is called partial pressure. Mathematically, p1 = 760 mm Hg, V1= 600 mL pTotal = p1+p2+p3+......(at constant T, V) (5.23) T1 = 25 + 273 = 298 K V2 = 640 mL and T2 = 10 + 273 = 283 K where pTotal is the total pressure exerted by According to Combined gas law the mixture of gases and p1, p2 , p3 etc. are partial pressures of gases. p1V1 = p2V2 Gases are generally collected over water T1 T2 and therefore are moist. Pressure of dry gas can be calculated by subtracting vapour ⇒ p2 = p1 V1 T2 pressure of water from the total pressure of T1V2 the moist gas which contains water vapours (760 mm Hg) × (600 mL) × (283 K) also. Pressure exerted by saturated water ⇒ p2 = (640 mL) × (298 K) vapour is called aqueous tension. Aqueous tension of water at different temperatures is = 676.6 mm Hg given in Table 5.3. 5.6.1 Density and Molar Mass of a pDry gas = pTotal – Aqueous tension (5.24) Gaseous Substance Table 5.3 Aqueous Tension of Water (Vapour Ideal gas equation can be rearranged as follows: Pressure) as a Function of Temperature n= p V RT m Replacing n by M , we get m =p (5.20) M V RT d= p (where d is the density) (5.21) M RT Partial pressure in terms of mole fraction On rearranging equation (5.21) we get the relationship for calculating molar mass of a gas. Suppose at the temperature T, three gases, enclosed in the volume V, exert partial M = d RT (5.22) pressure p1, p2 and p3 respectively, then, p p1 = n1RT (5.25) V 5.6.2 Dalton’s Law of Partial Pressures p2 = n2RT (5.26) The law was formulated by John Dalton in V 1801. It states that the total pressure exerted 2020-21

STATES OF MATTER 147 p3 = n 3 RT (5.27) Number of moles of neon V 167.5 g where n1 n2 and n3 are number of moles of these = 20 g mol−1 gases. Thus, expression for total pressure will be pTotal = p1 + p2 + p3 = 8.375 mol Mole fraction of dioxygen = n1 RT + n2 RT + n3 RT = 2.21 V V V 2.21 + 8.375 RT (5.28) = 2.21 = (n1 + n2 + n3) V 10.585 On dividing p1 by ptotal we get = 0.21 p1 =  n1  RTV Mole fraction of neon = 8.375 ptotal  n1 +n 2  RTV 2.21 + 8.375 +n 3 = 0.79 n1 = n1 Alternatively, n1+n2 +n3 n = = x1 mole fraction of neon = 1– 0.21 = 0.79 where n = n1+n2+n3 Partial pressure = mole fraction × x1 is called mole fraction of first gas. of a gas total pressure Thus, p1 = x1 ptotal Similarly for other two gases we can write ⇒ Partial pressure = 0.21 × (25 bar) of oxygen = 5.25 bar p2 = x2 ptotal and p3 = x3 ptotal Partial pressure = 0.79 × (25 bar) Thus a general equation can be written as of neon = 19.75 bar pi = xi ptotal (5.29) where pi and xi are partial pressure and mole 5.7 KINETIC ENERGY AND MOLECULAR fraction of ith gas respectively. If total pressure SPEEDS of a mixture of gases is known, the equation Molecules of gases remain in continuous motion. While moving they collide with each (5.29) can be used to find out pressure exerted other and with the walls of the container. This by individual gases. results in change of their speed and redistribution of energy. So the speed and Problem 5.4 energy of all the molecules of the gas at any instant are not the same. Thus, we can obtain A neon-dioxygen mixture contains only average value of speed of molecules. If 70.6 g dioxygen and 167.5 g neon. If there are n number of molecules in a sample pressure of the mixture of gases in the and their individual speeds are u1, u2,…….un, cylinder is 25 bar. What is the partial then average speed of molecules uav can be pressure of dioxygen and neon in the calculated as follows: mixture ? uav = u1 + u 2 + .........un Number of moles of dioxygen n = 70.6 g 32 g mol−1 = 2.21 mol 2020-21