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

4.3 Bond Parameters c = 99 pm 1984.3.1 Bond Length r pmBond 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 rvdw= bond length (Fig. 4.1). In the case of a covalent 180 bond, the contribution from each atom is pm called the covalent radius of that atom. pm 360 The covalent radius is measured approximately as the radius of an atom’s Fig. 4.2 Covalent and van der Waals radii incore which is in contact with the core of an a chlorine molecule. The inner circles adjacent atom in a bonded situation. The correspond to the size of the chlorine covalent radius is half of the distance between atom (rvdw and rc are van der Waals and two similar atoms joined by a covalent bond covalent radii respectively). Reprint 2025-26 108 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 Covalent Bondmolecules are given in Table 4.3. Bond Type Length (pm) The covalent radii of some common O–H 96elements are listed in Table 4.4. C–H 107 4.3.2 Bond Angle N–O 136 C–O 143It is defined as the angle between the orbitals C–N 143 containing bonding electron pairs around the C–C 154 central atom in a molecule/complex ion. Bond C=O 121 angle is expressed in degree which can be N=O 122 experimentally determined by spectroscopic C=C 133 methods. It gives some idea regarding the C=N 138 distribution of orbitals around the central C≡N 116 atom in a molecule/complex ion and hence it C≡C 120 helps us in determining its shape. For Table 4.3 Bond Lengths in Some Commonexample H–O–H bond angle in water can be Moleculesrepresented as under : Molecule Bond Length (pm) H2 (H – H) 74 F2 (F – F) 144 4.3.3 Bond Enthalpy Cl2 (Cl – Cl) 199 It is defined as the amount of energy required Br2 (Br – Br) 228 to break one mole of bonds of a particular I2 (I – I) 267 type between two atoms in a gaseous state. N2 (N ≡ N) 109 The unit of bond enthalpy is kJ mol–1. For O2 (O = O) 121 example, the H – H bond enthalpy in hydrogen HF (H – F) 92 molecule is 435.8 kJ mol–1. HCl (H – Cl) 127 HBr (H – Br) 141H2(g) → H(g) + H(g); ∆aH = 435.8 kJ mol–1 HI (H – I) 160 Similarly the bond enthalpy for molecules Table 4.4 Covalent Radii, *rcov/(pm)containing multiple bonds, for example O2 and N2 will be as under : O2 (O = O) (g) → O(g) + O(g); ∆aH = 498 kJ mol–1 N2 (N ≡ N) (g) → N(g) + N(g); ∆aH = 946.0 kJ mol–1 It is important that larger the bond dissociation enthalpy, stronger will be the bond in the molecule. For a heteronuclear diatomic molecules like HCl, we have HCl (g) → H(g) + Cl (g); ∆aH = 431.0 kJ mol–1 In case of polyatomic molecules, the measurement of bond strength is more complicated. For example in case of H2O * The values cited are for single bonds, except where molecule, the enthalpy needed to break the otherwise indicated in parenthesis. (See also Unit 3 for two O – H bonds is not the same. periodic trends). Reprint 2025-26 Chemical Bonding And Molecular Structure 109 H2O(g) → H(g) + OH(g); ∆aH1 = 502 kJ mol–1 OH(g) → H(g) + O(g); ∆aH2 = 427 kJ mol–1 The difference in the ∆aH value shows that the second O – H bond undergoes some change because of changed chemical environment. This is the reason for some difference in energy of the same O – H bond in different molecules like C2H5OH (ethanol) and water. Therefore in polyatomic molecules the term mean or average bond enthalpy is used. It is obtained by dividing total bond dissociation enthalpy by the number of bonds broken as explained below in case of water molecule, 502 + 427 Fig. 4.3 Resonance in the O3 molecule Average bond enthalpy = 2 (structures I and II represent the two canonical = 464.5 kJ mol–1 forms while the structure III is the resonance hybrid) 4.3.4 Bond Order In the Lewis description of covalent bond, In both structures we have a O–O single the Bond Order is given by the number bond and a O=O double bond. The normal of bonds between the two atoms in a O–O and O=O bond lengths are 148 pm molecule. The bond order, for example in and 121 pm respectively. Experimentally H2 (with a single shared electron pair), in O2 determined oxygen-oxygen bond lengths in (with two shared electron pairs) and in N2 the O3 molecule are same (128 pm). Thus the (with three shared electron pairs) is 1,2,3 oxygen-oxygen bonds in the O3 molecule are respectively. Similarly in CO (three shared intermediate between a double and a single electron pairs between C and O) the bond bond. Obviously, this cannot be represented order is 3. For N2, bond order is 3 and its by either of the two Lewis structures shown is 946 kJ mol–1; being one of the highest above. for a diatomic molecule. The concept of resonance was introduced Isoelectronic molecules and ions have to deal with the type of difficulty experienced identical bond orders; for example, F2 and in the depiction of accurate structures of O22– have bond order 1. N2, CO and NO+ have molecules like O3. According to the concept bond order 3. of resonance, whenever a single Lewis structure cannot describe a molecule A general correlation useful for accurately, a number of structures withunderstanding the stablities of molecules is that: with increase in bond order, similar energy, positions of nuclei, bonding bond enthalpy increases and bond length and non-bonding pairs of electrons are decreases. taken as the canonical structures of the hybrid which describes the molecule 4.3.5 Resonance Structures accurately. Thus for O3, the two structures It is often observed that a single Lewis structure shown above constitute the canonical is inadequate for the representation of a structures or resonance structures and molecule in conformity with its experimentally their hybrid i.e., the III structure represents determined parameters. For example, the the structure of O3 more accurately. This is ozone, O3 molecule can be equally represented also called resonance hybrid. Resonance is by the structures I and II shown below: represented by a double headed arrow. Reprint 2025-26 110 chemistry Some of the other examples of resonance structures are provided by the carbonate ion and the carbon dioxide molecule. Fig. 4.5 Resonance in CO2 molecule, I, II Problem 4.3 and III represent the three canonical Explain the structure of CO32– ion in forms. terms of resonance. Solution In general, it may be stated that The single Lewis structure based on • Resonance stabilizes the molecule as the the presence of two single bonds and energy of the resonance hybrid is less one double bond between carbon than the energy of any single cannonical and oxygen atoms is inadequate to structure; and, represent the molecule accurately as it • R e s o n a n c e a v e r a g e s t h e b o n d represents unequal bonds. According characteristics as a whole. to the 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 with resonance and the same need to be dispelled. You should remember that : • The cannonical forms have no real existence. • The molecule does not exist for a certain fraction of time in one cannonical form and for other fractions of time in other Fig. 4.4 Resonance in CO32–, I, II and III cannonical forms. represent the three canonical • There is no such equilibrium between forms. the cannonical forms as we have Problem 4.4 between tautomeric forms (keto and enol) in tautomerism. Explain the structure of CO2 molecule. • The molecule as such has a single Solution structure which is the resonance The experimentally determined carbon hybrid of the cannonical forms and to oxygen bond length in CO 2 is which cannot as such be depicted by 115 pm. The lengths of a normal a single Lewis structure. carbon to oxygen double bond (C=O) and carbon to oxygen triple bond (C≡O) 4.3.6 Polarity of Bonds are 121 pm and 110 pm respectively. The existence of a hundred percent ionic or The carbon-oxygen bond lengths in covalent bond represents an ideal situation. CO2 (115 pm) lie between the values for C=O and C≡O. Obviously, a single In reality no bond or a compound is either Lewis structure cannot depict this completely covalent or ionic. Even in case of position and it becomes necessary to covalent bond between two hydrogen atoms, write more than one Lewis structures there is some ionic character. and to consider that the structure of When covalent bond is formed between CO2 is best described as a hybrid of two similar atoms, for example in H2, O2, the canonical or resonance forms I, II and III. Cl2, N2 or F2, the shared pair of electrons is equally attracted by the two atoms. As a result Reprint 2025-26 Chemical Bonding And Molecular Structure 111 electron pair is situated exactly between the In case of polyatomic molecules the dipole two identical nuclei. The bond so formed is moment not only depend upon the individual called nonpolar covalent bond. Contrary to dipole moments of bonds known as bond this in case of a heteronuclear molecule like dipoles but also on the spatial arrangement HF, the shared electron pair between the two of various bonds in the molecule. In such atoms gets displaced more towards fluorine case, the dipole moment of a molecule is the since the electronegativity of fluorine (Unit 3) vector sum of the dipole moments of various is far greater than that of hydrogen. The bonds. For example in H2O molecule, which resultant covalent bond is a polar covalent has a bent structure, the two O–H bonds are bond. oriented at an angle of 104.50. Net dipole As a result of polarisation, the molecule moment of 6.17 × 10–30 C m (1D = 3.33564 possesses the dipole moment (depicted × 10–30 C m) is the resultant of the dipole below) which can be defined as the product of moments of two O–H bonds. the magnitude of the charge and the distance between the centres of positive and negative charge. It is usually designated by a Greek letter ‘µ’. Mathematically, it is expressed as follows : Dipole moment (µ) = charge (Q) × distance of separation (r) Dipole moment is usually expressed in Net Dipole moment, µ = 1.85 D Debye units (D). The conversion factor is = 1.85 × 3.33564 × 10–30 C m = 6.17 ×10–30 C m 1 D = 3.33564 × 10–30 C m The dipole moment in case of BeF2 is zero.where C is coulomb and m is meter. This is because the two equal bond dipoles Further dipole moment is a vector quantity point in opposite directions and cancel the and by convention it is depicted by a small effect of each other. arrow with tail on the negative centre and head pointing towards the positive centre. But in chemistry presence of dipole moment is represented by the crossed arrow ( ) put on Lewis structure of the molecule. The In tetra-atomic molecule, for example in cross is on positive end and arrow head is on BF3, the dipole moment is zero although thenegative end. For example the dipole moment B – F bonds are oriented at an angle of 120o of HF may be represented as : to one another, the three bond moments give a net sum of zero as the resultant of any two H F is equal and opposite to the third. 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 Let us study an interesting case of NH3 his work on X-ray diffraction and and NF3 molecule. Both the molecules have dipole moments. The magnitude pyramidal shape with a lone pair of electrons of the dipole moment is given in Debye units in order to honour him. on nitrogen atom. Although fluorine is more electronegative than nitrogen, the resultant Reprint 2025-26 112 chemistry dipole moment of NH3 (4.90 × 10–30 C m) is • The smaller the size of the cation and the greater than that of NF3 (0.8 × 10–30 C m). This larger the size of the anion, the greater is because, in case of NH3 the orbital dipole the covalent character of an ionic bond. due to lone pair is in the same direction as • The greater the charge on the cation, the the resultant dipole moment of the N – H greater the covalent character of the ionic bonds, whereas in NF3 the orbital dipole is in bond. the direction opposite to the resultant dipole • For cations of the same size and charge,moment of the three N–F bonds. The orbital the one, with electronic configurationdipole because of lone pair decreases the effect (n-1)dnnso, typical of transition metals, isof the resultant N – F bond moments, which more polarising than the one with a noble results in the low dipole moment of NF3 as gas configuration, ns2 np6, typical of alkalirepresented below : and alkaline earth metal cations. The cation polarises the anion, pulling the electronic charge toward itself and thereby increasing the electronic charge between the two. This is precisely what happens in a covalent bond, i.e., buildup of electron 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 Dipole moments of some molecules are cent covalent character of the ionic bond. shown in Table 4.5. Just as all the covalent bonds have 4.4 The Valence Shell Electron some partial ionic character, the ionic Pair Repulsion (VSEPR) Theory bonds also have partial covalent character. As already explained, Lewis concept is unable The partial covalent character of ionic to explain the shapes of molecules. This bonds was discussed by Fajans in terms of theory provides a simple procedure to predict the following rules: the shapes of covalent molecules. Sidgwick Table 4.5 Dipole Moments of Selected Molecules Dipole Type of Molecule Example Geometry Moment, µ(D) Molecule (AB) HF 1.78 linear HCl 1.07 linear HBr 0.79 linear Hl 0.38 linear H2 0 linear Molecule (AB2) H2O 1.85 bent H2S 0.95 bent CO2 0 linear Molecule (AB3) NH3 1.47 trigonal-pyramidal NF3 0.23 trigonal-pyramidal BF3 0 trigonal-planar Molecule (AB4) CH4 0 tetrahedral CHCl3 1.04 tetrahedral CCl4 0 tetrahedral Reprint 2025-26 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 For the prediction of geometrical shapes atoms. It was further developed and redefined of molecules with the help of VSEPR theory, by Nyholm and Gillespie (1957). it is convenient to divide molecules into The main postulates of VSEPR theory are two categories as (i) molecules in which as follows: the central atom has no lone pair and • The shape of a molecule depends upon (ii) molecules in which the central atom the number of valence shell electron pairs has one or more lone pairs. (bonded or nonbonded) around the central Table 4.6 (page114) shows the atom. arrangement of electron pairs about a • Pairs of electrons in the valence shell repel central atom A (without any lone pairs) and one another since their electron clouds are geometries of some molecules/ions of the type negatively charged. AB. Table 4.7 (page 115) shows shapes of some simple molecules and ions in which the central• These pairs of electrons tend to occupy atom has one or more lone pairs. Table 4.8 such positions in space that minimise (page 116) explains the reasons for the repulsion and thus maximise distance distortions in the geometry of the molecule. between them. As depicted in Table 4.6, in the• The valence shell is taken as a sphere compounds of AB2, AB3, AB4, AB5 and AB6, with the electron pairs localising on the the arrangement of electron pairs and the spherical surface at maximum distance B atoms around the central atom A are : from one another. linear, trigonal planar, tetrahedral, • A multiple bond is treated as if it is a single trigonal-bipyramidal and octahedral, electron pair and the two or three electron respectively. Such arrangement can be seen pairs of a multiple bond are treated as a in the molecules like BF3 (AB3), CH4 (AB4) and single super pair. PCl5 (AB5) as depicted below by their ball and • Where two or more resonance structures stick models. can represent a molecule, the VSEPR model is applicable to any such structure. The repulsive interaction of electron pairs decrease in the order: Lone pair (lp) – Lone pair (lp) > Lone pair (lp) – Bond pair (bp) > Bond pair (bp) – Bond pair (bp) Fig. 4.6 The shapes of molecules in which central atom has no lone pair Nyholm and Gillespie (1957) refined the VSEPR model by explaining the important The VSEPR Theory is able to predict difference between the lone pairs and bonding geometry of a large number of molecules, pairs of electrons. While the lone pairs are especially the compounds of p-block elements localised on the central atom, each bonded accurately. It is also quite successful in pair is shared between two atoms. As a result, determining the geometry quite-accurately the lone pair electrons in a molecule occupy even when the energy difference between more space as compared to the bonding pairs possible structures is very small. The of electrons. This results in greater repulsion theoretical basis of the VSEPR theory regarding between lone pairs of electrons as compared the effects of electron pair repulsions on to the lone pair - bond pair and bond pair - molecular shapes is not clear and continues bond pair repulsions. These repulsion effects to be a subject of doubt and discussion. Reprint 2025-26 114 chemistry Table 4.6 Geometry of Molecules in which the Central Atom has No Lone Pair of Electrons Reprint 2025-26 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). Reprint 2025-26 116 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 repulsion. So the angle is reduced to 119.5° from 120°. AB3E 3 1 Trigonal Had there been a bp in place of lp the shape would have pyramidal 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 AB2E2 2 2 tetrahedral if there were all bp 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 4 1 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 tetrahedron, a folded square (More stable) or a see-saw. Reprint 2025-26 Chemical Bonding And Molecular Structure 117 Molecule No. of No. of Arrangement Shape Reason for the type bonding lone of electrons shape acquired pairs pairs AB3E2 3 2 T-shape 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).

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