9.4 Alkynes 1 2 3 4 5 Like alkenes, alkynes are also unsaturated II. H3C–C≡ C– CH2– CH3 Pent–2-yne hydrocarbons. They contain at least one triple 4 3 2 1 bond between two carbon atoms. The number III. H3C–CH–C≡ CH 3-Methyl but–1-yne |of hydrogen atoms is still less in alkynes as CH3compared to alkenes or alkanes. Their general Structures I and II are position isomers formula is CnH2n–2. and structures I and III or II and III are chain The first stable member of alkyne series isomers. is ethyne which is popularly known as acetylene. Acetylene is used for arc welding Problem 9.13 purposes in the form of oxyacetylene flame Write structures of different isomers obtained by mixing acetylene with oxygen corresponding to the 5 th member of gas. Alkynes are starting materials for a large alkyne series. Also write IUPAC names of number of organic compounds. Hence, it all the isomers. What type of isomerism is interesting to study this class of organic is exhibited by different pairs of isomers? compounds. Solution 9.4.1 Nomenclature and Isomerism th 5 member of alkyne has the molecular In common system, alkynes are named as formula C6H10. The possible isomers are: derivatives of acetylene. In IUPAC system, they Table 9.2 Common and IUPAC Names of Alkynes (CnH2n–2) Value of n Formula Structure Common name IUPAC name 2 C2H2 H-C≡CH Acetylene Ethyne 3 C3H4 CH3-C≡CH Methylacetylene Propyne 4 C4H6 CH3CH2-C≡CH Ethylacetylene But-1-yne 4 C4H6 CH3-C≡C-CH3 Dimethylacetylene But-2-yne Reprint 2025-26 Hydrocarbons 315 (a) HC ≡ C – CH2 – CH2 – CH2 – CH3 Hex-1-yne (b) CH3 – C ≡ C – CH2 – CH2 – CH3 Hex-2-yne (c) CH3 – CH2 – C ≡ C – CH2– CH3 Hex-3-yne 3-Methylpent-1-yne 4-Methylpent-1-yne 4-Methylpent-2-yne Fig. 9.6 Orbital picture of ethyne showing (a) sigma overlaps (b) pi overlaps. orbitals of the other carbon atom, which undergo lateral or sideways overlapping to 3,3-Dimethylbut-1-yne form two pi (π) bonds between two carbon atoms. Thus ethyne molecule consists of one Position and chain isomerism shown by C–C σ bond, two C–H σ bonds and two C–C different pairs. π bonds. The strength of C≡C bond (bond enthalpy 823 kJ mol -1) is more than those 9.4.2 Structure of Triple Bond of C=C bond (bond enthalpy 681 kJ mol –1) Ethyne is the simplest molecule of alkyne and C–C bond (bond enthalpy 348 kJ mol–1). series. Structure of ethyne is shown in The C≡C bond length is shorter (120 pm) Fig. 9.6. than those of C=C (133 pm) and C–C (154 pm). Electron cloud between two carbon Each carbon atom of ethyne has two sp atoms is cylindrically symmetrical about thehybridised orbitals. Carbon-carbon sigma (σ) internuclear axis. Thus, ethyne is a linear bond is obtained by the head-on overlapping molecule. of the two sp hybridised orbitals of the two carbon atoms. The remaining sp hybridised 9.4.3 Preparation orbital of each carbon atom undergoes 1. From calcium carbide: On industrial overlapping along the internuclear axis with scale, ethyne is prepared by treating the 1s orbital of each of the two hydrogen calcium carbide with water. Calcium atoms forming two C-H sigma bonds. carbide is prepared by heating quick lime H-C-C bond angle is of 180°. Each carbon with coke. Quick lime can be obtained byhas two unhybridised p orbitals which are heating limestone as shown in the followingperpendicular to each other as well as to the reactions:plane of the C-C sigma bond. The 2p orbitals of one carbon atom are parallel to the 2p CaCO3 ∆ CaO + O2 (9.55) Reprint 2025-26 316 chemistry CaO + 3C CaC2 + CO (9.56) the sp hybridised carbon2 atoms whereas they are attached to sp hybridised carbon Calcium 3 atoms in ethene and sp hybridised carbons carbide in ethane. Due to the maximum percentage of CaC2 + 2H2O Ca(OH)2 + C2H2 (9.57) s character (50%), the sp hybridised orbitals of carbon atoms in ethyne molecules have2. From vicinal dihalides : Vicinal dihalides highest electronegativity; hence, these attract on treatment with alcoholic potassium the shared electron pair of the C-H bond of hydroxide undergo dehydrohalogenation. ethyne to a greater extent than that of the One molecule of hydrogen halide is 2 sp hybridised orbitals of carbon in ethene eliminated to form alkenyl halide which 3 and the sp hybridised orbital of carbon in on treatment with sodamide gives alkyne. ethane. Thus in ethyne, hydrogen atoms can be liberated as protons more easily as compared to ethene and ethane. Hence, hydrogen atoms of ethyne attached to triply bonded carbon atom are acidic in nature. You may note that the hydrogen atoms attached to the triply bonded carbons are acidic but not all the hydrogen atoms of alkynes. HC ≡ CH + Na → HC ≡ C–Na++ 1/2 H2 9.4.4 Properties Monosodium Physical properties ethynide Physical properties of alkynes follow the same (9.59) trend of alkenes and alkanes. First three HC ≡ C– Na + Na → Na+ C–Na+ ≡ C–Na++ 1/2 H2members are gases, the next eight are liquids and the higher ones are solids. All alkynes Disodium ethynide are colourless. Ethyene has characteristic (9.60)odour. Other members are odourless. Alkynes are weakly polar in nature. They are lighter CH3 – C ≡ C – H + Na+ NH–2 than water and immiscible with water but ↓ soluble in organic solvents like ethers, carbon CH3 – C ≡ C– Na+ + NH3 tetrachloride and benzene. Their melting Sodium propynide (9.61) point, boiling point and density increase with These reactions are not shown by alkenesincrease in molar mass. and alkanes, hence used for distinction Chemical properties between alkynes, alkenes and alkanes. What Alkynes show acidic nature, addition reactions about the above reactions with but-1-yne and and polymerisation reactions as follows : but-2-yne ? Alkanes, alkenes and alkynes A. Acidic character of alkyne: Sodium follow the following trend in their acidic metal and sodamide (NaNH2) are strong behaviour : bases. They react with ethyne to form sodium i) CH ≡ CH > H2C – CH2 > CH3 –CH3acetylide with the liberation of dihydrogen gas. These reactions have not been observed ii) HC ≡ CH > CH3 –C≡ CH >> CH3 –C≡C–CH3in case of ethene and ethane thus indicating that ethyne is acidic in nature in comparison B. Addition reactions: Alkynes contain a to ethene and ethane. Why is it so ? Has triple bond, so they add up, two molecules of it something to do with their structures dihydrogen, halogen, hydrogen halides etc. and the hybridisation ? You have read that Formation of the addition product takes place hydrogen atoms in ethyne are attached to according to the following steps. Reprint 2025-26 Hydrocarbons 317 The addition product formed depends upon stability of vinylic cation. Addition in unsymmetrical alkynes takes place according to Markovnikov rule. Majority of the reactions of alkynes are the examples of electrophilic addition reactions. A few addition reactions (9.66)are given below: (i) Addition of dihydrogen (iv) Addition of water Pt/Pd/Ni H2 Like alkanes and alkenes, alkynes are alsoHC≡CH+H2 [H2C = CH2] CH3–CH3 immiscible and do not react with water. (9.62) However, one molecule of water adds to alkynes on warming with mercuric sulphate CH3–C≡CH + H2 Pt/Pd/Ni [CH3–CH=CH2] and dilute sulphuric acid at 333 K to form Propyne Propene carbonyl compounds. ↓H2 CH3–CH2–CH3 Propane (9.63) (ii) Addition of halogens (9.67) (9.64) Reddish orange colour of the solution of bromine in carbon tetrachloride is decolourised. This is used as a test for unsaturation. (iii) Addition of hydrogen halides (9.68) Two molecules of hydrogen halides (HCl, HBr, (v) Polymerisation HI) add to alkynes to form gem dihalides (in (a) Linear polymerisation: Under suitable which two halogens are attached to the same conditions, linear polymerisation of ethyne carbon atom) takes place to produce polyacetylene or H–C≡C–H+H–Br [CH2 = CH–Br]→ CHBr2 polyethyne which is a high molecular Bromoethene weight polyene containing repeating units of CH3 (CH = CH – CH = CH ) and can be represented 1,1-Dibromoethane as —(CH = CH – CH = CH)n— Under special (9.65) conditions, this polymer conducts electricity. Reprint 2025-26 318 chemistry Thin film of polyacetylene can be used as but in a majority of reactions of aromatic electrodes in batteries. These films are good compounds, the unsaturation of benzene ring conductors, lighter and cheaper than the is retained. However, there are examples of metal conductors. aromatic hydrocarbons which do not contain a (b) Cyclic polymerisation: Ethyne on benzene ring but instead contain other highly unsaturated ring. Aromatic compoundspassing through red hot iron tube at 873K containing benzene ring are known asundergoes cyclic polymerization. Three benzenoids and those not containing amolecules polymerise to form benzene, which benzene ring are known as non-benzenoids.is the starting molecule for the preparation of Some examples of arenes are givenderivatives of benzene, dyes, drugs and large below:number of other organic compounds. This is the best route for entering from aliphatic to aromatic compounds as discussed below: Benzene Toluene Naphthalene (9.69) Biphenyl Problem 9.14 How will you convert ethanoic acid into 9.5.1 Nomenclature and Isomerism benzene? The nomenclature and isomerism of aromatic Solution hydrocarbons has already been discussed in Unit 8. All six hydrogen atoms in benzene are equivalent; so it forms one and only one type of monosubstituted product. When two hydrogen atoms in benzene are replaced by two similar or different monovalent atoms or groups, three different position isomers are possible. The 1, 2 or 1, 6 is known as the ortho (o–), the 1, 3 or 1, 5 as meta (m–) and the 1, 4 as para (p–) disubstituted compounds. A few examples of derivatives of benzene are given below:

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).

2.1 Two charges 5 × 10–8 C and –3 × 10–8 C are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.