"B B AAB E CIF 32 3 2 T – shaped 3 B Trigonal bi-pyramidal B B B A Square AB E BrF 5 5 1 5 B B pyramidalOctahedral B B A Square AB E XeF 42 4 2 4 B B planarOctahedral Example 13. Why the bond angle of H – C – H in methane (CH ) is 109° 28’ while 4 H – N – H bond angle in NH is 107° though both carbon and nitrogen are3 3 sp hybridizedSolution: In CH there are 4 bond pair of electrons while in NH are 3 bond pair of4 3 electrons and 1 lone pair of electrons. Since bond pair bond pair repulsion is lessthan lone pair bond pair repulsion, in NH bond angle is reduced from 109°28‘ 3 to 107°.Example 14. Why bond angle in NH is 107° while in H O it is 104.5°?3 2 Solution:In NH , central nitrogen atom bears only one lone pair of electrons whereas in3 H O central oxygen atom bears two lone pair of electrons. 2Since the repulsion between lone pair-lone pair and lone pair – bond pair is morethan that between bond pair-bond pair, the repulsion in H O is much greater than2 that in NH which results in contraction of bond angle from 109°28? to 104.5° in3 water while in NH contraction is less i.e. from 109°28? to 107°. 3 “If the electronegativity of the peripheral atoms is more, then the bond angle will be less”. For example if we consider NH and NF , F – N – F bond angle will be lower than H – N – H bond3 3 angle. This is because in NF the bond pair is displaced more towards F and in NH it is displaced3 3 more towards N. So accordingly the b.p. – b.p. interaction is less in NF and more in NH .3 3 Example 15. The bond angle of H O is 104° while that that of F O is 102°.2 2 Solution: Both H O and F O have a lone pair of electrons. But fluorine being highly2 2 electronegative, the bond pair of electrons are drawn more towards F in F O,2 whereas in H O it is drawn towards O. So in F O the bond pairs being displaced2 2 away from the central atom, has very little tendency to open up the angle. But inH O this opening up is more as the bond pair electrons are closer to each other.2 So bond ? of F O is less than H O.2 2 OOrepulsion moreHFHFRepulsion lessRESONANCE16 There may be many molecules and ions for which it is not possible to draw a single Lewisstructure.For example we can write two electronic structures of O .3 (+)O (+) O ( ?)( ?)O O O O (A) (B) In (A) the oxygen ? oxygen bond on the left is a double bond and the oxygen ?oxygen bond on theright is a single bond. In B the situation is just opposite. Experiment shows however, that the twobonds are identical. Therefore neither structure A nor B can be correct.One of the bonding pairs in ozone is spread over the region of all the three atom rather thanassociated with particular oxygen ?oxygen bond. This delocalised bonding is a type of bonding inwhich bonding pair of electrons is spread over a number of atoms rather than localised betweentwo.O O O (C) Structures (A) and (B) are called resonating or canonical structures and C is the resonancehybrid. This phenomenon is called resonance, a situation in which more than one plausiblestructure can be written for a species and in which the true structure cannot be written at all.Some other examples2– (i) CO ion3 O O O O ?O O O O O O O O(ii) Carbon ?oxygen bond lengths in carboxylate ion are equal due to resonance.- O O O R R R - O O O(iii) Benzene (iv) Vinyl ChlorideH C H C 2 2 + Cl ClDifference in the energies of the canonical forms and resonance hybrid is called resonancestabilization energy and provides stability to species.Rules for writing resonating structures? Only electrons (not atoms) may be shifted and they may be shifted only to adjacent atoms or bond positions.? The number of unpaired electrons should be same in all the canonical form.? The positive charge should reside as far as possible on less electronegative atom andnegative charge on more electronegative atom. 17 ? Like charge should not reside on adjacent atom? The larger the number of the resonating structures greater the stability of species.? Greater number of covalency add to the stability of the molecule.Example 17. Out of the following resonating structures for CO molecule, which are important2 for describing the bonding in the molecule and why?2 O C O O C O O C O O C O (I) (III) (II) (IV)Solution:Out of the structures listed above, the structure (III) is wrong since the number of electronpairs on oxygen atoms are not permissible. Similarly, the structures (II) has very little contribution towards thehybrid because one of the oxygen atoms (electronegative) is show to have positive charge. Carbon dioxide isbest represented by structures (I) and (IV).FACTORS GOVERNING POLARIZATION AND POLARISABILITY (FAJAN’S RULE)? Cation Size: Smaller is the cation more is the value of charge density ( ?) and hencemore its polarising power. As a result more covalent character will develop. Let us taketheexample of the chlorides of the alkaline earth metals. As we go down from Be to Ba thecation size increases and the value of ? decreases which indicates that BaCl is less2covalent i.e. more ionic. This is well reflected in their melting points. Melting points of BeCl = 405°C and BaCl = 960°C.2 2 ? Cationic Charge: More is the charge on the cation, the higher is the value of ? andhigher is the polarising power. This can be well illustrated by the example already given,NaBr and AlBr . Here the charge on Na is +1 while that on Al in +3, hence polarising3 power of Al is higher which in turn means a higher degree of covalency resulting in alowering of melting point of AlBr as compared to NaBr.3 ? Noble Gas vs Pseudo Noble Gas Cation: A Pseudo noble gas cation consists of anoble gas core surrounded by electron cloud due to filled d-subshell. Since d-electrons provide inadequate shielding from the nuclei charge due to relatively lesspenetration of orbitals into the inner electron core, the effective nuclear charge (ENC) isrelatively larger than that of a noble gas cation of the same period. NaCl has got amelting point of 800°C while CuCl has got melting point of 425°C. The configuration of+ 10 + Cu = [Ar] 3d while that of Na = [Ne]. Due to presence of d electrons ENC is more and– therefore Cl is more polarised in CuCl leading to a higher degree of covalency and lowermelting point. ? Anion Size: Larger is the anion, more is the polarisability and hence more covalentcharacter is expected. An e.g. of this is CaF and CaI , the former has melting2 2 – – point of 1400°C and latter has 575°C. The larger size of I ion compared to F causesmore polarization of the molecule leading to a lowering of covalency and increasing inmelting point. ? Anionic Charge: Larger is the anionic charge, the more is the polarisability. A wellillustrated example is the much higher degree of covalency in magnesium nitride ++ 3– ++ — (3Mg 2N ) compared to magnesium fluoride (Mg 2F ). This is due to higher chargeof nitride compare to fluoride. These five factors are collectively known as Fajan‘s Rule.+ Example 18. The melting point of KCl is higher than that of AgCl though the crystal radii of Ag+ and K ions are almost same.Solution:Now whenever any comparison is asked about the melting point of thecompounds which are fully ionic from the electron transfer concept it means that 18 the compound having lower melting point has got lesser amount of ioniccharacter than the other one. To analyse such a question first find out thedifference between the 2 given compounds. Here in both the compounds theanion is the same. So the deciding factor would be the cation. Now if the cation isdifferent, then the answer should be from the variation of the cation. Now in the+ above example, the difference of the cation is their electronic configuration. K =+ 10 [Ar]; Ag = [Kr] 4d . This is now a comparison between a noble gas core andpseudo noble gas core, the analysis of which we have already done.So try tofinish off this answer.Example19. AlF is ionic while AlCl is covalent.3 3 – Solution:Since F is smaller in size, its polarisability is less and therefore it is having moreionic character. Whereas Cl being larger in size is having more polarisability andhence more covalent character.Example 20. Which compound from each of the following pairs is more covalent and why? (a) CuO or CuS (b) AgCl or AgI ‘ (c) PbCl or PbCl(d) BeCl or MgCl2 4 2 2 Solution:(a) CuS (b) AgI(c) PbCl (d) BeCl4 2 DIPOLE MOMENTDifference in polarities of bonds is expressed on a numerical scale. The polarity of a molecule isindicated in terms of dipole moment ? ? ? . To measure dipole moment, a sample of the substanceis placed between two electrically charged plates. Polar molecules orient themselves in theelectric field causing the measured voltage between the plates to change. The dipole moment is defined as the product of the distance separating charges of equalmagnitude and opposite sign, with the magnitude of the charge. The distance between thepositive and negative centres called the bond length. Thus,= ? ? electric charge ?bond length qd ? ?10 ?8 ?18 As q is in the order of 10 esu and d is in the order of 10 cm, ? is the order of .10 esu cm Dipole moment is measured in =Debye‘ unit (D)?? 18 30 1D ? 10 esu cm ? 3.33 ?10 coulomb metre Note:(i)Generally as electronegativity difference increase in diatomic molecules, polarity of bondbetween the atoms increases therefore value of dipole moment increases. (ii)Dipole moment is a vector quantity Cl (iii) A symmetrical molecule is non- polar even though it contains polarbonds. For example, CO , BF , CCl etc because summation of all2 3 4 Cbond moments present in the molecules cancel each other.Cl ClCl 19(iv)Unsymmetrical non-linear polyatomic molecules have net value of dipole moment. Forexample, H O, CH OH, NH etc. 2 3 3 Calculation of Resultant Bond Moments ? B ? Let AB and AC are two polar bonds inclined at an angle ? their? 1 dipole moments areand .? ? 1 2 ? ? A R ? ? Resultant dipole moment may be calculated using vectorial method.? 2 ? C ?22 ? ? ? ? ? ? 2 ? ? cos ?R 1 2 1 2 o180when ? = 0 the resultant is maximum?? ? ? ? ?R 1 2 when,? = 180 ?, the resultant is minimum o 180? ? ? ? ??R 1 2Example 21. The compound which has zero dipole moment is (A) CH Cl (B) NF2 2 3(C) PCl F (D) ClO 3 2 2Solution:(C)Example 22. Sketch the bond moments and resultant dipole moment in (i) SO(ii) Cis ?C H Cl and (iii) trans ?C H Cl2 2 2 2 2 2 2 Solution:(i) S O O Cl Cl (ii) C C H HCl H (iii) C C H ClResultant ? = 0Example 23. CO has got dipole moment of zero. Why?2O = C = O Solutions:The structure of CO is . This is a highly symmetrical structure with a2 plane of symmetry passing through the carbon. The bond dipole of C–O isdirected towards oxygen as it is the negative end. Here two equal dipoles actingin opposite direction cancel each other and therefore the dipole moment is zero.Example 24.Dipole moment of CCl is zero while that of CHCl is non zero.4 3 Solution: Both CCl & CHCl have tetrahedral structure but CCl is symmetrical while4 3 4 CHCl is non-symmetrical. 320 "