"gives both bonding and anti-bonding orbitals, and the 2 electrons occupy the bonding M.O.leaving the anti-bonding MO empty.NO Molecule: The M.O. of NO is also quite complicated due to energy difference of the atomicorbitals of N and O.As the M.O.‘s of the heteronuclei species are quite complicated, so we should concentrate inknowing the bond order and the magnetic behaviour.Molecules/Ions Total No. of electrons Magnetic behaviourCO 14 DiamagneticNO 15 Paramagnetic+ NO 14 Diamagnetic– NO 16 DiamagneticCN 13 Paramagnetic–CN 14 DiamagneticINERT PAIR EFFECT Heavier p-block and d-block elements show two oxidation states. One is equal to group number2 2 and second is group number minus two. For example Pb(5s 5p ) shows two OS, +II and +IV.Here +II is more stable than +IV which arises after loss of all four valence electrons. Reason2 given for more stability of +II O.S. that 5s electrons are reluctant to participate in chemicalbonding because bond energy released after the bond formation is less than that required tounpair these electrons (lead forms a weak covalent bond because of greater bond length).Example 36. Why does PbI not exist?4 Solution: Pb(+IV) is less table than Pb(+II) due to inert pair effect and therefore Pb(+IV) is– – reduced to Pb(+II) by I which changes to I (I is a good reducing agent)2 MISCELLANEOUS EXERCISES + - Exercise 1: Compare the bond energies of N , N & N.2 2 2 Exercise 2: Though the electronegativities of nitrogen and chlorine are same, NH exists as3 liquid whereas HCl as gas. Why?Exercise 3: Explain giving reason:- + is linear, but the ionis bentClF ClF 2 2 Exercise 4: Explain with reason: Two different bond lengths are observed in PF but only one bond length5 observed in SF .6 Exercise 5: (a) Amongst BBr and BF which is a stronger acid and why?3 3 Exercise 6: Explain why the measured resultant dipole moment for FNO is 1.81 D, is so muchhigher than the value for nitryl fluoride FNO (0.47 D).231 H H O O OH Exercise7: Why is liquid at room temperature while is a high melting solid?(I) OH (II) Exercise 8: (i) Among the compounds CH COOH, NH , HF and CH in which maximum3 3 4 hydrogen bonding is present?(ii) Which one of the following has strongest bond? HF, HCl, HBr, HIExercise 9: Why BeF and BF are stable though Be and B have less than 8 electrons? Which2 3 one is more stable?Exercise10: Ether O and water O have same hybridisation at oxygen. Whatangle would you expect for them? R H R HANSWER TO MISCELLANEOUS EXERCISES Exercise 1: The configuration of N is2 * ?? ?? ? ? 2p ?? 2p ?? y *2 y 2 2 *2 2 ? ? ? ? ?1s ?? ?? 1s 2s 2s 2px * ? ?? ?? 2p ? ?? z 2p ?? z ?Nowmeans removal of an electron from a bonding M.O. This will decreaseN 2 the B.O.? ?B.O. of N = ½ (5 – 0) = 2.52 ? Now again forbond order is ½ (6–1) = 2.5N 2 ? ?So from the bond order it may seem that both& may have the same bondN N 2 2 energy. But removal of an electron from a diatomic species tend to decrease theinter electronic repulsion and thereby shortens the bond length. So the bond? energy becomes more than compared to N2 ? ? ?N > N > N2 2 2 Exercise 2: The size of nitrogen is less than the size of chlorine. Therefore, electron densityin nitrogen is more than that of chlorine. So, nitrogen forms hydrogen bondingleading to association of molecules. Hence, NH is liquid. Hydrogen bonding is3 not possible with chlorine.3 ? Exercise 3:In ClF central chorine atom involves sp d hybridisation, to have minimum2 electronic repulsion three lone pairs should be in equitorial position as follows;? giving linear shape to the ion. Whereas, in case ofion central atom ClClF 2 3 involves sp hybridisation having two lone pairs, resulting in bent shape for theion, (bond angle less than 109°28 ? due to repulsion of bond pair by lone pair)32F ??+120°ClFFBent shapeFLinear shape-bond angle-180°3 Exercise 4:PCl has trigonal pyramidal structure (sp d hybridisation of central atom) in which5 bond angles are 90° and 120° respectively and there are two types of bond axial3 2 and equatorial. In case of SF the structure is octahedral (sp d hybridization of6 the central atom - S) resulting only one type of bond, bond angle (90°) and onetype of bond length.FFFFFP(PF )5 (SF )6 FFFFBond angle - 90°FF(Bond angle 90° & 120°)Exercise 5: (a) BBr since back bonding is present in BF .3 3 Exercise 6: Molecular symmetry (in terms of bond angles) leads to lesser dipole moment.Exercise 7: (i) shows intramolecular hydrogen bonding while (ii) shows intermolecular hydrogen bonding.Exercise 8: (i) HF due to maximum electronegativity of fluorine.(ii) HFExercise 9: The stability is explained by symmetrical linear structure for BeF and triangular2 planar structure for BF . BeF is more stable because its greater bond angle3 2 o (180 ).o Exercise 10: In H O bond angle is less than 109 .28‘ due to lone pair and bond pair repulsion.2 But in ether, due to strong mutual repulsion between two alkyl groups bond angleo becomes greater than 109 .28‘. 33 SOLVED PROBLEMSSubjective: Board Type QuestionsProb 2. Arrange the bonds in order of increasing ionic character in the molecules: LiF, K O, N , SO and ClF2 2 2 3 Sol. N < ClF < SO < K O < LiF2 3 2 2 Prob 3. Arrange the following in order of increasing ionic character: C ?H, F ?H, Br ?H, Na ?I, K ?F and Li ?ClSol. C ?H < Br ?H < F ?H < Na ?I < Li ?Cl < K ?FIIT Level Questions Prob 6. Predict the shapes of the following ions- -(a) BeF(b) BF3 4 - -(c) IF (d) IBr4 2 Sol.(a) Triangular (b) Tetrahedral(c) Square planar(d) Linear + - 2- Prob 8. Arrange the following in increasing order of stability O , O , O , O 2 2 2 2 2- - + Sol. O < O < O < O2 2 2 2Calculate first the bond order which is as follows + - 2-O ? 2, O ? 2.5, O ? 1.5, O ?1 & then arrange according to increasing bond2 2 2 2 order.Prob 11. Arrange the following: (i)N , O , F , Cl in increasing order of bond dissociation energy.2 2 2 2(ii) Increasing strength of hydrogen bonding (X – H – X): O, S, F, Cl, NSol. (i) F < Cl < O < N(ii) Cl < S < N < O < F2 2 2 2 Prob 12. Explain the following o - hydroxy benzaldehyde is liquid at room temperature while p - hydroxy benzaldehydeis high melting solid.Sol.There is intramolecular H bonding in o - hydroxy benzaldehyde while intermolecularhydrogen bonding in p-hydroxy benzaldehyde 34 CH=O---H—O CH=O---H O CHO o-hydroxy benzaldehyde O—H----OCH OH- + Prob 14. Explain why ClF is linear but ClF is a bent molecular ion?2 2 3 Sol.Chlorine atom lies in sp d hybrid state. Three lone pairs are oriented along the cornersof triangular plane F Cl F ? ?? ClF 2 ??3Chlorine atom lies in sp hybrid state. Two lone pairs are oriented along two corners oftetrahedralCl F F + [ClF ] 2Prob 16. AlF is ionic while AlCl is covalent.3 3 – Sol. Since F is smaller in size, its polarisability is less and therefore it is having more ioniccharacter. Whereas Cl being large in size is having more polarisability and hence morecovalent character.Prob 18. Write down the resonance structure of nitrous oxide NO? ? 2 Sol.N N O N N O Prob 20. Explain why BeH molecule has zero dipole moment although the Be ?H bonds are2polar.o Sol. BeH is a linear molecule (H ?Be ?H) with bond angle equal to 180 . Although the Be ?H2 bonds are polar on account of electronegativity difference between Be and H atoms,the bond polarities cancel with each other. The molecule has resultant dipole momentof zero. 35 "