"In this context it can also be noted that in T.B.P. shape the bond lengths are not same. Theequatorial bonds are smaller than axial bonds. But in square bipyramidal shape, all bond lengthsare same.– 3- O O O 12. POTotal valence electrons : 32 4etc. Requirement : 4 ? bonds P P P 3 Hybridisation: sp O OOO OO O O OShape: tetrahedral– Here all the structures drawn are resonating structures with O resonating with double bondedoxygen.– 13. NO Total valence electron: 18 2Requirement : 2 ? bonds + 1 lone pairN 2Hybridisation: spO O Shape: angular 2– 14. CO Total valence electrons: 243Requirement = 3 ? bonds 2Hybrdisation = sp Shape: planar trigonal But C has 4 valence electrons of these 3 form ? bonds ? the rest will form a ? bond.– – O O O C C C OO OO OO In the structure one bond is a double bond and the other 2 are single. The position of the doublebonds keeps changing in the figure. Since peripheral atoms are isovalent, so contribution of theresonanting structures are equal. Thus it is seen that none of the bonds are actually single ordouble. The actual state is –2/3 O Bond order = 3/2 = 1.5 C –2/3 –2/3 O O 15. CO Total valence electrons : 16 O = C = O2Requirement: 2 ? bondsHybridisation: sp Shape: linear- – 16.Totalvalence electrons = 32BF 4 FRequirement= 4 ? bonds B 3 F FHybridisation: spFShape: Tetrahedral- 17.Total valence electron = 26ClO3 Requirement = 3 ? bond + 1 lone pair 3 Cl Cl ClHybridization: sp- - - O O O O O OShape: pyramidal O O O11 F 18. XeO F Total valence electrons : 342 2ORequirement: 4 ? bonds +1 lonepairs 3Hybridization : sp dXeShape: Distorted TBP (sea-saw geometry)O F 19. XeO Total valence electrons : 263 XeRequirement: 3 ? bonds + 1 lonepair3 O OHybridization: spOShape: PyramidalO 20. XeOF Total valence electrons : 424 F FRequirement: 5 ? bonds + 1 lonepairXe 3 2 FHybridization: sp dFShape: square pyramidal. Example 8.Predict the hybridization for the central atom in , , POCl OSF OIF 3 4 5 5?? 6 21 32 Solution:Total No. of V.E. = ?? 4POCl 3 88 3 So, hybridization = sp6?? 6 28 40 = ?? 5OSF 4 88 3 So, hybridization of s = dsp 6?? 7 35 48 ? ? ? 6OIF 5 88 23 So, hybridization of I = d sp Example 9. Out of the three molecules XeF , SF and SiF one which has tetrahedral structures4 4 4 is (A) All of three (B) Only SiF4(C) Both SF and XeF (D) Only SF and XeF4 4 4 4 3 2 3 3 Solution:Hybridization of XeF = sp d , SF = sp d, SiF = sp 4 4 4Hence (B) is correct. Example 10. Among the following compounds in which case central element uses d ?orbital tomake bonds with attached atom (A)(B) BeF XeF 2 2(C)(D) SiF BF 4 3 3 Solution:In . Xe atom has sp d hybridisation.XeF 2Hence (B) is correct.Example 11. When NH is treated with HCl, state of hybridisation on central nitrogen3 3 2 (A) Changes from sp to sp (B) Remains unchanged3 3 3 (C) Changes from sp to sp d (D) Changes from sp to sp 12 + 3 Solution:On NH state of hybridisation on central nitrogen atom is sp as 4 in NH .3 + H N H H H Hence (B) is the correct answer.Exercise 4.Among the following which has bond angle very near to 90° (A) NH (B) XeF3 4(C) BF(D) H O3 2 Exercise 5.Homolytic fission of C – C bond in hexafluoroethane gives an intermediate in whichhybridization state of carbon is2 3(A) sp(B) sp (C) sp(D) cannot be determinedExercise 6.A molecule XY contains two ?, two ? bonds and one lone pair of electrons in valence2 shell of X. The arrangement of lone pair as well as bond pairs is (A) Square pyramidal (B) Linear (C) Trigonal planar (D) UnpredictableExercise 7. Draw the structure the following indicating the hybridisation of the central atom.–(i) SOF ,(ii) SO ,(iii) POCl ,(iv) I2 2 3 3 Exercise 8. The type of hybridisation of orbitals employed in the formation of SF molecule is6 ………….. Exercise 9. The angle between two covalent bonds is maximum for……………(CH , H O, CO )4 2 2 Exercise 10. 2 ?The bond angle in SO ion is ………………..4 Maximum Covalency 3p 3s 3dElements which have vacant d-orbitalIn ground state?? ? ? ? ?can expand their octet by transferringelectrons, which arise after unpairing,In excited state? ? ? ? ? ?to these vacant d-orbital e.g. in sulphur. 13 In excited state sulphur has six unpaired electrons and shows a valency of six e.g. in SF . Thus6 an element can show a maximum covalency equal to its group number e.g. chlorine showsmaximum covalency of seven.VALENCE SHELL ELECTRON PAIR REPULSION (VSEPR) THEORY (SHAPES ANDGEOMETRY OF MOLECULES)Molecules exist in a variety of shapes. A number of physical and chemical properties ofmolecules arise from and are affected by their shapes. For example, the angular shape of thewater molecules explains its many characteristic properties while a linear shape does not.The determination of the molecular geometry and the development of theories for explaining thepreferred geometrical shapes of molecules is an integral part of chemical bonding. The VSEPRtheory (model) is a simple treatment for understanding the shapes of molecules.Strictly speaking VSEPR theory is not a model of chemical bonding. It provides a simple recipefor predicting the shapes of molecules. It is infact an extension of the Lewis interpretation ofbonding and is quite successful in predicting the shapes of simple polyatomic molecules. The basic assumptions of the VSEPR theory are that:Pairs of electrons in the valence shell of a central atom repel each other 1. These pairs of electrons tend to occupy position in space that minimize repulsions and thusmaximize distance between them.2. The valence shell is taken as a sphere with the electron pairs localizing on the sphericalsurface at maximum distance from one another.3. A multiple bonds are treated as a single super pair.4. Where two or more resonance structures can depict a molecule, the VSEPR model isapplicable to any such structures For the prediction of geometrical shapes of molecules with the help of VSEPR model, it isconvenient to divide molecules into two categoriesRegular GeometryMolecules in which the central atom has no lone pairs Irregular GeometryMolecules in which the central atom has one or more lone pairs, the lone pair of electrons inmolecules occupy more space as compared to the bonding pair electrons. This causes greaterrepulsion between lone pairs of electrons as compared to the bond pairs repulsions. Thedescending order of repulsion (lp – lp)> (lp – bp)> (bp – bp)where lp-Lone pair; bp-bond pairRegular GeometryNumber of Arrangement of electronsMolecular geometry Examples electron pairs ? 180 BeCl ,HgCl2 B – A – B22 LinearA Linear14 B BF ,AlCl333 A A B B Trigonal planar Trigonal planar ? = 120 ?? = 120 ?? CH ,NH SiF 44 4 4 0 ' 0 '109 28 109 28 A A PCl , PF555 0 0 90 0 90 0 120 120 trigonal bipyramidal trigonal bipyramidal B SF66 B B A B A 90 ? B B Octahedral Irregular GeometryMolecule No. of No. of Arrangement of ShapeExamplesType Bonding pairs lone pair electrons pairs (Geometry)A AB E SO ,O2 1 Bent2 23 B B Trigonal planar A Trigonal AB E NH 3 3 1 3 B B pyramidalB Tetrahedral AAB E HO 22 2 2 Bent 2 B B Tetrahedral B B AAB E SF 4 4 1 See saw 4 B B Trigonal bi-pyramidal15 "