Chemical bonding part 1


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Chemical bonding part 1

  1. 1. 29/11/2010 TOPIC TWO CHEMICAL BONDING 29/11/2010 mov 1 CHEMICAL BOND• This is the force of attraction that binds two or more atoms together.• Types of chemical bonds 1) Ionic 2) Covalent 3) Metallic 4) Coordinate 5) Van-derwaal’s 6) Hydrogen 29/11/2010 mov 2 1
  2. 2. 29/11/2010 Bond formation • Why do atoms form chemical bonds ? so that the system can achieve the lowest possible potential energy • Example covalent bonding in H2 H••H 29/11/2010 mov 3 Ionic bond• This is the electrostatic attraction between oppositely charged ions. It involves complete transfer of one or more electrons from a highly electropositive to a highly electronegative element.• The atom that gives out electron becomes positively charged, whereas the one that gains electrons becomes negatively charged Cation anion• Formed between elements with the biggest difference in electronegativity Example: 29/11/2010 mov 4 2
  3. 3. 29/11/2010 Coulomb’s Law• Especially prevalent in compounds formed between group 1A and 2A elements with group 6A and 7A elements.• Ionic bond formation is influenced by the of electrostatic force of attraction (F), which is directly proportional to product of charge and inversely proportional to distance between the ions F = k Q1xQ2/r2 k = (2.31 x 10-19 J nm) therefore, strong lattices are favoured when the ions have a high charge to size ratio29/11/2010 mov 5Conditions for formation of ionic bond1) Low ionisation potential of the metal –the lower the ionisation potential the greater the ease to form cation2) High electron affinity of the non-metal-the higher the electron affinity, the greater the ease to firm anion3) High lattice energy – the greater the lattice energy the stronger the bond formed29/11/2010 mov 6 3
  4. 4. 29/11/2010 Characteristics of ionic compounds1) They have high melting and boiling points due to strong electrostatic force of attraction between the ions in the solid2) They are non-volatile due to high melting and boiling points3) Electrical conductivity: 1) They are poor conductors of electricity in solid state due to strong electrostatic force of attraction-hence ions are immobile 2) Good conductors of electricity in aqueous state-ions are mobile 3) Good conductors of electricity in molten or fused state4) Soluble in polar solvents but in soluble in organic solvents 29/11/2010 mov 7 Formation of Ionic Compounds • Electron affinity: is the energy change that occurs when an electron is accepted by an atom in the gaseous state (kJ/mol) X(g) + e- X-(g) • The higher the electron affinity, the greater the tendency of the atom to accept an electron • Chlorine has the greatest electron affinity of any element: Cl(g) + e- Cl-(g) ∆H = -349 kJ 29/11/2010 mov 8 4
  5. 5. 29/11/2010 Formation of Ionic Compounds• Lattice energy: this is the energy required to completely separate one mole of a solid ionic compound into gaseous ions.• or the energy released when an ionic solid forms from its ions• It has a negative sign (-)• Calculation of lattice Energy is based on two methods 1) Born-lande Equation (Theoretical method) 2) The Born-Haber Cycle (Experimental method) 29/11/2010 mov 9 Born-lande Equation (Theoretical method)• It is based on the coulombic interaction within the ionic crystal: 1) Electrostatic force of attraction between oppositely charged ions (cation and anion) 2) Repulsive interaction due to interpenetration of electron charge clouds• Based on these two interactions, Born-Lande theoretically derived the equation for calculating lattice energy.• The equation is called Born-Lande Equation and is given by:• U = -NoAZ+Z-e2 (1-1/n) ro 29/11/2010 mov 10 5
  6. 6. 29/11/2010 Born-lande Equation Cont.• Where, No = Avogadro’s number = 6.022x1023 atom-1 A = Madlung constant Z+ = Charge on cation Z- = charge on anion e- = Electronic charge ro = Sum of radii of cation and anion n = Born exponent >1• Significance of Born-Lande Equation1) From U α Z+Z-, the higher the charge on cation and anion, the greater the magnitude of lattice energy e.g. U (LiF) < U(CaF2) < U(MgS) 29/11/2010 mov 11 Born-lande Equation Cont. 2) U α 1/ro The smaller the size of ions, the higher the lattice energy e.g. U (NaF) > U(NaCl) > U(NaBr) > U(NaI) Exercise: 1) Zinc oxide, ZnO, is a very effective sun screen. How would the lattice energy of ZnO compare with that of NaCl 2) The precious gem ruby is aluminium oxide, Al2O3, containing traces of Cr3+. The compound Al2Se3 is used in the fabrication of some semiconductor devices. Which of the two has a larger lattice energy? 29/11/2010 mov 12 6
  7. 7. 29/11/2010 Born-Haber Cycle• Direct experimental determination of lattice energy is difficult, hence determined indirectly by a cyclic process called Born- Haber Cycle.• Born-Haber cycle relates lattice energies of ionic compounds with other thermodynamic data such as sublimation energy (SE), ionization energies (IE), electron affinities (Ea), dissociation energy (DE) and other atomic and molecular properties• The cycle is based on Hess’s law that total amount of heat evolved or absorbed in a chemical reaction is constant whether the reaction is carried out in a single step or multiple steps 29/11/2010 mov 13 Born-Haber Cycle e.g. ∆Hf = ∆Hs + IP + ½ ∆HD + Ea + U Ea IP ½ ∆HD U ∆Hs ∆Hf 29/11/2010 mov 14 7
  8. 8. 29/11/2010 Ionic Bond Formation• requires Coulombic attractive energy (lattice energy) to be sufficiently large to overcome ionization energy of the element that forms the cation.• balance between energy input (ionization energies) and stability gained from formation of the solid.• The main impetus for the formation of an ionic compound rather than a covalent compound results from the strong mutual attraction among the ions 29/11/2010 mov 15 Ionic Bond Formation cont. • Bonding in MgCl2 is ionic; Ionization energies ( I.E.) Mg(g) Mg2+(g) + 2ebonding ∆H = I.E.1 +I.E.2 = 753 +1435 = +2180 kJ/mol • bonds in AlCl3 are polar covalent Al(g) Al3+(g) + 3e- ∆H = I.E.1 + I.E. 2 + I.E. 3 = 580 +1815 + 2740 = +4125 kJ/mol 29/11/2010 mov 16 8
  9. 9. 29/11/2010 Ionic Bond Formation cont.• Bonds in AlCl3 are polar covalent• energy input (ionization energies) out weighs stability gained from formation of an ionic solid Al (g) Al3+(g) + 3e- ∆H = I.E.1 + I.E. 2 + I.E.3 = 580 +1815 + 2740 = +4125 kJ/mol29/11/2010 mov 17 Application of Born-Haber Cycle1) To determine lattice energy of unknown crystals eg. Calculate the lattice energy for formation of NaCl crystal based on the data below 1) ∆Hf (NaCl) = -99 Kcal/mol 2) ∆Hs (Na) = 26 Kcal/g atom 3) IP (Na) = 117 Kcal/atom 4) ∆HD (Cl2) = 54 Kcal/mol 5) Ea (Cl) = -84 Kcal/mol (Ans = -185 Kcal/mol)29/11/2010 mov 18 9
  10. 10. 29/11/2010 Application of Born-Haber Cycle cont.2) To determine electron affinity of elements which are difficult to determine by other methodseg. Determine electron affinity of iodine given the following thermodynamic data: 1) ∆Hf (Nal) = -68.8 Kcal/mol 2) ∆Hs (Na) = 25.9 Kcal/atom 3) IP (Na) = 118.4 Kcal/atom 4) ∆HD (l2) = 25.5 Kcal/mol 5) U (Nal) = -165.4 Kcal/mol (Ans = -73.2 Kcal/atom)29/11/2010 mov 19 Application of Born-Haber Cycle cont.• Explain why group IIA oxides and chlorides are stable in higher oxidation state despite the fact that the formation of Mg2+, Ca2+ etc require more energy than Mg+ and Ca+ Ans. the higher lattice energy involved compensates for the energy required for formation of Mg2+, Ca2+ making MgCl2, CaCl2, Al2O3 more stable.29/11/2010 mov 20 10
  11. 11. 29/11/2010 Valence Bond Theory • It is based on linear combination of atomic orbitals. • Main features: 1) A covalent bond is formed by the overlap of half filled atomic orbitals of different atoms 2) Overlapping atomic orbitals must have electrons with opposite spins 3) The bonded electron pair is localized between the two linked atoms 4) The stability of covalent bond is due to exchange of valence electrons between participating atoms, which lowers the potential energy of the bonded atoms 5) Each atom of the covalent compound tends to acquire a noble gas configuration by sharing electrons • s 29/11/2010 mov 21 Bonding Types• Two types of bonds result from orbital overlap:• sigma σ bonds • from head-on overlap • lie along the bond axis • account for the first bond • Can freely rotate around bond 29/11/2010 22 11
  12. 12. 29/11/2010 Bonding Types• pi π bonds • from lateral overlap by adjacent p or d orbitals • pi bonds are perpendicular to bond axis • account for the second and third bonds in a multiple bond • Cannot undergo rotation around bond 29/11/2010 mov 23 Valence Bond Theory cont. • One pair of electrons can occupy this overlapping area • Electron density is maximizes in overlapped region • Example: H2 bonds form because atomic valence orbitals overlap • Each hydrogen contributed 1s orbital 1s 1s 29/11/2010 mov 24 12
  13. 13. 29/11/2010 Valence Bond Theory• HF involves overlaps between the s orbital of H and the 2p orbital of F 1s 2s 2p29/11/2010 mov 25 VB Theory And H2S• Assume that the unpaired e- in S and H are free to form a paired bond• We may assume that the H-S bond forms between an s and a p orbital29/11/2010 mov 26 13
  14. 14. 29/11/2010 VB Theory and NH3 cont.According to Valence Bond Theory:Which orbitals overlap in the formation of NH3?• Ground state of nitrogen 2s ↑↓ 2p _↑ ↑ _↑__ mov 27 Difficulties With VB Theory• Most experimental bond angles do not support those predicted by mere atomic orbital overlap• For example: C 1s22s22p2 and H 1s1• Experimental bond angles in methane are 109.5° and all are the same• p orbitals are 90° apart, and not all valence e- in C are in the p orbitals• How can multiple bonds form?29/11/2010 mov 28 14
  15. 15. 29/11/2010 Hybridization• The mixing of atomic orbitals to allow formation of bonds that have realistic bond angles• The new shapes that result are called “hybrid orbitals”• The number of hybrid orbitals required = the number of bonding domains + the number of non-bonding domains on the atom 29/11/2010 mov 29 Hybrid between s and p Orbitals Two sp Orbitals in Linear Arrangement Formed by Hybridization of a single s and a single p Orbital 29/11/2010 mov 30 15
  16. 16. 29/11/2010 Hybrid orbitals • Naming of the hybrid orbital is based on the combination of the orbitals used to form the new hybrid • The name shows what type of atomic orbitals, and how many of each were used in formation of hybrid orbitals e.g. sp, sp2, sp3 etc. • One atomic orbital is used for every hybrid formed (orbitals are conserved)29/11/2010 mov 31 Hybrids From s & p Atomic Orbitals explain VSEPR Geometry Hybrid Atomic Orbitals Electron Geometry Used sp3 s + p x + py + pz Tetrahedral, bond angles 109.5˚ mov 32 16
  17. 17. 29/11/2010 Hybrids From s & p Atomic Orbitals explain VSEPR Geometry Hybrid Atomic Orbitals Electron Geometry Used sp2 s + p x + py Trigonal planar, bond angles 120 ˚29/11/2010 mov 33 Hybrids From s & p Atomic Orbitals explain VSEPR Geometry Hybrid Atomic Orbitals Electron Geometry Used sp s + px Linear, bond angles 180 ˚29/11/2010 mov 34 17
  18. 18. 29/11/2010 Hybrid Orbitals in BeH2• Ground state of Be 2s↑↓ 2p _ ___ [He]2s2 No half filled orbitals available for bonding• For hybrid sp hybrid orbitals sp ↑ ↑ 2p ___• Now bond can form between 1s orbital of hydrogen and sp hybrid orbital or berylllium sp ↑↓ ↑↓ 2p ___29/11/2010 mov 3529/11/2010 mov 36 18
  19. 19. 29/11/2010 Consider the CH4 molecule• Ground state for carbon 2s ↑ ↓ 2p ↑ _ ↑ ___ [He]2s2 2p2• Form sp3 hybrid orbitals ↑ ↑_ ↑ ↑ .• Each sp3 hybrid can now overlap with 1s orbital of hydrogen ↑↓ ↑↓_ ↓↑ ↑↓ .29/11/2010 mov 37 Bonding in CH4• The 4 hybrid orbitals are evenly distributed around the C• The H s-orbitals overlap the sp3 hybrid orbitals to form the bonds.29/11/2010 mov 38 19
  20. 20. 29/11/2010 Bonding in NH3• The 4 hybrid orbitals are evenly distributed around the N• The H s-orbitals overlap the sp3 hybrid orbitals to form three bonds bonds• The remaining lone pair occupies the last hybrid orbital29/11/2010 mov 39 Hybridization for form sp2 orbitals29/11/2010 mov 40 20
  21. 21. 29/11/2010 Ethene and Double Bonds Sigma Bonds σ : Direct overlap of orbitals between the two nuclei Direct overlap of atomic orbitals is not affected by rotation around that bond C--C29/11/2010 mov 41 C=C double bond consists of one sigma σ bond and one pi π bond29/11/2010 mov 42 21
  22. 22. 29/11/2010 Orbitals used for bonding in Ethene Sigma bond: sp2 overlaps sp2 Pi bond: unhybridized p orbital overlaps unhybridized p orbital29/11/2010 mov 43 Formation of sp Hybrid Orbitals29/11/2010 mov 44 22
  23. 23. 29/11/2010 Hybrid Orbitals is CO2 molecule First bond is sigma bond Second bond is pi bond 29/11/2010 mov 45 CO2 Molecule• Unhybridized p orbitals are used to form pi bonds between carbon and oxygen• sp hybrid orbitals of carbon overlap sp2 hybrid orbitals from oxygen to form sigma bonds 29/11/2010 mov 46 23
  24. 24. 29/11/2010 • Triple bond in N2 molecule consists of one sigma bond and 2 pi bonds • Sigma bond stems from sp hybrid overlap • Pi bonds come from unhybridized p orbital overlap29/11/2010 mov 47 Expanded Octet Hybridization• Can be predicted from the geometry as well• In these situations, d orbitals are be needed to provide room for the extra electrons• One d orbital is added for each pair of electrons in excess of the standard octet29/11/2010 mov 48 24
  25. 25. 29/11/2010 Expanded Octet hybridization• dsp3 hybridization gives rise to trigonal bipyramid geometry of PCl529/11/2010 mov 49 d2sp3 hybridization gives rise to octahedral geometry29/11/2010 mov 50 25
  26. 26. 29/11/2010 Consider SF6• Ground state for sulfur 3s ↑ ↓ 3p ↑ ↓ ↑ ↑_ 3d _ _ _ _ _• Six hybrid orbitals needed sp3d2 ↑ ↑ ↑ ↑ ↑ ↑. 3d _ _ _• Each sp3d2 hybrid can now overlap with 2p orbital of fluorine ↑↓ ↑↓_ ↓↑ ↑↓ . ↓↑ ↓↑ 29/11/2010 mov 51 Bonding is XeF4 • Placing lone pairs at axial positions lets them be as far as possible from one another • Square planar geometery 29/11/2010 mov 52 26
  27. 27. 29/11/2010 • Hybrid orbital can also hold nonbonding electrons • Usually results in polar molecules29/11/2010 mov 53 Consider the SF4 molelcule F Lewis Structure : F S F F• Four bonding + 1 nonbonding pairs around sulphur • Five hybrid orbitals needed sp3d ↑↓ ↑ ↑ ↑ ↑ . 3d _ _ _ _ • Four half filled orbitals available to overlap with 2p orbital of fluorine sp3d ↑↓ ↓↑ ↓ ↑ ↓↑ ↓ ↑ . 3d _ _ _ _29/11/2010 mov 54 27
  28. 28. 29/11/2010 Geometry of SF4 F • sp3d requires trigonal bipyramid geometry • Nonbonding pair goes on equatorial : position S F F • Distorted tetrahedron geometry F29/11/2010 mov 55 Bonding in Ethene C2H4• Carbon forms sp2 hybrid orbitals, and one unhybridized p orbital29/11/2010 mov 56 28
  29. 29. 29/11/201029/11/2010 mov 57 O Sigma and Pi Bonding C H H29/11/2010 mov 58 29
  30. 30. 29/11/2010H−C≡C −H• Each C has a triple bond and a single bond• Requires 2 hybrid orbitals, sp• unhybridize d p orbitals used to form the pi bond29/11/2010 mov 59 Types of bonds in Acetylene29/11/2010 mov 60 30
  31. 31. 29/11/2010 Summary of Multiple Bonds• Molecular skeleton held together by σ bonds. First bond between two atoms always σ.• Hybrid orbitals are used to form σ bonds, and to hold nonbonding electrons • Number of hybrid orbitals needed = # atoms bonded + # of nonbonding pairs• π bonds are formed using non-hybridized p or d orbitals• Double bond is one σ and one π bond• Triple bond consists of one σ and two π bonds29/11/2010 mov 61 Molecular Orbital Theory • Modification of VB theory, considers that the orbitals may exhibit interference. • Waves may interfere constructively or destructively • Bonding orbitals stabilize, antibonding destabilize.29/11/2010 mov 62 31
  32. 32. 29/11/2010 Molecular Orbital Theory concept1) Atomic orbitals having same energy and symmetry combine to form molecular orbitals by linear combination of atomic orbitals (LCAO)• If ΨA and ΨB are the wave functions of atoms A and B, then by LCAO, Ψ = ΨA ± ΨB2) Molecular orbital formed by addition of two atomic orbital wave functions is called Bonding Molecular Orbital i.e. Ψb = ΨA + ΨBProbability of finding electron in BMO is given by Ψb2 = ΨA2 + ΨB2 + 2 ΨAΨB> ΨA2 + ΨB2 The energy of bonding molecular orbital is less than the energy of individual atomic orbitals 29/11/2010 mov 63 Molecular Orbital Theory concept 3) Molecular orbital formed by subtraction of two atomic orbital wave functions is called Anti-Bonding Molecular Orbital i.e. Ψa = ΨA + ΨB Probability of finding electron in ABMO is given by Ψa2 = ΨA2 + ΨB2 - 2 ΨAΨB< ΨA2 + ΨB2 The energy of anti-bonding molecular orbital is higher than the energy of individual atomic orbitals 29/11/2010 mov 64 32
  33. 33. 29/11/2010 Molecular Orbital Theory concept4) Molecular Orbitals that do not participate in bonding are called Non Bonding molecular orbitals. They occur at same energy as individual atomic orbitals5) Atomic orbitals are monocentric, whereas molecular orbitals are polycentric6) The Stability of the bond is expressed in terms of bond order- defined as one half the difference between the number of electrons in bonding molecular orbitals (Nb) and atibonding molecular orbitals (Na) i.e. Bond Order = (Nb – Na)/2 29/11/2010 mov 65 Molecular Orbital Theory concept • Bond order my be zero, fraction or positive but not negative • The greater the bond order, the stronger the bond formed, hence the greater the stability of the molecule BO α Bond Strength α Stability α -1/bond length α 1/reactivity 29/11/2010 mov 66 33
  34. 34. 29/11/2010 Molecular orbital Energy Diagram• This is a potential energy diagram showing the atomic orbitals combining and molecular orbitals formed.• Electrons fill molecular orbitals in order of increasing energy levels• Sequence of molecular orbital energies is based on light (H to N) and heavy (≥ O )molecules• Light molecules: σ1s< σ*1s< σ2s< σ*2s< π2px = π2py< σ2pz< π*2px = π*2py<σ*2pz• Heavy Molecules: σ1s< σ*1s< σ2s< σ*2s< σ2pz< π2px = π2py< π*2px = π*2py<σ*2pz 29/11/2010 mov 67Differences between bonding and anti- bonding molecular orbitals BMO ABMO 1. BMO is formed by the addition of 1. ABMO is formed by the subtraction of overlapping atomic orbitals i.e. overlapping atomic orbitals Ψmo= ΨA+ΨB Ψ*mo= ΨA-ΨB 2. It has greater electron density in the 2. It has lower electron density in the region between the two nuclei of bonded region between the two nuclei of the atoms atoms 3. Electrons in BMO contribute to 3. Electrons in ABMO contribute to attraction between the two atoms repulsion between the two atoms 4. It possesses lower energy than 4. It possesses higher energy than associated atomic orbitals associated atomic orbitals 29/11/2010 mov 68 34
  35. 35. 29/11/2010 MO diagram for H2 • Show atomic energy level diagram for each atom • Show molecular orbitals (bonding and antibonding*) • 1 MO for each Atomic orbital. • Show electron occupancy of the orbitals.29/11/2010 mov 69 Filling MO diagrams1. Electrons fill the lowest-energy orbitals that are available.2. No more than two electrons, with spins paired, can occupy any orbital.3. Electrons spread out as much as possible, with spins unpaired, over orbitals that have the same energy. (# bonding e - ) − (# antibonding e - ) Bond Order = 2 electrons/bond29/11/2010 mov 70 35
  36. 36. 29/11/2010 H2 vs He2 2−0 H 2 Bond Order = =1 2 2−2 He 2 Bond Order = =0 229/11/2010 mov 71 Molecular Orbitals Using p Orbitals Two boron atoms have one set of p orbitals that can directly overlap to for sigma bond. Two parallel p orbitals can form pi bonds29/11/2010 mov 72 36
  37. 37. 29/11/2010 • Two px orbitals overlap for form sigma bonding and antibonding molecular orbitals 29/11/2010 mov 73• Two p orbitals overlap to form pi bonding and anti- bonding orbitals• Can happen both to py pair and to pz pair, resulting in two bonding and two anti-bonding orbitals 29/11/2010 mov 74 37
  38. 38. 29/11/2010 Molecular Orbital Diagram for B2 Total Bonding - Total antibondin gBond Order = 2 4-2 = =1 2 B2 should be a stable molecule 29/11/2010 mov 75 Diatomic MO diagrams differ by group Second period used s and A) I - V p orbitals B) VI-VIIIA 29/11/2010 mov 76 38
  39. 39. 29/11/2010Molecular Orbitals Explains Paramagnetic O2• Paramagnetic; weakly attracted to magnetic field • Usually a result of unpaired electron • Simple Lewis structure has no unpaired electrons • However, MO treatment shows two unpaired electrons in π* orbitals29/11/2010 mov 77 Molecular Orbital Diagrams for B2 to F229/11/2010 mov 78 39
  40. 40. 29/11/2010 MO Diagram for Group I-VDraw the MOdiagram for N2 π∗2p σ∗2p 2p σ2p 2p π2p σ∗2s 2s 2s σ2s 29/11/2010 mov 79 MO Diagram for Group VI-VIIIDraw the MOdiagram for O2 σ∗2p π∗2p 2p 2p π2p σ2p σ∗2s 2s 2s σ2s 29/11/2010 mov 80 40
  41. 41. 29/11/2010Draw the MO MO also works fordiagram for NO heteronuclear diatomics 2p 2p 2s 2s nitrogen29/11/2010 mov oxygen 81 MO’s and Free Radicals NO σ∗2p Free radicals are molecules with π∗2p an unpaired electron 2p σ2p 2p π2p σ∗2s 2s 2s σ2s oxygen nitrogen29/11/2010 mov 82 41
  42. 42. 29/11/2010 Resonance Structures for O3 and NO329/11/2010 mov 83 Figure (a) benzene molecule (b) two resonance structures for benzene molecule29/11/2010 mov 84 42
  43. 43. 29/11/2010 Delocalized Electrons • Lewis structures use resonance to explain that the actual molecule appears to have several equivalent bonds, rather than different possible structures • MO theory shows the electrons being delocalized in the structure29/11/2010 mov 85Calculation of Hybridization in a molecule• Steps:1) Calculate the total number of valence electrons2) Calculate the number of duplet or octate Duplet = Total valence electrons/2 Octet = Total valence electrons/83) Calculate the number of lone pairs of electrons Lone pairs = (Total valence electrons – 2xNumber of duplets)/2 Lone pairs = (Total valence electrons – 8xNumber of duplets)/84) Calculate the number of orbitals used = No. Of duplets + No. Of lone pairs or electrons5) If there are no lone pairs of electrons, the structure and shape are ideal, else distortion in shape occurs29/11/2010 mov 86 43
  44. 44. 29/11/2010 Calculation of Hybridization in a molecule 1: Calculate the type of hybridisation, geometry and shapes of the following molecules: H2O, SO42- ions H2O i) Total No. Electrons = 1x2+6 = 8 ii) Required number of orbitals = 2 iii) Required electrons for duplets = 2x2 = 4 iv) No. Of lone pair electrons = (i)-(iii)/2 = 8-4/2 = 2 v) Number of orbitals = (ii) + (iv) = 2+2 = 4 • Thus, oxygen exhibits SP3 hybridisation and its structure is tetrahedral with distortion bond angle due to presence of lone pairs 29/11/2010 mov 87 Calculation of Hybridization in a molecule SO42- i) Total No. Electrons = 6+6x4+2 = 32 ii) Required number of orbitals = 4 iii) Required electrons for octet = 4x8 = 32 iv) No. Of lone pair electrons = ((i)-(iii))/2 = (32-32)/2 = 0 v) Number of orbitals = (ii) + (iv) = 4+0 = 4• Thus, SO42- involves SP3 hybridisation and its structure is a regular tetrahedron 29/11/2010 mov 88 44
  45. 45. 29/11/2010 Calculating the percentage s- and p- character• It is based on the equation: Cosθ = s/s-1 = p-1/s, where θ = bond angle Example: Calculate the percentage s-orbital and p-orbital character of a hybrid orbital if the bond angle between the hybrid orbital is 105o Cosθ = s/s-1 = p-1/s, where θ is the bond angle s/s-1 = cos 105o = cos (90 + 15) = -sin 15o = -0.2588 s = 0.2588/1.2588 = 0.2056 similarly, p-1/p = cos 105o = -0.2588 p = 0.7944 29/11/2010 mov 89 Calculating the number σ and π bonds• Steps:1) The number of π electrons in a noncyclic molecule which does not contain hydrogen atom is given by 6n+2-v, where n = total number of atoms of molecules and v is the sum of the valence electrons in the molecule2) The number of π electrons in a noncyclic molecule containing hydrogen atoms is given by 6n-6q+2-v, where q = the number of hydrogen atoms in the molecule3) If the molecule is of Ax n type, and none of the them is a hydrogen atom, then, Number of σ bonds = Total No. Valence electrons/84) If the molecule contains a hydrogen atom, then, Number of σ bonds = total no. Valence electrons/2 29/11/2010 mov 90 45
  46. 46. 29/11/2010Calculating the number σ and π bonds• Example: Calculate the number of π and σ bonds in the following molecules: CO2, CO32-, H2O, NH3, C2H2, C2H4CO2No. Of π electrons = 6x3+2-(4+2x6) = 4No. π bonds = 4/2 = 2No. σ bonds = (4+2x6)/8 = 2C2H4No. Of π electrons = 6x6-6x4+2-(2x4+4) = 2No. π bonds = 2/2 = 1No. σ bonds = (2x4+4x1)/2 = 6 29/11/2010 mov 91Calculating the number σ and π bonds • Calculate the number of π and σ bonds in the following molecules: CO2, CO32-, H2O, NH3, C2H2, C2H4 H2O No. Of π electrons = 6x3+6x2+2-8 = 0 No. π bonds = 0 No. σ bonds = (2+6)/2 = 4 CO32- No. Of π electrons = 6x4+2-(4+3x6+2) = 2 No. π bonds = 2/2 = 1 No. σ bonds = (4+3x6+2)/8 = 3 29/11/2010 mov 92 46