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  1. Chemistry


  3. Session Opener

  4. Session Objectives

  5. Session Objectives • Introduction • Octet rule • Different types of bonding • Lewis theory • VSEPR theory and shape of molecules

  6. What is Chemical bonding? Na+ Cl- Chemical bonds Force of attraction holdinggroup(s) of atoms But why bonds are formed ?? Better stability against chemical reagents

  7. Octet rule two electrons in the valenceshell (1s2) + Cl Na Na Cl - 2 8 7 2 8 2 8 1 2 8 8 Ne Very reactive Very reactive Ar Atoms noble gas configuration attain betterstability.

  8. Limitation of octet rule In SF6, ‘S’ has twelve electron in itsvalence shell, leads to minimisation of energy.Other examples are: PCl5, BF3

  9. Questions

  10. Illustrative Problem The molecule that deviates from octet rule is (a) NaCl (b) BeCl2 (c) MgO (d) NH3

  11. Solution The no. of valence electrons in different central atoms is: Na+ 8 Be+2 2 Mg+2 8 N atom in NH3 (covalent compounds) 8 Hence, the answer is (b).

  12. Ionic Covalent Sigma bond Pi bond Co-ordinate or dative Metallic Bonding

  13. Formation of ionic bond

  14. Covalent bond Triple bond Double bond Formed by mutual sharing of electrons Covalent bonds

  15. Formation of covalent bond non-polar covalent bond between two carbon atomspolar covalent bond between carbon and hydrogen atoms.

  16. Question

  17. Illustrative Problem electrostatic force of attraction. Ionic bond overlap of atomic orbitals covalent bond Covalent bonds are called directional while ionic bonds are called non-directional -explain Solution: p and d-orbitals generate directional covalent bond.

  18. Types of covalent bonds s-s s-p p-p + + + Strength of these sigma bonds is in the order: p-p > s-p >s-s sigma bond forms due to end-to-end or head-on overlap

  19. Types of covalent bonds or + This is formed by lateral or sideways overlap which is possible for p or d-orbitals. Sigma bond is stronger than pi bond due to greater extent of overlap.

  20. Formed by head-on overlapping of s-s or s-p or p-p or any hybrid orbital Formed by side ways overlapping of unhybridised p-orbital First bond between any two atoms is always sigma Rest are p bonds In plane of molecule Perpendicular to plane of molecule Stronger as compared to p bond Weaker as compared to s bond Difference between sigma and pi bonds

  21. Difference between ionic and covalent compound. Ionic compound(NaCl) Covalent compound (CHCl3) Very high Volatile liquid Almost insoluble Highly soluble Highly Insoluble Insoluble Except s-soverlap all are directional Non-directional MP/BP H20 solubility Benzene solubility Directional nature

  22. Question

  23. Illustrative Problem The structure of benzene molecule is How many sigma and pi-bondsare present in a benzene molecule? Solution: no. of pi bonds are 3 [C=C] no. of sigma bonds are 12 [C-C and C-H]

  24. Coordinate covalent bond H H O : H+ H N : H+ H H [NH4]+ H3O+ • Single atom donating lone pair • Shared by two atoms involved

  25. Question

  26. Illustrative Problem NH3 and BF3 form an adduct readily-explain. Solution: N- atom in NH3 have one lonepair and BF3is electron deficient. They form an adductthrough coordinate bond, so BF3 can complete its octet. The adduct.

  27. Formation of metallic bond Metals lose their valence electrons to form cation in the pool of electrons. This is the Electron SeaModel for metallic bond.

  28. Characteristics of Metallic bond Regular close packed structures Excellent electricaland thermal conductivity

  29. Question

  30. Illustrative Problem • Which of the following has other • type of bonding with covalent • bonding? • CCl4 (b)AlI3 (c) NH4Cl (d) HCl Solution: covalent bonding is between N and three H-atomsCo-ordinate bond is present between N and one H atom ionic bond is there between NH4+ and Cl– ions. Hence, the answer is (c).

  31. Lewis theory iii) Multiple bonds complete the octet of atoms. Important aspects • Central atom. lesselectronegative atom [Exception: NH3, H2O more electronegative central atoms.] ii) Formal charge on ‘each atom’= (valence electron in atom) – (no. of bonds) – (no. of unshared electrons)

  32. Structure and bonding in CH4 n1=4+4 =8; n2=2x4+8x1 =16; n3=n2-n1 =8; no. of bonds= n3/2=4 no. of non-bonding electron= n4=(n1-n3)=0 no. of lone pairs= 0

  33. Limitations of Lewis theory Odd electron species NO,NO2 Electron-deficient species BF3 ,BeCl2 Electron-rich species PCl5 ,SF6 It cannot explain

  34. VSEPR theory Electro-negativity of central atom Electro-negativity of other atoms • Order of repulsion • lp-lp > lp-bp > bp-bp 2.lp-lp repulsion • Decreasing order of repulsion, Triple bond > double bond > single bond.

  35. Linear Planar Tetrahedral Trigonal bipyramidal Octahedral

  36. Application of VSEPR theory PCl5 central atom is P. therefore, V= 5+5= 10 V/2=5 Shape will be trigonal bipyramidal.

  37. Shape of SF6 Central atom is S V= 6 + 6 = 12 V/2=6 No. of atoms attached to central atom is six.Hence, shape is octahedral.

  38. Question

  39. Illustrative Problem • The shape of CH3+ is likely to be • Pyramidal (b) tetrahedral • (c) linear (d) planar Solution: According to VSEPR,N = 4 +3 –1=7N/2=3the shape should be planar. Hence, the answer is (d).

  40. Limitations Cannot determine the shape Multiple bonded species CO2,SO4-2

  41. Question

  42. Illustrative Problem The shape of NH3 is very similar to (a) CH4 (b) CH3– (c) BH3 (d) CH3+

  43. Solution shape is pyramidal shape is tetrahedral shape is planar shape is pyramidal shape is planar Hence, the answer is (b).

  44. Class Test

  45. Class Exercise - 1 Pi bond formation involves ______ overlap. (a) s-p head-on (b) p-p head-on (c) s-s head-on (d) p-p sideways Solution: Pi bond formation involves only sideways overlap of pand d-orbitals. Hence the answer is (d)

  46. What is the formal charge on ‘N’ atom of ? Class Exercise - 2 Solution: According to Lewis theory, n1 = 5 + 1 + 6 × 3 = 24 n2 = 2 × 0 + 8 × 4 = 32 n3 = n2 – n1; number of bonds = Number of lone pairs = Formal charge on ‘N’ atom = 5 – 4 – 0 = +1

  47. For Class Exercise - 3 Molecular structures of SF4 and XeF4 are (a) the same, with 2 and 1 lone pairs respectively (b) the same, with 1 lone pair each (c) different, with 0 and 2 lone pairs respectively (d) different with 1 and 2 lone pairs respectively Solution: Lone pair is one and the structure is trigonal bipyramidal.

  48. For Solution Lone pairs are two and the structure is octahedral. Hence, answer is (d)

  49. Class Exercise - 4 Predict the geometry of H3O+ based on VSEPR theory. Solution: For H3O+, central atom is ‘O’ Since 3 atoms are attached to the central atom,geometry will be of pyramidal according toVSEPR to minimize lp-bp repulsion.

  50. Class Exercise - 5 • Among Ca metal and Ca+2 the more • reactive will be (Atomic No. of Ca is • 20) • Calcium metal • (b) Calcium ion • (c) both are equally reactive • (d) Cannot be predicted Solution: Electronic configuration of Ca metal is 2, 8, 8, 2. While the configuration for Ca+2 is 2, 8, 8 which is a stable noble gas configuration. Hence the answer is (a)