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Chapter 10 Molecular Structure and Bonding Theories

Chapter 10 Molecular Structure and Bonding Theories. VSEPR. Valence-Shell Electron-Pair Repulsion Model (VSEPR) predicts shape from Lewis Structures. VSEPR Rule 1 : A molecule has a shape that minimizes electrostatic repulsions between valence-shell electron pairs.

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Chapter 10 Molecular Structure and Bonding Theories

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  1. Chapter 10 Molecular Structure and Bonding Theories

  2. VSEPR Valence-Shell Electron-Pair Repulsion Model (VSEPR) predicts shape from Lewis Structures. • VSEPR Rule 1: A molecule has a shape that minimizes electrostatic repulsions between valence-shell electron pairs. • Minimum repulsion results when the electron pairs are as far apart as possible.

  3. Steric Number • Steric number = (number of lone pairs on central atom) + (number of atoms bonded to central atom) • The steric number is determined from the Lewis structure. • Steric number determines the bonded-atom lone-pair arrangement, the shape that maximizes the distances between the valence-shell electron pairs.

  4. Geometric Arrangements

  5. Geometric Arrangements

  6. Steric Number = 2 • In the Lewis structure of BeCl2, beryllium has two bonded atoms and no lone pairs, steric number = 2. • A linear geometry places the two pairs of electrons on the central beryllium atom as far apart as possible.

  7. Molecules with Multiple Bonds • The Lewis structure of HCN (H-Cº N:) shows that the carbon atom is bonded to two atoms and has no lone pairs, steric number = 2. • The bonded-atom lone-pair arrangement is linear. • The number of bonded atoms, not the number of bonds, determines the steric number.

  8. Steric Number = 3 • The Lewis structure of BF3 shows the boron atom has a steric number = 3; the bonded-atom lone-pair arrangement is trigonal planar.

  9. Steric Number = 4 • The Lewis structure of CH4 shows the carbon atom has a steric number = 4; the bonded-atom lone-pair arrangement is tetrahedral.

  10. Steric Number = 5 • The phosphorus atom in PF5 has a steric number = 5; the bonded-atom lone-pair arrangement is trigonal bipyramidal.

  11. Steric Number = 6 • The sulfur atom in SF6 has a steric number = 6; the bonded-atom lone-pair arrangement is octahedral.

  12. O H H Central Atoms with Lone Pairs • The Lewis structure of H2O is • Steric number = 4, 2 bonded atoms and 2 lone pairs. • The bonded-atom lone-pair arrangement is tetrahedral.

  13. Molecular Shape of H2O • Molecular shapeis the arrangement of the atoms in a species. • The bonded-atom lone-pair arrangement of H2O is tetrahedral (top); the molecular shape is bent or V-shaped (bottom).

  14. Molecular Shape of NH3 • What is the electron pair geometry and molecular shape of NH3?

  15. Electron Pair Repulsions • The measured bond angle in H2O (104.5o) is smaller than the predicted angle (109.5o) • Explanation: repulsions vary lone pair-lone pair > lone pair-bonding pair > bonding pair-bonding pair

  16. Location of Lone Pair in SF4 • The favored structure for a trigonal bipyramid minimizes 90o lone pair interactions – the one on the right. Two structures are possible:

  17. Lone Pairs in Trigonal Bipyramids • Lone pairs always occupy the equatorial positions in a trigonal bipyramid so that lone pair-lone pair repulsions are oriented at 120o.

  18. Location of Lone Pairs in XeF4 • The structure on right has no 90o lone pair-lone pair interactions and is favored.

  19. Test Your Skill • What is the steric number, the bonded-atom lone-pair arrangement, and the molecular shape of ClF3?

  20. Multiple Central Atoms • The geometry of each central atom is determined separately. • The CH3 carbon in CH3CN has tetrahedral geometry and the other carbon has linear geometry.

  21. Shapes of Molecules H H N S H • What are the bonded-atom lone-pairarrangements and the shapes about each central atom in NH2SH? • Draw the Lewis structure. • The bonded-atom lone-pairarrangements of both are tetrahedral, the nitrogen shape is trigonal pyramidal and sulfur is “V” shaped.

  22. Overall Shape of C2H4 • Ethylene, C2H4 , could be planar (left) or nonplanar (right). The VSEPR model does not predict which is preferred.

  23. Polarity of Molecules • The bond dipoles in CO2 cancel because the linear shape orients the equal magnitude bond dipoles in exactly opposite directions.

  24. Polarity of Molecules • The bond dipoles do not cancel in COSe; they are oriented in the same direction and are of unequal length. They do not cancel in OF2 because the V-shape of the molecule does not orient them in opposite directions.

  25. Polarity of Molecules • The bond dipoles in BCl3 and CCl4 cancel because of the regular shape and equal magnitude.

  26. Polarity of Molecules • The bond dipoles in BCl2F and CHCl3 do not cancel because they are not of the same magnitude.

  27. Test Your Skill • Are the following molecules polar or nonpolar: H2S, SiF4, CH2Cl2?

  28. Valence Bond Theory • Valence bond theory describes bonds as being formed by overlap of partially filled valence orbitals.

  29. Test Your Skill • Identify the orbitals that form the bond in HCl.

  30. Bonding in NH3 • The observed bond angles of 107.5o in NH3 are not consistent with the angles of90o expected if the bonds formed from N 2p orbitals.

  31. Hybrid Orbitals • Hybrid orbitalsare orbitals obtained by mixing two or more atomic orbitals on the same central atom. • Appropriate hybrid orbitals formed by mixing one s and xp atomicorbitals make bonds at either 180o (x = 1), 120o (x = 2), or 109.5o (x = 3).

  32. Analogy for Hybrid Orbitals

  33. sp Hybrid Orbitals

  34. Shape of Hybrid Orbitals • For clarity, hybrid orbitals are pictured as elongated with the small lobe omitted.

  35. Bonding in BeCl2 • The bonds in BeCl2 arise from the overlap of two sp hybrid orbitals on the beryllium atom with the 3p orbitals on the two chlorine atoms.

  36. sp2 Hybrid Orbitals

  37. Bonding in BF3 • The bonds in BF3 arise from the overlap of three sp2 hybrid orbitals on the boron atom with 2p orbitals on the three fluorine atoms.

  38. sp3 Hybrid Orbitals

  39. Bonding in CH4 • The bonds in CH4 arise from the overlap of four sp3 hybrid orbitals on the carbon atom with 1s orbitals on the four hydrogen atoms.

  40. Lone Pairs and Hybrid Orbitals • Hybrid orbitals can hold lone pairs as well as make bonds.

  41. Hybridization with d Orbitals • Hybrid orbitals of central atoms with steric numbers of 5 or 6 involve d orbitals.

  42. Hybrid Orbitals

  43. Test Your Skill • Identify the hybrid orbitals on the central atoms in SiH4 and HCN.

  44. Types of Bonds: Sigma Bonds • Sigma bonds (s): the shared pair of electrons is symmetric about the line joining the two nuclei of the bonded atoms.

  45. Bonding in C2H4 • The C-C sigma bond in C2H4 arises from overlap of sp2 hybrid orbitals and the four C-H sigma bonds from overlap sp2 hybrid orbitals on C with 1s orbitals on H. • The second C-C bond forms from sideways overlap of p orbitals.

  46. Types of Bonds: Pi Bonds • Pi bonds (p) places electron density above and below the line joining the bonded atoms – they form by sideways overlap of p orbitals.

  47. Bonding in C2H4 • The double bond in C2H4 is one sigma bond and one pi bond – each bond is of similar strength.

  48. Proof of Pi Bonds: Shape of C2H4 • C2H4 isplanar(A) because pi overlap is at a maximum. Rotation of one end by 90o (B) reduces pi overlap to zero.

  49. Triple Bonds • The triple bond in C2H2 is one sigma bond and two pi bonds between the sp hybridized carbon atoms.

  50. Sigma Bonds in Benzene • Each carbon atom in benzene, C6H6, forms three sigma bonds with sp2 hybrid orbitals.

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