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Bonding and Molecular Structure

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  1. Bonding and Molecular Structure Goals: Understand the differences between ionic and covalent bonds. Draw Lewis electron dot structures for small molecules and ions. Use the valance shell electron-pair repulsion theory (VSEPR) to predict the shapes of simple molecules and ions and to understand the structures of more complex molecules. Use electronegativity to predict the charge distribution in molecules and ions and to define the polarity of bonds. Predict the polarity of molecules.

  2. Chemical Bonding Problems and questions — How is a molecule or polyatomic ion held together? Why are atoms distributed at strange angles? Why are molecules not flat? Can we predict the structure? How is structure related to chemical and physical properties? There are 2 extreme forms of connecting or bonding atoms: Ionic—complete transfer of ________________from one atom to another. Covalent—some valence electrons _________ between atoms Most bonds are somewhere in between!

  3. Chemical Bonding • Wherever two atoms or ions are strongly attached to each other : chemical bond between them. • Ionic, covalent, metallic. • Ionic bond: electrostatic force between ions of opposite charge. • Covalent bond: sharing of electrons between two atoms.

  4. Lewis Symbols • Electrons involved in chemical bonding are the _______________ (those in the outermost occupied shell of an atom). • Lewis electron-dot symbols: simple way of showing the valance electrons. • Lewis symbol: • Chemical symbol for the element + a dot for each valance electron. • Dots are placed on the four sides of the atomic symbol. Each side can accommodate up to two electrons. • The number of valance electrons in any representative element is the same as the group number of the element. Sulfur: [Ne]3s23p4 •• S • • ••

  5. The Octet Rule • Atoms tend to gain, lose, or share electrons until they are surrounded by eight valance electrons. • They have the same number of electrons as the ________ closest to them in the period table – stable electron arrangements (high ionization energies, low affinity for additional e-s, general lack of chemical reactivity. • An octet of electrons consists of full s and p subshells in an atom: • ns2np6 • There are some exceptions…

  6. Ionic Bonding Na (s) + 1/2 Cl2(g) ---> NaCl (s) DHfo = -410.9 kJ Metal of low IE Nonmetal of high EA

  7. Ionic Bonding Na (s) + 1/2 Cl2(g) ---> NaCl (s) DHfo = -410.9 kJ • An electron has been lost by a sodium atom and gained by a chlorine atom – electron transfer from the Na atom to the Cl atom. • NaCl is a typical ionic compound: it consists of a metal of low ionization energy and a nonmetal of high electron affinity. • Lewis electron-dot representation:

  8. Energetics of Ionic Bond Formation • The principal reason that ionic compounds are stable is the attraction between ions of unlike charge. • This attraction draws the ions together, releasing energy and causing the ions to form a solid array – lattice. • A measure of how much stabilization results from arranging oppositely charged ions in an ionic solid: • Lattice energy – is the energy required to completely separate a mole of a solid ionic compound into its gaseous ions. • NaCl  Na+ (g) + Cl- (g) DHlattice = + 788 kJ/mol • Highly endotermic process – the reverse process – the coming together is highly exothermic. • Eel = kQ1Q2 / d For a given arrangement of ions, the lattice energy increases as the charges on the ions increase and their radii decrease.

  9. Arrange the following ionic compounds in order of increasing lattice energy: NaF, CsI, and CaO.

  10. Covalent Bonding • The bond arises from the mutual attraction of 2 nuclei for the same electrons. • Electron sharingresults. Bond is a balance of attractive and repulsive forces.

  11. Bond Formation • A bond can result from a “head-to-head” overlap of atomic orbitals on neighboring atoms. Overlap of H (1s) and Cl (2p) Note that each atom has a single, unpaired electron.

  12. •• H Cl • • •• lone pair (LP) shared or bond pair G. N. Lewis 1875 - 1946 Electron Distribution in Molecules • Electron distribution is depicted with Lewis electron dot structures • Valence electrons are distributed as shared or ______ PAIRS and unsharedor ______ PAIRS.

  13. Valance Electrons Electrons are divided between core and valence electrons: Valance e-: e- in the outermost shell B 1s2 2s2 2p1 Core = [He] , valence = 2s2 2p1 Br [Ar] 3d10 4s2 4p5 Core = [Ar] 3d10 , valence = 4s2 4p5

  14. The Octet Rule • Each atom (except H) has a share in four pairs of electrons, so each has achieved a noble gas configuration. • Each atom is surrounded by an octet of e-s. • Octet Rule: The tendency of molecules and polyatomic ions to ______________________________________ _______________. • No. of valence electrons of an atom = group number • For Groups 1A-4A, no. of bond pairs = group number • For Groups 5A -7A, BP’s = 8 – group number • Except for H (and sometimes atoms of 3rd and higher periods), BP’s + LP’s = 4

  15. Building an Electron Dot Structure Ammonia, NH3 1. Decide on the central atom: Central atom is atom of lowest affinity for electrons; never H. Therefore, N is central 2. Count total valence electrons H = 1 and N = 5 Total = (3 x 1) + 5 = 8 electrons / 4 pairs The trick is to count the e-s! Students should be familiar with writing Lewis structures.

  16. H H N H •• H H N H Ammonia, NH3 3. Form a single bond between the central atom and each surrounding atom. Each line represent 1 pair of shared e-. 4. Remaining electrons form LONE PAIRS to complete octet as needed. N: 3 BOND PAIRS and 1 LONE PAIR. Note that N has a share in 4 pairs (8 electrons), while H shares 1 pair. Practice with the Lab Experiment # 11

  17. Sulfite ion, SO32- Step 1. Central atom = S Step 2. Count total valence electrons Notice charge in ion: add or substract e- according to charge! Negative charge = 2 e- more

  18. Sulfite ion, SO32- Step 3. Form bonds Step 4. Assign as lone pairs to outside atoms (never to H), if some left, then to the center atom. Step 5. Check all atoms have an octet. If the central atom as fewer than eight e-, move one or more of the lone pairs on the terminal atoms and form multiple bonds. – See next example.

  19. Carbon Dioxide, CO2 Step 1. Central atom = C Step 2. Count total valence electrons Step 3. Form bonds Step 4. Assign as lone pairs to outside atoms (never to H), if some left, then to the center atom. Step 5. Check all atoms have an octet.

  20. Carbon Dioxide, CO2 Step 5. Check all atoms have an octet. Form multiple bonds if needed. So that C has an octet, we shall form DOUBLE BONDS between C and O. The second bonding pair forms a pi (π) bond.

  21. Sulfur Dioxide, SO2 Step 1. Central atom = S Step 2. Count total valence electrons Step 3. Form bonds Step 4. Assign as lone pairs to outside atoms (never to H), if some left, then to the center atom. Step 5. Check all atoms have an octet.

  22. Sulfur Dioxide, SO2 Step 5. Check all atoms have an octet. Form double bond so that S has an octet — but note that there are two ways of doing this. This leads to the following structures. These equivalent structures are called RESONANCE STRUCTURES. The true electronic structure is a HYBRID of the two.

  23. Formal Atom Charges • Atoms in molecules often bear a charge (+ or -). • The predominant resonance structure of a molecule is the one with charges as close to 0 as possible. • Formal charge = Number of valance e- (group number) – 1/2 (no. of bonding electrons) - (no. of LP electrons)

  24. Formal Atom Charges Formal charge = Valance e- – 1/2 (BP e-) – LP e- For C, = 4 – ½ (8) – 0 = 0 For O, = 6 – ½ (4) – 4 = 0 Yellow= negative&red= positive Relative size = relative charge

  25. • S C N • • • • • • • • • • S C N • • • • • • S C N • • • • • • Formal Atom Charges Formal charge = Group number – 1/2 (BP e-) – LP e- For S, 6 - ½ (2) - 6 = -1 For C, = 4 – ½ (8) – 0 = 0 For N, 5 - ½ (6) - 2 = 0 Thiocyanate Ion, SCN- Which is the most important resonance form? All atoms are negative, but most negative is S.

  26. Violations of the Octet Rule Usually occurs with B and elements of higher periods. BF3 SF4

  27. Calc’d partial charges in BF3 Boron Trifluoride, BF3 The B atom has a share in only 6 pairs of electrons (or 3 pairs). B atom in many molecules is electron deficient. F is negative and B is positive

  28. Sulfur Tetrafluoride, SF4 5 pairs around the S atom. A common occurrence outside the 2nd period.

  29. Which of the following species will have a Lewis structure most like that of a sulfate ion, SO42-? Assume that the Lewis structure has no double bonds. a) NH3b) CBr4c) SO3d) H2CO

  30. Which of the following species are isoelectronic to SO4-2? CO32- ; PO4-3 ; NH4+

  31. Free Radicals • Free radicals – atomic or molecular specie with ______ ____________. • Free radicals are generally _____________. • H, Cl • O2 - Oxygen itself is a free radical, but one with two unpaired electrons; reduction by adding one electron to give superoxide involves the pairing of two of the electron spins, leaving one unpaired O2 + 1e-  O2·- • NO – vascular system (11 valance electrons – odd number) • Vitamin C, Vitamin E – Antioxidants • They produce very stable radicals (stabilized by resonance structures) Vit E

  32. MOLECULAR GEOMETRY VSEPR • Valence Shell Electron Pair Repulsion theory. • Most important factor in determining geometry is ________________________ electron pairs. • Molecule adopts the shape that ___________ the electron pair repulsions.

  33. Exp 11: Molecular Geometry • I suggest you practice with: 3. SO2 13. CS2 29. NO2+ 7. NO3- 22. CH3- 36. PO33- 9. NH4+ 27. H3O+ 43. H2O 10.CH2Cl2 28. O3 46. CH3+ 48. HCN 1. Give the total number of valance electrons. 2. Draw the correct Lewis dot structure. 3. Give the geometry using VSEPR theory. 4. Give the polarity of the MOLECULE or ION.

  34. Electron Pair Geometries

  35. •• H H N H Structure Determination by VSEPR Ammonia, NH3 1. Draw electron dot structure 2. Count BP’s and LP’s = 4 3. The 4 electron pairs are at the corners of a tetrahedron. The ELECTRON PAIR GEOMETRY is tetrahedral.

  36. Structure Determination by VSEPR Ammonia, NH3 The electron pair geometry is tetrahedral. The MOLECULAR GEOMETRY — the positions of the atoms — is PYRAMIDAL.

  37. Structure Determination by VSEPR Water, H2O 1. Draw electron dot structure 2. Count BP’s and LP’s = 4 3. The 4 electron pairs are at the corners of a tetrahedron. The electron pair geometry is TETRAHEDRAL.

  38. Structure Determination by VSEPR • Water, H2O The electron pair geometry is tetrahedral. The molecular geometry is BENT.

  39. Geometries for Four Electron Pairs Students should be familiar with predicting molecular geometries.

  40. The Hydronium Ion H3O+ Draw the Lewis dot structure and predict the geometry.

  41. Molecular Geometries for Five Electron Pairs All based on trigonal bipyramid.

  42. 9 0 o F F • 1 2 0 o S • F F Sulfur Tetrafluoride, SF4 Lone pair is in the equator because _____ __________________.

  43. Molecular Geometries for Six Electron Pairs All are based on the 8-sided octahedron

  44. O c t a h e d r o n 9 0 o F F F S 9 0 o F F F Sulfur Hexafluoride, SF6 6 electron pairs

  45. Electronegativity • Concept proposed by Linus Pauling (1901-1994). The only person to receive two unshared Nobel prizes (for Peace and Chemistry). Chemistry areas: bonding, electronegativity, protein structure.

  46. Electronegativity •  Is the ability of _________________to ____________________. • It is related to its ionization energy and electron affinity (properties of isolated atoms). • Electronegativity values are unitless. • ____________, the most electronegative, has a value of 4.0 • __________, the least electronegative has a value of 0.7. • The values for all other elements lie between these two extremes.

  47. Electronegativity • Used for atoms in molecules. • It is related to its ionization energy and electron affinity (properties of isolated atoms).

  48. Bond Polarity HCl is ______ because it has a positive end and a negative end. Cl has a greater share in bonding electrons than does H. Cl has slight negative charge (-d) and H has slight positive charge (+ d)

  49. Bond Polarity Due to the bond polarity, the H—Cl bond energy is __________ than expected for a “pure” covalent bond. BOND ENERGY “pure” bond 339 kJ/mol calc’d real bond 432 kJ/mol measured Difference = 92 kJ. This difference is proportional to the difference in ELECTRONEGATIVITY,.

  50. Bond Polarity • Three molecules with polar, covalent bonds. • Each bond has one atom with a slight negative charge (-d) and and another with a slight positive charge (+ d) • Relative values ofdetermine BOND POLARITY (and point of attack on a molecule). • Dipole moment – _________ ______________________ ______________________ ______________________. • They are added as VECTORS (in math)