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Concepts of Chemical Bonding

Concepts of Chemical Bonding. Brown, LeMay Ch 8 AP Chemistry Monta Vista High School. 8.1: Types of “Inter-Atomic” Bonds. Ionic : electrostatic attraction between oppositely charged ions Covalent : sharing of e- between two atoms (typically between nonmetals)

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Concepts of Chemical Bonding

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  1. Concepts of Chemical Bonding Brown, LeMay Ch 8 AP Chemistry Monta Vista High School

  2. 8.1: Types of “Inter-Atomic” Bonds • Ionic: electrostatic attraction between oppositely charged ions • Covalent: sharing of e- between two atoms (typically between nonmetals) • Metallic: “sea of e-”; bonding e- are relatively free to move throughout the 3D structure IncreasingDiff. of EN

  3. Lewis symbols Valence e-: • e- in highest energy level and involved in bonding; all elements within a group on P.T. have same # of valence e- Lewis symbol (or electron-dot symbol): • Shows a dot only for valence e- of an atom or ion. • Place dots at top, bottom, right, and left sides and in pairs only when necessary (Hund’s rule). • Primarily used for representative elements only (Groups 1A – 8A) Ex: Draw the Lewis symbols of C and N. Gilbert N. Lewis(1875 – 1946) • •C • • • : N • •

  4. The Octet Rule • Atoms tend to gain, lose, or share e- until they are surrounded by 8 valence e- (have filled s and p subshells) and are thus energetically stable. • Exceptions do occur (and will be discussed later.) http://hyperphysics.phy-astr.gsu.edu/hbase/chemical/eleorb.html#c1

  5. [ ]1- ••: F : •• 8.2: Ionic Bonding • Results as atoms lose or gain e- to achieve a noble gas e- configuration; is typically exothermic. • The bonded state is lower in energy (and therefore more stable). • Electrostatic attraction results from the opposite charges. • Occurs when diff. of EN of atoms is > 1.7 (maximum is 3.3: CsF) • Can lead to interesting crystal structures (Ch. 11). • Use brackets when writing Lewis symbols of ions. Ex: Draw the Lewis symbol of fluoride. • http://web.jjay.cuny.edu/~acarpi/NSC/5-bonds.htmhttp://chemmovies.unl.edu/ChemAnime/LICLD/LICLD.html

  6. Lattice Energy • Measurement of the energy of stabilization present in ionic solids DHlattice = energy required to completely separate 1 mole of solid ionic compound into its gaseous ions • Electrostatic attraction (and thus lattice energy) increases as ionic charges increase and as ionic radii decrease. http://ibchem.com/IB/ibnotes/full/ene_htm/15.2.htm Ex: Which has a greater lattice energy? NaCl or KCl NaCl or MgS

  7. Transition metals typically form +1, +2, and +3 ions. • It is observed that transition metal atoms first lose both “s” e-, even though it is a higher energy subshell. • Most lose e- to end up with a filled or a half-filled subshell.

  8. 8.4 - 8.5: Covalent Bonding • Atoms share e- to achieve noble gas configuration that is lower in energy (and therefore more stable). • Occurs when diff. of EN of atoms is ≤ 1.7 • Polar covalent: 0.3 < diff. of EN ≤ 1.7 (e- pulled closer to more EN atom) • Nonpolar covalent: 0 ≤diff. of EN ≤ 0.3 (e- shared equally) http://www.youtube.com/watch?v=QqjcCvzWwww (ionic vs. covalent bonding in youtube video)

  9. Cl-Cl Bond in Cl2 molecule http://www.youtube.com/watch?v=BCYrNU-7SfA&feature=related

  10. Lewis Structures • Lewis structures are used to depict bonding pairs and lone pairs of electron in the molecule.  • Step 1 • Total number of valence electrons in the system: Sum the number of valence electrons on all the atoms . Add the total negative charge if you have an anion. Subtract the charge if you have a cation. • Example: CO32-

  11. Step 2 • Number of electrons if each atom is to be happy: Atoms in our example will need 8 e (octet rule) or 2 e ( hydrogen). So, for the ex. • Step 3 • Calculate number of bonds in the system: Covalent bonds are made by sharing of e. You need 32 and you have 24. You are 8 e deficient. If you make 4 bonds ( with 2 e per bond) , you will make up the deficiency. Therefore, • # of bonds= ( e in step 2- e in step 1)/2 =(32-24)/2= 4 bonds

  12. Step 4 • Draw the structure: The central atom is C ( usually the atom with least electro negativity will be in the center). The oxygens surround it . Because there are four bonds and only three atoms, there will be one double bond. • Step 5 • Double check your answer by counting total number of electrons. http://www.youtube.com/watch?v=PoJoIbAJkvQ&feature=fvsr

  13. Level 1 Practice for Lewis Structurs Draw the Lewis structures for the following using above steps. Show work! • A.Cl2 • B. CH2Cl2 • C. NH3

  14. Level 2 practice on Lewis Structures SO42- HCN H2O2 CNS1-

  15. 8.8: Exceptions to the Octet Rule • Odd-electron molecules:Ex: NO or NO2 (involved in breaking down ozone in the upper atmosphere) • Incomplete octet: H2 He BeF2 BF3 NH3 + BF3 → NH3BF3 (Lewis acid/base rxn)

  16. Expanded octet: occurs in molecules when the central atom is in or beyond the third period, because the empty 3d subshell is used in hybridization (Ch. 9) PCl5 SF6

  17. 8.6: Formal Chargehttp://wps.prenhall.com/wps/media/objects/4974/5093961/emedia/ch09/FormalChargesMovie.html • For each atom, the numerical difference between # of valence e- in the isolated atom and # of e- assigned to that atom in the Lewis structure. To calculate formal charge: • Assign unshared e- (usually in pairs) to the atom on which they are found. • Assign one e- from each bonding pair to each atom in the bond. (Split the electrons in a bond.) • Then, subtract the e- assigned from the original number of valence e-. #VALENCE e- in free atom – #NON-BONDING e- – ½(#BONDING e-) FC

  18. Used to select most stable (and therefore most likely structure) when more than one structure are reasonable according to “the rules”. • The most stable: • Has FC on all atoms closest to zero • Has all negative FC on most EN atoms. • FC does not represent real charges; it is simply a useful tool for selecting the most stable Lewis structure.

  19. Examples: Draw at least 2 Lewis structures for each, then calculate the FC of each atom in each structure. SCN1- N2O BF3

  20. 8.7: Resonance Structures • Equivalent Lewis structures that describe a molecule with more than one likely arrangement of e- • Notation: use double-headed arrow between all resonance structures. Ex: O3 • Note: one structure is not “better” than the others. In fact, all resonance structures are wrong, because none truly represent the e- structure of the molecule. The “real” e- structure is an “average” of all resonance structures.

  21. : : O=O : : Bond Order • An indication of bond strength and bond length • Single bond: 1 pair of e- shared Ex: F2 Longest, weakest •• •• :F-F: •• •• • Double bond: 2 pairs of e- shared Ex: O2 • Triple bond: 3 pairs of e- shared Ex: N2 Shortest, strongest :N ≡ N:

  22. Bond Order & Resonance Structures • To determine bond orderwith resonance structures: • Determine the bond order at one position in one resonance structure, and add it to the bond orders at the same bond position in all other resonance structures. • Divide the sum by the number of resonance structures to find bond order.

  23. Examples: draw the Lewis structure and determine the bond S-O, C-C, and C-H bond orders SO3C6H6

  24. 8.9: Bond enthalpy: • DH/mol to break a particular bond of substance (g) Ex: CH4 (g) + Cl2 (g) → CH3Cl (g) + HCl (g) DHrxn = ? • 1 C-H & 1 Cl-Cl bond are broken (per mole) • 1 C-Cl & 1 H-Cl bond are formed (per mole) Hrxn ≈  (Hbonds broken) -  (Hbonds formed) Note: this is the “opposite” of Hess’ Law where Hrxn = DHproducts – Dhreactants http://yteach.com/page.php/resources/view_all?id=bond_enthalpy_dissociation_molecule_charge_resonance_energy_t_page_4&from=search http://www.webchem.net/notes/how_far/enthalpy/enthalpy_diagrams.htm

  25. Ex: CH4 (g) + Cl2 (g) → CH3Cl (g) + HCl (g) DHrxn = ? BondAve DH/molBondAve DH/mol C-H 413 Cl-Cl 242 H-Cl 431 C-Cl 328 C-C 348 C=C 614 Hrxn ≈  (Hbonds broken) -  (Hbonds formed) Hrxn ≈[(1(413) + 1(242)] – [1(328) + 1(431)] Hrxn ≈-104 kJ/mol Hrxn = -99.8 kJ/mol (actual) Note: 2 C-C ≠ 1 C=C 2(348) = 696 kJ ≠ 614 kJ

  26. Ex: CH4(g) + Cl2(g) → CH3Cl(g) + HCl(g) DHrxn=? *CH3(g) + H(g) + 2 Cl(g) Absorb E, break 1 C-H and 1 Cl-Cl bond Release E, form 1 C-Cl and 1 H-Cl bond H CH4(g) + Cl2(g) CH3Cl (g) + HCl(g) DHrxn Hrxn =  (Hbonds broken) +  (- Hbonds formed) Hrxn =  (Hbonds broken) -  (Hbonds formed)

  27. 23.5: Metallic bonding • Metallic elements have low I.E.; this means valence e- are held “loosely”. • A metallic bond forms between metal atoms because of the movement of valence e- from atom to atom to atom in a “sea of electrons”. The metal thus consists of cations held together by negatively-charged e- "glue.“ • This results in excellent thermal & electrical conductivity, ductility, and malleability. • A combination of 2 metals is called an alloy.

  28. Free e- move rapidly in response to electric fields, thus metals are excellent conductors of electricity. http://www.uwgb.edu/dutchs/EarthSC202Notes/minerals.htm Free e- transmit kinetic energy rapidly, thus metals are excellent conductors of heat. Layers of metal atoms are difficult to pull apart because of the movement of valence e-, so metals are durable. However, individual atoms are held loosely to other atoms, so atoms slip easily past one another, so metals are ductile.

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