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Concepts of Chemical Bonding

Brown, LeMay Ch 8 AP Chemistry Monta Vista High School. Concepts of Chemical Bonding. 8.1: Types of “Inter-Atomic” or Intra molecular Bonding a.k.a bonding. Ionic : electrostatic attraction between oppositely charged ions. Ex. NaCl, K 2 SO 4. Generally solids.

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Concepts of Chemical Bonding

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

  2. 8.1: Types of “Inter-Atomic” or Intra molecular Bonding a.k.a bonding • Ionic: electrostatic attraction between oppositely charged ions. Ex. NaCl, K2SO4. Generally solids. • Covalent: sharing of e- between two atoms (typically between nonmetals). Ex. CO2, SO2. Generally gases and liquids • Metallic: “sea of e-”; bonding e- are relatively free to move throughout the 3D structure Ex. Fe, Al, generally solids IncreasingDiff. of EN 4. Covalent Network: large number of atoms/molecules bonded in a network through covalent bonding. Ex. SiO2, Si, Ge, Diamond, Graphite

  3. [ ]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. Electrostatic attraction (force) determines the strength of ionic bond, just like electro negativity difference determines the strength of covalent bond. • 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. Animation of Ionic/Covalent BondingAnimation of LiCl Crystal

  4. 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. Animation showing formation of a lattice

  5. The lattice energy of NaCl is the energy given off when Na+ and Cl- ions in the gas phase come together to form the lattice of alternating Na+ and Cl- ions in the NaCl crystal (in this case lattice energy is negative) or it may be defined as the amount of energy required to break 1 mole of solid NaCl into its ions in gaseous state (in this case lattice energy will be positive). • Na+(g) + Cl-(g)  NaCl(s) ΔHo = -787.3 kJ/mol

  6. The lattice energies for the alkali metal halides is therefore largest for LiF (smallest size, smallest d) and smallest for CsI ( largest size, largest d), as shown below. • Values of Lattice Energies of Alkali Metals Halides (kJ/mol) F- Cl-Br-I- Li+ 1036 853 807 757 Na+ 923 787 747 704 K+ 821 715 682 649 Rb+ 785 689 660 630 Cs+ 740 659 631 604 Data taken from Purdue website

  7. Understanding Check Ex: Which has a greater lattice energy? Why? NaCl or KClNaCl or MgS

  8. 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) (ionic vs. covalent bonding in youtube video) • Coordinate Covalent: Shared pair contributed by only one of the two sharing species. Ex. Lewis acids and bases

  9. Cl-Cl Bond in Cl2 molecule Animation of bonding in Chlorine Molecule

  10. 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.

  11. 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.

  12. Graphite Diamond SiO2

  13. The Octet Rule • Atoms tend to gain, lose, or share e- until they are surrounded by 8 e- in their outermost energy level (have filled s and p sub shells) and are thus energetically stable. • Exceptions do occur (and will be discussed later.) Visualizing atomic orbitals

  14. 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 • •

  15. 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. Cr2+, Cr3+ • Most lose e- to end up with a filled or a half-filled subshell. Ex. Cu+ ion

  16. 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-

  17. 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

  18. 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. Drawing Lewis Structures (youtube video)

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

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

  21. 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)

  22. 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

  23. 8.6: Formal Charge Movie on Formal Charge • 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

  24. 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.

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

  26. 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.

  27. : : 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:

  28. 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.

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

  30. 8.9: Bond enthalpy: • Amount of energy required to break a particular bond between two elements in gaseous state. Given in kJ/mol. Remember, breaking a bond always requires energy! • Bond enthalpy indicates the “strength” of a bond. • Bond enthalpies can be used to figure out Hrxn . 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 Bond Enthalpy link

  31. 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

  32. 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)

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