Ex. 1) F F bond in F = 0 < % covalent, no transfer of electrons

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1 #52 Notes Unit 7: Bonding Ch. Bonding I. Bond Character Bonds are usually combinations of ionic and covalent character. The electronegativity difference is used to determine a bond s character. Electronegativity Difference Covalent < 1.67 < Ionic shared electrons 50% ionic transferred electrons weaker bonds 50% covalent stronger bonds Ex. 1) F F bond in F = 0 < % covalent, no transfer of electrons Ex. 2) Na Cl in NaCl = 2.1 > 1.67 mainly ionic Na Cl is polar. There is a difference in charge. δ(+) δ(-) Na wants to lose electrons, Cl wants to gain electrons. Ex. 3) S Cl in SCl = 0.5 < 1.67 mainly covalent S and Cl both want to gain electrons, but Cl (higher electronegativity) wants electrons more than S. The bond is still polar, but not as much as NaCl. ** The farther apart the elements on the periodic table, the larger the electronegativity difference, the more ionic the bond, the more polar the bond and the stronger the bond. Most Polar (most ionic) Least Polar (least ionic) Na Cl > S Cl > F F (farthest apart on periodic table) (closest together) A) For ionic bonds the energy can be calculated, using Coulomb s Law. E = 2.31 X10-19 J nm (Q 1 Q 2 ) r Q 1,Q 2 = the charges of the ions r = the bond length For NaCl: E = 2.31 X10-19 J nm (+1)(-1) = X10-19 J (0.276 nm) the (-) shows that is an attractive force **The larger the charges, the stronger the bond, the greater the attraction.

2 B) But covalent bonds share electrons, so the Repulsion Energy must be measured. On the left: the H atoms are too close together and the nuclei will repel each other. ( High repulsion energy) On the right: the H atoms are too far apart and there is no attraction between atoms. (Zero repulsion energy) At the bond length of 0.73 Å: the H atoms are at the best distance, so that electrons are attracted to both nuclei and the electron repulsion is minimized. The atoms will bond. (negative repulsion energy would equal positive attraction energy) These bond energies are measured and found on the Bond Energy Table in textbook. When reactions occur energy is gained (+) by atoms to break bonds and released (-) by atoms as new more stable bonds form. This calculates ΔH for the reaction. ΔH reaction = Σ E bonds broken Σ E bonds formed reactants products 4 C H cancels 4 C H 1 C=C 2 C F 1 F F 1 C C ΔH reaction = [(1 C=C) ( 614 kj) + (1 F F ) ( 154 kj) ] [ (2 C F) (485 kj) + (1 C C ) ( 347 kj) ] = 614 kj kj 970 kj 347 kj = -549 kj favorable, since (-).

3 #53 Notes II. Lattice Energy (for ionic compounds) -is the change in energy that takes place when separated gaseous ions are packed together into a solid. (energy is released, exothermic). M + (g) + X - (g) MX (s) Lattice energy = k (Q 1 Q 2 ) r a modified version of Coulomb s Law **The most exothermic lattice energy =the most favorable = the largest attraction ** The largest attraction will have a more concentrated charge or a larger charge difference. Ex. 1a) Which has the most exothermic lattice energy? NaCl or CsCl NaCl: Na is a smaller ion, so its charge will be more concentrated, most favorable lattice energy. Ex. 1b) MgCl 2 or NaCl MgCl 2 : Mg 2+ Cl -1 has a larger charge difference (more concentrated) than Na +1 Cl -1 The bond energies of ionic compounds can be calculated without Coulomb s Law, using the lattice energy with other energies. Ex. 2) Find the bond energy of LiF. Li (s) + ½ F 2(g) LiF (s) Various energies will be provided. Li needs to be a gas: Li (s) Li (g) 157 kj/mol (enthalpy of sublimation) Li needs to be a (+) ion: Li (g) Li +1 (g) + e kj/mol (loses e -, ionization energy) F2 needs to be broken to F: F 2(g) 2 F (g) X 1/2 152 kj/mol 2 (bond energy) F needs to be a (-) ion: F (g) + e - F -1 (g) -328 kj/mol (gains e -, electron affinity) The ions need to form a compound: Li +1 (g) + F -1 (g) LiF (s) kj/mol (lattice energy) Li (s) + ½ F 2(g) LiF (s) -611 kj/mol

4 III. Localized Electron Model A molecule is composed of atoms that are bound together by sharing pairs of electrons, using the atomic orbitals of the bound atoms. Bonding Pairs: pairs of e - found in the space between atoms Lone Pairs: nonbonding pairs of e - found elsewhere on an atom A) Lewis Structures -show how the valence (outer shell) electrons are arranged around the atoms in a molecule. Octet Rule: atoms try to achieve a noble gas configuration of 8 electrons. Ex. 1) C 1s 2 2s 2 2p 2 4 e - in outer layer (valence e - ) O 1s 2 2s 2 2p 4 6 outer e- O 2-6 outer e e - for the charge (2-) Cl Cl -1 Na Na +1 Mg Mg 2+

5 #54 Notes B) VSEPR Model (Valence Shell Electron Pair Repulsion) The structure around an atom is determined principally by minimizing electron pair repulsions. Ex. 1) NaCl linear, 180 o Ex. 2) MgBr 2 linear, 180 o Ex. 3) AlCl 3 trigonal planar, 120 o Ex. 4) CH 4 tetrahedral, o

6 Ex. 5) PCl 3 trigonal pyramidal, 107 o Ex. 6) H 2 O bent, o Ex. 7) ClO 4 - tetrahedral, o Oxygen will take two e - to have a noble gas configuration, O is greedy! Ex. 8) CO 2 linear, 180 o

7 Ex. 9) NaCN linear, 180 o

8 #55 Notes B) continued Ex. 10) PBr 5 trigonal bipyramidal, 90 o & 120 o Ex. 11) SF 4 **e - must be around the middle seesaw, <90 o & < 120 o (irregular tetrahedran, bisphenoid) Ex. 12) SF 6 octahedral, 90 o

9 Ex. 13) ICl 3 T-shaped, <90 o (e - must be around middle) Ex. 14) ClF 5 square pyramidal, <90 o Ex. 15) XeCl 4 square planar, 90 o ** e- must be across from each other linear, 180 o

10 IV. Resonance -is when more than one Lewis structure is possible. NO 3 - trigonal planar, 120 o The electrons in the bonds are delocalized, making this more stable than all single bonds. **Each N O bond is 2/3 single and 1/3 double in character.

11 #56 Notes V. Formal Charge Formal Charge = # valence electrons on lone atom (outer e - from the periodic table) - # valence electrons on the atom in the compound ( all lone electrons plus one electron per bond) N: F.C.= 5 e e - = +1 (5e - from periodic table 4 e - { 1 e- per bond}) single bond O: F.C. = 6e - - 7e - = -1 (6e - per. table {1 e - from bond + 6 lone e - }) double bond O: F.C. = 6e - - 6e - = 0 (6e - per. table {2 e - from bonds + 4 lone e - }) Don t count the stick, it only shows position of the electrons on P. (only 2 e - total) P F.C. = 5 e - - 5e - = 0 (5e - per. table {3 e - from bonds + 2 lone e - }) Official Lewis Dot Diagrams only show electrons (no bonds):

12 VI. Polarity Polar means (+)/(-) ends on the molecule. The ends are different! ** unsymmetrical (unbalanced) = {(+) and (-) ends to the molecule} = polar **symmetrical (balanced, it will not have (+)/(-) ends, the ends are the same) = nonpolar [ 2 prefixes not together ( un & non) ] (see notes #54-55 for structures) Ex. 1) NaCl polar Ex. 2) MgBr 2 nonpolar Ex. 3) AlCl 3 nonpolar Ex. 4) CH 4 nonpolar Ex. 5) PCl 3 polar Ex. 6) H 2 O polar - Ex. 7) ClO 4 nonpolar Ex. 8) CO 2 nonpolar Ex. 9) NaCN polar Ex. 10) PBr 5 nonpolar Ex. 11) SF 4 polar Ex. 12) SF 6 nonpolar Ex. 13) IBr 3 polar Ex. 14) ClF 5 polar Ex. 15) XeCl 4 nonpolar

13 #57 Notes Ch. Bonding Orbitals I. Hybridization (a mixture of orbitals) -is a modification to the localized electron model to account for the observation that atoms will continue and modify orbitals when bonding. Compound # of bonds orbitals used hybrid Ex. 1) H 2 1 s none H H H _ The electron from the other H can go into this 1s orbital to make the bond. Ex. 2) BeCl 2 2 s, p sp Be 2s 2p Be _ 2s 2p Now that the electrons are separated, the 2 Cl s can each move in one electron to make a bond. Two equivalent orbitals (1/2 s, 1/2 p). sp sp Ex. 3) BF 3 3 s, p, p sp 2 B 2s 2p B _ _ 2s 2p Now the electrons are separated and each F can move in one electron to make a bond. sp 2 Three equivalent orbitals (1/3 s, 2/3 p).

14 Ex. 4) CH 4 4 s, p, p, p sp 3 C _ 2s 2p C _ _ _ _ 2s 2p Now each H can move in one electron to bond. sp 3 4 equivalent orbitals (1/4 s, 3/4 p). Ex. 5) NH lone pair s, p, p, p sp 3 N _ _ _ 2s 2p Each H can move in one electron (2 e - in the 2s are the lone pair of electrons) to bond. sp 3 4 equivalent orbitals (1/4 s, 3/4 p). Ex. 6) H 2 O lone pairs s, p, p, p sp 3 O _ _ 2s 2p Each H can move in one electron to bond.

15 sp 3 4 equivalent orbitals (1/4 s, 3/4 p). Ex. 7) PF 5 5 s, p, p, p, d dsp 3 P _ _ 3s 3p 3d P _ _ _ _ 3s 3p 3d Now each F can move in an electron to bond. dsp 3 5 equivalent orbitals (1/5 s, 3/5 p, 1/5 d). Ex. 8) SF 6 6 s, p, p, p, d, d d 2 sp 3 S _ 3s 3p 3d S _ _ _ _ _ 3s 3p 3d Now each F can move in one electron to bond. d 2 sp 3 6 equivalent orbitals (1/6 s, 3/6 p, 2/6 d). =1/6 s, 1/2 p, 1/3 d On fingers do: SF 4 (4 bonds and 1 e - pair = 5 positions/orbitals needed), so dsp 3, XeF 4 (4 bonds and 2 e - pairs = 6 positions/orbitals needed), so d 2 sp 3

16 #58 Notes II. Multiple Bonds and Hybridization Double Bond = 1 σ (sigma) bond & 1 π (pi) bond {stronger than single bond} 1 hotdog (σ) + 1 bun (2 parts of π bond) Triple Bond = 1 σ (sigma) bond & 2 π (pi) bonds {stronger than double bond} 1 hotdog (σ) + 2 buns (π) **They become successively stronger, less flexible, and more reactive. III. Molecular Orbital Theory -is a way of determining whether a compound will have single, double or triple bonds. A) Homonuclear Diatomic Molecules 1) Simple Molecules with s-orbitals only. (H 2, He 2 )

17 Ex. 1) H 2 The 2 old orbitals will make 2 new orbitals. (1 bonding and 1 anti-bonding) Each H has 1 e - in the 1s orbital, so H 2 has 2 e - total. Bond Order = # bonding e - - # anti-bonding e - = 2 0 = 1 favorable (single bond) 2 2 Any Bond Order > 0 is favorable (0 is non-bonding). Ex. 2) He 2 Each He has 2 e - in the 1s, so He 2 has a total of 4 e -. Ex. 3) H 2 - B.O. = 2-2 = 0 unfavorable, no bonding 2 Each H has 1e - in the 1s, so H 2 has 2 e -. The (-) tells us to add 1 e -, so 3 e - total. B.O. = 2-1 = 1/2 favorable 2

18 2) Molecules with p-orbitals. Ex. 1) O 2 Each O has 2 e - in the s orbital and 4 e - in the p orbital. So O 2 has 4e - in the s orbital and 8 e - in the p orbital. B.O. = 2-2 = 0 Non bonding 2 Since the s-orbitals will always be full and the bond order will be zero, it can be ignored. For the p-orbitals: each O has 3 p-orbitals, making a total of 6 p-orbitals. These will make 6 new orbitals. Remember: Triple Bond has 1 σ & 2 π Each O has 4 e - in the p, so 8 e - total. B.O. = 6-2 = 2 favorable (double bond) 2

19 Ex. 2) O 2 2- Each O has 4 e - in the p, so O 2 has 8 e - in the p, but the (2-) is 2 extra electrons = 10 e - B.O. = 6-4 = 1 favorable 2 Ex. 3) N 2 2+ Each N has 3e - in the p, so N 2 has 6e - in the p, but the (2+) is 2 fewer electrons = 4 e - B.O. = 4-0 = 2 favorable 2 *End of Notes* (Assignments #59-60 are Review Assignments. There are no notes for these assignments.)

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