AP CHEMISTRY: BONDING THEORIES REVIEW KEY p. 1

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1 AP CHEMISTRY: BONDING THEORIES REVIEW KEY p. 1 1) a) O-H PC b) Cs-Cl I c) H-Cl PC d) Br-Br NPC 2) differences in electronegativity determines amount of ity O3 0, P8 0, NO.5, CO2 1.0, CH4.4, H2S.4 answer (for those that contain covalent ) CH4 = H2S < NO < CO2 3) a) b) c) d) no dipole shared equally 4) a) b) c) answer b has zero dipole moment for the molecule since all the bond dipole moments will cancel out since the molecule is symmetrical. In KI, the overall dipole moment is the same as the bond. In H2Se, the overall dipole moment will point towards the Se as in the diagram. 19 Q1 Q2 19 ( 1)( 1) 5) E 2.31x 10 J nm 2.31x 10 J nm x J r nm x J ( 1 kj 1000 J x 1023 ) ( mol ) = -433 kj mol 6) a) 2C2H6 + 7O2 4CO2 + 6H2O C C 2(347 kj) = 694 kj C = 0 8(799 kj) = 6392 kj C H 12(413 kj) = 4956 kj O H 12(467 kj) = 5604 kj O = O 7(495 kj) = 3465 kj 9115 kj kj ANSWER: 9115 kj kj = kj b) HCN + 2H2 CH3NH2 C H 1(413 kj) = 413 kj C H 3(413 kj) = 1239 kj C N 1(891 kj) = 891 kj C N 1(305 kj) = 305 kj H H 2(432 kj) = 864 kj N H 2(391 kj) = 782 kj 2168 kj 2326 kj ANSWER: 2168 kj 2326 kj = -158 kj

2 AP CHEMISTRY: BONDING THEORIES REVIEW KEY p. 2 6) c) HC CH + H2 H2C=CH2 C C 1(839 kj) = 839 kj C H 4(413 kj) = 1652 kj C H 2(413 kj) = 826 kj C = C 1(614 kj) = 614 kj H H 1(432 kj) = 432 kj 2097 kj 2266 kj ANSWER: 2097 kj 2266 kj = -169 kj 7) The reactants undergo 3 changes: turn into a gas if not already, be single atoms if not already, and form the ion. step 1 sublimation of cation Na(s) Na(g) endothermic step 2 ionization of cation Na(g) Na + (g) endothermic step 3 vaporization of anion NA since F2 already gas endothermic step 4 dissociation of anion ½ F2(g) F(g) endothermic step 5 ionization of anion (electron affinity) F(g) F (g) exothermic (for 1 st e ) step 6 lattice forms (lattice energy) Na + (g) + F (g) NaF exothermic The reaction is exothermic because so the lattice energy is so large it overcomes all the endothermic processes itself. 8) Li(s) + ½ Br2(g) LiBr(s) for problem ionization energy for Li = +520 kj/mol +520 kj/mol electron affinity for Br = 342 kj/mol 342 kj/mol energy of sublimation for Li = +161 kj/mol +161 kj/mol lattice energy = 787 kj/mol 787 kj/mol bond energy of Br2 = +193 kj/mol ½ (+193 kj/mol) o H f = -352 kj/mol 9) The major difference between MgO and NaF when forming is the amount of lattice energy given off. The difference Q in lattice energy is related to 1 Q 2 where Q1 and Q2 are the ionic charges of the ions. With the (+2)(-2) for MgO r and the (+1)(-1) for NaF, the lattice energy of MgO is almost 4 times larger (the difference in r is insignificant for the problem as all other processes involving energy in the formation of the compounds). 10) a) MgO since Mg is smaller than Ca. b) BaCl2 since Ba has a larger charge than Na. 11) The higher the bond order, the higher the bond energy and the shorter the bond length. 12) a) b) O = C = O c) B.O. = 1 B.O. = 2 B.O. = 1⅓ 13) C, N, and O (sometimes P and S if bonded with C, N, or O)

3 AP CHEMISTRY: BONDING THEORIES REVIEW KEY p. 3 14) S has a 3d sublevel in its valence shell. Upon bonding, 2 of the electrons in S promote to the 3d sublevel giving it 6 half-filled orbitals to form 6. Thus it could have up to 12 electrons. C (and other period 2 elements) does not have a d sublevel in its valence shell to allow such promotion. Therefore, it can only promote within its 2s-2p sublevels (if possible) and have at most 8 electrons. 15) a) 1 b) 4 c) 1 d) 6 16) Elements form compounds to lower their potential energy and become more stable. 17) Resonance is the sharing of multiple throughout the molecule (the multiple are delocalized). It occurs whenever there are more than 1 possible Lewis structure for the molecule. 18) a) octahedral all 90 b) teeter totter axis < 90, equator < 120 c) tetrahedral all d) <109.5 (~107 ) 19) NH4 + and BF4 20) a) b) ) CONTINUED NEXT PAGE

4 AP CHEMISTRY: BONDING THEORIES REVIEW KEY p. 4 22) LEWIS STRUCTURE ATOM CN LP HYBRIDIZATION σ π ELECTRONIC MOLECULAR N 2 1 sp planar bent DRAWING OF POLARITY OF NO2 -O 1 3 sp tetrahedral end =O 1 2 sp planar end 23) IF3 I 3 2 dsp T-shape 24) ClO3 Cl 3 1 sp tetrahedral O 1 3 sp tetrahedral end 25) C 3 0 sp planar planar H2CO O 1 2 sp planar end 26) C 2 0 sp 2 2 linear linear ICN N 1 1 sp 1 2 linear end I 1 3 sp tetrahedral end 27) SnCl5 Sn 5 0 dsp bipyramdal Cl 1 3 sp tetrahedral end non 28) AsF5 As 5 0 dsp bipyramdal non 29) C6H6 C 3 0 sp planar planar non 30) 4 0 sp tetrahedral tetrahedral 3 0 sp planar planar H3CCOOH -O- 2 2 sp tetrahedral bent =O 1 2 sp planar end

5 AP CHEMISTRY: BONDING THEORIES REVIEW KEY p. 5 31) SeF4 LEWIS STRUCTURE ATOM CN LP HYBRIDIZATION σ π Se 4 1 dsp ELECTRONIC MOLECULAR teeter totter DRAWING OF POLARITY OF 32) CO2 C 2 0 sp 2 2 linear linear O 1 2 sp planar end non 33) P 5 0 dsp F must be on axis PCl3F2 34) AsF3 35) XeF4 36) ICl4 Cl 1 3 sp tetrahedral end As 3 1 sp tetrahedral Xe 4 2 d 2 sp octahedral square planar I 4 2 d 2 sp octahedral square planar Cl 1 3 sp tetrahedral end non non 37) S 3 1 sp tetrahedral H2SO3 -O- 2 2 sp tetrahedral bent -O 1 3 sp tetrahedral end 38) S 3 1 sp tetrahedral OSCl2 O 1 3 sp tetrahedral end Cl 1 3 sp tetrahedral end

6 AP CHEMISTRY: BONDING THEORIES REVIEW KEY p. 6 LEWIS STRUCTURE ATOM CN LP HYBRIDIZATION σ π ELECTRONIC MOLECULAR DRAWING OF POLARITY OF 39) 4 0 sp tetrahedral tetrahedral H2NCH2COCH3 3 0 sp planar planar O 1 2 sp planar end N 3 1 sp tetrahedral 40) NO2 C6H6 41) Hybrid orbitals form by the mixing of atomic orbitals when the molecule goes to bond. 42) They are degenerate they are alike in all ways but spatial orientation. 43) Lone pair of electrons require more room than bonded electrons. The equator positions on the molecule has more room than the axis positions. 44) The more electronegative elements have a tighter hold on their electrons and therefore require less room than the other elements of lesser electronegativity. Thus, these elements will take the axial positions since they need less room. 45) a) C2H2 b) C2H6 c) C2H6 d) C2H2 46) 20 σ, 8 π 47) Molecular orbitals form by the mixing of atomic orbitals upon the formation of a molecule. Orbitals are conserved the same number of molecular orbitals formed equal the number of atomic orbitals that make them. 48) Molecular orbitals are similar to atomic orbitals in many ways such as getting larger as the valence shell gets larger and that orbitals of the same level are degenerate. Two most important aspects are the molecular orbitals also can hold 2 electrons, with opposite spins, and the wave functions for the molecule indicates the electron probability for the orbitals just like atomic orbitals. 49) a) πp b) σs c) σp* d) σs e) πp* f) σs* 50) a) b) c) d) e) f)

7 AP CHEMISTRY: BONDING THEORIES REVIEW KEY p. 7 51) Bonding MO s have less energy than the atomic orbitals from which they come and antibonding MO s have more energy than the atomic orbitals from which they come. 52) Bonding MO s favor molecule formation since they have less energy than the original atomic orbitals. Since nature spontaneously goes in the direction of less energy, and therefore a more stable situation, the molecule will form. Antibonding MO s favor the separation of free atoms since they have more energy than the original atomic orbitals this means the formation of a molecule will not lead to a more stable substance and therefore won t happen on its own. 53) C2 O2 Cl2 Ne2 BO = 2 BO = 2 BO = 1 BO = 0 54) Can use either electron configuration or energy diagram to determine this. With the configurations, if the orbitals are full there are no unpaired e s whereas if they are not full then there is unpaired e (s). Li2 (σ2s) 2 diamagnetic N2 (σ2s) 2 (σ2s*) 2 (π2p) 4 (σ2p) 2 diamagnetic He2 + (σ2s) 2 (σ2s*) 1 paramagnetic H2 (σ1s) 2 diamagnetic O2 (σ2s) 2 (σ2s*) 2 (σ2p) 2 (π2p) 4 (π2p*) 2 paramagnetic Be2 (σ2s) 2 (σ2s*) 2 neither doesn t exist (bond order is 0) 55) N2 N2 + N2 N2 2 BO = 3 BO = 2.5 BO = 2.5 BO = 2 diagmagnetic paramagnetic paramagnetic paramagnetic 56) From the diagram above, N2 is diamagnetic and therefore is not attracted to a magnetic field. Liquid N2 will pour between the magnet s poles.

8 AP CHEMISTRY: BONDING THEORIES REVIEW KEY p. 8 57) CO (σ2s) 2 (σ2s*) 2 (π2p) 4 (σ2p) 2 BO = 3 diamagnetic CO (σ2s) 2 (σ2s*) 2 (π2p) 4 (σ2p) 2 (π2p*) 1 BO = 2.5 paramagnetic 58) B 1 C 2 N 3 O 2 F 1 Ne 0 59) IBr (σs) 2 (σs*) 2 (σp) 2 (πp) 4 (πp*) 4 BO = 1 diamagnetic 60) CN (σ2s) 2 (σ2s*) 2 (π2p) 4 (σ2p) 2 diamagnetic CN (σ2s) 2 (σ2s*) 2 (π2p) 4 (σ2p) 1 CN + (σ2s) 2 (σ2s*) 2 (π2p) 4 diamagnetic 61) THEORY STRENGTH WEAKNESS VSEPR determining shapes of molecules determining ity of molecules dealing with unpaired electrons and the properties derived from them dealing with delocalized electrons Hybrid Orbital Molecular Orbital explaining how are identical determining the type of bond between two atoms (sigma or pi) determining arrangement of electron pairs around an atom dealing with unpaired electrons and the properties derived from them dealing with delocalized electrons dealing with unpaired electrons determining the shapes of molecules beyond the basic shapes determining shapes of molecules and the properties derived from the shapes very complicated with polyatomic molecules

Name. CHM 115 EXAM #2 Practice KEY. a. N Cl b. N F c. F F d. I I e. N Br. a. K b. Be c. O d. Al e. S

Name. CHM 115 EXAM #2 Practice KEY. a. N Cl b. N F c. F F d. I I e. N Br. a. K b. Be c. O d. Al e. S Name CHM 115 EXAM #2 Practice KEY Circle the correct answer. (numbers 1-8, 2.5 points each) 1. Which of the following bonds should be the most polar? a. N Cl b. N F c. F F d. I I e. N Br 2. Choose the