5.111 Second Hour Exam - Answer Key

1. (30 points) Lewis structures and VSEPR theory Draw the most stable Lewis structure for each of the following molecules, subject to the information given for each. Be sure to include the lone pairs and, if applicable, draw any resonance forms that are equal in energy. Indicate any nonzero formal charges. (a) (i) (6 points) Draw the Lewis structure of POCl3. Include any relevant resonance forms, and indicate any nonzero formal charges. 32 valence electrons 40 electrons needed to fill valence shells 8 bonding electrons after valence shell expansion, 10 bonding electrons and 22 lone pair electrons 24 lone-pair electrons (eliminating all formal charges) O Cl P Cl Cl (ii) (2 points) Name the geometry around the phosphorus atom. tetrahedral (b) (8 points) Draw the Lewis structure of (NCNH)-1 (atom order as indicated). Include any relevant resonance forms, and indicate any nonzero formal charges. 16 valence electrons 26 electrons needed to fill valence shells 10 bonding electrons 6 lone-pair electrons -1 N C N H -1 N C -1 Formal charges are circled on atoms where the FC is not zero. N -1 H (c) (i) (6 points) Draw the Lewis structure of (SO3)-2. Include any relevant resonance forms, and indicate any nonzero formal charges. 26 valence electrons 32 electrons needed to fill valence shells 6 bonding electrons after valence shell expansion, 8 bonding electrons and 18 lone pair electrons 20 lone-pair electrons (eliminating all formal charges) -2 O S -2 -2 O O -1 -1 S O O S -1 O O -1 -1 O -1 O (ii) (2 points) Name the geometry around the sulfur atom. trigonal pyramidal (iii) (2 points) Circle the one value that best describes the O-S-O bond angle in (SO3)-2. < 90°; 90°; > 90°; < 109.5°; 109.5°; > 109.5°; < 120°; 120°; > 120°; < 180°; 180°; > 180° (iv) (2 points) Is (SO3)-2 a polar or a non-polar molecule? polar 2. (12 points) Ionic bonds KF has an ionic bond with a bond length of 0.217 nm. Calculate the ∆E, in kJ/mol, for the formation of a KF bond from the neutral atoms K and F. For this calculation, assume that the potassium and fluorine ions are point charges. IE and EA information for K and F is provided in the table below. Ionization energy (kJ/mol) Electron affinity (kJ/mol) potassium (K) 418 48 fluorine (F) 1680 328 Overall: K + F → KF First calculate the ΔE for the formation of an ionic bond from the two ions: K+ + F- → KF U(r) = z1z2e2 = 4πεor (-1) (1) (1.602 x 10-19 C)2 4π (8.854 x 10-12 C2J-1m-1)(0.217 x 10-9 m) U(r) = -1.063 x 10-18 J U(r) = -1.063 x 10-18 J x __ kJ___ x _6.022 x 1023 = -640.1 kJ/mol 1000 J mol U(r) = -640.1 kJ/mol ΔEtotal = IEK - EAF + U(r) K → K+ + e- ≡ IEK 418 kJ/mol F + e- → F- ≡ -EAF -328 kJ/mol K+ + F- → KF ≡ U(r) -640.1 kJ/mol K + F → KF -550.1 kJ/mol -550. kJ/mol or -5.50 x 10-2 kJ/mol 3. (20 points) Hybridization (a) (12 points) The structure of the amino acid histidine is provided below. For the indicated bonds, a-c, write the symmetry of each bond, and give the hybrid or atomic orbitals (with their principal quantum numbers) that overlap to form each of the bonds. Where appropriate, include the x, y, or z designations with the orbitals. a π(N2px, C2px) NH C H2C C H b CH O H2N N-C bond a: σ(N2sp2, C2sp2) π(N2py, C2py) or CH N N-C bond b: σ(N2sp3, C2sp3) c C H O-H bond c: σ(O2sp3, H1s) O (b) (8 points) (i) For the molecule below, indicate the symmetry in the C-Cl bond (labeled a), and give the hybrid or atomic orbitals (with their principal quantum numbers) that overlap to form the bond. If appropriate, include the x, y, or z designations with the orbitals. a F2C C Cl C-Cl bond a: σ(C2sp2, Cl3pz) C Cl (ii) Do the chlorine atoms in the F2C=C=CCl2 molecule above lie in the same plane as the fluorine atoms or in a perpendicular plane to the fluorine atoms? Briefly explain your answer (with words or a picture). perpendicular plane. The p-orbitals in the two sp2-hybridized carbon atoms are perpendicular to each other (py and px), meaning the F-C-F bonds and the Cl-C-Cl bonds also lie in perpendicular planes. F F Cl Cl going into and out of the page 4. (27 points) Molecular orbital theory (a) (21 points) (i) (9 points) Draw an energy correlation diagram for the molecular orbitals of the valence electrons in CN. Label the atomic and molecular orbitals, including the x, y and z designations where appropriate. The relative ordering of the energies of the states must be correct. Use the full space available to spread out your energy levels so that the labels for the orbitals fit easily. σ2pz* π2px* π2py* E 2px 2py 2pz 2pz 2py 2px σ2pz π2px π2py σ2s* 2s 2s σ2s C N (ii) (2 points) Of the CN molecular orbitals occupied by valence electrons, name the orbitals that have a nodal plane along the internuclear (bond) axis. π2px and π2py (iii) (4 points) Determine the bond order of the cyanide molecule, CN, and the cyanide ion, CN-1. ½(# of bonding electrons-# of anti-bonding electrons) BO of CN: ½(7-2) = 2.5 BO of CN-1: ½(8-2) = 3 (iv) (4 points) Below is an energy diagram of the CN covalent bond in a neutral CN molecule. On the same graph, plot the energy vs. internuclear distance, r, of the CN covalent bond in a CN-1 ion. Indicate the equilibrium bond distances with arrows. The relative values of the bond distances and energies must be correct, but no numbers are needed. Note: this question is graded based on your answer to part (iii), since it depends on bond order. Energy 0 CN internuclear distance, r CN-1 (v) (2 points) Which of the following are radical species: CN, CN-1, both, or neither? CN is the only radical species. (b) (6 points) Write the valence electron configuration for O2. (12 valence electrons) (σ2s)2(σ2s*)2(σ2pz)2(π2px)2(π2py)2(π2px*)1(π2py*)1 5. (11 points) Thermochemistry (a) (7 points) Consider the reaction below for the conversion nitrogen dioxide to nitric oxide and O2. 2NO2(g) → 2NO(g) + O2(g) ΔHf° (kJ/mol) NO2(g) +33.18 NO(g) +90.25 Calculate ΔHo (per mol of O2 formed) for the reaction at 298 K. ΔHr° = ΣΔHf°(products) - Σ ΔHf°(reactants) ΔHr° = [2ΔHf°(NO) + ΔHf°(O2)] - [2ΔHf°(NO2)] ΔHr° = [2(90.25 kJ/mol) + (0)] – [2(33.18 kJ/mol)] ΔHr° = 114.1 kJ or 114.14 kJ (also ok to have kJ/mol, since the questions says per mol of O2) (b) (4 points) Using the table of mean bond enthalpies provided, predict the bond enthalpy (in kJ/mol) for the CO bond marked with an arrow in the molecule below. -1 O H C O Note: the C-O bond is a 1½ bond (not a single bond) due to resonance. 360 kJ/mol < bond enthalpy < 743 kJ/mol It is also fine if a student calculated a number, as long as it is above 400 and less than 700 kJ/mol (ie. 552 kJ/mol). Bond C-H C-C C=C C-O C=O Mean Bond Enthalpy (in kJ/mol) 412 348 612 360 743 c = 2.99792 x 108 m/s h = 6.62608 x 10-34 J s Na = 6.02214 X 1023 mol-1 1 eV = 1.60218 x 10-19 J me = 9.10939 x 10-31 kg e = 1.60218 x 10-19 C U(r) = (z1z2e2)/(4πε0r) ε0 = 8.8542 x 10-12 C2/(Jm) Electronegativity = (IE + EA)/2 ΔG = ΔH - TΔS Image by MIT OpenCourseWare 9