O P O O. This structure puts the negative charges on the more electronegative element which is preferred. Molecular Geometry: O Xe O

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1 hemistry& 141 lark ollege Exam 4 olution 1. Draw the Lewis structures for the following molecules and ions. Include formal charges and resonance structures, where appropriate. Fill out the table for the central atom in each molecule. [18] The phosphite ion P e - total P This structure puts the negative charges on the more electronegative element which is preferred. teric number: 4 ybridization: sp 3 Ideal bond angle: Electronic Tetrahedral Molecular Pyramidal Xe 2 F 2 34e - total F Xe F Double bonds between Xe and eliminate all formal charge in this molecule. teric number: 5 ybridization: sp 3 d Ideal bond angle: 120 and 90 Electronic Trigonal bipyramidal Molecular ee-saw Exam 4 Fall 2008 Page 1 of 5

2 hemistry& 141 lark ollege 2. Two possible Lewis structures are shown for sulfine ( 2 ), which is a metabolite of sulfoxide drugs and is found in plasma: Formal harge = # valence e - - # lone pair electrons - # bonds a. ompute formal charges for the atoms and list them on structures directly. [4] b. f the two structures shown, which is better? Explain your selection. [4] The first structure is better, as the negative formal charge is on the more electronegative element. c. Draw a best Lewis structure for sulfine, for which all atoms have a formal charge of zero. [2] ulfur can have an expanded octet, which makes this structure possible! 3. omplete the Lewis structure for peroxyacetyl nitrate ( 2 3 N 5, PAN), which is a smog irritant. omplete the structure by filling in all remaining bonds, lone pairs and formal charges. ow many valence electrons should there be in the structure? [6] N N PAN actually exists as a pair of resonance structures. This set has the fewest formal charges. Number of Valence Electrons = 46 valence e - total Exam 4 Fall 2008 Page 2 of 5

3 hemistry& 141 lark ollege 4. The first ionization energy for phosphorus is 1012 kj/mol and the first ionization energy for sulfur is 1005 kj/mol. a. Does this follow the general trend for ionization energies? [2] No. The general trend is that ionization energies get bigger going across the periodic table. b. It is EAIER ARDER to remove an electron from phosphorus than from sulfur. [2] (circle one) c. Explain why the ionization energy for phosphorus is greater than the ionization energy for sulfur. [6] onsider the electron configurations for the parent atom and the -1 ion: P = [Ne]3s 2 3p 3 = [Ne]3s 2 3p 4 P -1 = [Ne]3s 2 3p 2-1 = [Ne]3s 2 3p 3 The ionization energy is the energy required to remove an electron from the valence shell, so the higher it is, the harder it is to remove the electron. Phosphorus has a beginning electron configuration that is semi-stable, as the p-orbitals are half-filled; removing an electron from phosphorus would decrease the stability of the system. owever, removing an electron from sulfur brings it to the half-filled, semi-stable state, so it is easier to remove the electron from sulfur. 5. Element 116 (discovered in 1999) is in group 6A and should have the electron configuration of [Rn]7s 2 5f 14 6d 10 7p 4. Element 118 (discovered in 2002) is in group 8A and should have the electron configuration of [Rn]7s 2 5f 14 6d 10 7p 6. a. Which atom should have the greater (more negative) electron affinity? Explain your response. [6] Element 116 should have the greater electron affinity, as it needs to gain two electrons to achieve a noble gas electron configuration. Element 118 is a noble gas and has a completely filled set of orbitals, so it does not want to gain an electron and little energy would be released if an electron were added to the system. b. Relative to elements 116 and 118, where would you expect the electron affinity of element 117 (not yet discovered) to fall? Explain your response. [4] Element 117 would have the highest electron affinity of all, as it needs only one electron to attain a noble gas configuration, so it would release a great deal of energy upon gaining an electron. Exam 4 Fall 2008 Page 3 of 5

4 hemistry& 141 lark ollege 6. Write the electron configurations for the following elements or ions. Give the number of valence electrons. Is the element paramagnetic or diamagnetic? If so, how many unpaired electrons are there? [10] onfiguration Valence electrons? Paramagnetic or diamagnetic? a) Br [Ar]4s 2 4p 5 7 Paramagnetic 1 Unpaired electrons? b) Mo [Kr]5s 1 4d 5 (Mo switches to have half-filled orbitals) 6 Paramagnetic 6 c) Zr +2 [Kr]4d 2 Remember to remove e - from the highest n level first! 2 Paramagnetic 2 d) P -3 [Ar] 8 Diamagnetic 0 7. An electron transitions from the n = 2 state to the n = 4 state in a B +4 ion. a. Does this process require energy or release energy? [1] b. What energy is associated with this transition, in kj/mol? [6]!E = E f - E i = E 4 - E 2 Energies are from the Bohr model " 1!E = -5 2 R - 1 % # $ & '!E = 1.02 x J x 1 kj 1000 J x x 1023 photons 1 mol = 6.15 x10 3 kj/mol Requires energy 6.15 kj/mol c. What wavelength of light (in m) is associated with this transition? [3]!E = hc " " = hc!e = hc 1.02 x J = 1.95 x 10-8 m 1.95 x 10-8 m d. What type/color of light is this? (e.g. gamma radiation, microwave radiation, purple light)? [2] Ultraviolet region Exam 4 Fall 2008 Page 4 of 5

5 hemistry& Which set of quantum numbers cannot occur to specify a specific orbital? [3] a. n = 2, l = 1, m l = -1 c. n = 3, l = 1, m l = 0 b. n = 3, l = 3, m l = 2 d. n = 4, l = 3, m l = 1 l can only be as large as n-l lark ollege 9. An electron in which orbital will be closer (on average) to the nucleus? [3] a. 2s b. 3s c. 2p d. 4s 10. Which type of orbital contains two axial nodes? [3] a. s b. p c. d d. f 11. All of these species have the same electron configuration (isoelectronic). Rank them in order increasing radius from smallest to largest. [4] F -, Ne, -2, Mg +2, Na + mallest Mg +2 < Na + < Ne < F - < -2 Largest ations are always smaller! 12. elect the orbital below that represents a p-orbital. [3] 13. onsider the two generic molecules below. In both situations, B is more electronegative than A. B A B B A B molecule A molecule B a. Is the A B bond polar? If so, which way does the bond dipole point? [3] Yes, since B is more electronegative than A, the bond is polar and the dipole points towards B. b. ne molecule is polar, and one molecule is non-polar. Which molecule is polar? Explain your choice. [5] Molecule A is non-polar, and molecule B is polar. Molecule A is linear, so the two bond dipoles cancel each other out, whereas the dipoles do not cancel due to the bent shape of molecule B. Exam 4 Fall 2008 Page 5 of 5

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