3. The energy for a H H bond is 436kJ/mol. What is the smallest frequency of light capable of breaking a H H molecule?

Transcription

1 EMITRY 103 Practice Problems #3 hapters 57 (7.17.5) Do the topics appropriate for your lecture Prepared by Dr. Tony Jacob (Resource page) uggestions on preparing for a chemistry exam: 1. rganize your materials (quizzes, notes, etc.). 2. Usually, a good method to prepare for a chem exam is by doing lots of problems. Rereading a section of a chapter is fine, but rereading entire chapters takes up large amounts of time that generally is better spent doing problems. 3. ld exams posted by your instructor should be completely worked though. ld exams give you a sense of how long the exam will be, the difficulty of the problems, the variability of the problems, and the style of your instructor. Quizzes written by your instructor are also a good resource. You might not necessarily want to do these problems in the order written. Good Luck! APTER 5 1. Which of the following statements is incorrect? a. The photoelectric effect occurs when light with sufficient frequency causes electrons to be ejected from a metal. b. The Bohr model predicted the spectrum only of hydrogen atoms. c. Electrons can absorb or emit energy and will then move from one permitted energy level to another. d. Light has properties of both waves and particles but moving bodies show only properties of particles. e. Bohr postulated that electrons in orbits were restricted to specific quantized energy levels. 2. A photon of light has a frequency of 5.50Mz. a. What is the energy (J) of this light? b. What is the wavelength (m) of this light? c. What is the energy of one mole of this light? 3. The energy for a bond is 436kJ/mol. What is the smallest frequency of light capable of breaking a molecule? 4. A green laser emits light with a wavelength of 532nm. ow many photons are needed to obtain a total energy of 1.15mJ? 5. A laser emits light with a wavelength of 634nm. ver a 50.0s period a detector found the total energy from the laser was 0.270J. ow many photon per second is the laser emitting? 6. Given that the Bohr radius is 0.529Å if the radius of the electron orbit is 52.9Å what energy level does the electron reside on? a. 1 b. 2 c. 5 d. 10 e What is the change in energy if an electron in the 6th energy level falls to the 2nd energy level in a atom? a. 1 / 3 Rhc b. 2 / 9 Rhc c. 1 / 4 Rhc d. 1 / 36 Rhc e. 1 / 32 Rhc 8. ow much energy does it take to ionize one mole of atoms if all the electrons reside in the 5 th energy level? 9. An electron in the 4th energy level can undergo 6 different transitions of which 5 are shown below. Which transition shown below will have the longest wavelength? a. 4 3 b. 4 2 c. 4 1 d. 3 1 e. 2 1 (nap time)

2 10. If an e has a wavelength of 1.00 x 10 2 nm associated with it (mass = 9.11 x g), how fast is it traveling? a x 10 9 m/s b x 10 6 m/s c x 10 3 m/s d e x 10 3 m/s 11. If an atom moving 7.50 x 10 5 m/s has a wavelength of 9.53 x m associated with it, what element is it? a. Re b. e c. e d. Ne e. Be 12. If incident light of 275nm wavelength is used in a photoelectric effect experiment with titanium that has a work function of 6.94 x J, what is the kinetic energy of the emitted electrons? a. 0J b x J c x J d x J e. need more info 13. I. Which of the following sets of quantum numbers (n, l, m l, m s ) is a valid set of quantum numbers? a. (2, 3, 2, 1 / 2 ) b. (4, 4, 2, 1 / 2 ) c. (4, 3, 4, 1 / 2 ) d. (4, 3, 2, 1 1 / 2 ) e. (7, 3, 2, 1 / 2 ) II. What is a possible set of quantum numbers (n, l, m l, m s ) for the last electron added to In in the ground state? a. (5, 2, 2, 1 / 2 ) b. (5, 1, 2, 1 / 2 ) c. (5, 1, 1, 1 / 2 ) d. (4, 1, 1, 1 / 2 ) e. (3, 0, 0, 1 / 2 ) 14. What subshell is described by n = 5 and l = 2? a. 2h b. 5h c. 5p d. 5s e. 5d 15. Which statement below is correct? a. The 2s subshell will penetrate closer to the nucleus than the 2p subshell. b. A maximum of 7 electrons can be placed into the fsubshell. c. The spin quantum number, m s, can take on values of ±1. d. The subshell that fills after 4s is 4p. e. The probability of finding an electron in the 1s subshell is zero at a point far away from the nucleus. 16. Which of the following statements about the 4p subshell is incorrect? a. The number of spherical nodes is 2. b. The 4p subshell is higher in energy than the 4d subshell. c. The number of surface (planar) nodes is 1. d. There are 3 orbitals within the subshell. e. The value of m l for the 4p x orbital can be I. What is the maximum #electrons that can have these quantum numbers (not a multiple choice question). a. n = 4 b. n = 5, m s = +1/2 c. n = 3, l = 2 d. n = 2, l = 1 e. n = 4, l = 3, m s = 1 / 2 II. What is the maximum #orbitals that can have these quantum numbers (not a multiple choice question). a. n = 3 b. n = 4, l = 2 c. n = 5, l = 1 d. n = 5, l = 5 e. n = 3, l = 2, m l = Which subshell penetrates closest to the nucleus? a. 3p b. 2p c. 3d d. 2s e. 4d 19. Which orbital below represents the p z orbital? a. b. c. d. e. (coffee time)

3 20. a. What neutral atom has the electron configuration in the ground state: 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 6? b. What 2 ion has the electron configuration in the ground state: 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 4? c. What +2 ion has the electron configuration in the ground state: 1s 2 2s 2 2p 6 3s 2 3p 6 3d 5? 21. Which ion/molecule is isoelectronic with P 3? a. 3 b. N 3 c. 4 d. 2 e. none of the above 22. Which ion/molecule is diamagnetic? a. b. Ga +3 c. u d. +2 e. none are diamagnetic 23. Draw the orbital box electron configuration diagram for e in the ground state. 24. Which orbital box diagram corresponds with o +2 in the ground state? nly electrons past [Ar] are shown. a. b. c. d. e. 25. Which atom or ion in the ground state has the following electron configuration, 1s 2 2s 2 2p 6 3s 2 3p 6 3d 5? a. Mn b. V c. Ti +2 d. Mn +2 e. o Which atom or ion is most paramagnetic in the ground state? a. Ni +2 b. V +2 c. a d. b +5 e. Ne 27. Place the following atoms in order of decreasing atomic radii: l, e, Ge, n a. e > Ge > n > l b. n > Ge > e > l c. l > e > Ge > n d. l > Ge > e > n e. None of the above are correct. 28. ort these atoms/ions from smallest to largest in size (i.e., atomic radii)., e 2, 2,, Ne (smallest) < < < < (largest) 29. I. Elements increase in atomic radii (size) going down a column of the periodic table. Explain why this occurs. II. Elements increase in atomic radii (size) going right to left on a row of the periodic table. Explain why this occurs. III. The 1st ionization energy increases going left to right on a row of the periodic table. Explain why this occurs. 30. Which atom or ion is smallest? a. b 3 b. Te 2 c. I d. Xe e. Xe Which of the following reactions corresponds to the third ionization energy reaction for bromine? a. 2 (g) + e 3 (g) b. 3 (g) (g) + 3e c. (g) +3 (g) + 3e d (g) +3 2 (g) + e e. +2 (g) +3 (g) + e 32. Rank the atoms from lowest to highest second ionization energy? a, Rb, K a. a < Rb < K b. Rb < a < K c. Rb < K < a d. K < Rb < a e. a < K < Rb (eat an apple)

4 33. Place the following atoms, N,,, in order of increasing first electron affinity (more exothermic/more negative values on the right). a. N < < b. N < < c. < < N d. < < N e. < N < 34. Which atom has the lowest third ionization energy? a. b. c. Mg d. P e. all have same IE3 35. Which atom has the highest next ionization energy? a. Al +2 b. Al c. Na + d. N e. B Which reaction represents the first electron affinity for fluorine? a. 2 (g) + e 2 (g) b. (g) + (g) + e c. 2 (g) + 2e 2 (g) d. (g) + e 2 (g) e. (g) + e (g) 37. Which material would be expected to be the least dense? a. sodium(s) b. aluminum(s) c. magnesium(s) d. need additional info 38. Which process is the most exothermic? a. e(g) e + (g) + e b. e +2 (g) e +3 (g) + e c. (g) + e 2 (g) d. (g) + e (g) e. Boiling 100g 2 (l) at Which atom would most likely align with the following ionization energies: IE 1 = 578kJ/mol IE 2 = 1817kJ/mol IE 3 = 2745kJ/mol IE 4 = 11,577kJ/mol a. Li b. Mg c. d. Al e. none of the above 40. Which reaction represents the lattice energy reaction for All 3 (s)? a. Al(s) + 3 / 2 l 2 (g) All 3 (s) b. Al(g) +2 Al(g) +3 + e and l(g) + e l(g) c. Al +3 (g) + 3l (g) All 3 (s) d. All 3 (s) Al +3 (aq) + 3l (aq) e. none of the above APTER Draw the Lewis dot structures for the following molecules. a. N 3 b. 3 3 c. P 3 3 d. l 2 e. 2 f. N 2 g. 2 l 2 h. N 2 i. 2 2 j. l 2 k. 3 2 l. N 3 m. K 2 n. a o. B p. q. B 3 r. Be 2 s. (not ) t. I 4 u. Kr 4 +2 v. I Which molecule violates the octet rule? a. B 4 b. il 4 c. AsI 3 d. 4 e. none 43. onsider the following covalent single bonded pairs of atoms. Which bond will be the shortest? a. N b. P l c. N l d. P e. none of the above 44. Which of the following molecules will have resonance structures? i. N 2 ii. 2 3 iii. N 3 iv. 3 a. ii b. ii and iii c. ii, iii, and iv d. i e. iv (nap time)

5 45. onsider the following bond lengths: single bond: 1.43 Å double bond: 1.23 Å triple bond: 1.09 Å In the carbonate ion, 3 2, a reasonable average bond length would be: a Å b Å c Å d Å e Å 46. What is the formal charge on the in 3? a. 2 b. 1 c. 0 d. +1 e I. onsider the following list of compounds. Rank the molecules from smallest to largest N bond length. N 2 N 3 N + N a. N + < N < N 2 < N 3 b. N 3 < N 2 < N < N + c. N 3 < N + < N 2 < N d. N + < N < N 2 = N 3 e. N 3 = N 2 < N < N + II. Using the same compounds above, which compound will have the greatest N bond enthalpy? 48. As indicated by Lewis structures which molecule could not exist? If all can exist then select choice e. a. P 5 b. N 2 2 c. Pl 3 d. N 5 e. all can exist 49. Identify each of the following bonds as nonpolar covalent, polar covalent, or ionic. a. I b. c. ii d. a e. 50. Which of the following statements about the bond in 2 is correct? a. The bond is nonpolar and there is a partial negative charge on the. b. The bond is nonpolar and there is a partial negative charge on the. c. The bond is polar and there is a partial negative charge on the. d. The bond is polar and there is a partial negative charge on the. e. The bond is ionic and there is a negative charge on the. 51. Which pairs of atoms will form the most ionic compound? a. nitrogen and oxygen b. chlorine and fluorine c. oxygen and oxygen d. sodium and oxygen e. phosphorus and oxygen 52. hown below are four possible Lewis dot structures for 2 3 without resonance structures drawn. Decide which structure is best based on formal charges. Explain your answer. 2 I II III IV 53. Draw the 3 Lewis dot resonance structures for thiocyanate, N ( is in the middle). Based on formal charges, which resonance structure would be the better structure? The electronegativity values (χ) are: ΕΝ = 2.3, ΕΝ = 2.4, and ΕΝ N = 3.0. (take a walk)

6 54. I. In the molecule shown, 2, choose the structure with the correct locations of the δ + or δ symbols. δ + δ δ δ b. δ δ + δ δ δ c. δ + δ + a. d. II. Would the bond be nonpolar covalent, polar covalent, or ionic? δ δ δ δ+ δ+ 55. alculate the change in enthalpy for the reaction using the bond energies (kj/mol) listed below Bond Bond enthalpy (kj/mol) Bond Bond enthalpy (kj/mol) 356 = = What is the l l bond dissociation energy given the bond energies (kj/mol) listed below. 2 (g) + 2l 2 (g) l 2 (g) + 2l(g) Δ rxn = 208kJ Bond Bond enthalpy (kj/mol) Bond Bond enthalpy (kj/mol) Bond Bond enthalpy (kj/mol) = l l 431 = 695 a. 484kJ b. 242kJ c. 26.5kJ d. 53.0kJ e. 208kJ APTER A central atom with 2 lone pairs and 3 bonding pairs of e will have a molecular geometry of: a. linear b. triangular pyramidal c. triangular planar d. Tshape e. triangular bipyramidal 58. In Lewis dot structures, which electron interactions repel the most? a. bonding pair e bonding pair e b. bonding pair e lone pair e c. lone pair e lone pair e d. since these are all electrons there is equivalent repelling 59. Which of the following compounds have tetrahedral molecular geometry? a. Pl 4 b. e 4 +2 c. I 4 + d. Xe 2 2 e. all do 60. Which of the following statements is false? a. A molecule with individual polar bonds can be nonpolar. b. A molecule with a central atom with 2 atoms and 2 lone pairs of electrons around it is a bent molecule. c. Repulsion between core electrons pairs is used to determine the shape of the molecule. d. The angle between two atoms in 2 e is expected to be less than e. Ammonia, N 3, will absorb microwave radiation. 61. Which of the following molecules are planar (i.e., which have a molecular geometry that is flat)? I. Bl 2 II. Nl 3 III. 2 e IV. e 4 V. Xe 4 a. I b. I, II, IV c. I, III d. I, III, V e. I, III, IV, V (eat a nicker s bar)

7 62. Which structure best represents the molecular geometry for 4 +? a. b. c. d. e. Images: Public Domain from Wikipedia.org ( 63. or each molecule below, draw the Lewis dot structure (LD). Draw the electron region geometry (ERG), and under the ERG name the ERG. Draw the molecular geometry (MG), and under the MG name the MG. In a separate diagram, draw the vectors representing the bond dipoles, and draw the net vector representing the net dipole if the molecule is polar otherwise write no dipole if the molecule is nonpolar and there is no net vector. a. 2 b. N 3 c. Xe 2 d. P ow many of these molecules have a tetrahedral electron region geometry around the central atom? l 4 4 il 4 2 ei 2 Kr 2 a. 1 b. 2 c. 3 d. 4 e Which molecule will have an angle of ~120? a. N 3 b. 6 c. 4 d. e 4 e. none 66. Which molecule has a net molecular dipole moment? a. P 4 + b. Gel 4 2 c. il 4 d. Xe 4 e. none 67. Which molecule will heat up in a microwave oven? a. I 3 b. 6 c. Xe 6 +2 d. I 4 e. none 68. There are two possible structures, one polar and one nonpolar, for each chemical given. Draw the two possible electron regional geometry structures, one polar and one nonpolar, for each chemical (i.e., two electron regional geometry structures for a and two for b ). a. Pl 2 3 b. Xe 2 l General short answer questions. a. Is the B bond nonpolar covalent, polar covalent, or ionic? b. ow many lone pairs of electrons exist on the central atom in el 4 2? c. Which bond is shorter: or N? d. What is the formal charge on l in l 4? e. Which bond is longer: the bond in or the bond in 2? f. What is the lattice energy reaction for Nal(s)? g. Name the electron region geometry for the central atom in I 4. h. Name the molecular geometry for the central atom in I 4. i. Is I 4 a polar or nonpolar molecule? j. Will 6 heat up in a microwave oven? k. Which bond is stronger: N or N? (almost done)

8 70. Which of the following statements is incorrect? a. Molecules can have polar bonds and while the molecule can be nonpolar. b. To reduce lone pairlone pair repulsions in a trigonal bipyramidal electron region geometry, the lone pairs are found in the equatorial positions. c. The four large lobes of the sp 3 hybridized orbitals point towards the vertices of a tetrahedral. d. A covalent bond forms when two orbitals overlap and two electrons with the same spin reside in the overlap region. e. The two atoms found above the plane in a trigonal bipyramidal electron region geometry are axial atoms. (The next 2 problems contain hybridization questions, sec 7.3 and 7.4; skip those parts if not on Exam 3. Q71. a, e, g, j; Q72. c, f, h, i) 71. Answer the questions below about the structure and bonding in this molecule. a. Draw in any lone pairs needed to complete octets. c a b. What is the hybridization on the N atom marked a? c. What is the bond order for the bond marked c? d. What is the angle for the with the atom marked d? N e. What orbitals overlap to form the bond marked f? e f. What is the bond order for the bond marked b? d g. ow many σ and π bonds are in the molecule? h. What is the electron region geometry at the N atom marked a? i. What is the molecular geometry of the atom marked d? j. What orbitals overlap to form the bond marked e? f b 72. Answer the questions below about the structure and bonding in this molecule. a. Draw in any lone pairs needed to complete octets. b. What is the bond order for the bond marked a? f c. What is the hybridization on the atom marked b? d. What is the molecular geometry around the atom marked c? e. What is the angle on the group marked d? f. What is the hybridization on the atom marked e? g. What is the bond order for the bond marked f? c h. What orbitals overlap to form the bonds marked a? i. ow many σ and π bonds are in the molecule? (yea done!) e d b a ANWER 1. d {light and moving bodies show both wave and particle properties} 2. a x J {E photon = hv = (6.626 x Js)(5.50 x 10 6 s 1 ) = x J = 3.64 x J} b. 54.5m {ν = c/λ; λ = (3.00 x 10 8 m/s)/[5.50 x 10 6 s 1 )] = 54.5m} c J {3.644 x J/photon x (6.022 x photons/1mol photons = J/mol photons = J} x s 1 {436kJ/mol x (1000J/1kJ) x (1mol 2 /6.022 x molecules) = x J/molecule; E = hν; ν = E/h = x /6.6s6 x = 1.09 x s 1 } x photons {convert wavelength in J: E = hc/λ; l = 532nm x (1m/1 x 10 9 nm) = 5.32 x 10 7 m; E = [(6.626 x Js)(3.00 x 10 8 m/s)]/(5.32 x 10 7 m) = x J/photon; 1.15mJ x (1J/1000mJ) = 1.15 x 10 3 J; #photons = (1.15 x 10 3 J)/(3.736 x J/photons) = x photons = 3.08 x photons} x photons/s {find energy of the light: E = hc/λ = [(6.626 x Js)(3.00 x 10 8 m/s)]/[(634nm)(1m/1 x 10 9 nm)] = x J/photon; E from the laser per second = 0.27J/50s = J/s; divide energy/photon by the energy/s to find the number of photons/s: [0.0054J/s]/[3.135 x J/photon] = x photons/s} 6. d {r n = a o n 2 ; 52.9Å = (0.529Å)n 2 ; n 2 = 100; n = 10} 7. b {ΔE = Rhc[ 1 /(n f ) 2 1 /(n i ) 2 ] = Rhc[ 1 /(2) 2 1 /(6) 2 ] = Rhc[ 1 /4 1 /36] = Rhc[ 9 /36 1 /36] = Rhc[ 8 /36] = Rhc[ 2 /9]}

9 kJ {ΔE = Rhc[ 1 /(n f ) 2 1 /(n i ) 2 ] where n f = ; ΔE= 2.18 x [ 1 /( ) 2 1 /(5) 2 ] = 2.18 x [0 1 /25] = 8.72 x J; this is for one atom; 8.72 x J/atom x (6.022 x atoms/1mol) = 52,512J/mol x (1kJ/1000J) = 52.5kJ} 9. a {as the energy levels have higher n values, they get closer together in energy so the 2 closest energy levels would be 4 and 3. The closer they are, the lower the energy difference the lower the ΔE the longer the wavelength since E 1/λ} 10. e {λ = h/mv v = h/mλ = (6.626 x Js)/[(9.11 x g x (1kg/1000g)][1 x 10 2 nm x (1m/1 x 10 9 nm)] = 7.27 x 10 3 m/s} 11. b {λ = h/(mv) m = h/(λv) = (6.626 x )/[(9.53 x )(7.5 x 10 5 )] = 9.27 x kg/atom; 9.27 x kg/atom x (1000g/1kg) x (6.022 x atoms/mol) = 55.83g/mol e} 12. b {E incident = work function + E KE e ; E incident = hc/λ = (6.626 x )(3.00 x 10 8 )/[(275nm)(1m/1 x 10 9 nm)] = 7.23 x J; work function = 6.94 x J; E KE e = 7.23 x J 6.94 x J = 2.88 x J} 13. I. e {(2, 3, 2, 1 / 2 ) and (4, 4, 2, 1 / 2 ): l value invalid; (4, 3, 4, 1 / 2 ): m l invalid; (4, 3, 2, 1 1 / 2 ): m s invalid} II. c {Last electron is 5p electron; n = 5 and l = 1 for 5p; if l = 1, m l = 1, 0, 1} 14. e {n = 5 5; l = 2 d; 5d} 15. a {b. 14 e not 7 e ; c ±1/2; d 3d fills after 4s; e the probability is always greater than 0} 16. b {the order of filling subshells corresponds to their relative energy; hence, the 4p subshell is lower in energy than the 4d subshell} 17. I. a. 32 {2n 2 = 2(4) 2 = 32 electrons} b. 25 {2n 2 = 2(5) 2 = 50 electrons; only half will have m s = 1 / 2 25 electrons} c. 10 {subshell = 3d 10 electrons} d. 6 {subshell = 2p 6 electrons} e. 7 {subshell = 4f 14 electrons; only half will have m s = 1 / 2 7 electrons} II. a. 9 {n 2 = 3 2 = 9 orbitals} b. 5 {subshell = 4d 5 orbitals} c. 3 {subshell = 5p 3 orbitals} d. 0 {when n = 5 l 5 0 orbitals} e. 1 {subshell = 3d; of the five 3d orbitals only one orbital has a m l = 1} 18. d {penetration: s > p > d and lower n values penetrate more; lower energy subshells penetrate more} 19. d 20. a. e b. Ge 2 c. Mn d {The number of electrons in P 3 is (1) = 18; only 2 has that number: 2(9) = 18e } 22. b {Ga = [Ar]4s 2 3d 10 4p 1 ; Ga +3 = [Ar]3d 10 all electrons are paired; note: remove the 4p 1 electron first and then Ga + resembles a transition metal so remove the 4s 2 electrons next; : [Ne]2s 2 2p 4 2 unpaired e ; u: [Ar]4s 1 3d 10 1 unpaired e ; +2 : [Ne]3s 2 3p 2 2 unpaired e } 23. 1s 2s 2p 3s 3p 3d 4s 4p 24. b {remove s electrons first for charged transition metals; o 0 = [Ar]4s 2 3d 7 ; o +2 = [Ar]3d 7 } 25. d {With charged transition metals, remove the s electrons first.} 26. b {most paramagnetic most unpaired e ; V: [Ar]4s 2 3d 3 ; V +2 : [Ar]3d 3 since s electrons are removed first} 27. b {Radii increase as you go to the left and down on the Periodic Table} 28. Ne < < < 2 < e 2 {Ne is farthest to the right and neutral so it is smallest; is larger than and by making both atoms a 1 charge the is still larger than ; as an ion becomes more anionic is becomes larger so 2 anion it will be larger than a 1 anion and yields: < 2 ; between 2 and e 2 since < e when both are neutral when both atoms become a 2 charge this yields: 2 < e 2 ; combining this information yields: Ne < < < 2 < e 2 }

10 29. I. As the n value increases, the size of the orbital increases. omparing Li to Na to K, the valence shells for these atoms are 2s 1, 3s 1, and 4s 1, respectively, and the 4s subshell is larger than the 3s subshell which is larger than the 2s subshell because the n value increases. II. Going from right to left the Z effective decreases. As Z effective decreases the outer most electrons are held less tightly allowing the electrons to pull away from the nucleus and increasing the size of the atom. omparing Li to, the Z effective for is greater than Z effective for Li. ence, the outer most electrons for (2s 2 2p 5 ) are pulled in more tightly as compared to Li and therefore has a smaller radius. III. The first ionization energy increases going left to right across a row because the Z effective increases left to right which implies that the electron is held more tightly by the nucleus and hence requires greater energy to remove it from the atom. The first ionization energy trend can also be explained by the atomic radius since the atoms are smaller on the right side of a periodic table row and if the atom is smaller then the electrons are closer to the nucleus and held more tightly. 30. e {ations < Anions Xe + } 31. e {the third ionization energy refers to the third electron being removed from a single atom in the gas phase} 32. a {Look at positions of the +1 ions, a +, Rb +, K + since the IE 2 involves removing the 2nd electron from the +1 ion. IE2 trend is higher as you go up and to the right on the Periodic Table. K + is in the Ar position.} 33. a {N has EA 0; EA increase (more negative) up a period but 3 rd period > 2 nd period} 34. d {IE increase as you go to the right and up on the Periodic Table; also, since we are considering the 3rd IE, think of the atoms as +2 ions: +2, +2, Mg +2, and P +2 ; assign these locations on the PT, i.e., +2 has 6 electrons so it is in the position and so on; P +2 is the only atom on the 3rd row and therefore has a lower IE 3 } 35. c {Na + is most up and to the right highest next IE; also removing an electron from Na + is breaking a noble gas configuration} 36. e {EA 1 : X(g) + e X(g) } 37. a {D = m/v; Na would be the largest in size and has the smallest mass so would be the least dense} 38. d { d is EA 1 and EA 1 is either 0 (zero) or negative (exothermic), in this case exothermic; a is IE 1 and b is IE 3 ionization energies are endothermic; c is EA 2 and EA 2 are always endothermic; e is endothermic; } 39. d {ince the IE 4 is so large as compared to the other IE, the removal of the 4 th electron breaks a noble gas configuration. nly Al breaks a noble gas configuration during IE 4 : Al +3 Al +4 + e } 40. c {Lattice energy means form 1 mol from its ions in the gas phase} 41. ee below for Lewis dot structures. 42. d { has 10 electrons around it} 43. a {this is an atomic size question; smaller atoms closer together, smaller bond length and stronger bond energy} b {N 2 : N N ; 2 3 : ; 3 : 45. b {rom Lewis dot structures, including the resonance structures, the bond is between a single and double bond 2 ; N 3 : N N N } B = 4 /3 bonds. (double bond).} Therefore, the bond length should be between 1.43 (single bond) and 1.23 Å 46. e { ; has resonance but this does not change for the atom; : 6 4 = 2}

11 47. I. a. {Longest bond has smallest bond order; N + : N 2 : N 3 : II. N + N N N N N + has N triple bond, N : N has N double bond, 1 has N with B = 1.5 and N 3 has a N B = 1.33; N 1 has smallest bond order longest bond length} {Greater B greater bond enthalpy; N + has N triple bond greatest B greatest bond enthalpy} P l P l 48. d {P 5 : ; N 2 2 : N N l ; Pl 3 : ; N 5 : you need to put 10 electrons around N; this is only allowed for elements in the 3rd period or later} 49. a. polar covalent b. nonpolar covalent c. nonpolar covalent d. ionic e. polar covalent 50. d {since the atoms are close to one another and are nonmetals the bond is a covalent one (eliminates choice e ); since the atoms are different the bond is polar (eliminates choices a and b ); since EN > EN the partial negative charge resides on (eliminates choice c )} 51. d {Most ionic means greatest ΔEN which means further apart from one another on the PT.} 52. tructure IV can be dropped because of the high. tructure I can be dropped because it has fewer zeros than tructures II or III. The difference between tructures II and III is a 1 on the atom and a of 0 on the (tructure II) and the reverse of that for tructure III. ince is more electronegative than the atom should have the more negative. Therefore tructure II is the best structure. (1) (1) (0) (0) (0) (+1) (0) (1) (2) (1) (1) (1) (1) (0) (0) (0) I II III IV Lower formal charges (closer to zero) are better. Two resonance structures have two 0 s and one 1 for formal charges. These are better structures than the structure with a 2 formal charge. To determine between these 2 resonance structures which is better, consider the electronegativity of and N. ince N has a greater EN, it should have the more negative formal charge. ence, the resonance structure with 2 double bonds is the best structure in which the 1 formal charge is on the N and not the. :.. N.. : (0) (0) (1).. :.. N: (1) (0) (0).. : Ṇ. : (+1) (0) (2) 54. I. b {compare atom pairs; the more EN atom in the pair gets a δ while the less EN atom gets a δ + } II. polar covalent kJ { + ; Δ r = bonds broken bonds formed = [1(=) + 4( ) + 2( )] [1( ) + 5( ) + 1( ) + 1( )] = [1(598) + 4(416) + 2(467)] [1(356) + 5(416) + 1(336) + 1(467)] = 43kJ} 56. b { + 2 l l l l + 2 l N ; Δ r = [2( ) + 1(=) + 2(l l)] [(2( l) + 1(=) + 2( l)]; 200 = [2(416) + 1(695) + 2(x)] [(2(327) + 1(695) + 2(431)]; x = 242} 57. d {from VEPR table} 58. c {repulsions: lone pair lone pair > bonding pair lonepair > bonding pair > bonding pair} l e : seesaw; e +2 4 : +2 l P l Xe 59. b {Pl 4 : l : tetrahedral; I + 4 : I : seesaw; Xe 2 2 : 60. c {valence electrons are used} 61. d {draw LD and then determine the molecular geometry; planar structures include: linear, triangular planar, bent, Tshaped, I I I + 2 : linear} and square planar; I. ; II. ; III. ; IV. ; V. }

12 + 62. d {draw LD; 1 lone pair and 4 bonding pairs around the central atom seesaw; } 63. a. b. c. d. 64. b {draw LD and then determine the electron domain geometry; ; ; ; ; } 65. d {triangular planar and triangular bipyramidal can have 120 angles; draw LD and then determine molecular geometry; N 3 : ~109.5 angles; 6 : 90 and 180 angles; 4 : angles; e 4 : 90, 120, and 180 angles} 66. b {a. P 4 + : ; b. Gel4 2 : ; c. il4 : ; d. Xe 4 : ; Note: the 4 0, etc., refers to the #atoms #lone pairs around the central atom; compare to VEPR table for molecular polarity} 67. e {all the molecules are nonpolar nonpolar molecules don t heat up in a microwave oven; 68. a. a. ; b. ; c. ; d. } l P l P l l Nonpolar Polar b. l Xe l Nonpolar l l Xe Polar

13 69. a. polar covalent {since and B electronegativity values are close to one another ΔEN is small polar covalent bond} b. 2 pairs of electrons {draw LD; 36 valence electrons; 32 electrons are tied up in single bonds and as lone pairs of electrons on the l atoms; hence, 2 pairs of electrons reside on the e atom} c. N {since atomic radius > atomic radius N and the atomic radius > atomic radius the bond length between the 2 shorter atoms (N and ) will be shorter} d. +3 { l = #valence electrons #electrons in LD; from LD: l : 7 4 = +3} e. in 2 {from LD of : the bond has a B = 3; from LD of 2 : the bond has a B = 2; smaller B longer bond length so in 2 is longer} f. Na + (g) + l (g) Nal(s) {lattice energy rxn: gaseous ions 1 mol of solid} g. octahedral {draw LD I 4 ; 36 valence electrons; 4 bonding atoms and 2 lone pairs on the central atom} h. square planar {draw LD I 4 ; 4 bonding atoms and 2 lone pairs on the central atom square planar} i. nonpolar molecule {4 atoms + 2 lone pairs on central atom MG = square planar if all 4 atoms around the central atom are the same nonpolar molecule} j. not heat up { 6 6 atoms on central atom MG = octahedral if all 6 atoms around central atom are the same nonpolar molecule nonpolar molecules don t heat up in microwaves} k. N {since atomic radius < atomic radius then the N bond is shorter; shorter bond stronger bonds} 70. d {electrons must have opposite spins} : : N 71. a. b. sp 3 c. 2 d e. sp 2 sp 2 f. 1.5 {B resonance = #bonds/#locations = 9/6 = 1.5} g. 23σ, 7π h. tetrahedral i. bent j. ssp 72. a. ee picture to right. f b. B = 3 {B = 1 for single bond; B = 2 for double bond; B = 3 for triple bond} e c. sp 3 {4 domains sp 3 } d. trigonal planar {3 domains trigonal planar} e. ~109.5 (or ) {tetrahedral electron region geometry around the atom} c f. 1s { atoms are not hybridized and remain 1s} g. B = 1.5 {resonance: ; B = # bonds ; B = 3/2 = 1.5} # locations h. σ = sp() sp(); π = p() p(); π = p() p() i. 23s bonds; 6p {single bond = σ bond; double bond = σ bond + π bond; triple bond = σ bond + 2π bonds} d b a

14 41.