2. 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. Fe c. He d. Ne e.

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1 EMTRY 103 ractice roblems #3 hapters 5 ( ), 6 (6.16.2, ), 7 (7.17.2d, ), and 6.12 **Do the appropriate topics for your lecture** repared 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. ractice exams posted by your instructor should be completely worked though as they 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! ATER 5 1. f 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 2. f 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. e e. Be 3. Draw the electron probability graph, Ψ 2 versus r, for each subshell. n each graph, label the axes, note on the graph where the spherical and planar (surface) nodes are (not all instructors go over where the nodes are on the plot; skip if not covered), and note how many surface nodes exist on the graph. a. 2s b. c. 5p 4.. 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 ). What is a possible set of quantum numbers (n, l, m l, m s ) for the last electron added to n 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 ) 5. What subshell is described by n = 5 and l = 2? a. 2h b. 5h c. 5p d. 5s e. 5d 6. Which statement below is correct? a. The 2s subshell will penetrate closer to the nucleus than the 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 subshell is zero at a point far away from the nucleus. 7. 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 planar (surface) nodes is 1. d. There are 3 orbitals within the subshell. e. The value of m l for the 4p x orbital can be 1. (coffee time)

2 8.. 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. 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 = 1 9. (ot all instructors cover subshell penetration; skip if not covered.) Which subshell penetrates closest to the nucleus? a. 3p b. c. 3d d. 2s e. 4d 10. Which orbital below represents the p z orbital? z y z y z y x x x a. b. c. d. e. 11. a. What neutral atom has the electron configuration in the ground state: 2 2s 2 6 3s 2 3p 6 4s 2 3d 6? b. What 2 ion has the electron configuration in the ground state: 2 2s 2 6 3s 2 3p 6 4s 2 3d 10 4p 4? c. What +2 ion has the electron configuration in the ground state: 2 2s 2 6 3s 2 3p 6 3d 5? 12. Which ion/molecule is isoelectronic with 3? a. 3 b. 3 c. 4 d. 2 e. none of the above 13. Which ion/molecule is diamagnetic? a. b. Ga +3 c. u d. +2 e. none are diamagnetic 14. Draw the orbital box electron configuration diagram for e in the ground state. 15. Which orbital box diagram corresponds with o +2 in the ground state? nly electrons past [Ar] are shown. a. 4s 3d b. 4s 3d c. 4s 3d d. 4s 3d e. 4s 3d 16. Which atom or ion in the ground state has the following electron configuration, 2 2s 2 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. i +2 b. V +2 c. a d. b +5 e. e 18. lace the following atoms in order of decreasing atomic radii:, e, Ge, n a. e > Ge > n > b. n > Ge > e > c. > e > Ge > n d. > Ge > e > n e. one of the above are correct. (eat an apple)

3 19. ort these atoms/ions from smallest to largest in size (i.e., atomic radii)., e 2, 2,, e (smallest) < < < < (largest) 20. Two elements are shown on the eriodic Table. Which statements are correct? B A. The n value for the valence shell for A is larger than the n value for the valence shell for B.. The effective nuclear charge for the valence electrons for A is less than the effective nuclear charge for the valence electrons of B.. The first ionization energy for A is larger than the first ionization energy for B. V. The effective nuclear charge for the electrons for A is greater than the effective nuclear charge for the electrons for B. a., V b., V c., d.,, V e.,,, V 21. Which atom or ion is smallest? a. b 3 b. Te 2 c. d. e 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 23. 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 24. lace the following atoms,,,, in order of increasing first electron affinity (more exothermic/more negative values on the right). a. < < b. < < c. < < d. < < e. < < 25. Which atom has the lowest third ionization energy? a. b. c. Mg d. e. all have same E3 26. Which atom has the highest next ionization energy? a. Al +2 b. Al c. a + d. e. B Which material would be expected to be the least dense? a. sodium(s) b. aluminum(s) c. magnesium(s) d. need additional info 28. 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/ion will require the most energy to remove an electron? a. a b. c. + d. + e. i (eat some pizza)

4 30. Which atom would most likely align with the following ionization energies: E 1 = 578kJ/mol E 2 = 1817kJ/mol E 3 = 2745kJ/mol E 4 = 11,577kJ/mol a. Li b. Mg c. d. Al e. none of the above 31. Determine the enthalpy for each reaction below using the data given on the right. (int: Write two reactions using the reaction given as a guide, such that when these two are added they generate the reaction desired!). (g) + (g) (g) + (g) E 1 for = 1681kJ/mol E 2 for = 3374kJ/mol E 1 for = 1251kJ/mol E 2 for = 2297kJ/mol. (g) + + (g) (g) +2 + (g). Mg(g) + + (g) Mg(g) + (g) E 3 for = 6050kJ/mol EA 1 for = 328kJ/mol E 1 for = 1681kJ/mol E 2 for = 3374kJ/mol E 3 for = 6050kJ/mol EA 1 for = 328kJ/mol E 1 for = 1251kJ/mol E 2 for = 2297kJ/mol E 3 for = 3826kJ/mol EA 1 for = 349kJ/mol E 3 for = 3826kJ/mol EA 1 for = 349kJ/mol E 1 for = 1314kJ/mol E 2 for = 3388kJ/mol E 3 for = 5300kJ/mol EA 1 for = 141kJ/mol E 1 for Mg = 738kJ/mol E 2 for Mg = 1451kJ/mol E 3 for Mg = 7732kJ/mol EA 1 for Mg = 0kJ/mol 32. (Do if lattice energy was covered.) Which reaction represents the lattice energy reaction for Al 3 (s)? a. Al(s) + 3 / 2 2 (g) Al 3 (s) b. Al(g) +2 Al(g) +3 + e and (g) + e (g) c. Al +3 (g) + 3 (g) Al 3 (s) d. Al 3 (s) Al +3 (aq) + 3 (aq) e. none of the above 33. (Do if Bornaber cycles were covered.) A Bornaber cycle is shown here for Mg(s). a. What does tep A represent? b. What does tep D represent? c. Which step represents the electron affinity for oxygen? d. Which step represents the heat of vaporization of magnesium? e. Which step represents the lattice energy of Mg(s)? f. What is the value of the lattice energy? ATER 6 1. Draw the Lewis dot structures for the following molecules. a. 3 b. 3 3 c. 3 3 d. 2 e. 2 f. 2 g. 2 2 h. 2 i. 2 2 j. 2 k. 3 2 l. 3 m. K 2 n. a o. B p. q. B 3 r. Be 2 s. (not ) t. 4 u. Kr 4 +2 v Which molecule violates the octet rule? a. B 4 b. i 4 c. As 3 d. 4 e. none 3. onsider the following covalent single bonded pairs of atoms. Which bond will be the shortest? a. b. c. d. e. none of the above 4. Which of the following molecules will have resonance structures? i. 2 ii. 2 3 iii. 3 iv. 3 a. ii b. ii and iii c. ii, iii, and iv d. i e. iv (watch some TV) Mg +2 (g) + (g) 844kJ 142kJ G Mg +2 (g) + (g) 299kJ Mg +2 (g) + 1 / 2 2 (g) 1451kJ E 738kJ D 601kJ A Mg + (g) + 1 / 2 2 (g) Mg(g) + 1 / 2 2 (g) 129kJ Mg(l) + 1 / 2 2 (g) 9kJ B Mg(s) + 1 / 2 2 (g) Mg(s) Mg +2 (g) + 2 (g)

5 5. onsider the following bond lengths: single bond: 1.43 Å double bond: 1.23 Å triple bond: 1.09 Å n the carbonate ion, 3 2, a reasonable average bond length would be: a Å b Å c Å d Å e Å 6. What is the formal charge on the in 3? a. 2 b. 1 c. 0 d. +1 e onsider the following list of compounds. Rank the molecules from smallest to largest bond length a. + < < 2 < 3 b. 3 < 2 < < + c. 3 < + < 2 < d. + < < 2 = 3 e. 3 = 2 < < +. Using the same compounds above, which compound will have the greatest bond enthalpy? 8. dentify each of the following bonds as nonpolar covalent, polar covalent, or ionic. a. b. c. ii d. a e. 9. 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. 10. 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 11. 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 V 12. Draw the 3 Lewis dot resonance structures for thiocyanate, ( 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 = As indicated by Lewis structures which molecule could not exist? f all can exist then select choice e. a. 5 b. 2 2 c. 3 d. 5 e. all can exist (B time)

6 14.. n 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.. Would the bond be nonpolar covalent, polar covalent, or ionic? δ δ δ δ+ δ+ 15. alculate the change in enthalpy for the reaction (Δ rxn ) using the bond energies (kj/mol) listed below Bond Bond enthalpy (kj/mol) Bond Bond enthalpy (kj/mol) 356 = = What is the bond dissociation energy given the bond energies (kj/mol) listed below. 2 (g) (g) 2 (g) + 2(g) Δ rxn = 208kJ Bond Bond enthalpy (kj/mol) Bond Bond enthalpy (kj/mol) Bond Bond enthalpy (kj/mol) = = 695 a. 484kJ b. 242kJ c. 26.5kJ d. 53.0kJ e. 208kJ ATER 7 1. A central atom with 2 lone pairs and 3 bonding pairs of e will have which molecular geometry? a. linear b. triangular pyramidal c. triangular planar d. Tshape e. triangular bipyramidal 2. n 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 3. Which of the following compounds have tetrahedral molecular geometry? a. 4 b. e 4 +2 c. 4 + d. 2 2 e. all do 4. 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, 3, will absorb microwave radiation. 5. Which of the following molecules are planar (i.e., which have a molecular geometry that is flat)?. B e V. e 4 V. 4 a. b.,, V c., d.,, V e.,, V, V (another nap)

7 6. Which structure best represents the molecular geometry for 4 +? a. b. c. d. e. mages: ublic Domain from Wikipedia.org ( 7. 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. n 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. 3 c. 2 d ow many of these molecules have a tetrahedral electron region geometry around the central atom? 4 4 i 4 2 e 2 Kr 2 a. 1 b. 2 c. 3 d. 4 e Which molecule will have an angle of ~120? a. 3 b. 6 c. 4 d. e 4 e. none 10. Which molecule has a net molecular dipole moment? a. 4 + b. Ge 4 2 c. i 4 d. 4 e. none 11. Which molecule will heat up in a microwave oven? a. 3 b. 6 c d. 4 e. none 12. There are two possible structures, one polar and one nonpolar, for each chemical given. Draw the two possible electron regional geometry structures in 3D, one structure polar and one structure nonpolar, for each chemical (i.e., draw two electron regional geometry structures for a and two for b ). a. 2 3 b General short answer questions. a. s the B bond nonpolar covalent, polar covalent, or ionic? b. ow many lone pairs of electrons exist on the central atom in e 4 2? c. Which bond is shorter: or? d. What is the formal charge on in 4? e. Which bond is longer: the bond in or the bond in 2? f. What is the lattice energy reaction for a(s)? g. ame the electron region geometry around the central atom in 4. h. ame the molecular geometry around the central atom in 4. i. s 4 a polar or nonpolar molecule? j. Will 6 heat up in a microwave oven? k. Which bond is stronger: or? (take a walk)

8 14. Which of the following statements is incorrect? a. Molecules can have polar bonds and 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. 15. Answer the questions below about the structure and bonding in this molecule. a. Draw in any lone pairs needed to complete octets. c b. What is the hybridization on the 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? 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? f h. What is the electron region geometry at the atom marked a? i. What is the molecular geometry of the atom marked d? j. What orbitals overlap to form the bond marked e? b a 16. 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? e d b a ATER 6.12 Molecular rbital Theory (skip these questions if M theory was not covered!) 1. Write the entire molecular electronic configuration including core electrons and determine the bond order for each molecule. a. Li 2 b Using molecular orbital theory, which molecule(s) could not exist?. e B V. Be 2 2 V. Li 2 2 a. b. c. V d., V e.,, V 3. Which of the following statements is incorrect? a. As the overlap between orbitals increase as a covalent bond forms, the energy of the interaction decreases; at this minimum energy the distance between the nuclei is the bond length. b. ingle bonds in molecular orbital theory are comprised of one sigma bond. c. A dipole is established when two opposite charges of equal magnitude are separated by a distance d. d. Molecular orbital theory explained why oxygen is paramagnetic. e. Molecular orbital theory explained why electrons follow the auli Exclusion rinciple while Valence Bond Theory did not. (almost done)

9 4. Using M theory, place the following in order of increasing bond order (smallest bond order on the left) a. < < b. < < c. < < d. < < e. < < 5. Based on M theory, which of the following are diamagnetic? e 2 2 V a., b., c.,, d.,, V e.,,, V 6. a. Draw a molecular orbital diagram for Be 2 2. Label all the molecular orbitals in the diagram. (ote: some instructors provide a blank M diagram for students to use.) b. What is the bond order? c. Write the molecular orbital electron configuration. d. s the molecular paramagnetic or diamagnetic? e. When an electron is added to this molecule to form Be 2 2 what happens to the bond length? 7. a. Draw a molecular orbital diagram for 2 2. Label all the molecular orbitals in the diagram. (ote: some instructors provide a blank M diagram for students to use.) b. What is the bond order? c. Write the molecular orbital electron configuration. d. s the molecular paramagnetic or diamagnetic? e. When an electron is added to this molecule what happens to the bond strength? 8. a. Draw a molecular orbital diagram for. Label all the molecular orbitals in the diagram. (ote: some instructors provide a blank M diagram for students to use. ote: ot all instructors cover heteronuclear diatomic molecules; skip if not covered!) b. What is the bond order? c. Write the molecular orbital electron configuration. d. s the molecular paramagnetic or diamagnetic? e. When an electron is removed from this molecule what happens to the bond length? 9. (ot all instructors cover drawing molecular orbitals; skip if not covered!) Below is two atomic orbitals as they form a molecular orbital. dentify the individual atomic and molecular orbitals. f there is a node, label it. a. + b (ot all instructors cover drawing molecular orbitals; skip if not covered!) Draw the formation of the molecular orbital π showing the two atomic orbitals (each labeled) and the molecular orbital (labeled) that is formed. f there is a node, label it. (yea done!)

10 AWER hapter 5 1. 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} 2. 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} 3. a. {graph is memorized; 2s subshell: n = 2; l = 0; spherical nodes = n l 1 = = 1 spherical node; all spherical nodes do not occur at the origin; planar (surface) nodes = l = 0 planar nodes} b. {graph is memorized; subshell: n = 2; l = 1; spherical nodes = n l 1 = = 0 spherical nodes; planar (surface) nodes = l = 1 planar nodes; all planar nodes occur at the original} c. {graph is memorized; 5p subshell: n = 5; l = 1; spherical nodes = n l 1 = = 3 spherical nodes; planar (surface) nodes = l = 1 planar nodes; all planar nodes occur at the original 4.. 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}. c {Last electron is 5p electron; n = 5 and l = 1 for 5p; if l = 1, m l = 1, 0, 1} 5. e {n = 5 5; l = 2 d; 5d} 6. a {b. 14 e not 7 e ; c ±1/2; d 3d fills after 4s; e the probability is always greater than 0} 7. b {the order of filling subshells corresponds to their relative energy; hence, the 4p subshell is lower in energy than the 4d subshell} 8.. 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 = 6 electrons} e. 7 {subshell = 4f 14 electrons; only half will have m s = 1 / 2 7 electrons}. 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} 9. d {penetration: s > p > d and lower n values penetrate more; lower energy subshells penetrate more} 10. d 11. a. e b. Ge 2 c. Mn d {The number of electrons in 3 is (1) = 18; only 2 has that number: 2(9) = 18e } 13. 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; : [e]2s unpaired e ; u: [Ar]4s 1 3d 10 1 unpaired e ; +2 : [e]3s 2 3p 2 2 unpaired e } 14. 2s 3s 3p 3d 4s 4p 15. b {remove s electrons first for charged transition metals; o 0 = [Ar]4s 2 3d 7 ; o +2 = [Ar]3d 7 } 16. d {With charged transition metals, remove the s electrons first.} 17. b {most paramagnetic most unpaired e ; V: [Ar]4s 2 3d 3 ; V +2 : [Ar]3d 3 since s electrons are removed first}

11 18. b {Radii increase as you go to the left and down on the eriodic Table} 19. e < < < 2 < e 2 {e 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: e < < < 2 < e 2 } 20. d { : True; since A is lower on the T than B, it will have a larger n value; : True; Z eff increases lefttoright on the T so Z eff (A) < Z eff (B); V: True; the electrons will not have any shielding so the p + e interaction strength will be determined the number of p + the electrons are interacting with; since A is further down the T it has more p + and therefore the electrons interaction will be stronger} 21. e {ations < Anions + } 22. e {the third ionization energy refers to the third electron being removed from a single atom in the gas phase} 23. a {Look at positions of the +1 ions, a +, Rb +, K + since the E 2 involves removing the 2nd electron from the +1 ion. E2 trend is higher as you go up and to the right on the eriodic Table. K + is in the Ar position.} 24. a { has EA 0; EA increase (more negative) up a period but 3 rd period > 2 nd period} 25. d {E increase as you go to the right and up on the eriodic Table; also, since we are considering the 3rd E, think of the atoms as +2 ions: +2, +2, Mg +2, and +2 ; assign these locations on the T, i.e., +2 has 6 electrons so it is in the position and so on; +2 is the only atom on the 3rd row and therefore has a lower E 3 } 26. c {a + is most up and to the right highest next E; also removing an electron from a + is breaking a noble gas configuration} 27. a {D = m/v; a would be the largest in size and has the smallest mass so would be the least dense} 28. d { d is EA 1 and EA 1 is either 0 (zero) or negative (exothermic), in this case exothermic; a is E 1 and b is E 3 ionization energies are endothermic; c is EA 2 and EA 2 are always endothermic; e is endothermic; } 29. c {removing an electron from a cation is more difficult than removing an electron from an anion or a neutral atom so the possible answers are + and + ; the smaller the cation the harder it is to remove an electron; since + is smaller than + (recall is smaller than ) it is harder to remove the next electron from + } 30. d {ince the E 4 is so large as compared to the other E, the removal of the 4 th electron breaks a noble gas configuration. nly Al breaks a noble gas configuration during E 4 : Al +3 Al +4 + e } kJ/mol {(g) (g) : EA 1 (); (g) (g) : EA 1 (); Δ r = EA 1 () + (EA 1 ()) = ((349)) = 21kJ/mol}. 3060kJ/mol {(g) (g) : EA 1 for ; (g) + (g) +2 : E 2 for ; Δ r = EA 1 () + E 2 () = = 3060kJ/mol}. 389kJ/mol {Mg(g) + Mg(g) : E 1 for Mg; (g) (g) : EA 1 for ; Δ r = E 1 (Mg) + (EA 1 ()) = ((349)) = 389kJ/mol} 32. c {Lattice energy means form 1 mol from its ions in the gas phase} 33. a. Δ f of Mg(s) {recall that Δ f is the heat require to form 1mol of substance from its elements in their natural state; for Mg(s) the Δ f reaction is: Mg(s) + 1 / 2 2 (g) Mg(s); tep A is: Mg(s) Mg(s) + 1 / 2 2 (g) which is the reverse of Δ f } b. E 1 of Mg {the only change in tep D is: Mg(g) Mg + (g); this is the first ionization energy of Mg, E 1 (Mg)} c. tep G {EA for : (g) + e (g); in tep G oxygen goes from (g) to (g)} d. tep {Δ vap is the process: liquid gas; in tep : Mg(l) Mg(g)} e. tep {lattice energy reaction: ions(g) ionic solid; tep : Mg +2 (g) + 2 (g) Mg(s)} f. 3929kJ {ess s Law: add all the steps and they total zero; tep A + tep B + tep + tep D + tep E + tep + tep G + tep + tep = 0; (142) tep = 0; tep = lattice energy = 3929kJ} hapter 6 1. ee below for Lewis dot structures. 2. d { has 10 electrons around it} 3. a {this is an atomic size question; smaller atoms closer together, smaller bond length and stronger bond energy}

12 b { 2 : ; 2 3 : ; 3 : 5. b {rom Lewis dot structures, including the resonance structures, the bond is between a single and double bond 2 ; 3 : } B = 4 /3 bonds. (double bond).} Therefore, the bond length should be between 1.43 (single bond) and 1.23 Å 6. e { 7.. a. {Longest bond has smallest bond order; + : 2 : 3 : ; has resonance but this does not change for the atom; : 6 4 = 2} + has triple bond, : has double bond, 1 has with B = 1.5 and 3 has a B = 1.33; 1 has smallest bond order longest bond length}. + {Greater B greater bond enthalpy; + has triple bond greatest B greatest bond enthalpy} 8. a. polar covalent b. nonpolar covalent c. nonpolar covalent d. ionic e. polar covalent 9. 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 E > E the partial negative charge resides on (eliminates choice c )} 10. d {Most ionic means greatest ΔE which means further apart from one another on the T.} 11. tructure V can be dropped because of the high. tructure can be dropped because it has fewer zeros than tructures or. The difference between tructures and is a 1 on the atom and a of 0 on the (tructure ) and the reverse of that for tructure. ince is more electronegative than the atom should have the more negative. Therefore tructure is the best structure. (1) (1) (0) (0) (0) (+1) (0) (1) (2) (1) (1) (1) (1) (0) (0) (0) V 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. ince has a greater E, 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 and not the. :.... : (0) (0) (1) 13. d { 5 :.. :.. : (1) (0) (0).. : Ṇ. : (+1) (0) ; 2 2 : ; 3 : (2) ; 5 : you need to put 10 electrons around ; this is only allowed for elements in the 3rd period or later} 14.. b {compare atom pairs; the more E atom in the pair gets a δ while the less E atom gets a δ + }. 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}

13 16. b { ; Δ r = [2( ) + 1(=) + 2( )] [(2( ) + 1(=) + 2( )]; 200 = [2(416) + 1(695) + 2(x)] [(2(327) + 1(695) + 2(431)]; x = 242} hapter 7 1. d {from VER table} 2. c {repulsions: lone pair lone pair > bonding pair lonepair > bonding pair > bonding pair} e 3. b { 4 : : seesaw; e +2 4 : : tetrahedral; + 4 : : seesaw; 2 2 : 4. c {valence electrons are used} 5. d {draw LD and then determine the molecular geometry; planar structures include: linear, triangular planar, bent, Tshaped, B and square planar;. triangular planar ;. triangular pyramidal ;. e bent ; V. e seesaw ; V. 6. d {draw LD; 1 lone pair and 4 bonding pairs around the central atom seesaw; 7. a. c. d. LD LD LD ERG = tetrahedral ERG = triangular bipyramid MG = bent ERG = triangular bipyramid polar MG = linear net dipole MG = seesaw b. vectors cancel; no dipole; nonpolar polar LD net dipole square planar } ERG = tetrahedral + + } MG = triangular pyramid : linear} net dipole polar 8. b {draw LD and then determine the electron domain geometry; tetrahedral ; bipyramidal ; tetrahedral ; 9. d {triangular planar and triangular bipyramidal can have 120 angles; draw LD and then determine molecular geometry; 3 : triangular pyramidal ~109.5 angles; 6 : 120, and 180 angles} 10. b {a. 4 + : same atoms nonpolar ; b. Ge 4 2 : octahedral 90 and 180 angles; 4 : Ge always polar ; c. i 4 : i 4 0 same atoms triangular nonpolar ; d. 4 : #atoms #lone pairs around the central atom; compare to VER table for molecular polarity} i 2 triangular bipyramidal ; tetrahedral angles; e 4 : e Kr triangular bipyramidal } e seesaw 90, 4 2 same atoms nonpolar ; ote: the 4 0, etc., refers to the

14 11. e {all the molecules are nonpolar nonpolar molecules don t heat up in a microwave oven; a. 2 3 same atoms nonpolar ; b. 6 0 same atoms nonpolar ; c. 6 0 same atoms nonpolar +2 ; d. 4 0 same atoms nonpolar } 12. a. onpolar olar b. onpolar olar 13. a. polar covalent {since and B electronegativity values are close to one another ΔE 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 atoms; hence, 2 pairs of electrons reside on the e atom} c. {since atomic radius > atomic radius and the atomic radius > atomic radius the bond length between the 2 shorter atoms ( and ) will be shorter} d. +3 { = #valence electrons #electrons in LD; from LD: : 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. a + (g) + (g) a(s) {lattice energy rxn: gaseous ions 1 mol of solid} g. octahedral {draw LD 4 ; 36 valence electrons; 4 bonding atoms and 2 lone pairs on the central atom} h. square planar {draw LD 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. {since atomic radius < atomic radius then the bond is shorter; shorter bond stronger bonds} 14. d {electrons must have opposite spins} : : 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 16. 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. { atoms are not hybridized and remain } g. B = 1.5 {resonance: ; B = # bonds ; B = 3/2 = 1.5} # locations h. σ = sp() sp(); π = p() p(); π = p() p() i. 23σ bonds; 6π bonds {single bond = σ bond; double bond = σ bond + π bond; triple bond = σ bond + 2π bonds} hapter 6.12 Molecular rbital Theory 1. a. (σ ) 2 (σ 2 (σ 2s ) 2 (σ 2s 1 ; B = 0.5 {B = (4 3)/2 = 1/2} b. (σ ) 2 (σ 2 (σ 2s ) 2 (σ 2s 2 (σ ) 2 (π ) 4 (π 3 ; B = 1.5 {B = (10 7)/2 = 3/2} 2. d {when B = 0 molecule doesn t exist; when total #e is 4, 8, or 20 B = 0; e 2 +2 = 2e exists; B2 2 = 12e exists; 2 2 = 4e doesn t exists as B = 0; Be2 2 = 10e exists; Li2 2 = 8e doesn t exists as B = 0} d b a

15 3. e {M Theory explained why oxygen was paramagnetic, why some chemicals were colored, and how electrons could be excited. either MT nor VBT address the auli Exclusion rinciple.} 4. e {#bond and #antibonding e determined from M e config; B 2 = 14e (σ ) 2 (σ 2 (σ2s ) 2 (σ 2s 2 (π ) 4 (σ ) 2 = (10 4)/2 = 6/2 = 3; B 2 +1 = 13e (σ ) 2 (σ 2 (σ2s ) 2 (σ 2s 2 (π ) 4 (σ ) 1 = (9 4)/2 = 5/2 = 2.5; B 2 2 = 16e (σ ) 2 (σ 2 (σ2s ) 2 (σ 2s 2 (π ) 4 (σ ) 2 (π 2 = (10 6)/2 = 4/2 = 2} 5. c { 2 +2 = 12e (σ ) 2 (σ 2 (σ2s ) 2 (σ 2s 2 (π ) 4 all e paired; 2 2 = 18e (σ ) 2 (σ 2 (σ2s ) 2 (σ 2s 2 (σ ) 2 (π ) 4 (π 4 all e paired; e 2 2 = 6e (σ ) 2 (σ 2 (σ2s ) 2 all e paired; 2 +2 = 10e (σ ) 2 (σ 2 (σ2s ) 2 (σ 2s 2 (π ) 2 2e unpaired;,, and are diamagnetic} M A Be Be s 2 1 σ * A Be 2 2s 2 1 π * π * σ π π 2s σ 2s * 2s σ 2s σ * σ 6. a. b. B = 1 {B = (6 4)/2 = 2/2 = 1} c. σ 2 σ *2 σ 2 2s σ *2 2s π 2 d. paramagnetic e. decreases {when an electron is added, the new M electron configuration becomes: σ 2 σ *2 σ2s 2 σ2s *2 π 3 and the new B = (7 4)/2 = 3/2 = 1.5; the B has increased the bond length decreases} M A 2 2 A 2 2s 2 3 σ * 2 2s 2 3 π * π * σ π π 2s σ 2s * 2s σ 2s σ * σ 7. a. b. B = 3 {B = (10 4)/2 = 6/2 = 3} c. σ 2 σ *2 σ 2 2s σ *2 2s π 4 σ 2 d. diamagnetic e. decreases {when an electron is added, the new M electron configuration becomes: σ 2 σ *2 σ2s 2 σ2s *2 π 4 σ 2 π *1 and the new B = (10 5)/2 = 5/2 = 2.5; the B has decreased the bond strength decreases}

16 M A A 2 2s 2 2 σ * 2 2s 2 4 π * π * σ π π 2s σ 2s * σ 2s 2s σ * σ 8. a. b. B = 3 {B = (10 4)/2 = 6/2 = 3} c. σ 2 σ *2 σ 2 2s σ *2 2s π 4 σ 2 d. diamagnetic e. decreases {when an electron is removed, the new M electron configuration becomes: σ 2 σ *2 σ2s 2 σ2s *2 π 4 σ 1 and the new B = (9 4)/2 = 5/2 = 2.5; the B has decreased the bond length increases} 9. a. f the horizontal axis = zaxis, then: + b. z z σ + (or 2s) (or 2s) σ * (or σ2s 10. f the horizontal axis = zaxis, then: + x or y x or y π

17 hapter 6 1.