GEERAL EMISTRY TPIAL: Bonding Test 1 Time: 22 Minutes* umber of Questions: 17 * The timing restrictions for the science topical tests are optional. If you are using this test for the sole purpose of content reinforcement, you may want to disregard the time limit.
MAT DIRETIS: Most of the questions in the following test are organized into groups, with a descriptive passage preceding each group of questions. Study the passage, then select the single best answer to each question in the group. Some of the questions are not based on a descriptive passage; you must also select the best answer to these questions. If you are unsure of the best answer, eliminate the choices that you know are incorrect, then select an answer from the choices that remain. Indicate your selection by blackening the corresponding circle on your answer sheet. A periodic table is provided below for your use with the questions. PERIDI TABLE F TE ELEMETS 1 1.0 2 e 4.0 3 Li 6.9 4 Be 9.0 5 B 10.8 6 12.0 7 14.0 8 16.0 9 F 19.0 10 e 20.2 11 a 23.0 12 Mg 24.3 13 Al 27.0 14 Si 28.1 15 P 31.0 16 S 32.1 17 35.5 18 Ar 39.9 19 K 39.1 20 a 40.1 21 Sc 45.0 22 Ti 47.9 23 V 50.9 24 r 52.0 25 Mn 54.9 26 Fe 55.8 27 o 58.9 28 i 58.7 29 u 63.5 30 Zn 65.4 31 Ga 69.7 32 Ge 72.6 33 As 74.9 34 Se 79.0 35 Br 79.9 36 Kr 83.8 37 Rb 85.5 38 Sr 87.6 39 Y 88.9 40 Zr 91.2 41 b 92.9 42 Mo 95.9 43 Tc (98) 44 Ru 101.1 45 Rh 102.9 46 Pd 106.4 47 Ag 107.9 48 d 112.4 49 In 114.8 50 Sn 118.7 51 Sb 121.8 52 Te 127.6 53 I 126.9 54 Xe 131.3 55 s 132.9 56 Ba 137.3 57 La * 138.9 72 f 178.5 73 Ta 180.9 74 W 183.9 75 Re 186.2 76 s 190.2 77 Ir 192.2 78 Pt 195.1 79 Au 197.0 80 g 200.6 81 Tl 204.4 82 Pb 207.2 83 Bi 209.0 84 Po (209) 85 At (210) 86 Rn (222) 87 Fr (223) 88 Ra 226.0 89 Ac 227.0 104 Rf (261) 105 a (262) 106 Unh (263) 107 Uns (262) 108 Uno (265) 109 Une (267) * 58 e 140.1 59 Pr 140.9 60 d 144.2 61 Pm (145) 62 Sm 150.4 63 Eu 152.0 64 Gd 157.3 65 Tb 158.9 66 Dy 162.5 67 o 164.9 68 Er 167.3 69 Tm 168.9 70 Yb 173.0 71 Lu 175.0 90 Th 232.0 91 Pa (231) 92 U 238.0 93 p (237) 94 Pu (244) 95 Am (243) 96 m (247) 97 Bk (247) 98 f (251) 99 Es (252) 100 Fm (257) 101 Md (258) 102 o (259) 103 Lr (260) G T TE EXT PAGE. 2 as developed by
Bonding Test 1 Passage I (Questions 1 7) If in the formation of a chemical bond, the electronegativities of the combining atoms are approximately the same, the electron density will be distributed evenly between the two atoms forming a covalent bond. If the combining atoms have different electronegativities, then the bonding electrons will be located around the more electronegative atom; the bond thus formed is known as a polar covalent bond. When the difference in electronegativity of the combining atoms is very large, a complete transfer of charge may occur, forming ionic species. The energy change that takes place in the formation of the ionic species can be attributed to the energy change when the electron is transferred and to the electrostatic energy between the ions. This energy change, the energy of the ionic bond, can be estimated with Equation 1, where D is the dissociation energy of the bond, I(A) is the ionization energy of atom A, E a (B) is the electron affinity of atom B, and Z A Z B e 2 /4πε 0 R is the oulombic potential energy of the two ions at a distance R with Z A and Z B being the quantized charge on ions A and B, respectively. (+1, -2, etc.) D = I(A) + E a (B) Z A Z B e 2 /4πε 0 R Equation 1 Tables 1 and 2 contain first ionization potentials and electron affinities of selected elements, respectively. Element K a Li Br Ar F Table 1 I/eV 4.34 5.14 5.39 11.81 12.97 15.76 17.42 Element Ar Li a K Br F Table 2 E a /ev 0.30 0.58 0.78 0.82 3.40 3.45 3.61 According to the valence-shell electron-pair repulsion (VSEPR) theory, molecules have definite shapes which result from a tendency of the atoms and nonbonded electron pairs attached to the central atom to get as far away from each other as possible. Table 3 contains the formulas, molecular structures, and bond angles of selected molecules. Formula Structure Bond Angle ai 2 K 2 3 Trigonal pyramidal 2 Bent S 2 Table 3 S Trigonal BF 3 planar P 5 P 107 G T TE EXT PAGE. KAPLA 3
MAT 1. According to Equation 1, at what point is it energetically favorable for Li and to form the ionic compound Li +? A. E a () Z Li Z e 2 /4πε 0 R I(Li) B. Z Li Z e 2 /4πε 0 R I(Li) + D + E a (). E a () I(Li) + Z Li Z e 2 /4πε 0 R D. I(Li) + D E a () Z Li Z e 2 /4πε 0 R 5. Which compound in Table 3 has two or more resonance forms? A. 3 B. P 5. S 2 D. a 2 6. What is the molecular structure of phosphorus pentachloride? 2. The bond distance in a(g) has been determined to be 250 10 12 m. According to Equation 1, what is the predicted bond energy of one molecule of a(g)? [ote: e 2 /4πε 0 R = 5.74 ev at an R of 250 pm.] A. 6.45 ev B. 4.21 ev. 4.21 ev D. 6.45 ev A. Trigonal pyramidal B. Linear. Tetrahedral D. Trigonal bipyramidal 7. Which of the following compounds has the strongest electrostatic interactions between the constituent particles? A. 3 B. P 5. a D. 2 3. In which of the following does ionic bonding predominate? I. 3 II. a 2 III. P 5 IV. K 2 A. I and III only B. II and III only. I and IV only D. II and IV only 4. Based on information presented in the passage, what would be the predicted bond angle in boron trifluoride? A. 120 B. 109. 90 D. 108 G T TE EXT PAGE. 4 as developed by
Bonding Test 1 Questions 8 through 11 are T based on a descriptive passage. 11. Which of the following is T a correct Lewis dot diagram? 8. Which of the following is a Lewis base? A. 4 + B. 4. P 3 D. 3 3 9. What is the formal charge on the nitrogen atom in 3? A. 1 B. +1. 0 D. +2 10. Which of the following is T a resonance structure of the others? A. B.. D. S B A. B.. D. 2 2 2 3 2 G T TE EXT PAGE. KAPLA 5
MAT Passage II (Questions 12 17) Some atoms have more power to draw electrons toward themselves than others. The stronger this tendency is, the more electronegative an element is said to be. Electronegativity can be quantified by using either of two scales: the Mulliken scale or the Pauling scale. As shown in Equation 1, the Mulliken electronegativity of an element is defined as the mean of the ionization potential, I, and the electron affinity, E a, of that element. X M = (I + E a ) 2 Equation 1 The actual values of the ionization potential and electron affinity that are of relevance, however, are not necessarily equal to those of the ground state element. This is because electronegativity is a characteristic of the atom when it is bonded to others in a molecule. The ionization energy and electron affinity to be used should be the ones for the atom in its correct hybridized state in a molecule. The ionization energy of carbon, for example, is the energy needed to remove an electron from its valence 2p orbital. In the calculation of its electronegativity in the molecule methane, however, what is meant by the ionization energy is the energy needed to remove an electron from the now sp 3 hybridized orbital. Because this orbital contains 25% s character, it is at a lower energy than a p orbital and thus the ionization energy would be greater. This also implies that the electronegativity of an element is dependent upon its hybridization in the molecule. The Pauling scale is approximately proportional to the Mulliken scale. The Mulliken and Pauling electronegativities of elements A and B are related with Equation 2, where X A M and X A P are the Mulliken and Pauling electronegativities of element A, respectively. 13. Which of the following is a true statement about a molecule of S 2? A. It is a polar molecule with polar bonds. B. It is a polar molecule with nonpolar bonds.. It is a nonpolar molecule with polar bonds. D. It is a nonpolar molecule with nonpolar bonds. 14. In the molecule shown below, which arrow indicates the direction of the dipole moment? 1 3 A. 1 B. 2. 3 D. The molecule has no overall dipole moment. 15. Which structure below has the partial charges correctly represented? A. δ + δ B. 2 δ δ + X A M X B M = 2.78(X A P X B P ) Equation 2. δ + δ D. δ δ + δ 12. Using Equation 2, if the Pauling electronegativity of phosphorus is 2.1, predict its Mulliken electronegativity. A. 2.1/2.78 B. 2.78/2.1. 0.68 D. The Mulliken electronegativity of phosphorus cannot be determined without more information. G T TE EXT PAGE. 6 as developed by
Bonding Test 1 16. Which of the following has the most polar bond? A. B.. D. F 17. In which of the following compounds will the nitrogen atom have the highest electronegativity on the Mulliken scale? A. 3 : B. ( 3 ) 3 :. D. : ED F TEST KAPLA 7
MAT ASWER KEY: 1. A 6. D 11. B 16. D 2. B 7. 12. D 17. D 3. D 8. 13. A 4. A 9. B 14. A 5. 10. 15. A 8 as developed by
Bonding Test 1 EXPLAATIS Passage I (Questions 1 7) 1. A The quickest way to obtain the answer is to recognize that bond dissociation is an endothermic process, and that the bond dissociation energy is the amount of energy one needs to supply to break the bond. A positive dissociation energy would indicate that the ionically-bonded state is more stable than the state of separated neutral atoms. An ionic bond forms, therefore, if D 0. Substituting in the terms on the left hand side of equation 1 for D and rearranging, one obtains choice A. To explain this more fully: the point at which bond formation becomes possible is that point where the energy released upon formation of the ionic bond is greater than or equal to the energy one needs to supply to generate the ionic species. The energy required to remove an electron from a neutral atom to form a cation is positive--in other words, this process costs energy. So, you may ask, why does an ionic bond form? It forms because energy is released in two ways: first, the electron affinity of a neutral atom is positive: i.e. energy is released as it accepts one more electron to form an anion; second, the oulombic attraction between the two oppositely-charged ions means that the system can lower its energy by coming close together (forming the bond). In order to make the process of forming an ionically-bonded species worthwhile, the energy one needs to invest in, the ionization energy, has to be less than or at most equal to the payback, that is, the energy released in the form of electron affinity and lower oulomb potential energy. hoice A describes this criterion: released energy that needs to be supplied = energy to ionize Li = I (Li) energy that one gets back = energy released from adding electron to + oulombic energy = electron affinity of + oulombic energy released What is the oulombic energy released? When the two ions are infinitely separated, the oulombic potential energy is zero, since R =. When the two are a distance R apart, this electrostatic potential energy is: U = Z Li Z e2 4πε 0 R as given in the passage. ote that Z Li and Z will have opposite signs, and so this term is negative. This makes sense as we would expect the energy to be lower when the two are closer than when they are infinitely far apart. (The ions are attracted to, not repelled by, each other.) The amount of oulombic potential energy released as the ionic bond forms, then, is U = Z Li Z e2 4πε 0 R which is positive since the opposite signs of Z Li and Z cancel the negative sign in front. The energy one gets back is therefore and this needs to be greater than or equal to I (Li). E a () Z Li Z e2 4πε 0 R 2. B This question requires you to use Equation 1 together with Tables 1 and 2. Looking at Equation 1, you can see that the missing values that you need to provide are the electron affinity and the ionization potential; the energy of interaction term--the 4πε ο part--has been given to you in the question stem as 5.74 ev. (Z a and Z b are +1 and -1 respectively and so do not affect the magnitude of the interaction term). Sodium chloride is a + and, and so we want the ionization potential of sodium and the electron affinity of chlorine. These values are 5.14 and 3.61, respectively. This gives: D = I(a) + E a () + 5.74 = 5.14 + 3.61 + 5.74 = 4.21 ev Remember, if the bond forms, then the bond energy (or bond dissociation energy) would have to be positive: bondformation is exothermic; bond-breaking is endothermic. KAPLA 9
MAT 3. D Ionic bonding predominates in those compounds which contain elements with large differences in electronegativity. The larger the difference, the more polar the bond. The easiest, and quickest, way to answer this question is to consult the periodic table. You should know that the most electronegative elements are located in the upper-right hand corner of the table (excluding the noble gases) and those elements that are the least electronegative are located in the lower left-hand corner. Looking at the answer choices, you can see that they are all two-item choices--in other words, look for the two compounds whose elements are the furthest apart in the periodic table. ompound I is ammonia, which you probably should know to be a covalent compound already. Since answer choices A and both contain roman numeral I, they can be eliminated. Roman numeral II, calcium chloride: calcium is located in the second column, fourth row; chlorine is located in the 17th column, third row--pretty far apart from each other---mostly ionic. You may also know that calcium tends to form the +2 cation, while chlorine forms the 1 anion to complete their octet. ppositely-charged ions would form ionic bonds. Roman numeral III, phosphorus pentachloride: in the periodic table, phosphorus and chlorine are only separated from each other by sulfur, making it mostly covalent. By elimination, choice D is the correct response. 4. A The easiest way to answer this question correctly is to go to Table 3 and see that the shape of BF 3 is listed as trigonal planar. If there wasn't any information in the passage, you could still answer this question correctly. From the periodic table you should be able to determine that boron has two nonvalence electrons in the 1s subshell and three valence electrons that are available for bonding; in other words, boron completes its valence by forming three bonds, and only three bonds. When boron forms these three bonds, there will be six bonding electrons surrounding it: there are no nonbonding electrons. Since there are no nonbonding electrons to worry about, the bonding electrons arrange themselves to be as far apart from each other as possible. As a result, a trigonal arrangement is adopted which has bond angles of 120. 5. In order for resonance structures to be possible in a molecule or polyatomic ion, it needs to have double and single bonds which can be interchanged. In resonance structures, bonds and nonbonding electrons may move, but atoms do not. In table 3, S 2 is the only one of these which has a double bond. The resonance structures are therefore: S S This implies that the 2 π electrons are actually delocalized over all three atoms. Each sulfur-oxygen bond has characteristics intermediate between a single and a double bond. 6. D This question is testing you on your understanding of the valence-shell electron-pair repulsion theory (VSEPR). The MAT requires you to make predictions of molecular shapes using the VSEPR theory. P 5 has a phosphorus atom at the center surrounded by 5 chlorine atoms; there are no lone pairs, thus giving it the formula AX 5 --trigonal bipyramidal, choice D. 7. In order to answer this question correctly, you need to remember that ionic bonding results from the attractions between opposite charges. Particles in ionic compounds therefore exhibit stronger electrostatic or oulombic interactions than atoms in covalent bonds. Looking at the answer choices, the only choice that is predominately ionic is choice, a. Discrete Questions 8. A Lewis base is a substance that can donate a pair of electrons. Among the answer choices, only P 3, choice, has a pair of nonbonding electrons, making it the correct response. The easiest way to determine if a substance is a Lewis base is to draw its Lewis dot structure and see if a lone electron pair exists: P 10 as developed by
MAT 9. B To answer this question, we need to write the Lewis dot diagrams and then use any one of several formulas to find the formal charge. ne such formula is Formal charge = Valence electrons - [number of bonds + number of nonbonding electrons]. The Lewis dot structure of 3 is as follows: Formal charge is equal to the valence electrons of nitrogen, which is 5, minus the sum of the number of bonds and the number of nonbonding electrons, which in this case is 4. So, the formal charge on the nitrogen is equal to 5 4 which is +1. 10. Remember that in resonance forms, only electrons (bonds and nonbonding electron pairs) can move. Examining the answer choices, you can see that the atomic linkages are all the same except for answer choice ; the carbon second from the left in choice now has a methyl group bonded to it, whereas the other choices have hydrogens in this position. Below are the resonance structures showing the actual way that the electrons can be shifted around: 2 2 2 It is always important to keep in mind, however, that shifting or moving electrons is just a figure of speech: the π electrons are really delocalized over all the bonds. 11. B The first thing you should do to verify a Lewis structure of a molecule is to make sure that all the valence electrons are accounted for. For choice A, acetylene, there are two carbons, each having 4 valence electrons, and 2 hydrogens, each having 1 valence electron. So, choice A needs to have 10 valence electrons, and indeed it has. For choice B, nitrogen dioxide, there is 1 nitrogen, which has 5 valence electrons, and 2 oxygens, each having six valence electrons--there should be a total of 17 electrons accounted for. ounting the electrons in choice B you can see that it has only 16 electrons--choice B is the correct answer. hoice, sulfur trioxide, should have a total of 24 valence electrons, 6 electrons from each element. ounting the valence electrons, you'll see that it has the required 24. hoice D, boron trichloride, should have and does have 24 electrons. Passage II (Qu estions 12 17) 12. D The relationship between the Pauling and Mulliken scales of electronegativity given in Equation 2 has four unknowns, though some may look similar at first glance. To solve such an equation we must be given, or be able to determine, three of these unknowns. In this question, only one of the unknowns (the Pauling electronegativity of phosphorus) is given--without more information there is no way to figure out the other two. To determine the equivalent Pauling or Mulliken electronegativity for an element A we must know both the Pauling and Mulliken electronegativities of the other element B. Likely mathematical errors might include the attempted algebraic cancellation of X B M with XB P from both sides of the equation. This is a common mistake and there are two reasons why this approach is incorrect. Firstly, the unknowns are not identical (though they look similar) and cannot be cancel each other out. Secondly, even if the two variables were identical (i.e. both X B M and X B P = X B ), cancellation would not work: the variables on the right side of the equation would have to be first multiplied by 2.78 (converting X B to 2.78X B ) before X B is added to both sides of the equation (leaving 1.78X B on the right side of the equation). The variable would still remain and the answer still could not be determined. 13. A nce again a question concerning VSEPR theory and vector addition. The correct VSEPR class for S 2 is AX 2 E; its VSEPR shape is angular. As can be seen from the drawing of S 2 in Table 3 of the first passage, the individual bonds are 11 as developed by
MAT polar and the vector sum of the bonds is nonzero, choice A is the correct response. If you had forgotten the shape of S 2, or it is not made available to you, you could have realized that it contains polar bonds, only homonuclear bonds are nonpolar, and eliminated choices B and D, giving you a 50/50 chance of selecting the correct answer. 14. A Another vector addition question--the vector sum of the three -- dipole moments is an arrow which goes straight up the middle of the molecule. You should know that a dipole vector is represented as an arrow, with the head of the arrow pointing to the partial negative charge. In this case, nitrogen would be the center of partial negative charge since it's more electronegative than hydrogen. Adding the three bond vectors together gives a dipole moment having the direction shown by arrow one: 15. A hoice A is the only one where the more electronegative atom has a partial negative charge. In this example, the nitrogen, being the more electronegative, has a small negative charge due to increased electron density and the electropositive carbon has a small partial positive charge. 16. D The most polar bond is the one that has the greatest difference in the electronegativities of the two elements. (Remember, electronegativity increases in going from the lower left to the upper right of the periodic table.) Among these choices, -F is the most polar since hydrogen and fluorine are the furthest apart in the periodic table. 17. D The key to this question lies in understanding the point made in the second paragraph. Depending on the hybridization of the atom, the ionization potential and the electron affinity values change and thus so does the Mulliken electronegativity. In particular, the higher the percentage of s character, the higher the electronegativity. hoice D has the nitrogen in an sp hybridized state. This means that each sigma orbital has 50% s character. This is the highest among the four shown and thus the nitrogen atom there has the highest electronegativity. hoices A and B both have the nitrogen in the sp 3 hybridized state. The nitrogen in choice is sp 2 hybridized. The sigma orbital thus has 33% s character. 12 as developed by