GENERAL CHEMISTRY TOPICAL: Electronic Structure and the Periodic Table Test Time: 22 Minutes* Number 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 MCAT DIRECTIONS: 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 PERIODIC TABLE OF THE ELEMENTS H 1.0 He 4.0 Li 6.9 Be 9.0 B 10.8 C 12.0 N 14.0 O 16.0 F 19.0 10 Ne 20.2 11 Na 23.0 12 Mg 24.3 13 Al 27.0 14 Si 28.1 15 P 31.0 16 S 32.1 17 Cl 35.5 18 Ar 39.9 19 K 39.1 20 Ca 40.1 21 Sc 45.0 22 Ti 47.9 23 V 50.9 24 Cr 52.0 25 Mn 54.9 26 Fe 55.8 27 Co 58.9 28 Ni 58.7 29 Cu 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 Nb 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 Cd 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 Cs 132.9 56 Ba 137.3 57 La * 138.9 72 Hf 178.5 73 Ta 180.9 74 W 183.9 75 Re 186.2 76 Os 190.2 77 Ir 192.2 78 Pt 195.1 79 Au 197.0 80 Hg 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 Ha (262) 106 Unh (263) 107 Uns (262) 108 Uno (265) 109 Une (267) * 58 Ce 140.1 59 Pr 140.9 60 Nd 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 Ho 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 Np (237) 94 Pu (244) 95 Am (243) 96 Cm (247) 97 Bk (247) 98 Cf (251) 99 Es (252) 100 Fm (257) 101 Md (258) 102 No (259) 103 Lr (260) GO ON TO THE NEXT PAGE as developed by Electronic Structure and the Periodic Table Test Passage I (Questions 1–6) In addition to kinetic energy, electrons also have potential energy due to their positions relative to the nucleus, other electrons, and other nuclei In a gaseous atom, the potential energy of an electron can be estimated from its ionization energy The first ionization energy of an element is defined as the minimum amount of energy required to remove an electron from the gaseous element in its ground state First ionization energies range from less than 400 kJ/mole for cesium to nearly 2400 kJ/mole for helium Table gives the first ionization energies of selected elements Table Element H He Li Be B C N F Na I (kJ/mol) 1312 2372 520 899 800 1086 1402 1680 496 Photoionization is a method by which first ionization energies can be conveniently measured A gaseous sample of an element is irradiated and the energy of absorbed photons is determined by the following equation: I = hν Equation where I is the ionization energy, ν is the frequency of the light absorbed, and h is Planck’s constant, 6.63 × 10–34 J • s Sodium has a first ionization energy of 496 kJ/mol What is the frequency of light absorbed per photon in the photoionization of a gas phase sodium sample? A B C D 1.24 × 1012 1.24 × 1015 7.48 × 1026 7.48 × 1038 Hz Hz Hz Hz The first ionization energy of aluminum is 577 kJ/mol while that of magnesium is 738 kJ/mol This difference can most likely be attributed to: A the higher energy of the 3p electron of aluminum as opposed to the lower lying 3s electrons of magnesium B the larger nuclear charge of magnesium compared to that of aluminum C the increased internuclear attraction between aluminum atoms D the electron-electron repulsion of aluminum 3p electrons The second ionization energy is defined as the minimum amount of energy needed to remove an electron from a gaseous +1 cation in its ground state Which of the following must be true for any element? A Second ionization energies are always higher than first ionization energies B Second ionization energies may be higher or lower than first ionization energies C Second ionization energies are usually lower than first ionization energies, but exceptions exist in the inner transition metals D Second ionization energies are always lower than first ionization energies due to the destabilization of the element brought about by ionization Which of the following elements has the smallest atomic radius? A B C D Potassium Calcium Cesium Barium GO ON TO THE NEXT PAGE KAPLAN MCAT When cesium is photoionized, an electron is removed from which of the following atomic orbitals? A B C D 4p 5p 5d 6s Why must the photoionization experiment in the passage be carried out on atoms in the gas phase? A Gas phase atoms are the only ones capable of absorbing visible light B The refractive index of most solids and liquids is greater than one and thus obscures the true frequency of absorption C The proximity of neighboring atoms in solids and liquids increases the potential energy of an electron, making ionization impossible D Factors such as lattice energy and van der Waals forces in the solid and liquid states alter the environment of the electron, thereby distorting the frequency of absorption GO ON TO THE NEXT PAGE as developed by Electronic Structure and the Periodic Table Test Passage II (Questions 7–12) The Aufbau, or Building-up, Principle states that the electronic configuration of an atom or molecule is the result of the placement of electrons in orbitals of increasing energy until the correct number of electrons has been accommodated This principle takes into consideration the attraction of the electron to the nucleus and inter-electron repulsion The electronic configuration of ions can be determined by subtracting electrons from the highest energy orbital for cations or by adding electrons according to the Building-up Principle There has been, however, experimental evidence that the Aufbau principle is an oversimplification of electronic structure The electronic configurations of transition elements and their ions often not correspond to those predicted by the Aufbau Principle These inconsistencies are due to electrons filling orbitals in such a way as to give the atom the lowest possible energy The total energy of an atom can be described as: E=F–G Equation where E is the total energy, F is the sum of the orbital energies of the atom, and G is the electron repulsion energy The F term increases with increasing values of the quantum numbers n and l, but the orbitals that minimize F may not necessarily minimize E As a result, the lowest total energy of some elements is achieved by placing electrons in orbitals with a higher energy (i.e., a higher principal quantum number) even though there are still unfixed subshells in the lower energy orbital This is especially true for the transition elements, where the s orbital of a higher electron shell fills before the d orbital with a lower n value Which of the following best explains why chromium exhibits a valence electron configuration of 3d54s1 instead of the predicted 3d44s2? A The 4s orbital has a higher energy and therefore fills last B After the first 3d electron is added to an atom, each successive electron will go into the 3d subshell until it is filled C Since the 3d subshell is half-filled, the total energy of the atom is minimized D Chromium exhibits multiple oxidation states and so must have many unpaired electrons Based on information in the passage, which of the following statements is true? A By nature of its principal quantum number alone, the 3d orbital has lower energy and fills before the 4s orbital B By nature of its principal quantum number alone, the 3d orbital has lower energy but fills after the 4s orbital because of electron repulsion energies C By nature of its principal quantum number alone, the 4s orbital has lower energy and fills before the 3d orbital D By nature of its principal quantum number alone, the 4s orbital has lower energy but fills after the 3d orbital because of electron repulsion energies What are the ground-state electronic configurations of the Fe2+ and Fe3+ ions respectively? A B C D [Ar]3d6 and [Ar]3d5 [Ar]3d6 and [Ar]3d44s1 [Ar]3d54s1 and[Ar]3d44s1 [Ar]3d44s2 and [Ar]3d34s2 What is the electronic configuration of indium (Z = 49)? A B C D 1s22s22p63s23p64s23dl04p65s24d105p1 1s22s22p63s23p64s23dl04p65s14d105p2 1s22s22p63s23p64s23dl04p65s14d95p3 1s22s22p63s23p64s23dl04p65s24d85p3 GO ON TO THE NEXT PAGE KAPLAN MCAT 1 What is the maximum number of electrons shell number can hold? A B C D 32 18 2 Which set of elements has electrons added to the 4f subshell as the atomic number increases? A B C D Metalloids Actinides Alkali metals Lanthanides GO ON TO THE NEXT PAGE as developed by Electronic Structure and the Periodic Table Test Questions 13 through 17 are NOT based on a descriptive passage All halogens have similar reactivity because: A they have the same number of protons B they have the same number of electrons C they have similar outer shell electron configurations D they have valence electrons with the same quantum numbers Which of the following would be different in a groundstate and an excited-state neon atom? A B C D The number of neutrons The number of electrons The atomic weight The electronic configuration END OF TEST Electron affinity is defined as: A the change in energy when a gaseous atom in its ground state gains an electron B the pull an atom has on the electrons in a chemical bond C the energy required to remove a valence electron from a neutral gaseous atom in its ground state D the energy difference between an electron in its ground and excited states The modern periodic table is ordered on the basis of A B C D atomic mass atomic radius atomic charge atomic number Two electrons with the same n, l, and ml values: A must be in different atoms B are in different orbitals of the same subshell C are in the same orbital of the same subshell with opposite spins D are indistinguishable from each other KAPLAN MCAT ANSWER KEY: B A A B D 8 10 D C B A A 11 12 13 14 15 A D C A D 16 17 C D as developed by Electronic Structure and the Periodic Table Test ELECTRONIC STRUCTURE AND THE PERIODIC TABLE TEST EXPLANATIONS Passage I (Questions 1–6) B In order to answer this question correctly you need to understand and use Equation The question-stem states that the ionization energy of sodium is 496 kJ/mole; in other words, 496 kilojoules of energy is required to remove one mole of electrons from one mole of sodium atoms Well, all you have to is plug the energy into Equation and divide by Planck's constant to get the frequency of the photon, right? Well, essentially, yes, but remember that 496 kJ is the amount of energy for one mole of these photons, not for an individual photon So if one mole of photons has an energy of 496 kJ, one photon has 496 kJ divided by 6.02 ∞ 1023, or 8.24 ∞ 10–22, kilojoules of energy So, converting that to joules to match units with Planck's constant and plugging into Equation 1, we get that the frequency of the photons is 8.24 ∞ 10–19 joules divided by 6.62 ∞ 10–34 joule seconds, or 1.24 ∞ 1015 hertz, answer choice B If you failed to recognize that the energy you were given was per mole, not per photon, you would have gotten answer choice D If you had forgotten that Planck's constant is in joule seconds and the energy was in kilojoules so there had to be a conversion, you would have gotten choice A Choice C represents a combination of calculation errors, including moving the decimal point the wrong way when converting kilojoules to joules A The first thing you may have noticed about this question is that in describing the first ionization energies of magnesium (atomic number 12) and aluminum (atomic number 13), the trend in the periodic table that shows increasing first ionization energies across a period doesn't apply Due to the increased nuclear pull, first ionization energies generally increase across a period, but here, the first ionization energy has decreased First of all, remember that the trends in the periodic table are not absolute Second, the question is essentially asking us to explain why these two elements don't follow the trend Why is the first ionization energy of aluminum lower than the first ionization energy of magnesium? To answer that, we have to look at the electronic structure of these two atoms Magnesium is in the second column of the periodic table, indicating that its two valence electrons are in the 3s subshell Aluminum, in column 13, has one of its three valence electrons in the 3p subshell and the other two in the 3s Aluminum's lone "p" electron has a higher energy than magnesium's "s" electrons and is, therefore, held less tightly by the nucleus Because it is held less tightly, it requires less energy to remove it A similar situation exists between beryllium and boron Choice B is wrong because aluminum, with one more proton than magnesium, has a greater nuclear charge Choice C is wrong because the two positively charged nuclei of neighboring atoms repel each other; they don't attract as indicated here D is also incorrect because there is only one "p" electron in aluminum; no electron-electron repulsions exist A As the question states, and is stated in the question, the second ionization energy is the energy required to remove an electron from the +1 cation; in essence, the energy needed to remove the second most energetic electron from an atom after the first is removed All atoms have as many ionization energies as they electrons, but we hardly ever deal with anything higher than the sixth; after all, when was the last time you had to deal with a +20 cation? Anyway, which of the statements is correct? Choice A says that the second ionization energy of an atom is always greater than the first ionization energy of the same atom That seems right After all, the second electron to be removed must have had a lower energy than the first to begin with, so it would take more energy to remove it Further, it requires more energy to remove an electron from a positively charged ion So choice A looks good, but we should check on the other three Choice B says that the second ionization energies can be lower or higher than the first The only way that the second ionization energy can be lower than the first is if the second electron to be removed has a greater energy than the first However, if the second electron had more energy, it would have been the first electron removed, so choice B is wrong Choice C is wrong because, as we've just discussed, second ionization energies are not lower than first ionization energies Choice D is wrong for the same reason, but also because ionization does not always destabilize a species Sodium tends to be found not as neutral Na, but as the Na+ cation because the cation is more stable than the neutral atom So choice A is the best answer B This question is only indirectly related to the passage The distance from the nucleus is a factor in ionization energy In general, the further an electron is from the nucleus, the greater its overall energy is and the easier it is to strip from the atom It should also be apparent to you that the further away from the nucleus the outermost electron is, the larger the atomic radius is The trend in atomic radius in the periodic table is just the opposite of that for ionization energy, that is, it generally decreases as you move across a period and increases as you move down a column The radius decreases from left to right across a period because the effective nuclear charge is increasing: the greater the effective nuclear charge, the stronger the electrons are KAPLAN MCAT attracted to the nucleus, the smaller the radius The radii increase in going down a group because in the higher periods, the value of n increases: the larger n is, the larger the orbitals are, and the further away the electrons are from the nucleus Applying this knowledge to the question, we can eliminate choices C and D pretty quickly since they are in the sixth period and the other choices are in the fourth We know that the atomic radii of potassium and calcium will be much smaller than those of cesium and barium Since calcium is to the right of potassium in the same period, we can say that choice B, calcium, has the smallest radius D This answer shouldn't come as a great surprise to you By now, you should have a handle on the fact that ionization, be it photoionization or some other method for removing an electron from an atom, takes an electron from the valence shell of an atom That means that this question essentially asks you what the valence shell of cesium is Cesium is element 55 It is the first element in the sixth period of the periodic table Here's where some knowledge of how the periodic table is set up comes in handy In the period table, the first two columns are the s-block elements Their valence electrons are in the s subshell of whatever energy level characterizes their outermost shell The next ten columns are the d-block elements, also called the transition elements The next six columns are the p-block elements and the 28 elements below in the lanthanide and actinide series, the inner transition metals, are the f-block elements Notice that the number of columns in a block is the same as the maximum number of electrons the subshell that block is named after can hold Now, don't think that the elements in any one block can only use the electrons in the subshell the block is named for as valence electrons Metals in the fourth period transition elements often use both 4s and 3d electrons in reactions and elements in the p-block often utilize the electrons in the s subshell too However, the blocks are named as such because each electron added in these blocks goes into the subshell the block is named for; the block name identifies the valence subshell Anyway, let's answer the question Since cesium is an sblock element, its valence electrons are in the s subshell, so the answer must be choice D Further, since cesium is in the sixth period, it must have valence electrons in the sixth shell Again, only choice D reflects electrons in the sixth shell So by both the shell and subshell identity, the only possible answer to this question is D D The best way to answer this question is by eliminating each incorrect choice In doing this, it forces you to read each answer choice in turn before deciding which one is the best So here we're looking for the best reason that photoionization, for the sake of determining the first ionization energy, should take place in the gas phase Choice A says that it is because only gas phase atoms can absorb visible light Well, this is untrue If solids or liquids were unable to absorb at least some visible light, there would be no colored solids or liquids So choice A must be wrong B says that the atoms must be in the gas phase because the refractive index is greater than one, so the frequency of the absorbed light is obscured This is untrue, the frequency of absorption can be measured; what effect the photon has on the material is another story entirely choice B is incorrect Choice C essentially says that solids and liquids cannot be ionized and so electrons cannot be removed or shifted in solids or liquids Not true! Think about electricity How could solid metal wires conduct electricity if their electrons couldn't be moved around? How could redox reactions of solids take place if electrons couldn't be gained or lost? That leaves choice D as the right answer It says that other energy factors on the electrons between the atoms of solids or liquids distorts the light frequency that an electron can absorb by altering the energy of the electron That makes sense After all, when bonds are formed, energy is released To remove the bonding electrons, energy must first be added to break the bond and then to release the electron from the nuclear pull When atoms are associated through some sort of intermolecular force like hydrogen bonds, energy must be added to break the intermolecular association and then to release the electron So only for elements in the gas phase, where the distances between atoms is so great that there is very little interatomic interactions, will photoionization give a measurement for the ionization energy that is not affected by other energy factors Passage II (Questions 7–12) C The answer to this question comes from the passage Electrons fill the orbitals around the nucleus of an atom in such a way as to give the lowest possible atomic energy That was right in the passage This knowledge, along with the fact that half-filled and fully filled subshells have increased stability, gives us the answer In the case of chromium, a lower energy is achieved when one of the 4s electrons is transferred to the 3d subshell, making it half-filled Choice C is the correct answer Other anomalies include molybdenum, which behaves similarly to chromium; copper, silver, and gold, all of which promote an "s" electron to give a fully filled "d" subshell All right, choice A is wrong because there would be no 4s electrons in this atom if the 4s orbital filled last Even with five 3d electrons, the subshell has room for more Choice B is wrong because it implies that the 4s electrons would have remained where they were while the 3d orbitals filled That doesn't seem to be the case here since a 4s electron has moved into the 3d subshell Finally, choice D is more of a consequence, not a cause, of the 10 as developed by Electronic Structure and the Periodic Table Test electronic configuration Because of the six different spatial orientations of the valence electron orbitals, there is more opportunity for reaction and so chromium exhibits a wide variety of oxidation states So choice C is the best answer B This question is a little confusing because of the oversimplifications that the Aufbau principle leads to However, you are told that just because the F term in Equation is minimized doesn't mean that the E term is also In fact, because of the electron repulsions, subshells with higher principal quantum numbers are sometimes filled before subshells with lower principal quantum numbers because that minimize the overall energy when electron repulsions are taken into account You are even told that this is the case for the d and s orbitals in transition elements A To find the electronic configurations of cations, all you need to is, starting with the highest value of n, remove electrons in the order np, then ns, then (n-1)d So, let's this for the +2 and +3 ions of iron The first thing that you need to is figure out the electronic configuration of neutral iron From the periodic table, the configuration is [argon core]3d64s2 To form the +2 ion you need to remove two electrons; following the rule I just stated, the configuration of the +2 ion is [Ar]3d6 The two 4s electrons are removed since they have the highest energy This makes answer choice A correct Answers B and C are choices that could have confused you if you had tried to impose the chromium atom electronic configuration on the iron II ion However, because electrons are removed from atoms to form ions in the order that I described earlier, this is not the case The configuration of the plus three state is [Ar]3d5 That leaves us with a half-filled 3d orbital The stability of this configuration makes +3 the preferred oxidation state of iron Choice D would have seemed correct if you had forgotten the higher energy of 4s electrons The destabilizing electron repulsions that caused the 4s subshell to fill before the 3d are countered by the greater nuclear charge (You may have expected the electronic configuration of Fe2+ to be [Ar]3d54s1 based on the stability of half-filled orbitals, but actually that is not the case here Even though you couldn’t have predicted when the anomaly applies, it is irrelevant here because [Ar]3d54s1 is not a choice that is accompanied by the correct configuration for the Fe3+ ion, which we know to be [Ar]3d5.) 10 A For this question you have to be able to recognize the electronic configuration of indium The first thing to when finding the electronic configuration of an element is to determine its atomic number and its atomic charge That will tell you how many electrons to put into the atom In this case, indium has an atomic number of 49 and an atomic charge of since it is neutral That means that you have to put 49 electrons into the correct orbitals If you look at the periodic table, you see that indium is in column 13, period That means that its electronic configuration will be built upon the fourth period noble gas, krypton, configuration In fact, you can see this by looking at the answer choices and seeing that they all start with 1s2-2s22p6-3s2-3p6-4s2-3d10-4p6 That takes care of the first 36 electrons, what about the next 13? 12 of them will fill the 5s and 4d orbitals, bringing us out of the transition elements The final one will go into the 5p orbital That gives a final valence configuration of 5s2-4d-105p1, choice A The other choices try to fool you by promoting electrons to the 5p orbital 11 A This question is just an outside knowledge question testing your memory of quantum numbers and subshells The best way to answer this question is to remember that the 4th electron shell has four subshells, an s, a p, a d and an f They can hold 2, 6, 10, and 14 electrons respectively So all you have to is add them up, which gives you 32, choice A You can either remember the subshells in the 4th electron shell and how many electrons each one holds as we just did, or you can remember that for n = 4, l can be 0, 1, 2, or Each l value had ml values ranging from l to -l, so there are 16 possible ml states in the 4th energy level Each ml value can hold electrons in the atom, so 16 ∞ is 32 12 D This is another outside knowledge question, although you can draw parallels to the passage The passage informs you that as the 3d electrons fill, you move across the fourth period transition elements So, as you fill in 4f electrons, you should move across another set of elements Well, first of all, the 4f subshell can hold 14 electrons, so whatever set of elements you are dealing with, there must be fourteen of them Further, you know that they must all be in the same period That limits the answers to one of the two periods of inner transition metals shown under the periodic table These are the lanthanide and actinide series, choices B and D The metalloids, choice A, are p-block elements that display properties intermediate to those of metals and nonmetals Since there are metalloids in a number of periods, choice A cannot be right since filling the subshell one electron at a time would not represent elements in different periods Choice C, the alkali metals that make up the first column of the periodic table, also suffers from this problem So since both choices B and D represent sets of elements that have f electrons added as the atomic number increases, you just have to decide which set you get when the 4f subshell is filled Well, the order of filling up to the 4f subshell is 1s2-2s2-2p6-3s2-3p6-4s2-3d10-4p6-5s2-4d10-5p6-6s2, which totals 56 electrons So the KAPLAN 11 MCAT next 14 elements above atomic number 56 will have the 4f subshell filling Element 57 is lanthanide, so this set of 4f elements is called the lanthanide series, or simply the lanthanides That's choice D Choice B, the actinides, are the elements that represent the set that fill the 5f orbital as atomic number increases Discrete Questions 13 C This question is basically another one that requires you to know something about the periodicity of the elements Basically, all this question asks you is why all the halogens behave so similarly in reactions What is it about the elements in a column that are the same? Not the number of protons, choice A If that were true, all these elements would be the same Not the number of electrons, choice B, since that is equal to the number of protons in a neutral atom Not the quantum numbers of the valence electrons, choice D Every period has higher valence quantum numbers than the last and every column contains elements from at least two periods That leaves choice C, the number of valence electrons Since the identity of the valence electrons is the same in each column, that is columns in the s-block have valence s electrons, columns in the p-block have valence s and p electrons, and columns in the d-block have valence s and d electrons, the only thing that can make them act similarly is the number of electrons in these shells That should make sense We discussed that the number of electrons in the valence shell affected the reactivity and stability of the shell The number of electrons affects the ionization energy and electron affinity, whether the atom will form cations or anions, and even the number of bonds the atom can participate in So choice C must be the correct answer 14 A This question really just boils down to definitions The electron affinity of an atom is defined as the change in energy that occurs when an electron is added to a gaseous neutral atom in its ground-state So electron affinities can be positive or negative, depending on whether energy is released when an atom spontaneously accepts an electron or energy is gained when an electron is forced onto an atom So choice A is the correct answer Electronegativity is a derived quantity, usually scaled for all atoms between and 4, that characterizes the pull an atom has for the electrons in a bond The electronegativity has nothing to with how likely an atom is to gain an electron, just how strong its pull on an electron in a bond is The "in a bond" part is very important The concept of electronegativity can only be applied to atoms that are already bonded It characterizes the polarity of the bond, not the likelihood of bond formation So choice B is the definition of electronegativity You probably recognize choice C as the definition of the first ionization energy This is kind of the opposite of electron affinity since in electron affinity, electrons are gained and in ionization, electrons are lost Choice D would give the energy of a photon released when an electron relaxed from an excited state to a lower lying state Since there are many quantities for this energy, depending on which excited state the electron is in, it hardly makes a good answer to this question 15 D When the periodic table was first being designed, it was thought that the periodicity of the elements could be explained on the basis of atomic mass Mendeleev discovered that when the elements were arranged in order of increasing atomic mass, certain chemical properties were repeated at regular intervals However, certain elements could not be fit into any group of a table based on increasing atomic mass It was the discovery of the nucleus and its components that led scientists to order the elements by increasing atomic number, the number of protons So, choice A is wrong and choice D is correct Choice B, atomic radius, is wrong because while the atomic radius does increase as the period increases, it decreases across a single period due to the increased nuclear charge acting on the same amount of electron shielding So using atomic radius as the basis of the periodic order doesn't make sense either Choice C might seem correct because each subsequent element has an additional proton, making each subsequent nuclear charge greater However, the choice is the atomic charge which, in neutral atoms, is always zero since one electron is added for each proton 16 C You know from the Pauli Exclusion Principle that no two electrons in the same atom can have identical quantum numbers So, with n, l, and ml the same, you may be tempted to say the two electrons must be in different atoms, choice A However, there is a fourth quantum number that could be different, so choice A is not necessarily true Let's review what we know The n value indicates that these electrons are in the same shell The value of n also limits the value of l, defining the possible subshells the electrons can occupy If they have the same l value, they must be in the same subshell if they are in the same atom The l value limits the ml value, which defines the specific spatial orientation of the orbital the electrons are in If the ml value is the same for two electrons, they are in the same orbital So, since the identical ml value puts these two electrons in the same orbital, choice B must be wrong So how we choose between C and D? Well, so far the electrons are indistinguishable in the same atom, but the Pauli principle says that we must be able to distinguish them, so choice D must be wrong Choice C is indeed correct because you know that the last quantum number, ms, must be different for these two 12 as developed by Electronic Structure and the Periodic Table Test electrons So one will have an ms value equal to +1/2 and the other will be -1/2 This simply denotes that one is spinning clockwise and the other is spinning counter-clockwise So the spins are opposite and choice C is correct 17 D Again, to answer this question, you simply need to know the definition of ground-state and excited-state The groundstate of an atom is when all the electrons in the atom are in their lowest energy state If any electron has absorbed energy and been promoted to a higher energy orbital in the atom without actually leaving the atom, the atom is said to be in an excitedstate An atom can have any number of excited states depending on how many electrons have been promoted and what orbitals they end up in Usually, excited states are unstable and the atom will release energy and return to the ground state So which of these choices explains the difference between these two definitions? Choice A says that a difference in the number of neutrons leads to the difference between the ground and excited states As we just said however, it is the electrons that we are concerned with here Atoms of the same element with different numbers of neutrons are called isotopes Choice B says the number of electrons Well, we just said that no electrons can leave an atom when going between ground and excited states, so this is wrong Atoms of the same element with different numbers of electrons are called ions Choice C says atomic weight This pretty much means a different number of protons and/or neutrons Well, we know a different number of neutrons is wrong and that if the number of protons was different, we'd have a different atom, so choice C is wrong That leaves choice D as correct When electrons gain energy and change which atomic orbital they're in, they change their quantum numbers which in turn changes the electronic configuration around the nucleus, so choice D is indeed correct KAPLAN 13 ... each other KAPLAN MCAT ANSWER KEY: B A A B D 8 10 D C B A A 11 12 13 14 15 A D C A D 16 17 C D as developed by Electronic Structure and the Periodic Table Test ELECTRONIC STRUCTURE AND THE PERIODIC. .. to the nucleus, the smaller the radius The radii increase in going down a group because in the higher periods, the value of n increases: the larger n is, the larger the orbitals are, and the. .. periodic table Here's where some knowledge of how the periodic table is set up comes in handy In the period table, the first two columns are the s-block elements Their valence electrons are in the