Atomic Structure and Periodic Trend
Compared to carbon, nitrogen has a: A. greater ionization energy and more negative electron affinity. B. smaller ionization energy and more positive electron affinity. C. smaller ionization energy and more negative electron affinity. D. greater ionization energy and more positive electron affinity.
A. Compared to carbon, nitrogen has a greater ionization energy and more negative electron affinity. This is a 2 x 2 question and is best approached by evaluating one of the arguments at a time to eliminate answer choices. Carbon and nitrogen are in the same period, and nitrogen is to the right of carbon. Ionization energy increases going left to right in a period due to greater effective nuclear charge, resulting in valence electrons being held more tightly. Thus, nitrogen will have a greater ionization energy, eliminating two answer choices. Evaluating the other argument, electron affinity becomes more negative (i.e., more favorable) going left to right in a period (once again due to the greater effective nuclear charge), making nitrogen more negative compared to carbon. Thus compared to carbon, nitrogen has a greater ionization energy and a more negative electron affinity. Periodic Trends: Ionization Energy/Electron Affinity
In organic solution, F- deprotonates dissolved HCl. Which of the following explains this observation? A. F- has a smaller radius than Cl-. B. F- is a stronger acid than Cl-. C. F- has a greater electronegativity than Cl-. D. F- has a lower ionization energy than Cl-.
A. Since F- has a smaller radius than does Cl-, it is less stable with excess charge, and therefore a better Bronsted base. It outcompetes Cl- for the proton. Periodic Trends: Atomic/Ionic Radius
The outermost shell of electrons in the noble gases (with the exception of helium) have which of the following electron configurations? A.ns2np6 B. ns2np4 C.ns2np2 D.ns2np8
A. The outermost shell of electrons in the noble gases (with the exception of helium) have an ns2np6 electron configuration. Noble-gas configurations are characterized by filled valence s and p subshells. Since the s holds 2 electrons and the p holds 6, the answer is ns2np6. Atomic Structure: Electron Configuration
The reactivity of alkali metals follows the trend Cs > Rb > K > Na > Li. Which of the following is the best explanation for this trend? A. The ionization energy increases from Cs to Li. B. The electron affinity increases from Li to Cs. C. The electronegativity decreases from Cs to Li. D. The atomic size decreases from Li to Cs.
A. The reactivity of alkali metals follows the trend Cs > Rb > K > Na > Li. The best explanation for this trend is that the ionization energy increases from Cs to Li. Cesium has the lowest ionization potential of the group and therefore it can give up an electron more easily than the other metals. For this reason, it is the most reactive of the group. Periodic Trends: Ionization Energy/Electron Affinity
Which of the following is the correct electron configuration for the iron ions present in rust, Fe2O3? A. [Ar] 3d5 B. [Ar] 4s23d3 C. [Ar] 4s23d4 D. [Ar] 3d6
A. [Ar] 3d5 is the correct electron configuration for the iron ions present in rust, Fe2O3. First, determine what the charge on the iron ion should be from the chemical formula for rust. Oxygen will always have a -2 charge in an ionic compound, so in order to make the formula neutral, each of the two iron ions will be Fe3+. The electron configuration for a neutral iron atom is [Ar] 4s23d6, and three electrons must be removed to yield Fe3+. Therefore, choices "[Ar] 4s23d4" and "[Ar] 3d6" can be eliminated since both show only two electrons have been removed. When transition metals lose electrons, they first lose them from the s subshell. The remaining electron that must be removed comes from the 3d subshell, making "[Ar] 3d5" the best answer.\ Atomic Structure: Electron Configuration
Which of the following appropriately ranks the atoms from smallest to largest atomic radius? A. Ge < Al < P < C B. C < P < Al < Ge C. C < Al < P < Ge D. Ge < P < Al < C
B. C < P < Al < Ge appropriately ranks the atoms from smallest to largest atomic radius. This is a ranking question, which is best approached by evaluating the extremes. For a given family, atomic radius increases as period increases due to greater shielding, which means that Ge must be larger than C. This eliminates choices Ge < P < Al < C and Ge < Al < P < C. Now, we must evaluate Al vs. P. These two atoms are in the same period. However, atomic radius decreases as you go left to right in the same period due to a greater effective nuclear charge, which means that P is smaller than Al. Thus, the atoms from smallest to largest atomic radius is C < P < Al < Ge. Periodic Trends: Atomic/Ionic Radius
Compared to the electronegativity of iodine, the electronegativity of chlorine is: A. less, because chlorine has a smaller nuclear charge which translates into a smaller force exerted on electrons. B. greater, because decreased nuclear shielding allows for a stronger pull on the valence electrons. C. greater, because chlorine has a greater nuclear charge which translates into a greater force exerted on electrons. D. less, because increased nuclear shielding results in a weaker pull on the valence electrons.
B. Compared to the electronegativity of iodine, the electronegativity of chlorine is greater, because decreased nuclear shielding allows for a stronger pull on the valence electrons. The periodic trend for electronegativity predicts that chlorine is more electronegative than iodine, which eliminates "less, because chlorine has a smaller nuclear charge which translates into a smaller force exerted on electrons" and "less, because increased nuclear shielding results in a weaker pull on the valence electrons". Since iodine (Z = 53) has more protons than chlorine (Z = 17), it has a greater nuclear charge, so "greater, because chlorine has a greater nuclear charge which translates into a greater force exerted on electrons" is false. Periodic Trends: Electronegativity
Which of the following could be the electron configuration of a boron atom in an excited state? A. 1s22s22p1 B. 1s22s12p2 C. 1s22s12p1 D. 1s22s22p2
B. The electron configuration of a boron atom in an excited state could be 1s22s12p2. A boron atom contains 5 electrons, so choices "1s22s22p2" and "1s22s12p1" are eliminated immediately, since they account for 6 and 4 electrons, respectively. The configuration in choice "1s22s22p1" is for a boron atom in the ground state, so the answer must be choice "1s22s12p2".
Give the electron configuration of a ground state Zn2+ ion. A. [Ar] 4s23d8 B. [Ar] 3d10 C. [Ar] 4s13d9 D. [Ar] 4s23d10
B. The electron configuration of a ground state Zn2+ ion is [Ar] 3d10. First, eliminate choices that involve the wrong number of electrons. Since an atom of zinc contains 30 electrons, a Zn2+ ion contains 28, eliminating [Ar] 4s23d10. Recall that when electrons are removed from a transition metal such as zinc, the electrons always come from the valence s orbital first. Therefore, the electron configuration, [Ar] 3d10 is the correct one. Atomic Structure: Electron Configuration
An excited electron drops down to its initial energy level within an atom and a photon is emitted. All of the following characteristics about photons and electron energy levels are true EXCEPT: A. the product of the wavelength and frequency of a photon is equal to 3.0 × 108 m/s. B. the energy of a photon is directly proportional to its wavelength. C. the difference in energy between adjacent energy levels decreases as the energy levels increase. D. the energy of a photon is directly proportional to its frequency.
B. The energy of electromagnetic radiation is directly proportional to its frequency and inversely proportion to its wavelength: E = hf = h(c/λ), where E is the energy of the photon, h is a constant, f is its frequency, λ is its wavelength, and c is the speed of light in a vacuum. Choice D is eliminated because photon energy and frequency are directly proportional. The relationship between f, λ, and c can also be expressed by: c = fλ = 3.0 × 108 m/s. Therefore, choice A is also a correct statement and is eliminated. As electron energy levels increase, the distance between adjacent levels and subsequently the difference in energy between adjacent levels decreases. Choice is C is another correct statement that can be eliminated. Atomic Structure: Electron Transitions Quantum Physics: Photoelectric Effect
Which one of the following correctly represents the electron configuration of sulfur in an excited state? A. 1s22s22p63s23p5 B. 1s22s22p63s23p44s1 C. 1s22s22p63s23p34s1 D. 1s22s22p63s23p4
C. 1s22s22p63s23p34s1 correctly represents the electron configuration of sulfur in an excited state. A sulfur atom contains 16 electrons, so the choices 1s22s22p63s23p5 and 1s22s22p63s23p44s1 are immediately eliminated since they each account for 17 electrons. The configuration in 1s22s22p63s23p4 is that of a ground-state sulfur atom, so the answer must be 1s22s22p63s23p34s1. Atomic Structure: Electron Configuration
A sample of excited hydrogen atoms emits a light spectrum of specific, characteristic wavelengths. The light spectrum is a result of: A. energy released as H atoms form H2 molecules. B. the light wavelengths which are not absorbed by valence electrons when white light is passed through the sample. C. excited electrons dropping to lower energy levels. D. particles being emitted as the hydrogen nuclei decay.
C. A sample of excited hydrogen atoms emits a light spectrum of specific, characteristic wavelengths. The light spectrum is a result of excited electrons dropping to lower energy levels. The line spectra of atoms are the result of the photons emitted when excited electrons drop to lower energy levels. Atomic Structure: Electron Transitions
Which of the following elements would be expected to exhibit the greatest ionization energy? A. Calcium B. Potassium C. Titanium D. Scandium
C. Of the elements listed, titanium would be expected to exhibit the greatest ionization energy. Ionization energy generally increases as we move to the right across a row in the periodic table. Therefore, of the choices given, titanium should have the highest ionization energy. Periodic Trends: Ionization Energy/Electron Affinity
Which of the following gives the ground-state electron configuration of a calcium atom once it acquires an electric charge of +2? A. [Ar] 4s23d2 B. [Ar] 4s2 C. [Ar] D. [Ar] 4s24p2
C. The ground-state electron configuration of a calcium atom once it acquires an electric charge of +2 is [Ar]. A calcium atom normally contains 20 electrons, so if it acquires a charge of +2, it must have lost 2 electrons, leaving just 18. Only [Ar] accounts for 18 electrons. Atomic Structure: Electron Configuration
What is the reactivity, as indicated by the tendency to lose an electron, of sodium compared to potassium? A. This cannot be determined from the information given. B. Potassium has a lower reactivity because potassium has more protons than sodium. C. Potassium has a higher reactivity because the valence electron on potassium is farther from the nucleus. D. Both metals have the same reactivity because the potassium and sodium valence electrons experience the same effective nuclear charge.
C. The reactivity of sodium compared to potassium is that potassium has a higher reactivity because the valence electron on potassium is farther from the nucleus. Reactivity of metallic elements increases down a column of the periodic table. This increase in reactivity corresponds with increased shielding from inner electron shells, increased atomic radius, and decreased ionization energy. Thus, electrons are easier to remove in potassium versus sodium. Atomic Structure: Electron Configuration
Which of the following is the correct electron configuration of Ge+? A. [Ar] 4s23d104p2 B. [Ar] 4s13d104p2 C. [Ar] 4s23d104p1 D. [Ar] 4s24d104p1
C. Ge+ contains 13 more electrons than Ar, eliminating choice A. In the ionization of germanium, the least stable valence electron will be lost. This corresponds to the electron in the highest energy level, which would be the 4p subshell, eliminating choice B. The correct value of n for the p subshell is found by subtracting 1 from the period number, eliminating choice D. Atomic Structure: Electron Configuration
An atom of beryllium is repelled by an externally produced magnetic field. Which of the following explains this phenomenon? A. The electrons of beryllium are all spin-paired and the atom is paramagnetic. B. The electrons of beryllium are not all spin-paired and the atom is paramagnetic. C. The electrons of beryllium are all spin-paired and the atom is diamagnetic. Correct Answer D. The electrons of beryllium are not all spin-paired and the atom is diamagnetic.
C. This is a two by two question. First, beryllium's electron configuration is 1s22s2, so all electrons are spin-paired, eliminating choices B and D. Atoms with all electrons spin-paired are diamagnetic, eliminating choice A. Diamagnetic atoms will be repelled by an externally produced magnetic field, and paramagnetic atoms (which have electrons that are not spin-paired) will be attracted to an externally produced magnetic field. Atomic Structure: Electron Configuration Magnetism: Magnetic Fields
Does a small atomic radius correspond to a high ionization energy? A. No, because the smaller space enhances electron-electron collisions, thus making it easier to remove these electrons. B. No, because the large concentration of negative charge in a smaller space causes increased repulsion, thus making it easier to remove these electrons. C. Yes, because atoms in lower periods exhibit an enhanced strong force on electrons. D. Yes, because the shorter distance between the positive nucleus and the negative electron enhances electrostatic attraction, and thus makes it difficult for these electrons to be removed
D. A small atomic radius does correspond to a high ionization energy because the shorter distance between the positive nucleus and the negative electron enhances electrostatic attraction, and thus makes it difficult for these electrons to be removed. The strategy to answer this question is to determine whether the question in the stem is true or not, so that we can eliminate half of the choices. Atomic radius increases and ionization energy decreases as period number increases. Thus, it is true that a small atomic radius corresponds to a high ionization energy. This eliminates the choices that begin with "No." Evaluating the final two choices, atoms in lower periods do not exhibit an enhanced strong force on electrons, because the strong force only pertains to holding the protons and neutrons in the nucleus together (and thus works over a very small distance which will never influence electrons). Thus, the choice "Yes, because atoms in lower periods exhibit an enhanced strong force on electrons" may be eliminated. The attractive force exhibited between charged particles such as protons in the nucleus and electrons is the electrostatic force, which increases in strength as the distance between the particles decreases. Periodic Trends: Ionization Energy/Electron Affinity
Compared to the atomic radius of S, the atomic radius of Al is: A. smaller, due to decreased nuclear charge. B. smaller, due to increased nuclear charge. C. larger, due to increased nuclear charge. D. larger, due to decreased nuclear charge.
D. Compared to the atomic radius of S, the atomic radius of Al is larger, due to decreased nuclear charge. Moving left to right across a period, nuclear charge increases due to increasing atomic number (number of protons). As the positive nuclear charge increases, effective nuclear charge increases and electrons are pulled closer to the nucleus, resulting in a decreased atomic radius. Aluminum is therefore larger than sulfur. This problem can also be treated as a two-by-two if you know the periodic trend for atomic radius, which decreases up and to the right. Based on this trend, it can be determined that Al is larger than S, eliminating the choices using "smaller." The distinction between the final two choices can then be determined based on the effect of nuclear charge. Periodic Trends: Atomic/Ionic Radius
Compared to the atomic radius of calcium, the atomic radius of gallium is: A. larger, because increased electron charge requires that the same force be distributed over a greater number of electrons. B. smaller, because gallium gives up more electrons, thereby decreasing its size. C. larger, because its additional electrons increase the atomic volume. D. smaller, because increased nuclear charge causes electrons to be held more tightly.
D. Compared to the atomic radius of calcium, the atomic radius of gallium is smaller, because increased nuclear charge causes electrons to be held more tightly. Atomic radius generally decreases as we move to the right across a row in the periodic table. Therefore, comparing calcium and gallium (both 4th period elements), gallium should have the smaller radius; this eliminates "larger, because increased electron charge requires that the same force be distributed over a greater number of electrons" and "larger, because its additional electrons increase the atomic volume". "Smaller, because gallium gives up more electrons, thereby decreasing its size" can be eliminated since there is no loss of electrons when we are simply comparing atomic size. "Smaller, because increased nuclear charge causes electrons to be held more tightly" is true and provides the correct explanation. Periodic Trends: Atomic/Ionic Radius
Which of the following is true of ionization energy? A. It increases with period, and the second ionization energy is greater than the first. B. It decreases with period, and the first ionization energy is greater than the second. C. It increases with period, and the first ionization energy is greater than the second. D. It decreases with period, and the second ionization energy is greater than the first.
D. Ionization energy decreases with period and the second ionization energy is greater than the first. This is a 2 x 2 question that is best approached by evaluating one of the arguments to eliminate answer choices. As period increases, ionization energy decreases due in part to the greater degree of shielding present. Thus, we can eliminate the two choices that begin with the choice "it increases." Evaluating the other argument, the second ionization energy is always greater than the first because electrons are more tightly bound after one is displaced (a more positively charged atom is less likely to give up an additional electron). Thus, ionization energy decreases with period and the second ionization energy is greater than the first. Periodic Trends: Ionization Energy/Electron Affinity
Which of the following is true of an electron in an excited state? A. It has absorbed a photon, and its energy has decreased. B. It has emitted a photon, and its energy has increased. C. It has emitted a photon, and its energy has decreased. D. It has absorbed a photon, and its energy has increased.
D. It is true that an electron in an excited state has absorbed a photon, and its energy has increased. Electrons become excited when they gain energy by absorbing a photon. Atomic Structure: Electron Transitions
Which of the following is the most stable ionization state for Sn? A. Sn+1 B. Sn+3 C. Sn+4 Your Answer D. Sn+2 Correct Answer
D. Sn+2 is the most stable ionization state for Sn. When Sn loses two electrons, the resulting configuration has a full and stable 4d10 subshell, making Sn2+ the most stable ionization state. Although losing two more electrons (from the 5s subshell) would also leave a full 4d10, the resulting Sn4+ has a greater reduction potential than Sn2+ due to its higher charge. Therefore, Sn4+ is less stable than Sn2+. Atomic Structure: Electron Configuration
Which of the following electron transitions (between energy levels, labeled by n) could account for the emission of photons of red, yellow, and blue light from a pure noble gas? A. n = 4 to n = 1, n = 4 to n = 2, and n = 4 to n = 3, respectively B. n = 1 to n = 0, n = 2 to n = 1, and n = 3 to n = 2, respectively C. n = 2 to n = 1, n = 3 to n = 2, and n = 4 to n = 3, respectively D. n = 4 to n = 3, n = 3 to n = 2, and n = 2 to n = 1, respectively
D. The electron transition (between energy levels, labeled by n) that could account for the emission of photons of red, yellow, and blue light from a pure noble gas is n = 4 to n = 3, n = 3 to n = 2, and n = 2 to n = 1, respectively. The energy of the photons satisfy red < yellow < blue. Therefore, the energy difference between the energy levels that produce these photons must follow the same order. Since the spacing between energy levels decreases with increasing n, then n = 4 to n = 3, n = 3 to n = 2, and n = 2 to n = 1 is best. Atomic Structure: Electron Transitions
Give the electronic configuration of vanadium, V. A. [Ar] 4s13d4 B. [Ar] 3d5 C. [Ar] 4s33d2 D. [Ar] 4s23d3
D. The electronic configuration of vanadium, V is [Ar] 4s23d3. Vanadium, one of the transition metals, contains 23 electrons, so its first 18 electrons have the configuration of the noble gas argon; the question is where to put the remaining 5 electrons. Since the 4s subshell fills before the 3d, the answer must be [Ar] 4s23d3. [Ar] 4s33d2 should have been eliminated right away, since any s subshell can never hold more than 2 electrons. Atomic Structure: Electron Configuration
In the ionization of titanium, which subshell will lose its electrons first and why? A. The 3d subshell because it has the least stable orbitals. B. The 4s subshell because it has the highest ionization energy. C. The 3d subshell because it has the highest energy. D. The 4s subshell because it has the least stable orbitals.
D. If a subshell has a high energy, then it is less stable, which means that choices A and C are the same and both must be wrong. Ionization energy is the amount of energy needed to remove an electron from an atom. The lower the ionization energy, the more easily an atom will be ionized (choice B is incorrect). In transition metals, such as titanium, the electrons in the (n)s subshell are ionized before the electrons in the (n-1)p subshell. When the (n-1)p subshell is occupied, it pushes the (n)s subshell to a higher and less stable energy level. Atomic Structure: Electron Configuration Atomic Structure: Orbitals
Rank the following atoms/ions in order of increasing atomic/ionic radius: I. Ne II. Ar III. S2- IV. Cs+ A. I < IV < II < III B. I < II < III < IV C. III < II < I < IV D. IV < I < II < III
The following atoms/ions in order of increasing atomic/ionic radius is I < II < III < IV. For ranking problems, start with the extremes. Cesium ion, despite its positive charge, possesses an additional two electron shells than any of the other answer choices and will be largest (eliminating answer choices "I < IV < II < III" and "IV < I < II < III"). Neon, with its large effective nuclear charge and limited number or shells, is the smallest (choice "I < II < III < IV" is correct). Sulfur and argon possess the same number of electrons, but argon has a greater effective nuclear charge which accounts for its smaller size. Periodic Trends: Atomic/Ionic Radius
