Cracking the AP Chemistry Exam

Part IV

Content Review for the AP Chemistry Exam

Chapter 3

Big Idea #1: Atoms, Elements, and the Building Blocks of Matter

CHAPTER 3 ANSWERS AND EXPLANATIONS

Multiple-Choice

  1. D  This element has five peaks, meaning a total of five subshells. The final peak, which would be located in the 3p subshell, is slightly higher than the 3s peak to the left of it. A full 3s peak has two electrons, therefore there must be at least three electrons in the 3p subshell. The element that best fits this is phosphorus.

  2. B  The peaks, in order, represent 1s, 2s, 2p, 3s, and 3p.

  3. D  The less ionization energy that is required to remove an electron, the more kinetic energy that electron will have after ejection.

  4. D  Valence electrons are those in the outermost energy level. In this case, that is the third level, which has five valence electrons in it (two in 3s and three in 3p).

  5. D  The ion has three more electrons than the neutral atom, meaning the overall repulsion will be greater. The electrons will “push” each other away more effectively, creating a bigger radius.

  6. B  In all of these transition metals, the 3d subshell is the last one to receive electrons (Aufbau’s Principle). A 3d subshell has 5 orbitals which can hold two electrons each (Pauli’s Exclusion Principle). The electrons will enter one by one into each orbital before any pair up (Hund’s Rule), meaning manganese will have all five of its 3d electrons unpaired.

  7. C  The smaller an atom is, the more the “pull” of the nucleus can be felt and the easier it will be for that element to attract more electrons; thus the best answer is chlorine. (B) is a tempting answer; however, the additional protons in bromine are much farther away from the outside of the atom, as it has one full energy level greater than chlorine.

  8. D  Zinc has 10 electrons in the 3d subshell, filling it completely.

  9. D  Potassium has the highest number of protons out of all the options, therefore it exerts a higher nuclear charge on the electrons, pulling them in closer and creating a smaller atomic radius.

10. B  Neutral sodium in its ground state has the electron configuration shown in choice (C). Sodium forms a bond by giving up its one valence electron (3s1) and becoming a positively charged ion with the same electron configuration as neon, which is option (B).

11. D  As we move from left to right across the periodic table within a single period (from sodium to chlorine), we add protons to the nuclei, which progressively increases the pull of each nucleus on its electrons. So chlorine will have a higher first ionization energy, greater electronegativity, and a smaller atomic radius.

12. C  The further an electron is from the nucleus, the less binding energy it has and the easier it is to eject. The electrons in 3p would have the lowest binding energy and thus the highest velocity (and kinetic energy). A 4s electron would be faster; however, silicon has no electrons in the 4ssubshell while in its ground state.

13. A  Via c = vλ, we can see that there is an inverse relationship between wavelength and frequency (as one goes up, the other goes down). So, ultraviolet radiation has a higher frequency, and via E = hv, also has more energy.

14. B  The atomic mass is the sum of the neutrons and protons in any atom’s nucleus. Since atomic number, which indicates the number of protons, is unique to each element we can subtract this from the weight to find the number of neutrons. The atomic number of B is 5; hence, the 11B has 6 neutrons, only 1 in excess. The atomic number of Cl is 17, so 37Cl has 20 neutrons, 3 in excess. The atomic number of Mg is 12, so 24Mg has the same number of neutrons as protons. The element Ga has an atomic number of 31, meaning there are 39 neutrons in the given nucleus.

15. D  Electronegativity describes how easy it is for an element to attract additional electrons. Because halogens are smaller than other elements in their period (save for the noble gases, which have full energy levels and are not likely to gain additional electrons), they tend to have the highest electronegativity values within their period.

16. D  Nonmetals appear on the right side of the periodic table, and so are closer to having a full energy level. Nonmetals are much more likely to gain electrons to fill their levels, and it is difficult to remove their electrons for that reason.

17. B  The ionization energy will show a large jump when enough electrons have been removed to leave a stable shell. In this case, the jump occurs between the second and third electrons removed, so the element is stable after two electrons are removed. Magnesium (Mg) is the only element on the list with exactly two valence electrons.

18. C  Radiation Energy = Binding Energy + Kinetic Energy

Radiation energy (photon) = 2.5 eV + 2.0 eV

Radiation energy (photon) = 4.5 eV

19. A  If the wavelength is decreased, that increases the frequency via c = λv (remember, the speed of light is constant). An increased frequency increases the amount of photon energy via ΔE = hv. As the binding energy of the electron would not change, the excess radiation energy would turn into kinetic energy, increasing the velocity of the electron.

20. D  The intensity of the light is independent of the amount of energy that the light has. Energy is entirely based on frequency and wavelength, and the brightness of the light would not change the amount of radiation energy. Thus, the amount of kinetic energy of the ejected electrons would not change either.

Free-Response

  1.  (a)  Ionization energy is the energy required to remove an electron from an atom. The outermost electron in Ca is at the 4s energy level. The outermost electron in Mg is at the 3s level. The outermost electron in Ca is at a higher energy level and is more shielded from the nucleus, making it easier to remove.

(b)  Calcium has two electrons in its outer shell. The second ionization energy will be larger than the first but still comparable because both electrons are being removed from the same energy level. The third electron is much more difficult to remove because it is being removed from a lower energy level, so it will have a much higher ionization energy than the other two.

(c)  The height of the peaks on a PES represent the relative number of electrons in each subshell. In carbon, all three subshells hold two electrons (1s22s22p2), and thus all peaks are the same height. In oxygen, the 2p subshell has four electrons, meaning its peak will be twice as high as the other two.

(d)  The valence electron to be removed from magnesium is located in the completed 3s subshell, while the electron to be removed from aluminum is the lone electron in the 3p subshell. It is easier to remove the electron from the higher-energy 3p subshell than from the lower energy (completed) 3s subshell, so the first ionization energy is lower for aluminum.

  2.  (a)  Electronegativity is the pull of the nucleus of one atom on the electrons of other atoms; it increases from P to S to Cl because nuclear charge increases. This is because as you move from left to right across the periodic table, atomic radii decrease in size. Increasing nuclear charge means that Cl has the most positively charged nucleus of the three and will exert the greatest pull on the electrons of other atoms.

(b)  Electronegativity is the pull of the nucleus of one atom on the electrons of other atoms; it decreases from Cl to Br to I because electron shells are added. The added electron shells shield the nucleus, causing it to have less of an effect on the electrons of other atoms. Therefore, iodine will exert the least pull on the electrons of other atoms.

(c)  Atomic radius increases from Li to Na to K because electrons are being added in higher energy levels, which are farther away from the nucleus; therefore, the K atom is the largest of the three.

(d)  Atomic radius increases from Al to Mg to Na because protons are being removed from the nucleus while the energy levels of the valence electrons remains unchanged. If there are fewer positive charges in the nucleus, the electrons of Na will be less attracted to the nucleus and will remain farther away.

  3.  (a)  Element 3 is chlorine. Chlorine and phosphorus would have the largest atomic radii as they both have three energy levels with electrons present. However, chlorine would be smaller than phosphorus because it has more protons (a higher effective nuclear charge). Additionally, chlorine would have a higher ionization energy than phosphorus due to its smaller size and greater number of protons.

(b)  Element 2 is phosphorus, and therefore the outermost energy level would be n = 3. Phosphorus has two electrons in 3s and three electrons in 3p for a total of five valence electrons.

(c)  Electronegativity increases as atomic radius decreases, so it is expected that element 4 (oxygen) would have the highest electronegativity value. Alternatively, electronegativity increases as an energy level comes close to being full, so it is possible that element 3 (chlorine) may have the highest electronegativity as it is only one electron away from filling its outermost energy level. (Either answer is acceptable with the proper justification.)

(d)  Element 4 is oxygen, so it would be expected to have three peaks in a PES, one for each subshell. The first two peaks would be the same height because there are two electrons each in the 1s and 2s subshells. The final peak would be twice the height of the others as there are four electrons in oxygen’s 2p subshell.