Ap Chemistry Free Response Answers Essay Example
1.
(a) I, III, and IV are correct. II is not correct. To explain III, de Broglie's equation states l = h/(mv), so nl = nh/(mv) = 2pi(r). Where: l = wavelength, v = velocity of electron, n = some positive integer, r = distance of electron from center, m = mass of electron. Solve, get mvr = L = nh/2pi.
(b) The current wave mechanical model for the atom states that there are an integer number of wavelengths in every standing integer number (n).
2.
(a) The first shell electrons in Lithium are the closest electrons to the nucleus. In addition, there are proportionally more protons to electrons. This pulls the electrons even closer to the nucleus. And in Potassium, the outer shell electrons are a substantial distance from the nucleus. There are a greater number of protons than electrons; however, the large number of electrons dissipates the effect. This is in addition to Lithium being a much smaller neutral atom than Potassium because of the difference in the outer shells.
(b) The outer shell for Cl is the same as Cl-; however, Cl- has more electrons being attracted by the same number of protons. This weakens the attraction per electron. Since the attraction is weaker, the electrons are farther from the nucleus. Since the attraction is stronger for Cl, the electrons are closer to the nucleus.
(c) Although the normal trend is for the ionization energy to increase going to the right in a period, aluminum has a lowered ionization energy and magnesium has a raised ionization energy due to the electron configurations of these two ionizations. This reverses the order of ionization energies.
(d) The ionization energy increases each time an electron is removed because there are fewer electrons attracted by the same number of protons while magnesium starts off at a relatively high value because it begins in one of the preferred forms. The second ionization energy is lowered because losing an electron forms a preferred form and because of this, this is a smaller
(a) I, III, and IV are correct. II is not correct. To explain III, de Broglie's equation states l = h/(mv), so nl = nh/(mv) = 2pi(r). Where: l = wavelength, v = velocity of electron, n = some positive integer, r = distance of electron from center, m = mass of electron. Solve, get mvr = L = nh/2pi.
(b) The current wave mechanical model for the atom states that there are an integer number of wavelengths in every standing integer number (n).
2.
(a) The first shell electrons in Lithium are the closest electrons to the nucleus. In addition, there are proportionally more protons to electrons. This pulls the electrons even closer to the nucleus. And in Potassium, the outer shell electrons are a substantial distance from the nucleus. There are a greater number of protons than electrons; however, the large number of electrons dissipates the effect. This is in addition to Lithium being a much smaller neutral atom than Potassium because of the difference in the outer shells.
(b) The outer shell for Cl is the same as Cl-; however, Cl- has more electrons being attracted by the same number of protons. This weakens the attraction per electron. Since the attraction is weaker, the electrons are farther from the nucleus. Since the attraction is stronger for Cl, the electrons are closer to the nucleus.
(c) Although the normal trend is for the ionization energy to increase going to the right in a period, aluminum has a lowered ionization energy and magnesium has a raised ionization energy due to the electron configurations of these two ionizations. This reverses the order of ionization energies.
(d) The ionization energy increases each time an electron is removed because there are fewer electrons attracted by the same number of protons while magnesium starts off at a relatively high value because it begins in one of the preferred forms. The second ionization energy is lowered because losing an electron forms a preferred form and because of this, this is a smaller