Posted: January 1st, 2020
Ap Chemistry Free Response Answers
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 than normal increase in ionization energy between the first and second ionization energies. The third ionization energy is increased the most because it starts in the most standard form. When you combine this with a lower than normal second ionization energy, you get a very large increase in ionization energy. 3. (a) As you go to the right of the period, there are more protons in the nucleus.
The greater attraction makes it more difficult to remove electrons and first ionization energy is the energy necessary to remove an electron from a neutral atom. (b) Although the general trend is to have Boron with a higher first ionization energy than Berylium, Boron’s ionization potential is lowered and Berylium’s ionization potential is raised, the order is reversed. (c) O loses one electron and makes it easier to remove the electron and lowers the ionization potential. For nitrogen, it more difficult to remove the electron and raises the ionization potential.
And since Oxygen’s ionization potential is lowered and Nitrogen’s ionization potential is raised, the order is reversed. (d) Na has a lower first ionization energy than Li and also a lower ionization energy than Ne. Ne has the second highest first ionization energy of all the elements. “1s2” is the most preferred electron configuration. “s2 p6” of other shells are also highly preferred. Ne has the second highest first ionization potential because it’s “2s22p6”. 4. (a) The type of decay expected for Carbon-11 would be positron emission. 116C -; 115B + 01e (b) The type of decay expected for Carbon-14 would be Beta Decay as well. 46C -; 147N + 0-1? (c) Gamma rays have no mass or charge, so they need not be shown in nuclear equations. (d) Measure the amount of Carbon-14 in the dead wood and compare with the amount of Carbon-14 in a similar living object. 5. (a) 23494Pu -; 23092U + 42? (b) The missing mass has been converted into energy (E = mc2). (c) A line should be drawn curving downward from the path of the dotted line. This will represent the path of the alpha particles which are repelled by the positive plate and attracted by the negative one. A second line should be drawn upward from the path of the dotted line.
This will represent the path of the beta particles which are repelled by the negative plate and attracted by the positive one. The line should curve more than the one for the alpha particles. A third line should be drawn as a continuation of the dotted line. This will represent the gamma rays. (d) Incineration is a chemical process. The only thing any chemical process can do is connect radioactive atoms to other atoms, which has no effect on the radioactivity. 6. (a) As you go down the column in the alkali metals, the outer shell electrons are farther from the nucleus.
The attraction for the outer shell electrons is decreased and because the attraction is decreased, therefore the melting point decreases. (b) Intermolecular forces determine boiling and melting points. Halogens are all diatomic, which means they bond with themselves. In these diatomic compounds, the only intermolecular force is London forces. The larger molecules can form temporary dipoles easier than small molecules. The larger molecules as you go down the column have a greater attractive force. This increases the melting point as you go down the column. 7. a) As radius increases the heat of reaction decreases. Which means less energy released by ionic attraction. (b) As ionization energy increases the heat of reaction decreases, which means more energy is required to form M2+ while other factors remain unchanged 8. Metals are good conductors of heat, generally malleable, and react by losing electrons to form cations. They tend to have “s1”, “s2”, “s2p1”, or “s2p2” as their outer shell. Most metals have just “s1” or “s2”. Nonmetals are poor conductors of heat, brittle, and gain electrons when reacting with metals to form anions.
Nonmetals have either 3, 4, 5, or 6 electrons in the p subshell in addition to s2 of the same shell number. When the last subshell is a d, the outer shell is s2 of the next shell. Occasionally there will be only 1 electron in the s subshell and this explains when the transition elements are metals. When the last subshell is a “f”, the outer shell is s2 of the second higher shell and this explains when the lanthanides and actinides are metals. This proves how more than half of the periodic table are metals. 9. (a) “you have not learned this one yet” (b) “F2” has the highest electronegativity and electron affinity.
Thus it has the greatest attraction for extra electrons. F2 + 2e -; 2F – 1 This makes the reaction more likely to occur. “I2” has the lowest electronegativity and electron affinity. Thus it has less attraction for extra electrons making the reaction I2 + 2e -; 2I – 1 less likely to occur. Because it can disperse the charge better, the reaction does occur. (c) The trend for alkali metals shows a very small variation in reducing strength without a real trend. Cesium has the lowest ionization potential and Lithium has the highest ionization potential. However, there is not a great difference in the alkali metals.
