Big Idea 1: “Atomic Theory and Structure” AP LEVEL ASSESSMENT
DUE: Mon 2/18
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
description is phosphorous.
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.
KEY
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 elections), 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.
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).
17. B The ionization energy will how a large jump when enough electrons have been
removed to leave a stable shell. In this case, the jump occurs between the
second and their 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.
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.
18. C Ephoton = IE + KE = 2.5 eV + 2.0 eV = 4.5 eV
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 bbe felt and the
easier it will be for that element to attract more electrons; this the best
answer is chlorine. (B) is a tempting answer; however, the additional
protons in bromine are much farther way 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
is exerts a higher nuclear charge on the electrons, attracting them to a
greater degree 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 PT 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 is 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 4s subshell while
in its ground state.
13. A Via c = νλ, we can see the 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=hν, also has more
energy.
14. A 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
6neutrons, 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.
19. A If the wavelength is decreased, that increases the frequency via c = νλ,
(remember, the speed of light is constant). An increased frequency
increases the amount of photon energy via ΔE = hν. As the binding energy
of the electron would not change, the excess radiation energy would turn
into KE, 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 KE of the ejected electrons would not change either.
21. (SFR) Explain each of the following in terms of atomic and molecular structures and/or forces. (Note, these are the key features each response should contain)
(a) The first ionization energy for magnesium is greater than the first ionization energy for calcium.
-Outermost e− in Ca 4s sublevel (subshell/energy level)
-Outermost e− in Mg 3s sublevel (subshell/energy level)
-Ca is at a higher energy level, increased e− shielding (or shielding effect) results in a decrease of Zeff (or Effective Nuclear Charge)
-Reducing the amount of energy required to induce e− removal.
(b) The first and second ionization energies for calcium are comparable, but the third ionization energy is much larger.
-Ca has 2 valence e−
-2nd IE will be larger than 1st IE but comparable as they are removed from same energy level
-3rd is being removed for an inner (non-valence) energy level. Increased Zeff (or Effective Nuclear Charge) as a result of decreased shielding effect
-Increased amount of energy required to induce 3rd e− removal.
(c) There are three peaks of equal height in the PES of carbon, but on the PES of oxygen the last peak has a height twice as high as all the others.
-Peak height on a PES indicates relative # of e− in each sublevel (subshell)
-In C, all 3 subshells hold two e−, 1s2 2s2 2p2, and are thus the same height
-In O, 1s2 2s2 2p4, the 2p4 has 4 e−, having twice the # of e− compared to the other peaks will result in a peak twice as large
(d) The first ionization energy for aluminum is lower than the first ionization energy for magnesium.
-Outermost e− in Mg is in the full 3s sublevel (subshell/energy level)
-Outermost e− in Al is the single e- occupied 3p sublevel (subshell/energy level)
-Easier to remove lone e- in p- sublevel (subshell) as p- e- do not penetrate the nuclear region as greatly as s- e- do and are therefore not as tightly held.
-Reducing the amount of energy required to induce e− removal.
22. (SFR) Use your knowledge of thee periodic table of the elements to answer the following questions.
(a) Explain the trend in electronegativity from P to S to Cl.
-P, S, Cl all in same period and EN increase from P to S, to Cl
-L to R across period atomi radii decrease in size due to increased nuclear charge and a constant e- shielding effect.
-Cl has most positively charge nucleus of 3 and will exert greatest attraction for e(b) Explain the trend in electronegativity from Cl to Br to I.
-Cl, Br, I all in same group and EN decrease from Cl to Br to I
-Down a group, electron shells (energy levels) are added, increasing the shielding effect and thus decreasing Zeff (or Effective Nuclear Charge)
-Iodine will exert least pull as it has the most energy level.
(c) Explain the trend in atomic radius from Li to Na to K.
-Li, Na, K all in same group and AR increased from Li to Na to K
-Down a group, electron shells (energy levels) are added, increasing the shielding effect and thus decreasing Zeff (or Effective Nuclear Charge)
-K atom is largest as has the most shells (energy levels, 4)
(d) Explain the trend in atomic radius from Al to Mg to Na.
-Al, Mg, Na all in same period and AR decreases from Na to Mg to Al
-L to R across period atomi radii decrease in size due to increased nuclear charge and a constant e- shielding effect.
-Decrease nuclear charge of Na will result in decreased valence e- attraction and thus an increased atomic radii
23. (LFR) The table below gives data on four different elements, in no particular order:
Carbon, Oxygen, Phosphorus, and Chlorine
Atomic radius
(pm)
First Ionization energy
( kJ/mol )
Element 1
170
1086.5
Element 2
180
1011.8
Element 3
175
1251.2
Element 4
152
1313.9
(a) Which element is element number 3? Justify your answer using both properties.
-Chlorine, Cl
-Cl and P would have largest AR as both have 3 energy levels and thus greatest shielding effect
-Cl has a smaller AR than P as its increased Z (nuclear charge) results in an increased Zeff (Effective Nuclear Charge)
-Cl has a higher IE than P due to its decreased AR and resulting increased Zeff (Effective Nuclear Charge)
(b) What is the outermost energy level that has electrons in element 2? How many valence electrons does element 2 have?
-Phosphorous, P
-Outermost, n = 3 (3rd energy level)
-5 valence electrons (2 in 3s and 3 in 3p)
(c) Which element would you expect to have the highest electronegativity? Why?
-Oxygen, O
-EN increase as AR decrease (partial credit) due an increased Zeff (Effective Nuclear Charge)
-O has smallest AR and thus most attraction for e−
(d) How many peaks would the PES for element 4 have and what would the relative heights of these peaks be to each other.
-Element 4 is Oxygen, O
-As EC is 1s2 2s2 2p4, would be expected to have 3 peaks (one for each subshell/sublevel)
-As 2p4 has 4 e− while 1s2 and 2s2 have 2 e−, the last peak will be twice as large as the other two as it has twice the # of electrons.