Periodicity Presenter APD Cover - UHS chemistry · 2018-09-07 · Periodicity ELECTRON AFFINITY −...

AP* is a trademark of the College Entrance Examination Board. The College Entrance Examination Board was not involved in the production of this material.

Copyright © 2008 Laying the Foundation®, Inc., Dallas, TX. All rights reserved. Visit: www.layingthefoundation.org

AP* CHEMISTRY

Periodicity Periodic Trends − What to Write to Wow `Em

Teacher Packet

AP* is a trademark of the College Entrance Examination Board. The College Entrance Examination Board was not involved in the production of this material.


Periodicity Periodic Trends and What to Write to Wow `EM

Objective To review the student on the concepts, processes and problem solving strategies necessary to successfully answer questions dealing with periodic trends. Standards Periodicity is addressed in the topic outline of the College Board AP Chemistry Course Description Guide as described below. AP Chemistry Exam Connections Periodic trends are tested every year on the multiple choice and in most years on the free response portion of the exam. The list below identifies free response questions that have been previously asked. These questions are available from the College Board and can be downloaded free of charge from AP Central http://apcentral.collegeboard.com.

Free Response Questions 2008 Question 5 (a−c) 2007 B Question 6 2006 Question 8 (c) 2006 B Question 7 (a & c) 2005 Question 7 (c) 2005 B Question 8 (c) 2002 Question 6 (a & b) 2003 B Question 7 (a & d) 2000 Question 7 (c)

I. Structure of Matter A. Atomic Theory and Atomic Structure

5. Periodic relationships including, for example, atomic radii, ionization energies, electron affinities, oxidation states


Periodicity Periodic Trends − What to Write to Wow `Em

What I Absolutely Have to Know to Survive the AP Exam

Atomic radii; ionic radii; electronegativity; ionization energy; electron affinity; comparison of properties between two elements, etc…

IMPORTANT

The trends on the periodic table for each property listed are just that, TRENDS. It is NEVER ok to use the trend as your justification for an answer. It is only a reminder that allows you to develop the correct explanation of the forces at work in the atom. Use the arguments below to get justification points in your answers...

Trends in the Periodic Table Justifying all of the trends on the periodic table can be simplified using these two generalizations.

Use Zeff (effective nuclear charge) to justify trends across a period. Use increased distance (greater value of n) to justify trends down a group (also can argue based on the shielding effect…)

ATOMIC RADIUS − A boundary beyond which the electron rarely strays. The “size” of the atom is affected by two things; the attraction the protons in the nucleus have towards the electron cloud and the number of energy levels of electrons (n) in the cloud.

As you move across a period… Atomic radii decrease… i.e. the atom becomes smaller

Why…? The greater the effective nuclear charge (Zeff) the greater the attraction the nucleus has towards the electrons; therefore, the electron cloud is pulled closer to the nucleus resulting in a smaller atomic radius.

As you move down a family… Atomic radii increase… i.e. the atom becomes larger

Why…? The distance from the nucleus to the valence level increases due to the addition of energy levels, n. The attractive forces of the nucleus dissipate with this increased distance making the atom larger − i.e. electrons at higher n values are freer to drift than those at lower n values as they are less attracted to the nucleus… it is also acceptable to argue that full energy levels provide some shielding between the nucleus and the outer electrons, decreasing their attraction to the nucleus.

4 of 15


Periodicity

IONIZATION ENERGY (IE) − The energy required to remove a single electron from an atom in the gas phase. This requires an input of energy.

X(g) + energy → e− + X+(g)

Removing each subsequent electron requires even more energy (second IE, third IE, etc.) A HUGE energy price is paid if the removal of the electron is from another principal E level (core electron).

As you move across a period… Ionization Energy (IE) increases… i.e. it becomes more difficult; i.e. it takes more energy to remove the electron

Why…? The increased effective nuclear charge (Zeff) means that the outer electrons are being held more tightly, thus more energy is required to remove that electron, and ionization energy increases.

Exceptions to the trend occur between groups 2 and 13 and 15 and 16; in spite of increasing Zeff the ionization energy actually decreases when comparing elements in both of these two groups. Why?

Between 2 and 13: Group 2 elements have a valence electron configuration of ns2 while those in Group 13 are ns2np1. This drop in energy required to remove the electron is due to the fact that you are removing a p electron rather than an s electron. The p’s are less tightly held because they do not penetrate the electron cloud toward the nucleus as well as an s electron. The s is held more tightly since it penetrates more than p, d, and f sublevels.

Between 15 and 16: Group 15 elements have a valence configuration of ns2 np3 while those in Group 16 are ns2np4. This drop in energy required to remove the electron is due to the fact that two of the p4 electrons are paired within one of the orbitals, whereas the p3 electrons are all unpaired. Thiscauses an increased repulsion between the 2 electrons, which lowers the amount of energy required to remove it! NEVER, EVER, NEVER, EVER say that the half filled sublevel is more stable!!!

As you move down a family… Ionization Energy (IE) decreases… i.e. becomes less difficult; i.e. it takes less energy to remove the electron Why…? The increased distance from the nucleus to the outer electrons (the ones being removed) and increased shielding by full principal energy levels (n) means it requires less energy to remove an electron, thus ionization energy decreases.

Atomic Number

Ioni

zatio

n En

ergy

s2p1 “glitch”: remember p is less penetrating…

p4 “glitch”: think added electron−electron repulsion

5 of 15


Periodicity

ELECTRON AFFINITY − Can be thought of as the energy associated with the addition of an electron to a neutral atom. The process can require or release energy.

X(g) + e− → + X−(g) If the addition of the electron is exothermic, then you will see a negative sign on the energy change. The converse is also true. The more negative the quantity the more energy is released. The exothermic values can be confusing since − 500 kJ represents a higher electron affinity than −100 kJ. Remember the negative sign is simply indicating the direction of energy flow.

As you move across a period… Electron Affinity increases… i.e. more energy is released with the addition of the electron Why…? The increased effective nuclear charge (Zeff) means that the outer electrons are being held more tightly, thus better attracting the electron that is being added. It is important to note that the electron-electron interactions wreak havoc on some of these trends.

As you move down a family… Electron Affinity decreases… i.e. less energy is released with the addition of the electron Why…? The increased distance from the nucleus to the outer electrons (where the electron will be added) and increased shielding by full principal energy levels (n) means the nucleus attracts said electron less effectively, thus less energy is released or the process becomes less negative from an energy point of view. IONIC RADIUS − The same “size trends” found with atomic radii apply when comparing cations to cations and anions to anions.

Cations v. Atoms When an atom loses an electron (forms a cation) its radius shrinks Why…? The loss of electrons means there is an increase in the effective nuclear charge (Zeff) as the number of protons (which remain unchanged) are attracting fewer electrons; therefore, the remaining electrons are closer and more tightly held. If a certain number of electrons is lost in the formation of the cation, an entire energy level, n, will be lost.

Anions v. Atoms When an atom gains an electron (forms an anion) its radius increases Why…? The additional electron lessens the effective nuclear charge (Zeff) as the ratio of electrons to protons increases. There is also enhanced electron-electron repulsion; therefore, the nucleus doesn’t attract or pull the electrons as closely, so the anion is larger than its atom.

6 of 15


Periodicity

ELECTRONEGATIVITY − The ability of atoms to attract a shared electron pair in a covalent bond. Think tug of war!

As you move across a period… Electronegativity increases… i.e. the atom has an increased attraction for the electrons Why…? The increased effective nuclear charge (Zeff) means that the outer electrons are more attracted, thus the atom is smaller and can better “pull” or attract the bonded electrons.

As you move down a family… Electronegativity decreases… i.e. the atom has less attraction for the electrons Why…? The increased distance from the nucleus to the outer electrons and increased shielding by full principal energy levels (n) means the attraction is less and its ability to attract the bonded electrons is less.

Final thoughts

First recognize what the question is asking, i.e. what property is in question and what elements are being compared. To best answer the question use the following three steps…

1. Locate both elements on the periodic table and note the energy level, n, and sublevel of their valence electrons.

2. Compare their n values! If same n: argue with Zeff If different n: argue n vs. n distances

3. ENERGY! Comment of whether the electrons are more attracted or less attracted; or whether it takes more energy or less energy to remove the electron, etc… NEVER FORGET the ENERGY relationship!

Be careful when comparing isoelectronic species − those that have the same electron configurations, such as Ca2+, K+, and Cl−. If they all have the same number of electrons in the same energy levels, n, then all the property differences can be explained using effective nuclear charge! Determine the number of protons and answer accordingly. 7 of 15


Periodicity

Free Response 1. Answer the following questions about nitrogen, oxygen, fluorine and iodine using principles of atomic and

molecular structure.

Element First Ionization Energy (kJ mol−1)

N 1402 O 1314 F 1681 I ???

(a) Explain why N has a smaller first ionization energy than F.

Both N and F have their valence electrons at energy level 2, n=2, however N has a less effective nuclear charge (Zeff) than F (7 protons to 9). This means N does not have as great an attraction for its valence electrons as F does; therefore, it takes less energy to remove an electron from N than F.

1 point for explaining that N and F have the same n value. 1 point for explaining that N has less Zeff than F thus a smaller first ionization energy.

(b) Explain why O has a smaller first ionization energy than N.

Both O and N have their valence electrons at energy level 2, n=2, and even though O has a greater effective nuclear charge (Zeff) than N (8 protons to 7), O has paired electrons in one of its p orbitals. These paired electrons experience electron-electron repulsions and therefore take less energy to remove, hence the lower first ionization energy.

1 point for explaining that both atoms have the same n value. 1 point for explaining that O has more repulsion because of paired electrons thus a lower first ionization energy.

(c) Would you predict the first ionization energy of atoms of iodine to be greater than, less than, or equal to

that of fluorine? Explain. The first ionization energy of I would be less than that of F. The electron being removed from the F atom is in a 2p orbital, whereas the electron being removed from atoms of I is in a 5p orbital. The 5p orbital is farther away from, and their electrons not as attracted to, the nucleus as in the 2p orbital, thus it takes less energy to remove it.

1 point for noting the IE for I would be greater than F. 1 point for a proper justification based on the energy level difference.

(d) Which of the atoms listed in the table above would have the largest atomic radius? Explain.

Atoms of I would have the largest atomic radius as the valence shell electrons are in a 5p orbital. The 5p orbital is farther away from, and not as attracted to, the nucleus as the 2p orbitals where the other elements have their valence electrons located, thus the atoms of I are larger and would have a bigger atomic radius.

1 point for noting atoms of I have the largest radius.

8 of 15


Periodicity

(e) When bonded with fluorine, nitrogen atoms form the molecule NF3; however, atoms of iodine can form

IF3 and IF5 molecules. Explain. Atoms of I have empty d orbitals thus can expand their valence shell to form trigonal bipyramidal structures whereas atoms of N do not have empty d orbitals, and thus have at most 4 regions of electron density.

1 point for noting atoms of I have empty d orbitals and can expand their valence shell whereas N cannot.

9 of 15


Periodicity

Free Response

2. Answer the following questions using principles of atomic and molecular structure. The elements in the table below (W, X, Y, and Z) are actual elements found in either period 2 or 3 in the periodic table.

Element First

Ionization Energy(kJ mol−1)

Second Ionization Energy

(kJ mol−1)

Third Ionization Energy

(kJ mol−1)

Fourth Ionization Energy

(kJ mol−1) W 520 7298 11815 −−−− X 900 1757 14850 21000 Y 801 2427 3660 25000 Z 496 4562 6910 9543

(a) Which of the elements listed above has a valence electron configuration of 3s1? Justify your answer.

Element Z has a valence electron configuration of 3s1, as shown by the large jump in ionization energy after the first electron is removed. Element W also shows a large jump in ionization energy after the first electron is removed; but since its first ionization energy is greater than that of W (520 > 496) it must come from a lower energy level than Z.

1 point for identifying Z. 1 point for correct justification stating evidence of both a high 2nd ionization energy and correctly comparing the 1st ionization energies of W and Z.

(b) Which of the elements listed above is an alkaline earth metal? Explain.

Element X is an alkaline earth metal. X shows a large jump in ionization energy after the 2nd electron is removed. This suggests an ns2 electron configuration which is indicative of a Group 2 element.

1 point for identifying X. 1 point for correct justification.

(c) Which of the elements listed above has the largest atomic radius? Explain.

Element Z has the largest radius. Both W and Z are Group 1 elements, thus would be the largest. Z is larger than W because its valence shell is at a higher energy level than that of W, as evidenced by the lower first ionization energy.

1 point for identifying Z. 1 point for correct justification.

(d) Which element, X or Y has more protons? Assume both have the same principal valance energy level, n.

Element Y has more protons. It loses 3 electrons from its valence level before experiencing a large jump in ionization energy whereas Element X only loses 2. This means Element Y is farther to the right of element X; therefore, it must have more protons.

1 point for identifying Y. 1 point for correct justification.

10 of 15


Periodicity

Free Response 3. Use appropriate principles of atomic structure to account for each of the following statements. Be sure to

discuss both substances in your response.

a) The radius of the fluoride ion, F−, is larger than atomic fluorine, F. Both F− and F have the same principal valence energy level, n=2 and the same number of protons; however; F− has one more electron than F, thus less effective nuclear charge. F− also has more electron-electron repulsion, due to the additional electron. Therefore, its electrons are less attracted to the nucleus making the ion larger than its atom.

1 point for identifying F has a more effective nuclear charge. 1 point for stating the more effective nuclear charge increases the attraction for the electrons thus decreasing the size.

b) The fluoride ion, F− and the aluminum ion, Al3+ both have the same electron configuration; however,

the fluoride ion, F− is larger than the aluminum ion, Al3+. Even though both have the same electron configuration, the aluminum ion has a more effective nuclear charge than fluoride ion (13 to 9) thus it better attracts the electrons and pulls them in closer making the aluminum ion smaller than the fluoride ion.

1 point for identifying aluminum ion has a greater nuclear charge. 1 point for stating the greater nuclear charge increases the attraction for the electrons thus decreasing the size.

The table below compares the elements lithium, sodium, potassium and rubidium.

Element Atomic Radius (pm)

Difference in Radii (pm)

Li 145 Na 180 35

K 220 Rb 235 15

c) Rubidium has a larger atomic radius than lithium.

Both Rb and Li have a valence electron configuration of s1; Rb is 5s1 and Li is 3s1. The higher energy level, n, for Rb means its valence shell is father from and less attracted to the nucleus. This lessened attraction means the electrons in Rb are farther from the nucleus and thus its atoms have a larger atomic radius than Li.

1 point for identifying the energy level, n, difference. 1 point for stating the greater the distance from the nucleus the less attraction, thus the larger the radius (size).

11 of 15


Periodicity

d) The difference between the atomic radii of lithium and sodium is relatively large compared to the

difference between the atomic radii of potassium and rubidium. The energy separation between higher energy levels is smaller than the energy separation between lower energy levels. The attractive force between protons and electrons dissipates with increased distance from the nucleus. Therefore the difference in radii between elements in higher energy levels is not as large as between elements in lower energy levels.

1 point for recognizing as n increases, the energy separations between energy levels become smaller and smaller. 1 point for recognizing that the attractive force the nucleus exerts on the valence electrons in the electron cloud varies inversely with the square of the distance according to the following equation:

218

22.178 10 J ZEn

− ⎛ ⎞= − × ⎜ ⎟

⎝ ⎠; where Z is the nuclear

charge.

12 of 15


Periodicity

Multiple Choice Questions 1 − 3 refer to the atoms below.

(A) O (B) Ba (C) Rb (D) Mg (E) P

1. Is the most electronegative

A Oxygen has the smallest radius thus the greatest attraction for its electrons. Thus, it will have a large attraction for other electrons.

2. Has the greatest first ionization energy

A Oxygen has the smallest radius thus the greatest attraction for its electrons. Thus, it will require the most energy to remove an electron.

3. Has the largest radius

B Barium has the highest occupied energy level, n=6, thus its attraction for its electrons is the weakest and the radius the largest.

4. Which lists species that are isoelectronic?

(A) Ba, Ra, Ca (B) Cl−, As3−, S2− (C) Cr, Mn, Fe (D) S2−, Cl−, K+ (E) Ba2+, Ca2+, Be2+

D S2−, Cl−, K+ all have an electron configuration that is the same as Argon.

5. In the periodic table, as the atomic number increases from 3 to 10, which statement best describes the

effect on the atomic radius?

(A) It increases. (B) It decreases. (C) It remains constant. (D) It decreases and then increases. (E) It increases and then decreases.

B The trend is that atomic radii generally decrease as you go across the periodic table since the effective nuclear charge increases.

13 of 15


Periodicity

6. Which of these sequences correspond to a correct trend in ionization energy?

I. Al < Si < P < Cl II. Be < Mg < Ca < Sr III. I < Br < Cl < F IV. Na+

< Mg2+ < Al3+

< Si4+

(A) I only (B) III only (C) I and II only (D) I and IV only (E) I, III, and IV only

E Ionization energy increases as effective nuclear charge increases in a given period and decreases as the number of energy levels increases within a given group.

7. Based on the information in the table below, what is the most likely charge of element X ?

Element First

Ionization Energy(kJ mol−1)

Second Ionization Energy

(kJ mol−1)

Third Ionization Energy

(kJ mol−1)

Fourth Ionization Energy

(kJ mol−1) X 900 1757 14850 21000

(A) 1− (B) 1+ (C) 2+ (D) 3+ (E) 4+

C A huge jump in ionization energy occurs at IE3 suggesting the 3rd electron is being removed from a core energy level.

8. The effective nuclear charge experienced by the valence shell electron of Na is different than that

experienced by the valence shell electron(s) of Ar. This difference best describes which of the following? Na has

(A) a smaller radius than Ar. (B) a higher melting point than Ar. (C) a greater electron affinity than Ar. (D) a lower first ionization energy than Ar (E) a higher neutron-to-proton ratio than Ar.

D

Na has a Zeff of 1 while Ar has a Zeff of 8. That means Ar has a much higher attractive force for its electrons than does Na. It also means that it takes less energy to remove an electron (first ionization energy) from Na than from Ar.

14 of 15


Periodicity

9. Which group of elements has most nearly the same atomic radius?

(A) F, Cl, I (B) Li, C, F (C) N, S, Br (D) Li, K, Cs (E) Mn, Fe, Co

E

Moving across the periodic table within a period, the atomic radius decreases due to an increase in effective nuclear charge (Zeff). Moving down the periodic table within a family, the atomic radius increases due to adding an entire principal energy level. Because they are all transition metals within the same period, moving across a given period within the transition metals adds 3d electrons, BUT the 4s electrons are also present, therefore the size of the radius stays relatively constant.

10. Which of the elements below is the least reactive?

(A) F (B) Li (C) Ne (D) Xe (E) Ra

C Ne is the least reactive. Xe is also a noble gas but does form certain compounds like XeF4 using empty d orbitals to expand its valence shell.

15 of 15

Periodicity Presenter APD Cover - UHS chemistry · 2018-09-07 · Periodicity ELECTRON AFFINITY −...

Documents

Transcript of Periodicity Presenter APD Cover - UHS chemistry · 2018-09-07 · Periodicity ELECTRON AFFINITY −...