08 silberberg 8e ISMChapter 8 PDF

Preview

Summary

Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.CHAPTER 8 ELECTRON CONFI...

Description

CHAPTER 8 ELECTRON CONFIGURATION AND CHEMICAL PERIODICITY FOLLOW–UP PROBLEMS 8.1A

Plan: The superscripts can be added to indicate the number of electrons in the element, and hence its identity. A horizontal orbital diagram is written for simplicity, although it does not indicate the sublevels have different energies. Based on the orbital diagram, we identify the electron of interest and determine its four quantum numbers. Solution: The number of electrons = 2 + 2 + 4 = 8; element = oxygen, atomic number 8. Orbital diagram:

1s

2s

2p

The electron in the middle 2p orbital is the sixth electron (this electron would have entered in this position due to Hund’s rule). This electron has the following quantum numbers: n = 2, l = 1 (for p orbital), ml = 0, ms = +1/2. By convention, +1/2 is assigned to the first electron in an orbital. Also, the first p orbital is assigned an ml value of –1, the middle p orbital a ml value of 0, and the last p orbital a ml value of +1. 8.1B

Plan: Use the quantum numbers to determine the orbital into which the last electron is added and whether the electron is the first or second electron added to that orbital. Draw the orbital diagram that matches the quantum numbers. Then use the orbital diagram and the periodic table to identify the element. Finally, write the electron configuration for the element. Solution: The last electron added to the atom has the quantum numbers n = 2, l = 1, ml = 0 and ms = +1/2. This electron is in the second 2p orbital (n = 2, l = 1 tells us that the orbital is 2p; ml = 0 tells us that the orbital is the 2nd of the three 2p orbitals). Additionally, because ms = +1/2, we know the electron is the first electron in that orbital. The orbital diagram that matches this description is: Orbital diagram:

1s

2s

2p

There are a total of 6 electrons in the orbital diagram. Carbon is the element with 6 protons and, in its neutral state, 6 electrons. Its electron configuration is: 1s22s22p2. 8.2A

Plan: The atomic number gives the number of electrons. The order of filling may be inferred by the location of the element on the periodic table. The partial orbital diagrams shows only those electrons after the preceding noble gas except those used to fill inner d and f subshells. The number of inner electrons is simply the total number of electrons minus those electrons in the partial orbital diagram. Solution: a) For Ni (Z = 28), the full electron configuration is 1s22s22p63s23p64s23d8. The condensed configuration is [Ar]4s23d8. The partial orbital diagram for the valence electrons is

Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

8-1

4s

3d

4p

There are 28 – 10(valence) = 18 inner electrons. b) For Sr (Z = 38), the full electron configuration is 1s22s22p63s23p64s23d104p65s2. The condensed configuration is [Kr]5s2. The partial orbital diagram is

5s

4d

5p

There are 38 – 2(valence) = 36 inner electrons. c) For Po (84 electrons), the full configuration 1s22s22p63s23p64s23d104p65s24d105p66s24f145d106p4. The condensed configuration is [Xe]6 s 2 4 f 1 4 5 d 1 0 6 p 4 . The partial orbital diagram represents valence electrons only; the inner electrons are those in the previous noble gas (Xe or 1s22s22p63s23p64s23d104p65s24d105p6) and filled transition (5d10) and inner transition (4f14) levels.

6s

6p

There are 84 – 6(valence) = 78 inner electrons. 8.2B

Plan: The partial orbital diagrams show the electrons that have been added after the preceding noble gas except for those used to fill inner d and f subshells (in most cases). Use the orbital diagrams and the periodic table to identify the elements. Write their full and condensed electron configurations. The number of inner electrons is simply the total number of electrons minus those electrons in the partial orbital diagram. a) As (Z = 33) has two 4s electrons and three 4p electrons. Its full electron configuration is 1s22s22p63s23p64s23d104p3 and its condensed electron configuration is [Ar] 4s23d104p3. There are 33 – 5(valence) = 28 inner electrons. b) Zr (Z = 40) has two 5s electrons and two 4d electrons. Its full electron configuration is 1s22s22p63s23p64s23d104p65s24d2 and its condensed electron configuration is [Kr] 5s24d2. There are 40 – 4(valence) = 36 inner electrons. c) I (Z = 53) has two 5s electrons and five 5p electrons. Its full electron configuration is 1s22s22p63s23p64s23d104p65s24d105p5 and its condensed electron configuration is [Kr] 5s24d105p5. There are 53 – 7(valence) = 46 inner electrons.

8.3A

Plan: Locate each of the elements on the periodic table. All of these are main-group elements, so their sizes increase down and to the left in the periodic table. Solution: a) Cl < Br < Se. Cl has a smaller n than Br and Se. Br experiences a higher Zeff than Se and is smaller. b) Xe < I < Ba. Xe and I have the same n, but Xe experiences a higher Zeff and is smaller. Ba has the highest n and is the largest.

8.3B

Plan: Locate each of the elements on the periodic table. All of these are main-group elements, so their sizes increase down and to the left in the periodic table.

Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

8-2

Solution: a) Cs > As > S. Cs has the largest n of the group. Additionally, it is further to the left of the table than the other two elements. Therefore, Cs is the largest. As has a larger n than S and is further to the left on the periodic table, so it is the next largest element of this group. b) K > P > F. K has the largest n of the group. Additionally, it is further to the left of the table than the other two elements. Therefore, K is the largest. P has a larger n than F and is further to the left on the periodic table, so it is the next largest element of this group. 8.4A

Plan: These main-group elements must be located on the periodic table. The value of IE1 increases toward the top (same column) and the right of the periodic table (same n). Solution: a) Sn < Sb < I. These elements have the same n, so the values increase to the right on the periodic table. Iodine has the highest IE, because its outer electron is most tightly held and hardest to remove. b) Cs < Na < Mg. Mg is farther to the right on the periodic table than either of the other two elements, so it has the largest IE. Cs and Na are in the same column, so the values decrease towards the bottom of the column. Because Cs is lower than Na in the column, it has a lower IE.

8.4B

Plan: These main-group elements must be located on the periodic table. The value of IE1 increases toward the top (same column) and the right of the periodic table (same n). Solution: a) O > As > Rb. Oxygen is the farthest to the right on the periodic table (and closer to the top of the table), so it has the largest IE. Similarly, As is farther to the right and closer to the top of the table than Rb and, thus, has a higher IE. b) Cl > Si > Sn. Cl and Si have the same n value and are closer to the top of the periodic table than Sn. Therefore, Sn has the lowest IE. Because Cl is closer to the right hand side of the periodic table than Si, Cl has the highest IE.

8.5A

Plan: We must look for a large “jump” in the IE values. This jump occurs after the valence electrons have been removed. The next step is to determine the element in the designated period with the proper number of valence electrons. Solution: The exceptionally large jump from IE3 to IE4 means that the fourth electron is an inner electron. Thus, Q has three valence electrons. Since Q is in period 3, it must be aluminum, Al: 1s22s22p63s23p1.

8.5B

Plan: A large jump in IE values occurs after the valence electrons have been removed. Thus, the highest IE3, for example, will be for an element that has 2 valence electrons because it will require much more energy to remove the third electron, an inner electron. Write the condensed electron configurations for the atoms, determine the number of valence electrons for each, and use this information to determine where the jump in IE values will occur. Solution: Rb: [Kr] 5s1 Sr: [Kr] 5s2 Y: [Kr] 5s24d1 Because Rb has one valence electron, a large jump in IE values will occur between IE1 and IE2. Therefore, IE2 will be large for Rb. Because Sr has two valence electrons, a large jump in IE values will occur between IE2 and IE3. Therefore, IE3 will be large for Sr. Plan: We need to locate each element on the periodic table. Elements in the first two columns on the left or the two columns to the left of the noble gases tend to adopt ions with a noble gas configuration. Elements in the remaining columns may use either their ns and np electrons, or just their np electrons. Solution: a) Barium loses two electrons to be isoelectronic with Xe: Ba ([Xe] 6s2) → Ba2+ ([Xe]) + 2e– b) Oxygen gains two e– to be isoelectronic with Ne: O ([He] 2s22p4 + 2e– → O2– ([Ne])

8.6A

Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

8-3

c) Lead can lose two electrons to form an “inert pair” configuration: Pb ([Xe]6s2 4f14 5d 10 6 p2 ) → Pb 2+ ([Xe]6s2 4 f14 5d 10 ) + 2e– or lead can lose four electrons form a “pseudo–noble gas” configuration: Pb ([Xe]6s 2 4 f 1 4 5 d 1 0 6 p 2 ) → Pb 4 + ([Xe]4f 1 4 5d 1 0 ) + 4e – 8.6B

Plan: We need to locate each element on the periodic table. Elements in the first two columns on the left or the two columns to the left of the noble gases tend to adopt ions with a noble gas configuration. Elements in the remaining columns may use either their ns and np electrons, or just their np electrons. Solution: a) Fluorine gains one electron to be isoelectronic with Ne: F ([He] 2s22p5) + e– → F– ([He] 2s22p6) b) Thallium loses 1 e– to form an “inert pair” configuration: Tl ([Xe] 6s24f145d106p1 → Tl + ([Xe] 6s24f145d10) + e– or thallium can lose 3 e– to form a “pseudo-noble gas” configuration: Tl ([Xe] 6s24 f14 5d 106p1 → Tl3+ ([Xe] 4f14 5 d10 ) + 3e – c) Magnesium loses two electrons to be isoelectronic with Ne: Mg ([Ne] 3s2) → Mg2+ ([Ne]) + 2e–

8.7A

Plan: Write the condensed electron configuration for each atom, being careful to note those elements which are irregular. The charge on the cation tells how many electrons are to be removed. The electrons are removed beginning with the ns electrons. If any electrons in the final ion are unpaired, the ion is paramagnetic. If it is not obvious that there are unpaired electrons, a partial orbital diagram might help. Solution: a) The V atom ([Ar]4s23d3) loses the two 4s electrons and one 3d electron to form V3+ ([Ar]3d2). There are two unpaired d electrons, so V3+ is paramagnetic. b) The Ni atom ([Ar]4s23d8) loses the two 4s electrons to form Ni2+ ([Ar]3d8). There are two unpaired d electrons, so Ni2+ is paramagnetic.

3d

c) The La atom ([Xe]6s25d1) loses all three valence electrons to form La3+ ([Xe]). There are no unpaired electrons, so La3+ is diamagnetic. 8.7B

8.8A

Plan: Write the condensed electron configuration for each atom, being careful to note those elements which are irregular. The charge on the cation tells how many electrons are to be removed. The electrons are removed beginning with the ns electrons. If it is not obvious that there are unpaired electrons, a partial orbital diagram might help. Solution: a) The Zr atom ([Kr]5s24d2) loses the two 5s electrons to form Zr2+ ([Kr]4d2). There are two unpaired d electrons. b) The Os atom ([Xe] 6s2 4f 14 5d 6 ) loses the two 6s electrons and one 5d electron to form Os3+ ([Xe] 4f145d5). There are five unpaired d electrons. c) The Co atom ([Ar] 4s23d7) loses the two 4s electrons to form Co2+ ([Ar] 3d7). There are three unpaired d electrons.

3d Plan: Locate each of the elements on the periodic table. Cations are smaller than the parent atoms, and anions are larger than the parent atoms. If the electrons are equal, anions are larger than cations. The more electrons added or removed; the greater the change in size.

Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

8-4

Solution: a) Ionic size increases down a group, so F – < Cl– < Br–. b) These species are isoelectronic (all have 10 electrons), so size increases with increasing negative charge: Mg2+ < Na+ < F – c) Ionic size increases as charge decreases for different cations of the same element, so Cr3+ < Cr2+. 8.8B

Plan: Locate each of the elements on the periodic table. Cations are smaller than the parent atoms, and anions are larger than the parent atoms. If the electrons are equal, anions are larger than cations. The more electrons added or removed; the greater the change in size. Solution: a) P3- > S2- > Cl- Anion size decreases with decreasing charge. b) Ionic size decreases up (toward the top of) a group, so Cs+ > Rb+ > K+. c) These species are isoelectronic (all have 54 electrons), so size decreases with increasing positive charge: I– > Cs+ > Ba2+.

END–OF–CHAPTER PROBLEMS 8.1

Elements are listed in the periodic table in an ordered, systematic way that correlates with a periodicity of their chemical and physical properties. The theoretical basis for the table in terms of atomic number and electron configuration does not allow for an “unknown element” between Sn and Sb.

8.2

Today, the elements are listed in order of increasing atomic number. This makes a difference in the sequence of elements in only a few cases, as the larger atomic number usually has the larger atomic mass. One of these exceptions is iodine, Z = 53, which is after tellurium, Z = 52, even though tellurium has a higher atomic mass.

8.3

Plan: The value should be the average of the elements above and below the one of interest. Solution: a) Predicted atomic mass (K) = 22.99 + 85.47 Na + Rb = = 54.23 amu (actual value = 39.10 amu) 2 2 b) Predicted melting point (Br2) = Cl2 + I2 −101.0 +113.6 = 6.3°C (actual value = –7.2°C) = 2 2

8.4

a) Predicted boiling point (HBr) = HCl + HI −84.9 + (− 35.4) = = –60.15 = –60.2°C 2 2 b) Predicted boiling point (AsH3) = PH3 + SbH3 −87.4 + (−17.1) = = –52.25 = –52.2°C 2 2

(actual value = –67.0°C)

(actual value = –55°C)

8.5

The allowed values of n: positive integers: 1, 2, 3, 4,...∞ The allowed values of l: integers from 0 to n – 1: 0, 1, 2, ... n – 1 The allowed values of ml: integers from –l to 0 to +l: –l, (–l + 1), ... 0, ... (l – 1), +l The allowed values of ms: –1/2 or +1/2

8.6

The quantum number ms relates to just the electron; all the others describe the orbital.

Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

8-5

8.7

The exclusion principle states that no two electrons in the same atom may have the same four quantum numbers. Within a particular orbital, there can be only two electrons and they must have opposing spins.

8.8

In a one-electron system, all sublevels of a particular level (such as 2s and 2p) have the same energy. In many electron systems, the principal energy levels are split into sublevels of differing energies. This splitting is due to electron-electron repulsions. Be3+ would be more like H since both have only one 1s electron.

8.9

Shielding occurs when inner electrons protect or shield outer electrons from the full nuclear attractive force. The effective nuclear charge is the nuclear charge an electron actually experiences. As the number of inner electrons increases, shielding increases, and the effective nuclear charge decreases.

8.10

Penetration occurs when the probability distribution of an orbital is large near the nucleus, which results in an increase of the overall attraction of the nucleus for the electron, lowering its energy. Shielding results in lessening this effective nuclear charge on outer shell electrons, since they spend most of their time at distances farther from the nucleus and are shielded from the nuclear charge by the inner electrons. The lower the l quantum number of an orbital, the more time the electron spends penetrating near the nucleus. This results in a lower energy for a 3p electron than for a 3d electron in the same atom.

8.11

Plan: The integer in front of the letter represents the n value. The l value designates the orbital type: l = 0 = s orbital; l = 1 = p orbital; l = 2 = d orbital; l = 3 = f orbital. Remember that a p orbital set contains 3 orbitals, a d orbital set has 5 orbitals, and an f orbital set has 7 orbitals. Any one orbital can hold a maximum of 2 electrons. Solution: a) The l = 1 quantum number can only refer to a p orbital. These quantum numbers designate the 2p orbital set (n = 2), which hold a maximum of 6 electrons, 2 electrons in each of the three 2p orbitals. b) There are five 3d orbitals, therefore a maximum of 10 electrons can have the 3d designation, 2 electrons in each of the five 3d orbitals. c) There is one 4s orbital which holds a maximum of 2 electrons.

8.12

a) The l = 1 quantum number can only refer to a p orbital, and the ml value of 0 specifies one particular p orbital, which holds a maximum of 2 electrons. b) The 5p orbitals, like any p orbital set, can hold a maximum of 6 electrons. c) The l = 3 quantum number can only refer to an f orbital. These quantum numbers designate the 4f orbitals, which hold a maximum of 14 electrons, 2 electrons in each of the seven 4f orbitals.

8.13

Plan: The integer in front of the letter represents the n value. The l value designates the orbital type: l = 0 = s orbital; l = 1 = p orbital; l = 2 = d orbital; l = 3 = f orbital. Remember that a p orbital set contains 3 orbitals, a d orbital set has 5 orbitals, and an f orbital set has 7 orbitals. Any one orbital can hold a maximum of 2 electrons. Solution: a) 6 electrons can be found in the three 4p orbitals, 2 in each orbital. b) The l = 1 quantum number can only refer to a p orbital, and the ml value of +1 specifies one particular p orbital, which holds a maximum of 2 electrons with the difference between the two electrons being in the ms quantum number. c) 14 electrons can be found in the 5f orbitals (l = 3 designates f orbitals; there are 7f orbitals in a set).

8.14

a) Two electrons, at most, can be found in any s orbital. b) The l = 2 quantum number can only refer to a d orbital. These quantum numbers designate the 3d orbitals, which hold a maximum of 10 electrons, 2 electrons in each of the five 3d orbitals. c) A maximum of 10 electrons can be found in the five 6d orbitals.

Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or ...