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NAME____________________________________ PER____________ DATE DUE____________
ACTIVE LEARNING IN CHEMISTRY EDUCATION
"ALICE"
CHAPTER 13
MODERNVIEW
OFATOMIC
STRUCTURE(PART 2)
NOTICE OF RIGHTS
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© 1997 A.J. Girondi, Ph.D.505 Latshmere DriveHarrisburg, PA 17109
Website: www.geocities.com/Athens/Oracle/2041
13-2 ©1997, A.J. Girondi
SECTION 13.1 The Valence Shell and Valence Electrons
Now that you know how to write electron configurations for atoms, we can examine chemicalfamilies in terms of electron configurations. You will recall that chemical families are the vertical columns onthe periodic table. One such family consists of the elements Li, Na, K, Rb, Cs, and Fr. These elementscomprise column 1A. They are called the alkali metals. Their electron configurations are listed in Table13.1 below.
Table 13.1Electron Configurations for Alkali Metals
At. No. Configuration Notation
3 Li 1s22s1
11 Na 1s22s22 p 63s1
19 K 1s22s22 p 63s23 p 64s1
37 Rb 1s22s22 p 63s23 p 64s23 d 104 p 65s1
55 Cs 1s22s22 p 63s23 p 64s23 d 104 p 65s24 d 105 p 66s1
87 Fr 1s22s22 p 63s23 p 64s23 d 104 p 65s24 d 105 p 66s24 f 145 d 106 p 67s1
There is an important similarity in all of these electron configurations. Can you see what it is? Circle theoutermost energy level (1,2,3,4,etc.) of each of the elements in Table 13.1. How many electrons arethere in the outermost energy level of each of these elements? {1}__________. The shell or energy levelthat contains the outermost electrons for an element is called the valence shell. The electrons in thatenergy level are called valence electrons. Because these electrons are among the ones that are farthestfrom the nucleus, they are relatively easy to remove. The chemical activity of an element is determined inpart by the number of valence electrons that it has.
The alkali metals all have one valence electron. This similarity in number of valence electronscauses the chemical behavior of these elements to be very similar. Sometimes hydrogen, which has a 1s1configuration, is also listed with this family since it also has one valence electron. However, it behavesquite differently from the other elements in the family because of its small size and the fact that there areno other electrons in levels beneath the valence shell, as there are in the rest of the family. On someperiodic tables, hydrogen is placed in a class by itself.
When you attempt to identify the valence electrons in an atom, be sure to count all the electronsin the entire outermost energy level, not just the outermost sublevel. For example, for germanium (#32)the configuration notation is 1s 22s22 p 63s23 p 64s23 d 104 p 2. How many valence electrons are there?The answer is 4. That's right, 4. The outermost energy level is 4 (the N shell), and there are 4 electronsthere. They are the 4s2 and the 4p2 electrons. Many students would make the mistake of thinking thatonly the 4p2 electrons are in the valence shell. All of the electrons in the outermost energy level (energylevel 4 in this case) must be included. Reread this paragraph until you are sure you understand.
Let's see if you've mastered this idea. How many valence electrons are there in the atom whoseconfiguration notation is listed below?
1s22s22 p 63s23 p 64s23 d 104 p 6 Number of valence electrons = {2}_____________
What is the configuration notation of only the valence electrons? {3}____________________________
13-3 ©1997, A.J. Girondi
Write the electron configuration notation for only the valence electrons in aluminum (#13): {4}__________
O.K., now write the valence shell electron configuration notation for Te (#52): {5}___________________
Problem 1. In Table 13.2, write the electron configuration notation for only the valence shell of theindicated families. Be careful, only the valence shell configuration is being asked for, but be sure toinclude all of the electrons in the valence shell!
Table 13.2Valence Shell Configuration for Families 2A & 3A
Family 2AElement At. No. Valence Shell Configuration
Be 4 _____________________
Mg 12 _____________________
Ca 20 _____________________
Sr 38 _____________________
Ba 56 _____________________
Ra 88 _____________________
Family 3AElement At. No. Valence Shell Configuration
B 5 _____________________
Al 13 _____________________
Ga 31 _____________________
In 49 _____________________
Tl 81 _____________________
The chemical activity of the elements depends on their electronic structure.Elements in the same group or family have the same valence shell structure. For example, in Group 3Athe element B has the valence shell: 2s22p1, while in the same family Al has the valence shell: 3s23p1.Mendeleev was not correct in arranging the elements according to increasing atomic masses, whileMoseley was correct in arranging them according to increasing atomic number. Why? Well, the atomicnumber indicates the number of protons in the nucleus as well as the number of electrons in a neutralatom. When arranged according to increasing atomic number, elements with the same valence shellstructure can be lined up in a vertical group (family). Thus, it is the valence shell configurations that allowus to arrange elements periodically in the table. Mendeleev did his work prior to the discovery of theinternal structure of atoms. It is a tribute to his insight that he developed his arrangement based onchemical similarities. Arranging the atoms according to increasing atomic masses was a very logical movegiven the state of knowledge at his time. The fact that he was incorrect does not detract from thecontribution he made to chemistry.
13-4 ©1997, A.J. Girondi
Problem 2. For further practice in working with valence shell electron configuration notation, completeTable 13.3 (show valence electrons only).
Table 13.3Electron Configuration and Valence Electrons of Selected Elements
Element At. No. Valence Shell Electron Configuration Notation
Si 14 __________________________________
Br 35 __________________________________
V 23 __________________________________
As 33 __________________________________
S 16 __________________________________
Y 39 __________________________________
At 85 __________________________________
Au 79 __________________________________
Cl 17 __________________________________
SECTION 13.2 The Use of "Core" Notation
The task of writing notations can get cumbersome for atoms that have a lot of electrons. Ashortcut method, core notation, has been developed that is quicker and more convenient. Suppose youwant to write the configuration and orbital notation for chlorine. To write core notation, find the noble gason the periodic table that has the atomic number closest to but less than that of chlorine. The atomicnumber of chlorine is 17. The closest noble gas is Neon (#10). Then write the notations of chlorine asfollows:
3s 3pChlorine: 17Cl = [Ne] 3s23p5 or: [Ne] (X) (X)(X)(/)
This indicates that chlorine atoms have all the electrons that neon has (the neon core) plus those shown inthe orbitals that follow. Here's lead:
Lead = 82Pb = [Xe] 6s24 f 145 d 106 p 2
6s 4f 5d 6p or: [Xe] (X) (X)(X)(X)(X)(X)(X)(X) (X)(X)(X)(X)(X) (/)(/)( )
Lead has all the electrons that xenon has (the xenon core) plus those shown in the notation. The coreform of configuration notation for 100 elements is shown in Table 13.4. This table follows the diagonalrule. There are a number of exceptions to the diagonal rule, but they are not shown in table 13.4. Thetable of electron configurations in Appendix B of your ALICE materials shows the actual electronconfigurations of the elements including exceptions to the diagonal rule.
Refer to Tables 13.4 and 13.5 as you study examples of the "core form" of orbital notation. Table13.5 shows the configuration of the last electron to enter each atom as orbitals fill up according to thediagonal rule. It really helps when writing core notation. Examine Table 13.5 for a moment now.
13-5 ©1997, A.J. Girondi
Element Orbital and Configuration Notation Using Core Notation
2s 2pLithium (#3) [He] (/) or [He] 2s1
boron (#5) [He] (X) (/)( )( ) or [He] 2s22 p 1
oxygen (#8) [He] (X) (X)(/)(/) or [He] 2s22 p 4
3s 3pmagnesium (#12) [Ne] (X) or [Ne] 3s2
silicon (#14) [Ne] (X) (/)(/)( ) or [Ne] 3s23 p 2
4s 3d 4pgallium (#31) [Ar] (X) (X)(X)(X)(X)(X) (/)( )( ) or [Ar] 4s23 d 104 p 1
bromine (#35) [Ar] (X) (X)(X)(X)(X)(X) (X)(X)(/) or [Ar] 4s23 d 104 p 5
Problem 3. Try writing the core form of configuration notation for the elements below. Check youranswers using Table 13.4.
Element Orbital Notation Using Core Notation
Sulfur (#16) ___________________________________________________________
Scandium (#21) ___________________________________________________________
Niobium (#41) ___________________________________________________________
Tellurium (#52) ___________________________________________________________
SECTION 13.3 Ionization Energy
You will recall that when energy is absorbed by an electron, it is said to be "excited." It is thevalence electrons that are easiest to "excite." Look at the two electron configurations below. There is anobvious difference between them. Both electron configurations represent eleven electrons. Both aresodium atoms (assuming the elements represented have 11 protons).
1s22s22 p 63s1 1s22s22 p 65s1 (sodium atom) (unstable sodium atom)
The 3s1 electron has been "excited"and jumped up to a higher energy level
Since sodium has an atomic number of 11, these configurations represent a sodium atom. What hashappened is that the electron in the 5s orbital is excited and is extremely unstable. It can emit that extraenergy as light to get back down to its ground state (the 3s level). But, when you excite the electron in thefirst place, what if the 3s1 electron absorbs more energy than it can handle? Indeed, this is a commonoccurrence. In this case, the 3s1 electron would simply leave the atom altogether. What is left is called asodium ion. An ion is an atom with an unequal number of electrons and protons. This means that all ionshave either a positive or negative charge. To see whether sodium would become a positive or negativeion, we must look at its electron configuration with and without its 3s1 electron.
Neutral Na Atom = 1s22s22 p 63s1 Na1+ Ion = 1s22s22 p 6
The sodium ion has a plus 1 (+1) charge. It is important for you to remember that ions can only be made byeither adding or removing electrons. The number of protons is never altered in a chemical change.
13-6 ©1997, A.J. Girondi
Table 13.4 "Core Notation" of 100 Elements According to the Diagonal Rule
1 H 1s1 51 Sb [Kr]5s24d105p3
2 He 1s2 52 Te [Kr]5s24d105p4
3 Li [He]2s1 53 I [Kr]5s24d105p5
4 Be [He]2s2 54 Xe [Kr]5s24d105p6
5 B [He]2s22p1 55 Cs [Xe]6s1 6 C [He]2s22p2 56 Ba [Xe]6s2 7 N [He]2s22p3 57 La [Xe]6s24f1 8 O [He]2s22p4 58 Ce [Xe]6s24f2 9 F [He]2s22p5 59 Pr [Xe]6s24f3 10 Ne [He]2s22p6 60 Nd [Xe]6s24f4 11 Na [Ne]3s1 61 Pm [Xe]6s24f5 12 Mg [Ne]3s2 62 Sm [Xe]6s24f6 13 Al [Ne]3s23p1 63 Eu [Xe]6s24f7 14 Si [Ne]3s23p2 64 Gd [Xe]6s24f8 15 P [Ne]3s23p3 65 Tb [Xe]6s24f9 16 S [Ne]3s23p4 66 Dy [Xe]6s24f10 17 Cl [Ne]3s23p5 67 Ho [Xe]6s24f11
18 Ar [Ne]3s23p6 68 Er [Xe]6s24f12 19 K [Ar]4s1 69 Tm [Xe]6s24f13 20 Ca [Ar]4s2 70 Yb [Xe]6s24f14 21 Sc [Ar]4s23d1 71 Lu [Xe]6s24f145d1
22 Ti [Ar]4s23d2 72 Hf [Xe]6s24f145d2
23 V [Ar]4s23d3 73 Ta [Xe]6s24f145d3
24 Cr [Ar]4s23d4 74 W [Xe]6s24f145d4
25 Mn [Ar]4s23d5 75 Re [Xe]6s24f145d5
26 Fe [Ar]4s23d6 76 Os [Xe]6s24f145d6
27 Co [Ar]4s23d7 77 Ir [Xe]6s24f145d7
28 Ni [Ar]4s23d8 78 Pt [Xe]6s24f145d8
29 Cu [Ar]4s23d9 79 Au [Xe]6s24f145d9
30 Zn [Ar]4s23d10 80 Hg [Xe]6s24f145d10
31 Ga [Ar]4s23d104p1 81 Tl [Xe]6s24f145d106p1 32 Ge [Ar]4s23d104p2 82 Pb [Xe]6s24f145d106p2
33 As [Ar]4s23d104p3 83 Bi [Xe]6s24f145d106p3
34 Se [Ar]4s23d104p4 84 Po [Xe]6s24f145d106p4
35 Br [Ar]4s23d104p5 85 At [Xe]6s24f145d106p5
36 Kr [Ar]4s23d104p6 86 Rn [Xe]6s24f145d106p6
37 Rb [Kr]5s1 87 Fr [Rn]7s1 38 Sr [Kr]5s2 88 Ra [Rn]7s2 39 Y [Kr]5s24d1 89 Ac [Rn]7s25f1 40 Zr [Kr]5s24d2 90 Th [Rn]7s25f2 41 Nb [Kr]5s24d3 91 Pa [Rn]7s25f3 42 Mo [Kr]5s24d4 92 U [Rn]7s25f4 43 Tc [Kr]5s24d5 93 Np [Rn]7s25f5 44 Ru [Kr]5s24d6 94 Pu [Rn]7s25f6 45 Rh [Kr]5s24d7 95 Am [Rn]7s25f7 46 Pd [Kr]5s24d8 96 Cm [Rn]7s25f8 47 Ag [Kr]5s24d9 97 Bk [Rn]7s25f9 48 Cd [Kr]5s24d10 98 Cf [Rn]7s25f10 49 In [Kr]5s24d105p1 99 Es [Rn]7s25f11 50 Sn [Kr]5s24d105p2 100 Fm [Rn]7s25f12
13-7 ©1997, A.J. Girondi
1s1 1s2
2s1
3s1
4s1
5s1
6s1
7s1
2s2 2p1 2p2 2p3 2p4 2p5 2p6
3p1
4p1
5p1
6p1
3p2
4p2
5p2
6p2
3p3
4p3
5p3
6p3
3p4
4p4
5p4
6p4
3p5
4p5
5p5
6p5
3p6
4p6
5p6
6p6
3s2
4s2
5s2
6s2
7s2
3d1 3d2 3d3 3d4 3d5 3d6 3d7 3d8 3d9 3d10
4d1
5d1
6d1
4d2
5d2
6d2
4d3
5d3
6d3
4d4
5d4
6d4
4d5
5d5
6d5
4d6
5d6
6d6
4d7
5d7
6d7
4d8
5d8
6d8
4d9
5d9
6d9
4d10
5d10
4f1 4f2 4f3 4f4 4f5 4f6 4f7 4f8 4f9 4f10 4f11 4f12 4f13 4f14
5f1 5f2 5f3 5f4 5f5 5f6 5f7 5f8 5f9 5f10 5f11 5f12 5f13 5f14
Table 13.5Last–Entering Electrons According to the Diagonal Rule*
*Since there are many exceptions to the diagonal rule, see Appendix B of your reference notebook for the actual electron distributions in atoms.
The energy required to remove electrons is called the ionization energy. The amounts ofionization energy required to remove the outermost electrons of sodium, magnesium, and aluminum arelisted in Table 13.6.
After the first electron is removed from an atom, the size of the atom decreases. The remainingelectrons are "pulled" closer to the nucleus and become even harder to remove. This is evident in Table13.6. In the table, valence electrons and the energies required to remove valence electrons are enclosedin boxes. Notice that after the valence electron(s) are removed from the atoms, the amount of energyrequired to remove any more electrons goes up dramatically. The reason for this is that after the valenceelectrons are removed, the atoms are left with the electron configuration of a noble gas and additionalelectrons would have to come from another energy level which is much closer to the nucleus. Suchelectrons are much more strongly attracted to the nucleus. Notice that to get a second electron out ofsodium (Na) we would have to move to another energy level which is closer to the nucleus.
Because the noble gas configurations are so stable, it is very difficult to remove more electronsfrom an atom after it has achieved an "octet" of eight electrons in the outermost energy level. In Table13.6, the biggest increase in ionization energy comes after which electron is removed from Al? (1st, 2nd,3rd, etc.): {6}_________ What accounts for this big increase in ionization energy in aluminum?{7}______________________________________________________________________________________
Notice in Table 13.6 that the big increase in ionization energy for Mg comes between the second and third
electrons. This means that it is much harder to remove the third electron from an atom of Mg than it was to
remove the first and second ones. This is evidence that the third electron in much closer to the nucleus.
We conclude that the third electron belongs to a different energy level.
13-8 ©1997, A.J. Girondi
Table 13.6Energy Needed to Remove Electrons
Element Configuration Ionization Energy (Kilojoules/mole)Notation 1st Elec. 2nd Elec. 3rd Elec. 4th Elec.
Na 1s22s22p63s1 498 4560
Mg 1s22s22p63s2 736 1445 7730
A l 1s22s22p63s23p1 577 1815 2740 11600
◊◊◊
◊◊◊
◊◊◊
◊◊◊ indicates a change in energy levels
So, when the big increase in ionization energy occurs after the first electron, the atom probably belongs to what family on the periodic table? {8}__________ When the big increase occurs after two electrons have been removed, the atom probably belongs to family {9}__________.
The valence shell arrangement of electrons in noble gases makes them very stable. They havewhat we call an octet of electrons in their outer (valence) shells. For neon the octet is in the 2nd energylevel: 2s22p6. For argon it is in the 3rd energy level: 3s23p6. Helium is the only noble gas that does nothave an octet. Instead, it has a full outer energy level (1s2) which is also very stable. (Remember, the firstenergy level holds a maximum of 2 electrons.)
The large "gaps" in ionization energies serve as great evidence for the existence ofenergy levels, sublevels, and the stability of certain electron arrangements.
Atoms gain or lose electrons in order to obtain a stable arrangement. The two most stable arrangementsare:
1. an octet (s2p6) in the outermost energy level (as seen in the noble gases except for helium)2. the full outermost energy level of helium (1s2)
Other arrangements that are "slightly" stable include:
3. a full sublevel (such as s2, or p6, or d10, etc.) 4. a half-full sublevel (such as s1, p3, d5, etc.)
In order to become more stable, atoms often lose, gain, or share electrons when they react witheach other. They lose, gain, or share as many electrons as needed to establish a stable arrangement(especially an octet). To accomplish this, atoms generally tend to lose or gain enough electrons toachieve the electron configuration of the closest noble gas. (By closest we mean according to atomicnumber.)
Sodium loses one electron and ends up with the arrangement of neon:
Na atom: 1s22s22 p 63s1 ---> Na1+ ion: 1s22s22 p 6
Magnesium, [Ne]3s2, loses two electrons and becomes like neon:
Mg atom: 1s22s22 p 63s2 ---> Mg2+ ion: 1s22s22 p 6
Aluminum, [Ne]3s23p1, loses three electrons to become like neon:
Al atom: 1s22s22 p 63s23 p 1 ---> Al3+ ion: 1s22s22 p 6
Let's go to row four and consider the elements potassium (19K) and calcium (20Ca). Have a periodic table infront of you as you read this. Element 19, potassium, becomes an ion by losing one electron to achievethe stable argon octet (3s23p6):
13-9 ©1997, A.J. Girondi
K atom: 1s22s22 p 63s23 p 64s1 ---> K1+ ion: 1s22s22 p 63s23 p 6
Calcium loses two electrons to achieve the argon octet:
Ca atom: 1s22s22p63s23p64s2 ---> Ca2+ ion: 1s22s22p63s23p6
SECTION 13.4 The Order in Which Atoms Lose Their Electrons
Before going on, a very important point has to be made. When electrons enter atoms, they fill theenergy levels and sublevels according to the diagonal rule (with exceptions). However, after the electronsare in an atom their distances from the nucleus do not follow the diagonal rule. For example, the order offilling for titanium (#22)is:
1s22s22 p 63s23 p 64s23 d 2
however, the actual distances of the sublevels from the nucleus is:
1s22s22 p 63s23 p 63 d 24s2
Another example is rhenium (#75) for which the order of filling is:
1s22s22 p 63s23 p 64s23 d 104 p 65s24 d 105 p 66s24 f 145 d 5
while the distance of the sublevels from the nucleus is:
1s22s22 p 63s23 p 63 d 104s24 p 64 d 104 f 145s25 p 65 d 56s2
Remember, the order of the energy levels and sublevels according to distance from the nucleus is:
1s,2s,2p,3s,3p,3d,4s,4p,4d,4f,5s,5p,5d,5f,6s,6p,6d,6f*,7s
*The 6f sublevel is never used, because there are only 111 elements as of the year 1995.
When atoms lose their electrons, they do not lose them in the reverseorder of the diagonal rule. Rather, they lose the ones that are farthestfrom the nucleus, first.
This is because the nuclear attraction is weaker at greater distances. The electrons in the highest energylevel of an atom may not be farthest from the nucleus. For example, electrons enter the 4s sublevelbefore they enter the 3d. The 3d is described as a "higher" sublevel because electrons in 3d orbitalscontain more energy than those in the 4s. The electrons in the 4s orbital, however, are farther from thenucleus.
Example: The configuration notation for 30Zn is 1s22s22 p 63s23 p 64s23 d 10
The orbital notation for 30Zn is:
1s 2s 2p 3s 3p 4s 3d (X) (X) (X)(X)(X) (X) (X)(X)(X) (X) (X)(X)(X)(X)(X)
The electrons in 30Zn are lost in the following order: 4s, 3d, 3p, 3s, 2p, 2s, and finally, 1s
In ordinary chemical reactions an atom normally loses electrons only from its valence shell. Forzinc those electrons would be in the 4s sublevel. Sometimes atoms can also lose some "d" electrons inthe sublevel immediately below the valence shell. For zinc those would be the electrons in the 3dsublevel. Zinc does not lose any of its 3d electrons, however, because the full 3d sublevel is too stable.
13-10 ©1997, A.J. Girondi
As a result, the only oxidation number that zinc has is +2, indicating that it can lose two electrons (the 4s2).
Many of the transition metals (21Sc through 30Zn) in row 4 can lose electrons and form ions in atleast three ways:
1. They may lose their 4s and 3d electrons and be left with the stable argon octet. For example, titanium(#21) can lose 4 electrons to achieve the stable argon octet:
Ti atom: 1s22s22 p 63s23 p 64s23 d 2 ---> Ti4+ ion: 1s22s22 p 63s23 p 6
2. They may lose only their 4s electrons and be left a highest energy level which is full: (3s23p63d10).Remember the third energy level has a capacity of 18. Zinc (30Zn) loses two electrons to achieve a highest(3rd) energy level which is full with 18 electrons (3s23p63d10):
Zn atom: 1s22s22 p 63s23 p 64s23 d 10 ---> Zn2+ ion: 1s22s22 p 63s23 p 63 d 10
3. Or, they may lose their 4s electrons and be left with arrangements such as half-full sublevels:
Manganese (25Mn) loses 2 electrons and is left with a half-full outer sublevel:
Mn atom: 1s22s22 p 63s23 p 64s23 d 5 ---> Mn2+ ion: 1s22s22 p 63s23 p 63 d 5
Manganese (25Mn) can also lose its 3d electrons in addition to its 4s electrons thus forming an Mn7+ ion:
Mn atom: 1s22s22 p 63s23 p 64s23 d 5 ---> Mn7+ ion: 1s22s22 p 63s23 p 6
It can also lose electrons in other ways resulting in the Mn6+ and the Mn4+ ions.
Remember! It is always the electrons in the valence shell that are lost first.
Let's consider element #31, gallium. It is a family 3A element. You should recall that elements inthis family have an oxidation number of +3. The elements above gallium in this family, B and Al, form ionsby losing three electrons to achieve stable octets. However, gallium is in row four and contains 3delectrons. It also loses three electrons, but that does not leave gallium with a stable octet, but it doesleave it with a full outermost energy level:
Ga atom: 1s22s22 p 63s23 p 64s23 d 104 p 1 ---> Ga3+: 1s22s22 p 63s23 p 63 d 10
You see, the 4s2 and 4p1 electrons are the ones that are lost first. Remember, again, that the outermostelectrons are the ones that go. In this case, the electrons in the highest sublevel (3d) are not members ofthe outermost (4th) energy level. Let's briefly get back to that third kind of stable arrangement thatinvolves half-full sublevels. For example, the electron configuration of copper is listed below according tothe diagonal rule. Also listed, is the actual electron arrangement in copper as determined by experiment.
Cu (from diagonal rule): 1s22s22 p 63s23 p 64s23 d 9
Cu (actual configuration) : 1s22s22 p 63s23 p 64s13 d 10
Why does the copper atom deviate from the trend represented by the diagonal rule? If one of copper's 4selectrons moves to the 3d level, the next-to-the-highest sublevel will be half-full (4s1), while the highestone will be full (3d10). The stability of these half-full and full sublevels makes this the preferred state forcopper. Similar reasoning explains the deviation of chromium from the diagonal rule as seen below.
Cr (from diagonal rule): 1s22s22 p 63s23 p 64s23 d 4
Cr (actual configuration): 1s22s22 p 63s23 p 64s13 d 5
13-11 ©1997, A.J. Girondi
Thus, you see that the information you get from the diagonal rule is not perfect. In fact, there are quite afew other deviations from the diagonal rule involving other elements. Some of them are not wellunderstood. If you want the best information regarding the electron configuration of electrons in an atom,you must check a reference source such as the tables found in Appendix B of your ALICE materials. Lookthere now, and find at least two other elements that deviate from the diagonal rule. Name them:
{10}_____________________________________________________________________________
SECTION 13.5 Trends in Ionization Energies and Atomic Radii
The amounts of ionization energy required to drive out the first electron from atoms is summarizedin Table 13.7. You should notice that certain generalizations can be made regarding trends in theionization energies as you go down a family or across a row.
Generally speaking, how do the values change as you go down a family?
{11}____________________________________________________________________________
Generally speaking, how do the values change as you go across a row or series from left to right?
{12}____________________________________________________________________________
As you move down a family, the atoms are adding energy levels and are getting larger. Theoutermost electrons are farther and farther from the nucleus, which makes them easier and easier toremove. As you follow the trends in Table 13.7, some of the values seem too large and out of place.These numbers reflect the extra stability that results from full and half-full outer sublevels such as the 1013kJ/mole for phosphorous (15P) resulting from its half-full p3 outer sublevel. Therefore, the ionizationenergies get smaller as you move down a family.
As you move from left to right across a row, the atoms are not adding energy levels, but they areadding sublevels and (you might think) gaining a little in size. However, protons are added to the nucleusas you move across a row, and the increasingly positive nuclei pull the electrons in tighter. So generallyspeaking, as you go across a row from left to right the atoms actually get a bit smaller. This puts electronscloser to the nucleus, and ionization energies tend to increase.
Table 13.8 contains data about atomic radii which will give you an idea of how the sizes of theatoms are changing. Look at that table now and answer the following:
Generally speaking, summarize how atomic sizes vary as you go down a family:
{13}____________________________________________________________________________
Why? {14}_______________________________________________________________________
Generally speaking, summarize how atomic sizes vary as you go across a row:
{15}____________________________________________________________________________
Why? {16}_______________________________________________________________________
13-12 ©1997, A.J. Girondi
H
1314
Li
519
Na
498
K
418
Rb
402
Cs
377
Fr
- - - -
Be
900
Ca
590
Sr
548
Ba
502
Ra
510
Sc
632
Y
615
La
540
Ac
665
Ti
657
Zr
661
Hf
674
Unq
V
649
Cr
653
Mn
715
Fe
757
Co
757
Ni
736
Cu
745
Zn
908
Ga
577
Ge
761
As
946
Se
941
Br
1138
Kr
1351
Nb
665
Mo
686
Tc
703
Ru
711
Rh
720
Pd
803
Ag
732
Cd
866
In
556
Sn
707
Sb
833
Te
870
I
1008
Xe
1172
Ta
761
W
770
Re
761
Os
841
Pd
803
Pt Au Hg
1008
Tl Pb
715
Bi
703
Po
820
At
- - - -
Rn
1038
Unp Unh Uns Uno Une Uun Uuu
Al
577
Mg
736
Si
787
P
1013
S
1000
Cl
1251
Ar
1519
B
799
C
1088
N
1402
O
1314
F
1682
Ne
2079
He
2372
Ce
527
Pr
523
Nd
531
Pm
536
Sm
544
Eu
548
Gd
594
Tb
565
Dy
573
Ho
582
Er
590
Tm
598
Yb
602
Lu
523
Th
665
Pa U Np Pu
561
Am
577
Cm Bk Cf Es Fm Md No Lr
Table 13.7First Ionization Energies of Atoms (kJ/Mole)
1A
2A 3A 4A 5A 6A 7A
8A
First ionization energies generally increase from left to right, and generally decrease from top to bottom.
---------------------------------------------------------------------------------------------------------------------
H
0.32
Li
1.23
Na
1.54
K
2.03
Rb
2.16
Cs
2.35
Fr
- - - -
Be
0.89
Ca
1.74
Sr
1.91
Ba
1.98
Ra
2.20
Sc
1.44
Y
1.62
La
1.69
Ac
2.0
Ti
1.32
Zr
1.44
Hf
1.44
Unq
V
1.22
Cr
1.18
Mn
1.17
Fe
1.17
Co
1.16
Ni
1.15
Cu
1.17
Zn
1.25
Ga
1.26
Ge
1.22
As
1.20
Se
1.17
Br
1.14
Kr
1.12
Nb
1.34
Mo
1.30
Tc
1.27
Ru
1.25
Rh
1.25
Pd
1.28
Ag
1.34
Cd
1.48
In
1.44
Sn
1.40
Sb
1.40
Te
1.36
I
1.33
Xe
1.31
Ta
1.34
W
1.30
Re
1.28
Os
1.26
Pd
1.27
Pt
1.30
Au
1.34
Hg
1.48
Tl
1.48
Pb
1.47
Bi
1.46
Po
1.46
At
1.45
Rn
- - - -
Unp Unh Uns Uno Une Uun Uuu
Al
1.18
Mg
1.36
Si
1.11
P
1.06
S
1.02
Cl
0.99
Ar
0.98
B
0.82
C
0.77
N
0.74
O
0.74
F
0.68
Ne
0.67
He
0.31
Ce
1.65
Pr
1.64
Nd
1.64
Pm
1.63
Sm
1.62
Eu
1.85
Gd
1.62
Tb
1.61
Dy
1.60
Ho
1.58
Er
1.58
Tm
1.58
Yb
1.70
Lu
1.56
Th
1.65
Pa U
1.42
Np Pu Am Cm Bk Cf Es Fm Md No Lr
Table 13.8Atomic Radii (Size) of Atoms (Å)
1A
2A 3A 4A 5A 6A 7A
8A
1 Å = 1 X 10-8 cm
The size of atoms generally decreases across a row from left to right, and increases down a family from top to bottom.
13-13 ©1997, A.J. Girondi
SECTION 13.6 Configuration and Orbital Notation of Ions
Problem 4. In Table 13.9 determine the charge of the most common ions formed by each element.The electron configuration of the ions should be the same as the nearest noble gas (according to atomicnumber) to each element. Some atoms gain electrons to become ions, while others will be losingelectrons. Some atoms can do both. Complete Table 13.9.
Table 13.9Electron Configurations of Selected Atoms and Ions
Element Electron Configuration Electron Configuration Charge of the Atom of the Ion on Ion
N ______________________ ______________________ ______
Na ______________________ ______________________ ______
Mg ______________________ ______________________ ______
Li ______________________ ______________________ ______
Be ______________________ ______________________ ______
O ______________________ ______________________ ______
F ______________________ ______________________ ______
C ______________________ ______________________ +4
Problem 5. Complete Table 13.10 below. Do not use core notation here.
Table 13.10Electron Configurations of Some Common Ions
Ion Electron Configuration No. of Protons No. of Electrons of Ion in Ion in Ion
Si4+ __________________________ ______ ______
C4- __________________________ ______ ______
N3- __________________________ ______ ______
Al3+ __________________________ ______ ______
Cl1- __________________________ ______ ______
K1+ __________________________ ______ ______
P3- __________________________ ______ ______
S2- __________________________ ______ ______
13-14 ©1997, A.J. Girondi
Problem 6. Table 13.11 below, asks you to write the orbital notation for the IONS for which you wroteconfiguration notation in Table 13.10. Do not use core notation here.
Table 13.11Orbital Notation of Some Common Atoms and Ions
Atom Ion Orbital Notation
1s 2s 2p 3s 3p 4s
Si ( ) ( ) ( )( )( ) ( ) ( )( )( ) ( )Si4+ ( ) ( ) ( )( )( ) ( ) ( )( )( ) ( )
C ( ) ( ) ( )( )( ) ( ) ( )( )( ) ( )C4- ( ) ( ) ( )( )( ) ( ) ( )( )( ) ( )
N ( ) ( ) ( )( )( ) ( ) ( )( )( ) ( )N3- ( ) ( ) ( )( )( ) ( ) ( )( )( ) ( )
Al ( ) ( ) ( )( )( ) ( ) ( )( )( ) ( )Al3+ ( ) ( ) ( )( )( ) ( ) ( )( )( ) ( )
Cl ( ) ( ) ( )( )( ) ( ) ( )( )( ) ( )Cl1- ( ) ( ) ( )( )( ) ( ) ( )( )( ) ( )
K ( ) ( ) ( )( )( ) ( ) ( )( )( ) ( )K1+ ( ) ( ) ( )( )( ) ( ) ( )( )( ) ( )
P ( ) ( ) ( )( )( ) ( ) ( )( )( ) ( )P3- ( ) ( ) ( )( )( ) ( ) ( )( )( ) ( )
S ( ) ( ) ( )( )( ) ( ) ( )( )( ) ( )S2- ( ) ( ) ( )( )( ) ( ) ( )( )( ) ( )
Problem 7. Knowing what you now do about how and why atoms become ions, can you see whyelements in the same family often form ions that have a common charge or charges? Look back over thework you have done and determine the most common negative and positive charge on ions for each ofthe families listed in Table 13.12.
Table 13.12Ion Charges in Families of Elements
Number of Most Common Most CommonFamily Valence Electrons Positive Charge Negative Charge
1A _______ ________ __XXXX__
2A _______ ________ ________
3A _______ ________ ________
4A _______ ________ ________
5A _______ ________ ________
6A _______ ________ ________
7A _______ ________ ________
8A _______ __XXXX__ __XXXX__
XXXX = these ions are either rare or nonexistant, so we will ignore them.
13-15 ©1997, A.J. Girondi
SECTION 13.7 Quantum Mechanics and Atomic Orbitals
Quantum mechanics is a set of laws which describe the behavior of very small particles such aselectrons. These laws provide a logical basis for the periodic arrangement of the elements on the periodictable. The similarities in numbers of valence electrons accounts for the existence of chemical families.The energy level diagrams and shorthand electron configurations we have examined are good"bookkeeping" devices for keeping track of electrons in elements.
Quantum mechanics also allows us to "draw pictures" of the probable locations of electronsaround the nucleus. The theory is based partially on the fact that small particles, electrons in this case,which are confined to a very small space exhibit very unusual behavior. The electrons in an atom do notfollow any predictable pathway. In fact, sometimes they behave like particles and sometimes they behavelike waves. When an electron has wave characteristics, it acts as though it is spread out over a large space.Because of their dual nature, electrons cannot be accurately represented as being little particles that orbitthe nucleus like planets orbit the sun. Such a picture ignores the wave characteristics of the electron.
What quantum mechanics theory does allow us to do is to construct three-dimensional shapesarranged about the nucleus of an atom to represent the space in which a particular electron spendspractically all of its time. The analogy below might be helpful in understanding what these three-dimensional shapes mean.
Suppose that you were to take a map of your city and use it to mark points on it to represent yourlocation at particular times during the day and night. Further, suppose that each hour you mark a point onyour map for your given location and that you do this for one month. At the end of the month, your mapwould show a large number of points clustered around the location of your school and your home. That is,there would be a large "density" of points at these particular locations. There would be other locations atwhich there might be a considerable density (number) of points such as a favorite shopping mall or eatingplace. However, the density of these locations would not be as great as that around your school andhome. It is likely that there would be locations on your map at which no points would be placed (city dump,police station, etc.).
Now suppose someone wanted to locate you for some reason. If they had a copy of your map,they could see where the "probable" places were in which they might locate you. They would probably goto the place with the highest density of points, and, if you were not there, they would travel to the nextmost probable location. This map would not tell them exactly where you were, but it certainly would savethem time in trying to find you. Such a map could be called a "probability density plot" for your locationover a long time period.
Quantum mechanics can give us similar information about the location of electrons. It allows us toconstruct probability density plots for the various electrons in an atom. The equations which have beendeveloped to do this are too complex to be presented here. Unlike your map, which is a two-dimensionalrepresentation, the probability density plots for electrons must be constructed in three dimensions.(Although we can only draw two- dimensional plots on these flat pages, we must hope that your mind willallow you to think in three dimensions.) The plots that we draw for electrons represent the space in whichparticular electrons spend almost all their time. These are orbitals. They do not represent the space inwhich electrons spend 100% of their time, because this would require plots with infinite extent. You mighttake a trip out of the city during the month and would not be able to place any points on your map torepresent your location, because your map only encloses the boundaries of your city. Your map hasbecome a probability density plot of your location for almost all of the time (but not 100%). Just as youmight spend some of your time out of town, electrons can and do spend some time outside of the three-dimensional spaces that we plot for them.
What do these three-dimensional probability density plots look like? Drawings appear in Figures13.1 through 13.3. The size and shape of these plots depend on the orbital that an electron occupiesabout the nucleus. There is some regularity among them. For example, all "s" orbitals (like 1s, 2s, 3s, etc.)
13-16 ©1997, A.J. Girondi
have the same basic shape - a hollow sphere. (Actually their description is not this simple, but it's goodenough for our purposes.) The radius for the sphere representing the 1s orbital is smaller than that for the2s orbital, which is smaller than that for the 3s orbital, and so on.
The three orbitals in the "p" sublevel for atoms all have the same shape. The "p" orbitals areshaped like a long balloon squeezed together in the middle. Since there are three of these in every major"p" sublevel, they are found to point along the directions of a set of x, y, z axes that intersect at thenucleus. Just as is the case for "s" orbitals, the 3p orbitals extend farther out from the nucleus than the2p, and so on. The shapes of the "s" and "p" orbitals are shown in Figures 13.1 & 13.2. The shapes thatappear in Figures 13.1 through 13.3 are actually depictions of three-dimensional waves. In a sense, thesurfaces of these figures correspond to the boundaries of your city on the map in the analogy that weused. The electrons in a particular orbital are very likely to be within the boundary surfaces most of thetime, just as you were very likely to be located within the boundaries of your city most of the time. In yourprobability density map there would be a concentration of points around those locations where you spendmost of your time. Within the boundaries of the probability density plots for electrons shown in figure11.1, there are regions in which the "s" electrons spend most of their time. The same thing is shown the"p" electrons in Figure 13.2.
The shape of "d" and "f" probability density plots become more complicated. The shape of justone of the "d" orbitals is shown in Figure 13.3. There are five "d" orbitals, since there are five "d"sublevels. Electrons are important in understanding chemical reactions and chemical bonding.Knowledge about quantum mechanics is helpful in understanding the kinds of ions formed by atoms andthe shapes of molecules that atoms form.
1s 2s 3s
Figure 13.1Depictions of 1s, 2s, and 3s Orbitals
px py pz
Figure 13.2Depictions of the Three 2p Orbitals
13-17 ©1997, A.J. Girondi
Figure 13.3One of the Five "d" Orbitals
The subject of atomic structures and electron behavior asrevealed by the use of quantum mechanics is a very difficultsubject. You have been given much information, but don't feelthat anything is wrong if this leaves you somewhat confused.You have only scratched the surface of the subject, and a realunderstanding would require much more study (perhaps incollege–level chemistry or physics). You have discovered thatelectrons do not behave in a manner that makes it easy to draw"pictures" of atoms. In this chapter, you have been "let in" on avery clever way that scientists have developed to deal withelectrons. While this way is at times confusing and difficult, it isalso very fascinating and quite useful, as you will see in the nextchapter.
ACTIVITY 13.8 Creating a Probability Density Plot
The position of an electron in an atom at a given moment cannot be predicted. The region ofspace in which the electron can most probably be found is, however, predictable. This region is oftencalled an "electron cloud" or "orbital" and is represented by a fuzzy shape with the nucleus at the center.The shape is determined by mathematical calculations using what is known as the wave mechanical modelof the atom. The electron cloud is not absolute. Sometimes, the electron may be found outside thecloud.
The purpose of this activity is to create a probability distribution of locations (a probability densityplot) around a central point. You will need a crayon and a paper "bull's eye" target. A copy of the targetcan be found within the next few pages. The procedure follows:
1. Place the target in front of you on your lab table or on the floor.
2. Obtain the crayon provided and hold it about 2 feet above the target. Drop the crayon onto the target,trying to hit the center. Repeat this at least 50 times. (See Figure 13.4 below.)
3. Count the number of marks in each numbered region of the target and record the numbers in data table13.13 below.
Figure 13.4Creating a Probability Density Plot
Table 13.13Distribution of Hits
Area Number of Hits
1
2
3
4
5
Out
13-18 ©1997, A.J. Girondi
Plot your results on the grid on the next page. Plot the number of hits on the Y axis and thenumbered areas on the X axis.
Based on your results, which numbered area of the target had the highest probability of a hit?_______
According to your data, the probability that a given drop of the crayon will hit region 1 is: 1 in _______. Of
the various shapes of plots shown in figures 13.1, 13.2, and 13.3 would you say that your plot most
resembles an s, p, or d orbital?_________
Have your lab partner make a copy of your plot by placing another copy of the target under yours,and holding it up to a window. Using a pen or crayon, duplicate the hit marks. Be sure to include yourtarget when submitting this chapter to your instructor for approval.
SECTION 13.9 Learning Outcomes
The learning outcomes for Chapter 13 are listed below. Check each one when you are certainthat you have mastered it. Arrange to take any quizzes or exams, and then move on to chapter 14.
_____1. Write the electron configuration of atoms and of ions.
_____2. Write the orbital notation for atoms and of ions.
_____3. Write the core notation (configuration and orbital) for atoms and ions.
_____4. Define valence electrons and determine the number of them in an atom.
_____5. Describe trends in ionization energy across a row and down a column on the periodic table.
_____6. Describe trends in the size of atoms across a row and down a column on the periodic table.
_____7. Explain what is meant by ionization energy.
_____8. List the common charges of ions of elements in columns 1A - 8A of the periodic table.
_____9. Be able to identify probability density plots of s, p, and d orbitals.
_____10. Define a probability density plot and explain how it is useful in describing the location of an
electron in an atom.
13-19 ©1997, A.J. Girondi
SECTION 13.10 Answers to Questions and Problems
Questions:
{1} one; {2} eight; {3} 4s24p6; {4} 3s23p1; {5} 5s25p4; {6} third; {7} Fourth electron must come from thesecond energy level which is closer to the nucleus; {8} 1A; {9} 2A; {10} copper, niobium, molybdenumand many more; {11} decrease; {12} increase; {13} get larger; {14} each successive one has one moreenergy level; {15} get smaller; {16} increased attraction for electrons by nuclei with more protons;
Problems:
1. 2s2, 3s2, 4s2, 5s2, 6s2, 7s2; 2s22p1, 3s23p1, 4s24p1, 5s25p1, 6s26p1
2. 3s23p2, 4s24p5, 4s2, 4s24p3, 3s23p4, 5s2, 6s26p5, 6s2, 3s23p5
3. See Table 13.4 for answers.4. N 1s22s22p3 1s22s22p6 -3
Na 1s22s22p63s1 1s22s22p6 +1Mg 1s22s22p63s2 1s22s22p6 +2Li 1s22s1 1s2 +1Be 1s22s2 1s2 +2O 1s22s22p4 1s22s22p6 -2F 1s22s22p5 1s22s22p6 -1C 1s22s22p2 1s2 +4
5. Si4+ 1s22s22p6 14 10C4- 1s22s22p6 6 10N3- 1s22s22p6 7 10Al3+ 1s22s22p6 13 10Cl1- 1s22s22p63s23p6 17 18K1+ 1s22s22p63s23p6 19 18P3- 1s22s22p63s23p6 15 18S2- 1s22s22p63s23p6 16 18
1s 2s 2p 3s 3p 4s
Si (X) (X) (X)(X)(X) (X) (/)(/)( ) ( )Si4+ (X) (X) (X)(X)(X) ( ) ( )( )( ) ( )
C (X) (X) (/)(/)( ) ( ) ( )( )( ) ( )C4- (X) (X) (X)(X)(X) ( ) ( )( )( ) ( )
N (X) (X) (/)(/)(/) ( ) ( )( )( ) ( )N3- (X) (X) (X)(X)(X) ( ) ( )( )( ) ( )
Al (X) (X) (X)(X)(X) (X) (/)( )( ) ( )Al3+ (X) (X) (X)(X)(X) ( ) ( )( )( ) ( )
Cl (X) (X) (X)(X)(X) (X) (X)(X)(/) ( )Cl1- (X) (X) (X)(X)(X) (X) (X)(X)(X) ( )
K (X) (X) (X)(X)(X) (X) (X)(X)(X) (/)K1+ (X) (X) (X)(X)(X) (X) (X)(X)(X) ( )
P (X) (X) (X)(X)(X) (X) (/)(/)(/) ( )P3- (X) (X) (X)(X)(X) (X) (X)(X)(X) ( )
S (X) (X) (X)(X)(X) (X) (X)(/)(/) ( )S2- (X) (X) (X)(X)(X) (X) (X)(X)(X) ( )
6.
7. 1A 1 +1 XXXX2A 2 +2 -63A 3 +3 -54A 4 +4 -45A 5 +5 -36A 6 +6 -27A 7 +7 -18A 8 XXX XXX
13-20 ©1997, A.J. Girondi
1
2
3
4
5
Figure 13.5Target for Probability Density Plot
13-21 ©1997, A.J. Girondi
SECTION 13.11 Student Notes
13-22 ©1997, A.J. Girondi