1. Periodic Relationships Among the Elements
Chapter 8
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2. When the Elements Were Discovered

3. Ground State Electron Configurations of the Elements
ns2np6
Ground State Electron Configurations of the Elements
ns1
ns2np1
ns2np2
ns2np3
ns2np4
ns2np5
ns2
d10 d1 d5 4f 5f

4. Classification of the Elements

5. Electron Configurations of Cations and Anions
Of Representative Elements
Na [Ne]3s1
Na+ [Ne]
Atoms lose electrons so that cation has a noble-gas outer electron configuration.
Ca [Ar]4s2
Ca2+ [Ar]
Al [Ne]3s23p1
Al3+ [Ne]
H 1s1
H- 1s2 or [He]
Atoms gain electrons so that anion has a noble-gas outer electron configuration.
F 1s22s22p5
F- 1s22s22p6 or [Ne]
O 1s22s22p4
O2- 1s22s22p6 or [Ne]
N 1s22s22p3
N3- 1s22s22p6 or [Ne]

6. Cations and Anions Of Representative Elements+1 +2 +3 -3 -2 -1

7. Isoelectronic: have the same number of electrons, and hence the same ground-state electron configuration
Na+: [Ne]
Al3+: [Ne]
F-: 1s22s22p6 or [Ne]
O2-: 1s22s22p6 or [Ne]
N3-: 1s22s22p6 or [Ne]
Na+, Al3+, F-, O2-, and N3- are all isoelectronic with Ne
What neutral atom is isoelectronic with H- ?
H-: 1s2
same electron configuration as He

8. Electron Configurations of Cations of Transition Metals
EXAMPLE 8.1
An atom of a certain element has 15 electrons. Without consulting a periodic table,
answer the following questions: (a) What is the ground-state electron confi guration of
the element? (b) How should the element be classifi ed? (c) Is the element diamagnetic
or paramagnetic?
Strategy (a) We refer to the building-up principle discussed in Section 7.9 and start
writing the electron confi guration with principal quantum number n 5 1 and continuing
upward until all the electrons are accounted for. (b) What are the electron confi guration
characteristics of representative elements? transition elements? noble gases? (c) Examine
the pairing scheme of the electrons in the outermost shell. What determines whether an
element is diamagnetic or paramagnetic?
Solution (a) We know that for n 5 1 we have a 1 s orbital (2 electrons); for n 5 2 we
have a 2 s orbital (2 electrons) and three 2 p orbitals (6 electrons); for n 5 3 we have a
3 s orbital (2 electrons). The number of electrons left is and these three
electrons are placed in the 3 p orbitals. The electron confi guration is 1 s 2 2 s 2 2 p 6 3 s 2 3 p 3 .
(b) Because the 3 p subshell is not completely fi lled, this is a representative element.
Based on the information given, we cannot say whether it is a metal, a nonmetal, or
a metalloid.
(c) According to Hund’s rule, the three electrons in the 3 p orbitals have parallel spins
(three unpaired electrons). Therefore, the element is paramagnetic.
Check For (b), note that a transition metal possesses an incompletely fi lled d subshell
and a noble gas has a completely fi lled outer shell. For (c), recall that if the atoms of
an element contain an odd number of electrons, then the element must be paramagnetic.
Practice Exercise An atom of a certain element has 20 electrons. (a) Write the
ground-state electron confi guration of the element, (b) classify the element, (c) determine
whether the element is diamagnetic or paramagnetic.
Electron Configurations of Cations of Transition Metals
When a cation is formed from an atom of a transition metal, electrons are always removed first from the ns orbital and then from the (n – 1)d orbitals.
Fe: [Ar]4s23d6
Mn: [Ar]4s23d5
Fe2+: [Ar]4s03d6 or [Ar]3d6
Mn2+: [Ar]4s03d5 or [Ar]3d5
Fe3+: [Ar]4s03d5 or [Ar]3d5

9. (a) What is the ground-state electron configuration of the element?
EXAMPLE 8.1
An atom of a certain element has 15 electrons. Without consulting a periodic table,
answer the following questions: (a) What is the ground-state electron confi guration of
the element? (b) How should the element be classifi ed? (c) Is the element diamagnetic
or paramagnetic?
Strategy (a) We refer to the building-up principle discussed in Section 7.9 and start
writing the electron confi guration with principal quantum number n 5 1 and continuing
upward until all the electrons are accounted for. (b) What are the electron confi guration
characteristics of representative elements? transition elements? noble gases? (c) Examine
the pairing scheme of the electrons in the outermost shell. What determines whether an
element is diamagnetic or paramagnetic?
Solution (a) We know that for n 5 1 we have a 1 s orbital (2 electrons); for n 5 2 we
have a 2 s orbital (2 electrons) and three 2 p orbitals (6 electrons); for n 5 3 we have a
3 s orbital (2 electrons). The number of electrons left is and these three
electrons are placed in the 3 p orbitals. The electron confi guration is 1 s 2 2 s 2 2 p 6 3 s 2 3 p 3 .
(b) Because the 3 p subshell is not completely fi lled, this is a representative element.
Based on the information given, we cannot say whether it is a metal, a nonmetal, or
a metalloid.
(c) According to Hund’s rule, the three electrons in the 3 p orbitals have parallel spins
(three unpaired electrons). Therefore, the element is paramagnetic.
Check For (b), note that a transition metal possesses an incompletely fi lled d subshell
and a noble gas has a completely fi lled outer shell. For (c), recall that if the atoms of
an element contain an odd number of electrons, then the element must be paramagnetic.
Practice Exercise An atom of a certain element has 20 electrons. (a) Write the
ground-state electron confi guration of the element, (b) classify the element, (c) determine
whether the element is diamagnetic or paramagnetic.
EXAMPLE 8.1
An atom of a certain element has 15 electrons. Without consulting a periodic table, answer the following questions:
(a) What is the ground-state electron configuration of the element?
(b) How should the element be classified?
(c) Is the element diamagnetic or paramagnetic?

10. The shielding Effect and Effective Nuclear Charge تأثير الحجب في الذرات متعددة الإلكترونات وشحنة النواة المؤثرة
In a many-electron atom, electrons are both attracted to the nucleus and repelled by other electrons.
The nuclear charge that an electron experiences depends on both factors.
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11. The shielding Effect and Effective Nuclear Charge
The effective nuclear charge, Zeff, is found this way:
Zeff = Z − S
where Z is the atomic number and S is a screening constant(الجزء من شحنة النواة المحجوب عن جذب الكترونات التكافؤ ) , usually close to the number of inner electrons.
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12. Effective nuclear charge (Zeff) is the “positive charge” felt by an electron.
Zeff = Z - s
0 < s < Z (s = shielding constant)
Zeff  Z – number of inner or core electrons
Zeff
Core Z
Radius (pm) Na Mg Al Si 11 12 13 14 10 1 2 3 4
186
160
143
132

13. Effective Nuclear Charge (Zeff)

14. What Is the Size of an Atom?
The bonding atomic radius is defined as one-half of the distance between covalently bonded nuclei.
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15. Atomic Radii
covalent radius
metallic radius

17. Sizes of Atoms Bonding atomic radius tends to…
…decrease from left to right across a row
(due to increasing Zeff).
…increase from top to bottom of a column
(due to increasing value of n).
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19. Trends in Atomic Radii

20. Comparison of Atomic Radii with Ionic Radii

21. Cation is always smaller than atom from which it is formed.
Anion is always larger than atom from which it is formed.

22. The Radii (in pm) of Ions of Familiar Elements

23. Sizes of Ions Ionic size depends upon: The nuclear charge.
The number of electrons.
The orbitals in which electrons reside.
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24. Sizes of Ions Cations are smaller than their parent atoms.
The outermost electron is removed and repulsions between electrons are reduced.
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25. Sizes of Ions Anions are larger than their parent atoms.
Electrons are added and repulsions between electrons are increased.
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26. Sizes of Ions Ions increase in size as you go down a column.
This is due to increasing value of n.
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27. Sizes of Ions
In an isoelectronic series, ions have the same number of electrons.
Ionic size decreases with an increasing nuclear charge.
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28. EXAMPLE 8.2 Referring to a periodic table, arrange the following atoms in order of increasing atomic radius: P, Si, N. Strategy What are the trends in atomic radii in a periodic group and in a particular period? Which of the preceding elements are in the same group? in the same period? Solution From Figure 8.1 we see that N and P are in the same group (Group 5A). Therefore, the radius of N is smaller than that of P (atomic radius increases as we go down a group). Both Si and P are in the third period, and Si is to the left of P. Therefore, the radius of P is smaller than that of Si (atomic radius decreases as we move from left to right across a period). Thus, the order of increasing radius is N , P , Si . Practice Exercise Arrange the following atoms in order of decreasing radius: C, Li, Be.
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29. Chemistry in Action: The 3rd Liquid Element?
117 elements, 2 are liquids at 250C – Br2 and Hg
223Fr, t1/2 = 21 minutes
Liquid?

30. Ionization energy is the minimum energy (kJ/mol) required to remove an electron from a gaseous atom in its ground state.
I1 + X (g) X+(g) + e-
I1 first ionization energy
I2 + X+(g) X2+(g) + e-
I2 second ionization energy
I3 + X2+(g) X3+(g) + e-
I3 third ionization energy
I1 < I2 < I3

32. Trends in First Ionization Energies
Generally, as one goes across a row, it gets harder to remove an electron.
As you go from left to right, Zeff increases.
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33. Trends in First Ionization Energies
However, there are two apparent discontinuities in this trend.
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34. Trends in First Ionization Energies
The first occurs between Groups IIA and IIIA.
In this case the electron is removed from a p-orbital rather than an s-orbital.
The electron removed is farther from nucleus.
There is also a small amount of repulsion by the s electrons.
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35. Trends in First Ionization Energies
The second occurs between Groups VA and VIA.
The electron removed comes from doubly occupied orbital.
Repulsion from the other electron in the orbital aids in its removal.
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36. General Trends in First Ionization Energies
Increasing First Ionization Energy
Increasing First Ionization Energy

37. Electron affinity is the negative of the energy change that occurs when an electron is accepted by an atom in the gaseous state to form an anion.
X (g) + e X-(g)
F (g) + e F-(g)
DH = -328 kJ/mol
EA = +328 kJ/mol
O (g) + e O-(g)
DH = -141 kJ/mol
EA = +141 kJ/mol

38. Trends in Electron Affinity
In general, electron affinity becomes more exothermic as you go from left to right across a row.
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39. Trends in Electron Affinity
There are again, however, two discontinuities in this trend.
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40. Trends in Electron Affinity
The first occurs between Groups IA and IIA.
The added electron must go in a p-orbital, not an s-orbital.
The electron is farther from nucleus and feels repulsion from the s-electrons.
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41. Trends in Electron Affinity
The second occurs between Groups IVA and VA.
Group VA has no empty orbitals.
The extra electron must go into an already occupied orbital, creating repulsion.
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43. Variation of Electron Affinity With Atomic Number (H – Ba)

45. Diagonal Relationships on the Periodic Table

46. Group 1A Elements (ns1, n  2)
M M+1 + 1e-
2M(s) + 2H2O(l) MOH(aq) + H2(g)
4M(s) + O2(g) M2O(s)
Increasing reactivity

47. Group 1A Elements (ns1, n  2)

48. Group 2A Elements (ns2, n  2)
M M+2 + 2e-
Be(s) + 2H2O(l) No Reaction
Mg(s) + 2H2O(g) Mg(OH)2(aq) + H2(g)
M(s) + 2H2O(l) M(OH)2(aq) + H2(g) M = Ca, Sr, or Ba
Increasing reactivity

49. Group 2A Elements (ns2, n  2)

50. Group 3A Elements (ns2np1, n  2)
4Al(s) + 3O2(g) Al2O3(s)
2Al(s) + 6H+(aq) Al3+(aq) + 3H2(g)

51. Group 3A Elements (ns2np1, n  2)

52. Group 4A Elements (ns2np2, n  2)
Sn(s) + 2H+(aq) Sn2+(aq) + H2 (g)
Pb(s) + 2H+(aq) Pb2+(aq) + H2 (g)

53. Group 4A Elements (ns2np2, n  2)

54. Group 5A Elements (ns2np3, n  2)
N2O5(s) + H2O(l) HNO3(aq)
P4O10(s) + 6H2O(l) H3PO4(aq)

55. Group 5A Elements (ns2np3, n  2)

56. Group 6A Elements (ns2np4, n  2)
SO3(g) + H2O(l) H2SO4(aq)

57. Group 6A Elements (ns2np4, n  2)

58. Group 7A Elements (ns2np5, n  2)
X + 1e X-1
X2(g) + H2(g) HX(g)
Increasing reactivity

59. Group 7A Elements (ns2np5, n  2)

60. Group 8A Elements (ns2np6, n  2)
Completely filled ns and np subshells.
Highest ionization energy of all elements.
No tendency to accept extra electrons.

61. Compounds of the Noble Gases
A number of xenon compounds XeF4, XeO3, XeO4, XeOF4 exist.
A few krypton compounds (KrF2, for example) have been prepared.

62. Comparison of Group 1A and 1B
The metals in these two groups have similar outer electron configurations, with one electron in the outermost s orbital.
Chemical properties are quite different due to difference in the ionization energy.
Lower I1, more reactive

63. Properties of Oxides Across a Period
basic
acidic

64. Chemistry in Action: Discovery of the Noble Gases
Sir William Ramsay