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Periodic Relationships Among the Elements Chapter 8 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

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Presentation on theme: "Periodic Relationships Among the Elements Chapter 8 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display."— Presentation transcript:

1 Periodic Relationships Among the Elements Chapter 8 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

2 2 The Modern Periodic Table Period Group Alkali Metal Noble Gas Halogen Alkali Earth Metal

3 3 Electron Configurations of Cations and Anions Na [Ne]3s 1 Na + [Ne] Ca [Ar]4s 2 Ca 2+ [Ar] Al [Ne]3s 2 3p 1 Al 3+ [Ne] Atoms lose electrons so that cation has a noble-gas outer electron configuration. H 1s 1 H - 1s 2 or [He] F 1s 2 2s 2 2p 5 F - 1s 2 2s 2 2p 6 or [Ne] O 1s 2 2s 2 2p 4 O 2- 1s 2 2s 2 2p 6 or [Ne] N 1s 2 2s 2 2p 3 N 3- 1s 2 2s 2 2p 6 or [Ne] Atoms gain electrons so that anion has a noble-gas outer electron configuration. Of Representative Elements

4 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. 4

5 Cations and Anions Of Representative Elements

6 6 Na + : [Ne] Al 3+ : [Ne] F - : 1s 2 2s 2 2p 6 or [Ne] O 2- : 1s 2 2s 2 2p 6 or [Ne] N 3- : 1s 2 2s 2 2p 6 or [Ne] Na +, Mg 2+,Al 3+, F -, O 2-, and N 3- are all isoelectronic with Ne What neutral atom is isoelectronic with H - ? H - : 1s 2 same electron configuration as He Isoelectronic: have the same number of electrons, and hence the same ground-state electron configuration Mg 2+ : 1s 2 2s 2 2p 6 or [Ne]

7 7 Electron Configurations of Cations of Transition Metals ( d-block elements) 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. Co: [Ar]4s 2 3d 7 Co 2+ : [Ar]4s 0 3d 7 or [Ar]3d 7 Co 3+ : [Ar]4s 0 3d 6 or [Ar]3d 6 Mn: [Ar]4s 2 3d 5 Mn 2+ : [Ar]4s 0 3d 5 or [Ar]3d 5

8 8 Atomic Radii metallic radius covalent radius

9 9

10 10 Cation is always smaller than atom from which it is formed. Anion is always larger than atom from which it is formed.

11 11 Ionization energy is the minimum energy (kJ/mol) required to remove an electron from a gaseous atom in its ground state. I 1 + X (g) X + (g) + e - I 2 + X + (g) X 2 + (g) + e - I 3 + X 2+ (g) X 3 + (g) + e - I 1 first ionization energy I 2 second ionization energy I 3 third ionization energy I 1 < I 2 < I 3

12 12 General Trends in First Ionization Energies Increasing First Ionization Energy

13 13 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 - X - (g) O (g) + e - O - (g)  H = -328 kJ/mol EA = +328 kJ/mol  H = -141 kJ/mol EA = +141 kJ/mol

14 14 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. Comparison of Group 1A and 1B Lower I 1, more reactive


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