Chapter 8

# Chapter 8 - CHAPTER 8 PERIODIC RELATIONSHIPS AMONG THE...

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CHAPTER 8 PERIODIC RELATIONSHIPS AMONG THE ELEMENTS 8.19 Hydrogen forms the H + ion (resembles the alkali metals) and the H ion (resembles the halogens). 8.20 Strategy: (a) We refer to the building-up principle discussed in Section 7.9 of the text. We start writing the electron configuration with principal quantum number n = 1 and continue upward in energy until all electrons are accounted for. (b) What are the electron configuration characteristics of representative elements, transition elements, and 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 = 1, we have a 1 s orbital (2 electrons). For n = 2, we have a 2 s orbital (2 electrons) and three 2 p orbitals (6 electrons). For n = 3, we have a 3 s orbital (2 electrons). The number of electrons left to place is 17 12 = 5. These five electrons are placed in the 3 p orbitals. The electron configuration is 1 s 2 2 s 2 2 p 6 3 s 2 3 p 5 or [Ne]3 s 2 3 p 5 . (b) Because the 3 p subshell is not completely filled, this is a representative element . Without consulting a periodic table, you might know that the halogen family has seven valence electrons. You could then further classify this element as a halogen . In addition, all halogens are nonmetals . (c) If you were to write an orbital diagram for this electron configuration, you would see that there is one unpaired electron in the p subshell. Remember, the three 3 p orbitals can hold a total of six electrons. Therefore, the atoms of this element are paramagnetic. Check: For (b), note that a transition metal possesses an incompletely filled d subshell, and a noble gas has a completely filled outer-shell. For (c), recall that if the atoms of an element contain an odd number of electrons, the element must be paramagnetic. 8.21 (a) and (d) ; (b) and (f) ; (c) and (e) . 8.22 Elements that have the same number of valence electrons will have similarities in chemical behavior. Looking at the periodic table, elements with the same number of valence electrons are in the same group. Therefore, the pairs that would represent similar chemical properties of their atoms are: (a) and (d) (b) and (e) (c) and (f) . 8.23 (a) 1 s 2 2 s 2 2 p 5 (halogen) (c) [Ar]4 s 2 3 d 6 (transition metal) (b) [Ar]4 s 2 (alkaline earth metal) (d) [Ar]4 s 2 3 d 10 4 p 3 (Group 5A) 8.24 (a) Group 1A (b) Group 5A (c) Group 8A (d) Group 8B Identify the elements.

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CHAPTER 8: PERIODIC RELATIONSHIPS AMONG THE ELEMENTS 221 8.25 There are no electrons in the 4 s subshell because transition metals lose electrons from the ns valence subshell before they are lost from the ( n 1) d subshell. For the neutral atom there are only six valence electrons. The element can be identified as Cr (chromium) simply by counting six across starting with potassium (K, atomic number 19).
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## This homework help was uploaded on 04/21/2008 for the course CHEM 112 taught by Professor Heglund during the Spring '08 term at SDSMT.

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Chapter 8 - CHAPTER 8 PERIODIC RELATIONSHIPS AMONG THE...

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