report4_soln - The Solution for the 4th Report Chapter 5...

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1 The Solution for the 4 th Report Chapter 5 5-5 (a) ns 2 np 3 , Group VA (b) ns 2 , Group IIA (c) ns 2 ( n – 1) d 0–2 ( n – 2) f 1–14 , lanthanides ( n = 6) and actinides ( n = 7) 5-11 (a) Ca 2+ = 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 (b) K + = 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 (c) O 2– = 1 s 2 2 s 2 2 p 6 Both Ca 2+ and K + have 18 electrons. They are isoelectronic. 5-17 Cl > S > Ar > K + . A neutral Ar atom would be smaller than the S atom due to the larger effective nuclear charge attracting the valence electrons. The K + ion is smaller than the Ar atom because both have 18 electrons, but the K + ion has one more proton to attract the valence electrons. Although the Cl atom is slightly smaller than the S atom due to effective nuclear charge, the Cl ion is larger than the S atom because there is an extra electron in the valence shell. The Cl ion has 17 protons and 18 electrons. 5-27 The dip at IIIA is due to the greater ease of removing a lone p electron than one of the pair of s electrons of Group IIA. Group VA is especially stable compared to its neighbors. The fourth
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This note was uploaded on 05/24/2011 for the course ENGINEERIN 1 taught by Professor Lee during the Spring '11 term at Sungkyunkwan.

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report4_soln - The Solution for the 4th Report Chapter 5...

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