Conceptual Problems 9.131 If six electrons rather than eight electrons led to a stable configuration, the electron configuration of the stable configuration would be ns2np4. a. A noble gas would have the electron configuration ns2np4. This could correspond to the O atom. b. A reactive nonmetal would have one less electron than the stable configuration. This would have the electron configuration ns2np3. This could correspond to the N atom. c. A reactive metal would have one more electron than the stable configuration. This would have the electron configuration of ns1. This could correspond to the Li atom. 9.132 Atom B would have the higher first ionization energy. Even though the effective nuclear charge is less, the outermost electron is closer to the nucleus and the potential energy becomes more negative with decreasing distance, making it harder to remove and requiring a larger ionization energy. 9.133 (a) An electron in a 3s orbital is more shielded than an electron in a 2s orbital. This is true True: because there are more core electrons below a 3s orbital. (b)
True:
An electron in a 3s orbital penetrates the region occupied by the core electrons more than electrons in a 3p orbital. Examine Figure 9.4, the radial distribution functions for the 3s, 3p, and 3d orbitals. You will see that the 3s electrons penetrate more deeply than the 3p electrons and more than the 3d electrons.
(c)
False:
(d)
True:
An electron in an orbital that penetrates closer to the nucleus will experience less shielding than an electron in an orbital that does not penetrate as far. An electron in an orbital that penetrates close to the nucleus will tend to experience a higher effective nuclear charge than one that does not. Because the orbital penetrates closer to the nucleus, the electron will experience less shielding and therefore a higher effective nuclear charge.
9.134 An electron in a 5p orbital could have any one of the following combinations of quantum numbers: (5, 1, − 1, + 1/2) (5, 1, − 1, − 1/2) (5, 1, 0, + 1/2) (5, 1, 0, − 1/2) (5, 1, 1, + 1/2) (5, 1, 1, − 1/2) An electron in a 6d orbital could have any one of the following combinations of quantum numbers: (6, 2,− 2,+ 1/2)
(6, 2,− 2,− 1/2)
(6, 2,− 1,+ 1/2)
(6, 2,− 1,− 1/2)
(6, 2, 0,+ 1/2)
(6, 2, 0,− 1/2)(6, 2, 1,+ 1/2)
(6, 2, 1,− 1/2)
(6, 2, 2,+ 1/2)
(6, 2, 2,− 1/2)
1=590 kJ/mol; they are valence electrons. The energetic cost for 9.135 The 4s electrons in calcium have relatively lowIE2=1145 ionizationkJ/mol) energiesbecause (IE to be lost is a core electron. Similarly, the electron affinity of fluorine to gain one electron (−328 kJ/mol) is highly exothermic because the added electron completes fluoride’s valence shell. The gain of a second electron by the negatively charged fluoride anion would not be favorable. Therefore, we would expect calcium and fluoride to combine in a 1:2 ratio.
Questions for Group Work 9.136 Shielding or screening occurs when one electron is blocked from the full effects of the nuclear charge so that the electron experiences only a part of the nuclear charge. Penetration occurs when an electron penetrates the electron cloud of the 1s orbital and now experiences the full effect of the nuclear charge. 9.137 The orbitals fill in order of increasing energy of the orbitals, which is 1s<2s<2p<3s<3p<4s<3d<4p<5s<4d<5p<6s. Keep in mind that s subshells can contain two electrons, p subshells can contain six electrons, and d subshells can contain ten electrons. 9.138
Extended Figure Description for Chapter 9 Problem 138 Figure 1 9.139 a. The Br− and Se2− ions are about the same size because the ions are an isoelectronic pair. b. The Br− ion is smaller than the Se2− ion because the Br− ion has a larger nuclear charge and fewer extra electrons than the Se2− ion. c. The Fr+(194 pm) ion is the singly charged cation that is closest in size to the Br− (195 pm) and Se2− (198 pm) ions because cations are much smaller than their corresponding atoms. 9.140 The atomic radius decreases as you move to the right across a period in the periodic table, as the effective nuclear charge increases. The atomic radius increases as you move down a column in the periodic table, as larger and larger shells are filled with electrons. The first ionization energy generally decreases as you move down a column in the periodic table because electrons in the outermost principal level become farther away from the positively charged nucleus and are therefore held less tightly. The first ionization energy generally increases as you move to the right across a period in the periodic table because electrons in the outermost principal energy level generally experience a greater effective nuclear charge. The metallic character (how easy it is to move electrons) decreases as you move to the right across a period in the periodic table, as electrons are held tighter and tighter. The metallic character increases as you move down a column in the periodic table, as the valence electrons are at increasing distances from the nucleus.
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