+] P. Bb Tests-CHE-100-01Y Spring 22 * Question Completion Status: QUESTION 9 Which is true regarding Se? The atoms lose 2 electrons to form a +2 ion The atoms lose 2 electrons to form a -2 ion The atoms gain 2 electrons to form a +2 ion O The atoms gain 2 electrons to form a -2 ion QUESTION 10 Is Al stable as an individual atom? Briefly explain why or why not. For the toolbar, press ALT+F10 (PC) or ALT+FN+F10 (Mac). Paragraph Arial 10pt x² X, 8 S 次。 QUESTION 11 How is Na attached to the other atoms in NaHCO3 (sodium bicarbonate)? Shared electrons Opposite charge Click Save and Submit to save and submit. Click Save All Answers to save all ansvers.

+] P. Bb Tests-CHE-100-01Y Spring 22 * Question Completion Status: QUESTION 9 Which is true regarding Se? The atoms lose 2 electrons to form a +2 ion The atoms lose 2 electrons to form a -2 ion The atoms gain 2 electrons to form a +2 ion O The atoms gain 2 electrons to form a -2 ion QUESTION 10 Is Al stable as an individual atom? Briefly explain why or why not. For the toolbar, press ALT+F10 (PC) or ALT+FN+F10 (Mac). Paragraph Arial 10pt x² X, 8 S 次。 QUESTION 11 How is Na attached to the other atoms in NaHCO3 (sodium bicarbonate)? Shared electrons Opposite charge Click Save and Submit to save and submit. Click Save All Answers to save all ansvers.

Chemistry

9th Edition

ISBN:9781133611097

Author:Steven S. Zumdahl

Publisher:Steven S. Zumdahl

Chapter20: The Representative Elements

Section: Chapter Questions

Problem 70E

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Slaters rules are a way to estimate the effective nuclear charge experienced by an electron. In this approach, the shielding constant, S, is calculated. The effective nuclear charge is then the difference between S and the atomic number, Z. (Note that the results in Table 7.2 and Figure 7.2 were calculated in a slightly different way.) Z = Z S The shielding constant, S, is calculated using the following rules: (1) The electrons of an atom are grouped as follows: (1s) (2s, 2p) (3s, 3p) (3d) (4s, 4p) (4d), and so on. (2) Electrons in higher groups (to the right) do not shield those in the lower groups. (3) For ns and np valence electrons a) Electrons in the same ns, np group contribute 0.35 (for 1s 0.30 works better). b) Electrons in n 1 groups contribute 0.85. c) Electrons in n 2 groups (and lower) contribute 1.00. (4) For nd and nf electrons, electrons in the same nd or nf group contribute 0.35, and those in groups to the left contribute 1.00. As an example, let us calculate Z for the outermost electron of oxygen: S = (2 0.85) + (5 0.35) = 3.45 Z = 8 3.45 = 4.55 Here is a calculation for a d electron in Ni: Z = 28 [18 1.00] [7 0.35] = 7.55 and for an s electron in Ni: Z = 28 [10 1.00] [16 0.85] [1 0.35] = 4.05 (Here 3s, 3p, and 3d electrons are in the (n 1) groups.) a) Calculate Z for F and Ne. Relate the Z values for O, F, and Ne to their relative atomic radii and ionization energies. b) Calculate Z for one of the 3d electrons of Mn, and compare this with Z for one of the 4s electrons of the element. Do the Z values give us some insight into the ionization of Mn to give the cation?
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