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Slater’s 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: (1 s ) (2 s , 2 p ) (3 s , 3 p ) (3 d ) (4 s , 4 p ) (4 d ), 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 1 s 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 3 s , 3 p , and 3 d 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 3 d electrons of Mn, and compare this with Z * for one of the 4 s 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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Abel Cruz
Abel CruzLv7
9 Dec 2020

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