Concept explainers
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
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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Chapter 7 Solutions
Chemistry & Chemical Reactivity, Hybrid Edition (with OWLv2 24-Months Printed Access Card)
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