Answered: Static electricity is observed when…

Static electricity is observed when whic part of the atom is transferred? a. Protons b. Neutrons I Electrons С. Naturally occurring static is observed in a. Thunder b. Lightning C. Wind

Introductory Chemistry For Today

8th Edition

ISBN:9781285644561

Author:Seager

Publisher:Seager

Chapter2: Atoms And Molecules

Section: Chapter Questions

Problem 2.20E

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Concept explainers

Atoms and Molecules

An atom is the smallest unit of an element that possesses all the properties of that particular element. Molecules are the smallest unit of a substance that possess all the properties of that particular substance.

Question

Static
Static electricity is observed when whic
part of the atom is transferred?
a.
Protons
b. Neutrons I
Electrons
С.
Naturally occurring static is observed in
a.
Thunder
b. Lightning
Wind
с.
speaker notes
areer investig
docs
Career inve stig
docx
Career Investig

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Def of ionization
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?