Find the sheilding constant of 2s or 2P
electon in F atom (Where z=9)
Answers
Answer::
First, write the electron configuration on the following way
(1s)(2s,2p)(3s,3p) (3d) (4s,4p) (4d) (4f) (5s,5p)
Electrons from higher orbitals of an atom do not contribute any shielding.
3. All other electrons in the same group as the electron of interest shield to an extent of 0.35 nuclear charge units.
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Shielding By S or P electrons
4. If the electron of interest is an s or p electron: All electrons with one less value of the principal quantum number shield to an extent of 0.85 units of nuclear charge.
All electrons with two fewer values of the principal quantum number shield to an extent of 1.00 units.
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5. If the electron of interact is a d or f electron: All electrons to the left shield to an extent of 1.00 units of nuclear charge.
Subtract the screening or shielding from the actual nuclear charge we can give an effective atomic number.
Calculate Z* for a valence electron in fluorine.
(1s2)(2s2,2p5)
Shielding or Screening = 0.35 · 6 + 0.85 · 2 = 3.8
Z* = 9 — 3.8 = 5.2 for a valence electron.
Calculate Z* for a 6s electron in Platinum.
(1s2)(2s2,2p6)(3s2,3p6) (3d10) (4s2,4p6) (4d10) (4f14) (5s2,5p6) (5d8) (6s2)
Shielding or Screening 0.35 · 1 + 0.85 · 16 + 60 · 1.00 = 73.95
Z* = 78–73.95 = 4.15 for a valence electron.
Inorganic Chemistry Topics
The main topics in inorganic chemistry include atoms