After teaching rules of Slater, our teacher gave some examples. Some of them were to calculate effective nuclear charge on an electron, but some of them involved calculating Z* of atoms and ions.

For example, he calculated screening constant of fluoride ion as follows. The configuration is (1s 2)(2s 2 2p 6). Screening constant for ion (not for any specific electron) = 2×0.85 (for 1s electrons) + 8×0.35 (for 2s and 2p electrons).

I asked what is the significance of this screening constant (what is fluoride ion screening? If it were to screen a 3s electron added later, the calculation will be different). The teacher could not provide any satisfactory answer.

Is there any reason for doing this calculation? This calculation actually evaluates the screening constant for an imaginary 7th 2p electron.

  • $\begingroup$ It's for screening a 2p electron in fluoride. $\endgroup$ – Tan Yong Boon Jan 6 '18 at 0:36
  • $\begingroup$ Then it should be 2×0.85 + 7×0.35, not 8×0.35 $\endgroup$ – Archisman Panigrahi Jan 6 '18 at 5:06
  • $\begingroup$ Right. That's true. My mistake... $\endgroup$ – Tan Yong Boon Jan 6 '18 at 5:49

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