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O ~ (1s)2 (2s, 2p)6

S = (0.35 * 5) + (0.85 * 2) = 3.45

Zeff = Z - S

= 8 - 3.45

= 4.55

The effective nuclear charge experienced by a valence electron on an O atom is + 4.55
ajeigbeibraheem

The Zeff for a valence electron in an oxygen atom is 4.55

The Zeff is the effective nuclear charge present in the number of protons existing within the nucleus.  It is given by the formula:

[tex]\mathbf{Z_{eff} = Z - \sigma}[/tex]

where;

  • Z = no of protons in nucleus
  • σ = screening constant which is estimated using slater's rule

The electronic configuration for an oxygen atom is 1s² 2s² 2p⁴

  • In the first shell n, there are two (2) electron
  • In the next shell m = (n -1), there are 6 electrons. i.e m = 6 - 1 = 5

By the application of Slater's rule which posits that the nth shell electrons add to the extent 0.35 for each n-shell electron and 0.85 for each (n-1) electron.

As such; For oxygen, the screening constant (σ)

σ = (1.75) + (1.7)

σ = 3.45

The Zeff of the valence electron in an oxygen atom is:

[tex]\mathbf{Z_{eff} = Z - \sigma}[/tex]

[tex]\mathbf{Z_{eff} = 8.0 - 3.45}[/tex]

[tex]\mathbf{Z_{eff} = 4.55}[/tex]

Therefore, we can conclude that the Zeff of the valence electron in an oxygen atom is 4.55

Learn more about valence electrons here:

https://brainly.com/question/18214726?referrer=searchResults

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