If Z_{eff} of B is 2.6,then the value of Z_{e | vedclass

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(C) The effective nuclear charge $Z_{eff}$ is calculated using Slater's rules: $Z_{eff} = Z - S$,where $Z$ is the atomic number and $S$ is the shieldi

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$4.55, 5.20$

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(C) The effective nuclear charge $Z_{eff}$ is calculated using Slater's rules: $Z_{eff} = Z - S$,where $Z$ is the atomic number and $S$ is the shielding constant.For Oxygen $(Z=8)$: The electronic configuration is $1s^2 2s^2 2p^4$. The shielding constant $S = (2 	imes 0.85) + (5 	imes 0.35) = 1.7 + 1.75 = 3.45$.Thus,$Z_{eff} = 8 - 3.45 = 4.55$.For Fluorine $(Z=9)$: The electronic configuration is $1s^2 2s^2 2p^5$. The shielding constant $S = (2 	imes 0.85) + (6 	imes 0.35) = 1.7 + 2.1 = 3.80$.Thus,$Z_{eff} = 9 - 3.80 = 5.20$.

If $Z_{eff}$ of $B$ is $2.6$,then the value of $Z_{eff}$ of oxygen and fluorine will be :-

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