If Z_{eff} of boron is x,then Z_{eff} of oxyg | vedclass

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(D) The atomic number of Boron is $5$. The electronic configuration is $1s^2 2s^2 2p^1$.Using Slater's rules,the shielding constant $\sigma$ for the $

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$x + 1.95$

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(D) The atomic number of Boron is $5$. The electronic configuration is $1s^2 2s^2 2p^1$.Using Slater's rules,the shielding constant $\sigma$ for the $2p$ electron is $(2 	imes 0.85) + (2 	imes 0.35) = 1.7 + 0.7 = 2.4$.Thus,$Z_{eff} = Z - \sigma = 5 - 2.4 = 2.6$. Let this be $x$.The atomic number of Oxygen is $8$. The electronic configuration is $1s^2 2s^2 2p^4$.Using Slater's rules,the shielding constant $\sigma$ for a $2p$ electron is $(2 	imes 0.85) + (5 	imes 0.35) = 1.7 + 1.75 = 3.45$.Thus,$Z_{eff} = 8 - 3.45 = 4.55$.Comparing the two values: $4.55 - 2.6 = 1.95$.Therefore,the $Z_{eff}$ of oxygen is $x + 1.95$.

If $Z_{eff}$ of boron is $x$,then $Z_{eff}$ of oxygen will be

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